S. Chand’s IIT Foundation Series

A Compact & Comprehensive Book of

Class - IX

S.Chand’s IIT Foundation Series

A Compact and Comprehensive Book of

IIT Foundation
Mathematics

CLASS - IX

S.Chand’s IIT Foundation Series

A Compact and Comprehensive Book of IIT Foundation Mathematics

CLASS - IX

S.K. GUPTA
ANUBHUTI GANGAL

EURASIA PUBLISHING HOUSE

(An imprint of $S$. Chand Publishing)

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C) 2014. S.K. Gupta, Anubhuti Gangal

All rights reserved. No part of this publication may be reproduced or copied in any material form (including photocopying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice.

Jurisdiction : All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, Tribunals and Forums of New Delhi, India only.

First Published in 2014

Reprints 2016 (Thrice)

ISBN : 978-93-837-4662-0

Code : 1014A 680

PREFACE AND A NOTE FOR THE STUDENTS

ARE YOU ASPIRING TO BECOME AN ENGINEER AND AN IIT SCHOLAR?

Here is the book especially designed to motivate you, to sharpen your intellect, to develop the right attitude and aptitude, and to lay a solid foundation for your success in various entrance examinations like IIT, EAMCET, WBJEE, MPPET, SCRA, J&K CET, Kerala PET, OJEE, Rajasthan PET, AMU, BITSAT, etc.

SALIENT FEATURES

Content based on the curriculum of the classes for CBSE, ICSE, Andhra Pradesh and Boards of School Education of Other States.
Full and comprehensive coverage of all the topics.
Detailed synopsis of each chapter at the beginning in the form of ‘Key Concepts’. This will not only facilitate thorough ‘Revision’ and ‘Recall’ of every topic but also greatly help the students in understanding and mastering the concepts besides providing a back-up to classroom teaching.
The books are enriched with an exhaustive range of hundreds of thought provoking objective questions in the form of solved examples and practice questions in practice sheets which not only offer a great variety and reflect the modern trends but also invite, explore, develop and put to test the thinking, analysing and problem-solving skills of the students.
Answers, Hints and Solutions have been provided to boost up the morale and increase the confidence level.
Self Assessment Sheets have been given at the end of each chapter to help the students to assess and evaluate their understanding of the concepts and learn to attack the problems independently.

We hope this book will be able to fulfil its aims and objectives and will be found immensely useful by the students aspiring to become top class engineers.

Suggestions for improvement and also the feedback received from various sources would be most welcome and gratefully acknowledged.

AUTHORS[^0]

Chapter 1. Logarithms ….. 1-1 to $1-15$

Chapter 2. Polynomials ….. 2-1 to $2-16$

Chapter 3. Quadratic Equations ….. 3-1 to 3-19

Chapter 4. Inequalities ….. 4-1 to $4-26$

Chapter 5. Relations ….. 5-1 to $5-13$

Chapter 6. Plane Geometry-Triangles ….. 6-1 to $6-22$

Chapter 7. Quadrilaterals ….. 7-1 to $7-26$

Chapter 8. Permutations and Combinations ….. $8-1$ to $8-30$

Chapter 9. Probability ….. 9-1 to $9-31$

Chapter 10. Trigonometry ….. $10-1$ to $10-25$

Chapter 11. Coordinate Geometry ….. $11-1$ to $11-28$

Chapter 12. Area and Perimeter ….. $12-1$ to $12-14$

Chapter 13. Volume and Surface Area of Solids ….. $13-1$ to $13-18$

Logarithms

KEY FACTS

Definition: If $a$ and $n$ are positive real numbers such that $a \neq 1$ and $x$ is real, then $a^{x}=n \Rightarrow x=\log _{a} n$.

Here $x$ is said to be the logarithm of the number $n$ to the base $a$.

Ex. $4^{3}=64 \Rightarrow \log _4 64=3, \quad 10^{-1}=\frac{1}{10}=0.1 \Rightarrow \log _{10} 0.1=-1, \quad 5^{x}=4 \Rightarrow x=\log _5 4$,

$ a^{0}=1 \Rightarrow \log _{a} 1=0, \quad a^{1}=a \Rightarrow \log _{a} a=1 $

2. Some Important Facts about Logarithms

$\log _{a} n$ is real if $n>0$
$\log _{a} n$ is imaginary if $n<0$
$\log _{a} n$ is not defined if $n=0$
The logarithm of 1 to any base $\boldsymbol{a}, a>0$ and $a \neq 1$ is zero. $\log _{\boldsymbol{a}} \mathbf{1}=\mathbf{0}$
The logarithm of any number $\boldsymbol{a}, a>0$ and $a \neq 1$, to the same base is 1 . $\log _{\boldsymbol{a}} \boldsymbol{a}=\mathbf{1}$
If $\boldsymbol{a}$ and $\boldsymbol{x}$ are positive real numbers, where $a \neq 1$, then $\boldsymbol{a}^{\log _{a} x}=\boldsymbol{x}$

Proof. Let $\log _{a} x=p$. Then, $x=a^{p}$ (By def.) $\Rightarrow x=a^{\log _{a} x}$ (Substituting the value of $p$ )

Ex. $3^{\log _3 7}=7,2^{\log _2 9}=9,5^{\log _5 x}=x$

For $a>0, a \neq 1, \log _{a} x_1=\log _{a} x_2 \Rightarrow x_1=x_2(x_1, x_2>0)$
If $a>1$ and $x>y$, then $\log _{a} x>\log _{a} y$.
If $0<a<1$ and $x>y$, then $\log _{a} x<\log _{a} y$

3. Laws of Logarithms

For $x>0, y>0$ and $a>0$ and $a \neq 1$, any real number $n$

$\log _{\boldsymbol{d}} x y=\log _{\boldsymbol{d}} x+\log _{\boldsymbol{a}} y \quad$ Ex. $\log _2(15)=\log _2(5 \times 3)=\log _2 5+\log _2 3$
$\log _{a}(x / y)=\log _{a} x-\log _{a} y$ Ex. $\log _2(\frac{3}{7})=\log _2 3-\log _2 7$
$\log _{\boldsymbol{a}}(\boldsymbol{x})^{\boldsymbol{n}}=\boldsymbol{n} \log _{\boldsymbol{a}} \boldsymbol{x} \quad$ Ex. $\log (2)^{5}=5 \log 2$,

$ \log (\frac{a^{3}}{b^{3}})=\log a^{3}-\log b^{3}=3 \log a-3 \log b $

$\log _{a} x=\frac{1}{\log _{x} a} \quad$ Ex. $\log _5 2=\frac{1}{\log _2 5}$
$\log _{a^{n}} x=\frac{1}{n} \log _{a} x \quad$ Ex. $\log _8 7=\log _2 3(7)=\frac{1}{3} \log _2 7, \log _{\sqrt{5}} 3=\log _{(5) \frac{1}{2}}(3)=\log _{5^{1 / 2}}(3)=\frac{1}{1 / 2} \log _5 3=2 \log _5 3$
$\log _{a^{n}} x^{m}=\frac{m}{n} \log _{a} x \quad$ Ex. $\log _{2^{5}} 5^{4}=\frac{4}{5} \log _2 5$

Base changing formula

$\log _{a} x=\log _{b} x \cdot \log b$ Ex. $\log _{12} 32=\log _{16} 32 . \log _{12} 16$. Old base New base
$x^{\log _{a} y}=y^{\log _{a} x}$

Ex. $3^{\log 7}=7^{\log 3}$

(It being understood that base is same)

[Proof. $x^{\log _{a} y} \to x^{\log _{x} y \cdot \log _{a} x}$ (Base changing formula)

$ \begin{matrix} =(x^{\log _{x} y})^{\log _{a} x} & (\text{ Using } n \log _{a} x=(\log _{a} x)^{n}) \\ =y^{\log _{a} x} & (\text{ Using } x^{\log _{x} y}=y .) \end{matrix} $

$\log _{a} b=\frac{\log b}{\log a} \quad$ (It being understood that base is same)
If $\log _{a} b=x$ for all $a>0, a \neq 1, b>0$ and $x \in R$, then $\log _{1 / a} b=-x, \log _{a} 1 / b=-x$ and $\log _{1 / a} 1 / b=x$

4. Some Important Properties of Logarithms

$a, b, c$ are in G.P. $\Leftrightarrow \log _{a} x, \log _{b} x, \log _{c} x$ are in H.P.
$a, b, c$ are in G.P. $\Leftrightarrow \log _{x} a, \log _{x} b, \log _{x} c$ are in A.P.

5. Natural or Naperian logarithm is denoted by $\log _{e} N$, where the base is $e$.

Ex. $\log _{e} 7, \log _{e}(\frac{1}{64}), \log _{e} b$, etc.

Common or Brigg’s logarithm is denoted by $\log _{10} N$, where the base is 10 .

Ex. $\log _{10} 5, \log _{10}(\frac{1}{81})$, etc.

$\log _{a} x$ is a decreasing function if $0<a<1$
$\log _{a} x$ is an increasing function if $a>1$.

6. Characteristic and Mantissa

Characteristic: The integral part of the logarithm is called characteristic.

(i) If the number is greater than unity and there are $n$ digits in integral part, then its characteristic $=(n-1)$

(ii) When the number is less than 1, the characteristic is one more than the number of zeroes between the decimal point and the first significant digit of the number and is negative. It is written as $(\overline{n+1})$ or $Bar(n+1)$.

Ex.

Number	Characteristic	Number	Characteristic
4.1456	0	0.823	$\overline 1$
24.8920	1	0.0234	$\overline 2$
238.1008	2	0.000423	$\overline 4$

7. Arithmetic Progression. A sequence $a_1, a_2, a_3, \ldots \ldots ., a_n$ is said to be in arithmetic progression, when $a_2-a_1=a_3-a_2=\ldots \ldots \ldots . .=a_n-a _{n-1}$, i.e., when the terms in the sequence increase or decrease by a constant quantity called the common difference. Ex. $1,3,5,7,9, \ldots$. $6,11,17,23, \ldots$ $-5,-2,1,4,7, \ldots$

Sum of first ’ $n$ ’ terms of an Arithmetic Progression

$S_n=\frac{n}{2}[2 a+(n-1) d]=\frac{n}{2}[a+l]$,

where $a=$ first term, $n=$ number of terms, $d=$ common difference, $l=$ last term.

Sum of first " $n$ " natural numbers.

$ S_n=1+2+3+\ldots \ldots \ldots \ldots . .+n=\frac{n(n+1)}{2} $

Also, written as $\Sigma \boldsymbol{n}=\frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})}{\mathbf{2}}$

Also, if $a, b, c$ are in A.P. then $2 b=a+c$

8. Geometric Progression : A sequence $a_1, a_2, a_3, \ldots \ldots \ldots . ., a_n$ is said to be in Geometric Progression when,

$ \frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\ldots \ldots \ldots \ldots=\frac{a_n}{a _{n-1}}=r(say) $

where $a_1, a_2, a_3, \ldots \ldots . .$. are all non zero numbers and $\boldsymbol{r}$ is called the common ratio.

Ex. 3 , 6, 12, 24, ………….. $\quad \boldsymbol{r}=\mathbf{2}$;

$64,16,4,1, \frac{1}{4}, \frac{1}{16} \cdot \frac{1}{64}, \ldots \ldots \ldots . . . \boldsymbol{r}=\frac{1}{4}$

Sum of first $n$ terms of a G.P. $S_n=\frac{a(r^{n}-1)}{(r-1)}$ if $r>1=\frac{a(1-r^{n})}{(1-r)}$ if $r<1=\frac{l r-a}{r-1}$

where, $a=$ first term, $r=$ common ratio, $l=$ last term

Sum of an infinite G.P. $S _{\infty}=\frac{\boldsymbol{a}}{\boldsymbol{1}-\boldsymbol{r}}$, where $a=$ first term, $r=$ common ratio.
For three terms $a, b, c$ to be in G.P., $\boldsymbol{b}^{2}=\boldsymbol{a} \boldsymbol{c}$

9. Harmonic Progression : A series of quantities $a_1, a_2, a_3, \ldots \ldots \ldots . ., a_n$ are said to be in H.P. when their reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \ldots \ldots, \frac{1}{a_n}$ are in A.P.

When three quantities $a, b, c$ are in H.P., then, $b=\frac{\mathbf{2 a c}}{\boldsymbol{a}+\boldsymbol{c}}$.

SOLVED EXAMPLES

Ex. 1 . If $\log _{a} 5+\log _{a} 25+\log _{a} 125+\log _{a} 625=10$, then find the value of $a$.

Sol. $\log _{a} 5+\log _{a} 25+\log _{a} 125+\log _{a} 625=10$

$\Rightarrow \log _{a}(5 \times 25 \times 125 \times 625)=10$

$\Rightarrow \log _{a}(5^{1} \times 5^{2} \times 5^{3} \times 5^{4})=10$

$\Rightarrow \log _{a} 5^{10}=10 \Rightarrow a^{10}=5^{10} \Rightarrow \boldsymbol{a}=\mathbf{5}$.

[Using $\log _{a} x=n \Rightarrow x=a^{n}$ ]

Ex. 2 . Solve for $x: \log _{10}[\log _2(\log _3 9)]=x$.

Sol. $\log _{10}[\log _2(\log _3 9)]=x$

$\Rightarrow \log _2(\log _3 9)=10^{x}$

$\Rightarrow \log _2(\log _3 3^{2})=10^{x}$

$\Rightarrow \log _2(2 \log _3 3)=10^{x}$

$\Rightarrow \log _2 2=10^{x} \Rightarrow 10^{x}=1=10^{0} \Rightarrow \boldsymbol{x}=\mathbf{0}$.

Ex. 3 . Find the value of $\log _{x} x+\log _{x} x^{3}+\log _{x} x^{5}+\ldots \ldots . .+\log _{x} x^{2 n-1}$.

Sol. $\log _{x} x+\log _{x} x^{3}+\log _{x} x^{5}+\ldots \ldots . .+\log _{x} x^{2 n-1}=\log _{x} x+3 \log _{x} x+5 \log _{x} x+\ldots \ldots \ldots+(2 n-1) \log _{x} x$

$=1+3+5+\ldots \ldots \ldots \ldots+(2 n-1)=\frac{n}{2}[1+(2 n-1)]=\boldsymbol{n}^{2} \quad[.$ Using $\log _{x} x=1$ and for A.P. $.S_n=\frac{n}{2}(a+l)]$

Ex. 4 . If $f(x)=\log (\frac{1+x}{1-x})$, show that $f(\frac{2 x}{1+x^{2}})=2 f(x)$.

Sol. $f(\frac{2 x}{1+x^{2}})=\log [\frac{1+\frac{2 x}{1+x^{2}}}{1-\frac{2 x}{1+x^{2}}}]=\log [\frac{1+x^{2}+2 x}{1+x^{2}-2 x}]=\log [\frac{(1+x)^{2}}{(1-x)^{2}}]=2 \log [\frac{1+x}{1-x}]=\mathbf{2} \boldsymbol{f}(\boldsymbol{x})$.

Ex. 5 . If $a=\log _{24} 12, b=\log _{36} 24, c=\log _{48} 36$, then prove that $1+a b c=2 b c$.

Sol. $\quad 1+a b c=1+\log _{24} 12 . \log _{36} 24 . \log _{48} 36=1+\log _{36} 12 . \log _{48} 36$

$ \begin{aligned} & =1+\log _{48} 12=\log _{48} 48+\log _{48} 12 \\ & =\log _{48}(48 \times 12)=\log _{48}(24 \times 24) \\ & =\log _{48}(24)^{2}=2 \log _{48} 24 \end{aligned} $

Also, $\quad 2 b c=2 \log _{36} 24 . \log _{48} 36=2 \log _{48} 24$

From $(i)$ and (ii), we have RHS $=$ LHS.

Ex. 6 . Solve $\log _{2 x+3}(6 x^{2}+23 x+21)=4-\log _{3 x+7}(4 x^{2}+12 x+9)$.

Sol. Given, $\log _{(2 x+3)}(6 x^{2}+23 x+21)=4-\log _{(3 x+7)}(4 x^{2}+12 x+9)$

$\Rightarrow \log _{(2 x+3)}(2 x+3)(3 x+7)=4-\log _{(3 x+7)}(2 x+3)^{2}$

$\Rightarrow \log _{(2 x+3)}(2 x+3)+\log _{(2 x+3)}(3 x+7)=4-2 \log _{(3 x+7)}(2 x+3)$

$\Rightarrow \log _{(2 x+3)}(3 x+7)+2 \log _{(3 x+7)}(2 x+3)=4-1=3$

[Since $\log _{2 x+3}(2 x+3)=1$ ]

$\Rightarrow \log _{(2 x+3)}(3 x+7)+\frac{2}{\log _{(2 x+3)}(3 x+7)}=3$

$[.$ Using $.\log _{a} x=\frac{1}{\log _{x} a}]$

Let $\log _{(2 x+3)}(3 x+7)=t$. Then,

$ \begin{aligned} t+\frac{2}{t} & =3 \Rightarrow t^{2}-3 t+2=0 \Rightarrow(t-1)(t-2)=0 \Rightarrow t=1,2 \\ t=1 & \Rightarrow \log _{(2 x+3)}(3 x+7)=1 \Rightarrow \log _{(2 x+3)}(3 x+7)=\log _{(2 x+3)}(2 x+3)[\text{ Replacing } 1 \text{ by } \log _{(2 x+3)}(2 x+3)] \\ & \Rightarrow 3 x+7=2 x+3 \Rightarrow x=-4 . \\ t=2 & \Rightarrow \log _{(2 x+3)}(3 x+7)=2 \Rightarrow \log _{(2 x+3)}(3 x+7)=\log _{(2 x+3)}(2 x+3)^{2} \\ & \Rightarrow(3 x+7)=(2 x+3)^{2} \Rightarrow 4 x^{2}+9 x+2=0 \Rightarrow(4 x+1)(x+2)=0 \Rightarrow x=-1 / 4,-2 \end{aligned} $

But $x=-4$ and -2 are extraneous solutions, so $x=-\frac{1}{\mathbf{4}}$.

Ex. 7 . If $\log _{x}(a-b)-\log _{x}(a+b)=\log _{x}(b / a)$, find $\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}$.

(CAT 2012)

Sol. Given, $\log _{x}(a-b)-\log _{x}(a+b)=\log _{x}(b / a) \Rightarrow \log _{x}[\frac{(a-b)}{(a+b)}]=\log _{x}(\frac{b}{a})$

$\Rightarrow a(a-b)=b(a+b) \Rightarrow a^{2}-a b=a b+b^{2}$

$\Rightarrow a^{2}-b^{2}=2 a b \Rightarrow a^{2}-2 a b-b^{2}=0 \Rightarrow(\frac{a}{b})^{2}-2(\frac{a}{b})-1=0$

This is a quadratic equation in $\frac{a}{b}$ and the product of the roots is -1 i.e, if $a / b$ is a root, then $(-\frac{b}{a})$ is the other root. Also, sum of its roots $=2$

$\therefore(\frac{a}{b})^{2}+(\frac{b}{a})^{2}=\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=[\frac{a}{b}+(-\frac{b}{a})]^{2}+2=2^{2}+2=\mathbf{6}$.

Ex. 8 . If $\log _{e} 2 \cdot \log _{b} 625=\log _{10} 16 . \log _{e} 10$, then find the value of $b$.

Sol. Given, $\log _{e} 2 . \log _{b} 625=\log _{10} 16 \cdot \log _{e} 10 \Rightarrow \log _{e} 2 . \log _{b} 5^{4}=\log _{10} 2^{4} \cdot \log _{e} 10$

$\Rightarrow \log _{e} 2.4 \log _{b} 5=4 \log _{10} 2 . \log _{e} 10$

$\Rightarrow \log _{b} 5 .=\frac{\log _{10} 2 \cdot \log _{e} 10}{\log _{e} 2}=\frac{\log _{e} 2}{\log _{e} 2}=1 \Rightarrow b^{1}=5 \Rightarrow b=\mathbf{5}$.

$[\because \log _{a} x \cdot \log _{x} b=\log _{a} b]$

Ex. 9 . If $(x^{4}-2 x^{2} y^{2}+y^{2})^{a-1}=(x-y)^{2 a}(x+y)^{-2}$, then the value of $a$ is

(a) $x^{2}-y^{2}$

(b) $\log (x y)$

(d) $\log (x-y)$

Sol. Given, $(x^{4}-2 x^{2} y^{2}+y^{2})^{a-1}=(x-y)^{2 a}(x+y)^{-2}$

$\Rightarrow[(x^{2}-y^{2})^{2}]^{a-1}=(x-y)^{2 a}(x+y)^{-2}$

$\Rightarrow(x-y)^{2(a-1)}(x+y)^{2(a-1)}=(x-y)^{2 a}(x+y)^{-2}$

$\Rightarrow \frac{(x-y)^{2(a-1)}}{(x-y)^{2 a}} \cdot \frac{(x+y)^{2(a-1)}}{(x+y)^{-2}}=1 \Rightarrow(x-y)^{-2}(x+y)^{2 a}=1$

$\Rightarrow \log [(x-y)^{-2}(x+y)^{2 a}]=\log 1 \Rightarrow-2 \log (x-y)+2 a \log (x+y)=\log 1$

$\Rightarrow 2 a \log (x+y)=2 \log (x-y) \Rightarrow a=\frac{\log (x-y)}{\log (x+y)}$.

[Since $\log 1=0]$

Ex. 10 . If $\log _{x} a, a^{x / 2}$ and $\log _{b} x$ are in GP, then $x$ is

(a) $\log _{a}(\log _{b} a)$

(b) $\log _{a}(\log _{e} a)+\log _{a}(\log _{e} b)$

(d) $\log _{a}(\log _{e} b)-\log _{a}(\log _{e} a)$

Sol. If $\log _{x} a, a^{x / 2}$ and $\log _{b} x$ are in GP, then $(a^{x / 2})^{2}=(\log _{b} x) \times(\log _{x} a)$

$\Rightarrow a^{x}=\log _{b} a \Rightarrow \log a^{x}=\log (\log _{b} a) \Rightarrow x \log a=\log (\log _{b} a) \Rightarrow x \log _{a} a=\log _{a}(\log _{b} a)$

$\Rightarrow x=\log _{a}(\log _{b} a)$.

Ex. 11 . What is the least value of the expression $2 \log _{10} x-\log _{x}(1 / 100)$ for $x>1$ ?

Sol. $2 \log _{10} x-\log _{x} \frac{1}{100}=2 \log _{10} x-\frac{\log _{10} 10^{-2}}{\log _{10} x}$

$[.$ Using $.\log _{a} b=\frac{\log _{x} b}{\log _{x} a}]$

$ =2 \log _{10} x+\frac{2}{\log _{10} x}=2(\log _{10} x+\frac{1}{\log _{10} x}) $

Given, $x>1 \Rightarrow \log _{10} x>0$

But since $AM \geq GM$

$\therefore[\frac{\log _{10} x+\frac{1}{\log _{10} x}}{2}] \geq \sqrt{\log _{10} x \times \frac{1}{\log _{10} x}}$

$\Rightarrow \log _{10} x+\frac{1}{\log _{10} x} \geq 2 \Rightarrow 2[\log _{10} x+\frac{1}{\log _{10} x}] \geq 4$

For $x=10,2[\log _{10} x+\log _{10} x] \geq 4$

Hence, the least value of $[\log _{10} x-\log _{x} \frac{1}{100}]$ is 4 .

Ex. 12 . If $\log _3 2, \log _3(2^{x}-5)$ and $\log _3(2^{x}-7 / 2)$ are in A.P., then what is the value of $x$ ?

Sol. Given, $\log _3 2, \log _3(2^{x}-5)$ and $\log _3(2^{x}-7 / 2)$ are in A.P.

$ \Rightarrow \quad 2[\log _3(2^{x}-5)]=\log _3 2+\log _3(2^{x}-\frac{7}{2}) $

$ \begin{aligned} & \Rightarrow \log _3(2^{x}-5)^{2}=\log _3[2 \times(2^{x}-7 / 2)] \\ & \Rightarrow(2^{x}-5)^{2}=(2^{x+1}-7) \Rightarrow 2^{2 x}-10.2^{x}+25=2.2^{x}-7 \\ & \Rightarrow 2^{2 x}-12.2^{x}+32=0 \Rightarrow y^{2}-12 y+32=0 \\ & \Rightarrow(y-8)(y-4)=0 \Rightarrow y=8 \text{ or } 4 \Rightarrow 2^{x}=8 \text{ or } 2^{x}=4 \Rightarrow x=\mathbf{3} \text{ or } 2 . \end{aligned} $

Ex. 13 . Let $u=(\log _2 x)^{2}-6(\log _2 x)+12$, where $x$ is a real number. Then the equation $x^{u}=256$ has :

(a) No solution for $x$

(b) Exactly one solution for $x$

(d) Exactly three distinct solutions for $x$

(CAT 2004)

Sol. Given, $u=(\log _2 x)^{2}-6(\log _2 x)+12=p^{2}-6 p+12$

(where $p=\log _2 x$ )

Also, given, $x^{u}=256$

Taking $\log$ to the base 2 of both the sides, we have

$u \log _2 x=\log _2 256=\log _2 2^{8}=8 \log _2 2 \Rightarrow u \log _2 x=8 \Rightarrow u=\frac{8}{\log _2 x}=8 / p$

From (i) and $(i i) \frac{8}{p}=p^{2}-6 p+12$

$\Rightarrow 8=p^{3}-6 p^{2}+12 \Rightarrow p^{3}-6 p^{2}+12 p-8=0$

$\Rightarrow(p-2)^{3}=0 \Rightarrow p=2$.

[Using $(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$ ]

$\therefore \log _2 x=2 \Rightarrow x=2^{2}=4$

Hence the equation $u^{4}=256$ has exactly one solution.

Ex. 14 . If $\log _{y} x=(a \cdot \log _{z} y)=(b \cdot \log _{x} z)=a b$, then which of the following pairs of values for $(a, b)$ is not possible?

(a) $(-2, \frac{1}{2})$

(b) $(1,1)$

(d) $(\pi, \frac{1}{\pi})$

(e) $(2,2)$

(CAT 2004)

Sol. Given, $\log _{y} x=(a \cdot \log _{z} y)=(b \log _{x} z)=a b$

$ \begin{aligned} & \Rightarrow a=\frac{\log _{y} x}{\log _{z} y} \text{ and } b=\frac{\log _{y} x}{\log _{x} z} \\ & \therefore a \times b=\frac{\log _{y} x}{\log _{z} y} \times \frac{\log _{y} x}{\log _{x} z}=\frac{[\frac{\log x}{\log y}]}{[\frac{\log y}{\log z}]} \times \frac{[\frac{\log x}{\log y}]}{[\frac{\log z}{\log x}]}=(\frac{\log x}{\log y})^{3}=(\log _{y} x)^{3}=(a b)^{3} \\ & \Rightarrow a b-a^{3} b^{3}=0 \Rightarrow a b(1-a^{2} b^{2})=0 \Rightarrow a b= \pm 1 \end{aligned} $

$\therefore$ Only option $(e)$ does not satisfy the condition, $\sin (2,2)$ is not a possible value of $(a, b)$.

PRACTICE SHEET

LEVEL-1

(i) The solution of $\log _{\pi}(\log _2(\log _7 x))=0$ is

(a) 2

(b) $\pi^{2}$

(d) None of these

(WBJEE 2008)

Similar questions

(ii) $\log _{27}(\log _3 x)=\frac{1}{3} \Rightarrow x=$

(a) 3

(b) 6

(d) 27

(EAMCET 2004) (iii) The solution of $\log _{99}(\log _2(\log _3 x))=0$

(a) 4

(b) 9

(d) 99

(BCECE 2006)

~~ 2. If $x=\log _{b} a, y=\log _{c} b, z=\log _{a} c$, then $x y z$ is

(a) 0

(b) 1

(d) $a+b+c$

(UPSEE 2003)

~~ 3. (i) $7^{2 \log _7 5}$ is equal to

(a) 5

(b) 25

(d) $\log _7 35$

(KCET 2007)

Similar question

(ii) The real roots of the equation $7^{\log _7(x^{2}-4 x+5)}$

(a) 1 and 2

(b) 2 and 3

~~ 4. $(\frac{1}{\log _3 12}+\frac{1}{\log _4 12})$ is

(DCE 2001)

(a) 0

(b) $\frac{1}{2}$

(d) 2

(WBJEE 2009)

~~ 5. If $a, b, c$ do not belong to the set ${0,1,2,3 \ldots 9}$, then $\log _{10}(\frac{a+10 b+10^{2} c}{10^{-4} a+10^{-3} b+10^{-2} c})$ is equal to

(a) 1

(b) 2

(d) 4

(EAMCET 2005)

~~ 6. Assuming that the base is 10 , the value of the expression $\log 6+2 \log 5+\log 4-\log 3-\log 2$ is

(a) 0

(b) 1

(d) 3

~~ 7. $\log \frac{a^{2}}{b c}+\log \frac{b^{2}}{a c}+\log \frac{c^{2}}{a b}$ equals

(a) -1

(b) $a b c$

(d) 0

~~ 8. If $\log _{r} 6=m$ and $\log _{r} 3=n$, then what is $\log _{r}(r / 2)$ equal to?

(a) $m-n+1$

(b) $m+n-1$

(CDS 2009)

~~ 9. The value of $25^{(-1 / 4 \log _5 25)}$ is

(a) $\frac{1}{5}$

(b) $-\frac{1}{25}$

(d) None of these

~~ 10. If $\log _{10} x-\log _{10} \sqrt{x}=\frac{2}{\log _{10} x}$, find the value of $x$.

(a) 10

(b) -1

(CAT 2004)

~~ 11. If $\log _4 2+\log _4 4+\log _4 x+\log _4 16=6$, then $x$ is equal to

(a) 4

(b) 8

(d) 64

~~ 12. If $2^{x} \cdot 3^{x+4}=7^{x}$, then $x$ is equal to

(KCET 2006)

(a) $\frac{3 \log _{e} 4}{\log _{e} 7-\log _{e} 6}$

(b) $\frac{4 \log _{e} 3}{\log _{e} 6-\log _{e} 7}$

(d) $\frac{4 \log _{e} 3}{\log _{e} 7-\log _{e} 6}$

(MPPET 2009)

~~ 13. (i) If $n=1000$ !, then the value of $\frac{1}{\log _2 n}+\frac{1}{\log _3 n}+\ldots+\frac{1}{\log _{1000} n}$ is

(a) 0

(b) 1

(d) $10^{3}$

(KCET 2009, Kerala PET 2006, DCE 2005)

Similar question

(ii) If $x=1999$ !, then $\sum _{x=1}^{1999} \log _{n} x$ is equal to

(a) -1

(b) 0

(d) $\sqrt[1999]{1999}$

(AMU 2003)

~~ 14. If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$, then the value of $x^{b+c} \cdot y^{c+a}$ .$z^{a+b}$ is

(a) 1

(b) 0

(d) $x y z$

(KCET 2011)

~~ 15. If $\log _{x} 484-\log _{x} 4+\log _{x} 14641-\log _{x} 1331=3$, then the value of $x$ is

(a) 1

(b) 3

(d) None of these

(DCE 2008)

LEVEL-2

~~ 16. If $\log _{\sqrt{3}} 5=a$ and $\log _{\sqrt{3}} 2=b$, then $\log _{\sqrt{3}} 300$ is equal to

(a) $a+b+1$

(b) $2(a+b+1)$

(d) $(a+b+4)$ (Kerala 2007)

~~ 17. If $\log _7 2=\lambda$, then the value of $\log _{49}(28)$ is

(a) $\frac{1}{2}(2 \lambda+1)$

(b) $(2 \lambda+1)$

(d) $\frac{3}{2}(2 \lambda+1)$

(WBJEE 2011)

~~ 18. The value of $x$ satisfying $\log _2(3 x-2)=\log _{\frac{1}{2}} x$ is

(a) -1

(b) $-\frac{1}{3}$

(d) 1

(AMU 2011)

~~ 19. $\log _3 2, \log _6 2, \log _{12} 2$ are in

(a) A.P.

(b) G.P.

(d) None of these

(Raj PET 2006, 2001)

~~ 20. (i) The value of $\frac{\log _3 5 \times \log _{25} 27 \times \log _{49} 7}{\log _{81} 3}$ is

(a) 1

(b) $\frac{2}{3}$

(d) 6

(ii) $\log _{\sqrt[4]{4^{2}}}(\frac{1}{1024})$ is equal to

(WBJEE 2010)

(a) -5

(b) -3

(d) 5

(COMEDK 2010)

~~ 21. If $2^{\log _{10} 3 \sqrt{3}}=3^{k \log _{10} 2}$ then the value of $k$ is :

(a) 1

(b) $1 / 2$

(d) $\frac{3}{2}$

~~ 22. $\frac{1}{(\log _{a} b c)+1}+\frac{1}{(\log _{b} a c)+1}+\frac{1}{(\log _{c} a b)+1}$ is equal to :

(a) 1

(b) 2

(d) $a b c$

~~ 23. What is the value of $(\log _{1 / 2} 2)(\log _{1 / 3} 3)(\log _{1 / 4} 4)$ ….. $(\log _{1 / 1000} 1000)$ ?

(a) 1

(b) -1

(d) 0

(CDS 2007)

~~ 24. The value of $\log _{10} \sqrt{10 \sqrt{10 \sqrt{10 \sqrt{10}}}} \ldots \ldots .$. to $\infty$ is

(a) 4

(b) 3

(d) 1

~~ 25. If $\log _{a} m=x$, then $\log _{1 / a}(\frac{1}{m})$ equals

(a) $\frac{1}{x}$

(b) $-x$

(d) $x$

~~ 26. Find the value of $x$ if the base is 10 :

$5^{\log x}-3^{\log x-1}=3^{\log x+1}-5^{\log x-1}$

(a) 1

(b) 0

(d) 10

~~ 27. If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$, then $a^{a} b^{b} c^{c}$ equals :

(a) -1

(b) 0

(d) 1

~~ 28. The value of $\log _{\sqrt{b}} a \log _{\sqrt[3]{c}} b \log _{\sqrt[4]{a}} c$ is :

(a) 1

(b) 10

(d) 0

~~ 29. Evaluate $x$ if $\log _3(3+x)+\log _3(8-x)-\log _3(9 x-8)$

(a) 2

(b) -2

(d) -4

$=2-\log _3 9$

~~ 30. If $x=\log _{a} b c, y=\log _{b} c a, z=\log _{c} a b$, then

(a) $x y z=x+y+z+2$

(b) $x y z=x+y+z+1$

(d) $x y z=1$

LEVEL-3

~~ 31. Given, $\log _{a} x=\frac{1}{\alpha}, \log _{b} x=\frac{1}{\beta}, \log _{c} x=\frac{1}{\gamma}$, then $\log _{a b c} x$ equals :

(a) $\alpha \beta \gamma$

(b) $\frac{1}{\alpha \beta \gamma}$

(d) $\frac{1}{\alpha+\beta+\gamma}$

~~ 32. If $y=\frac{1}{a^{1-\log _{a} x}}, z=\frac{1}{a^{1-\log _{a} y}}$ and $x=a^{k}$, then $k=$

(a) $\frac{1}{a^{1-\log _{a} z}}$

(b) $\frac{1}{1-\log _{a} z}$

(d) $\frac{1}{1-\log _{z} a}$

~~ 33. Solve for $x$ if $a>0$ and $2 \log _{x} a+\log _{a x} a+3 \log _{a^{2} x} a=0$

(a) $a^{3 / 2}$

(b) $a^{1 / 2}$

(d) $a^{-4 / 3}$

~~ 34. Find the value of $x$, if $\log _2(5.2^{x}+1), \log _4(2^{1-x}+1)$ and 1 are in A.P.

(a) $1+\log _5 2$

(b) $1-\log _2 5$

(d) $\log _2 5+1$

(AIEEE 2002)

~~ 35. If $\frac{1}{3} \log _3 M+3 \log _3 N=1+\log _{0.008} 5$, then

(a) $M^{9}=\frac{9}{N}$

(b) $N^{9}=\frac{9}{M}$

(d) $N^{9}=\frac{3}{M}$

(CAT 2003)

~~ 36. If $\frac{\log x}{a^{2}+a b+b^{2}}=\frac{\log y}{b^{2}+b c+c^{2}}=\frac{\log z}{c^{2}+c a+a^{2}}$, then $x^{a-b} \cdot y^{b-c} \cdot z^{c-a}=$

(a) 0

(b) -1

(d) 2

~~ 37. If $x, y, z$ are distinct positive numbers different from 1 , such that $(\log _{y} x \cdot \log _{z} x-\log _{x} x)+(\log _{x} y \cdot \log _{z} y-\log _{y} y)$ $+(\log _{x} z \cdot \log _{y} z-\log _{z} z)=0$ then $x y z$ equals

(a) 100

(b) -1

(d) 1

~~ 38. If $a, b, c$ be the $p^{\text{th }}, q^{\text{th }}, r^{\text{th }}$ terms of a GP, then the value of $(q-r) \log a+(r-p) \log b+(p-q) \log c$ is :

(a) 0

(b) 1

(d) $p q r$

~~ 39. If $1, \log _9(3^{1-x}+2)$ and $\log _3(4.3^{x}-1)$ are in A.P., then $x$ is equal to

(a) $\log _4 3$

(b) $\log _3 4$

~~ 40. What is the sum, of ’ $n$ ’ terms in the series:

$\log m+\log (\frac{m^{2}}{n})+\log (\frac{m^{3}}{n^{2}})+\log (\frac{m^{4}}{n^{3}})+\ldots \ldots$.

(a) $\log [\frac{n^{(n-1)}}{m^{(n+1)}}]^{n / 2}$

(b) $\log [\frac{m^{m}}{n^{n}}]^{n / 2}$

(d) $\log [\frac{m^{(1+n)}}{n^{(n-1)}}]^{n / 2}$

~~ 41. Find $x$, if $\log _{2 x} \sqrt{x}+\log _{2 \sqrt{x}} x=0$ :

(CAT 2003)

(a) $1,2^{-5 / 6}$

(b) $1,2^{-6 / 5}$

(d) None of these

~~ 42. The number of solutions satisfying the given equation $x^{[(\log _3 x)^{2}-\frac{9}{2} \log _3 x+5]}=3 \sqrt{3}$ for $x \in R$ are :

(a) 0

(b) 1

(d) 3

~~ 43. Solve the following equations for $x$ and $y$. $\log _{100}|x+y|=\frac{1}{2}, \log _{10} y-\log _{10}|x|=\log _{100} 4$

(a) $(\frac{8}{3}, \frac{16}{3}),(-8,-16)$

(b) $(\frac{10}{3}, \frac{20}{3}),(+10,20)$

(d) None of these

~~ 44. If $\log (a+c)+\log (a-2 b+c)=2 \log (a-c)$, then $a, b, c$ are in

(a) A.P.

(b) G.P.

(d) None of these

~~ 45. If $5^{3 x^{2} \log _{10} 2}=2^{(x+1 / 2) \log _{10} 25}$, then the value of $x$ is :

(a) -1

(b) 2

(d) $-\frac{1}{3}$

~~ 46. The number $\log _2 7$ is :

(a) a prime number

(b) a rational number

(d) an integer (DCE 2000)

~~ 47. If $x, y, z$ are in G.P. and $(\log x-\log 2 y),(\log 2 y-\log 3 z)$ and $(\log 3 z-\log x)$ are in A.P., then $x, y, z$ are the lengths of the sides of a triangle which is :

(a) acute angled

(b) equilateral

(d) obtuse angled

(Rajasthan PET 2006)

~~ 48. In a right-angled triangle, the sides are $a, b$ and $c$ with $c$ as hypotenuse and $c-b \neq 1, c+b \neq 1$. Then the value of $[\frac{\log _{c+b} a+\log _{c-b} a}{2 \log _{c+b} a \times \log _{c-b} a}]$ is

(a) -1

(b) $\frac{1}{2}$

(d) 2

(WBJEE 2010)

~~ 49. The value of $6+\log _{\frac{3}{2}}[\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}}}} \ldots.$ is

(a) $\frac{8}{3 \sqrt{2}}$

(b) $\frac{4}{3}$

(d) 4

(IIT 2012)

~~ 50. The equation $x^{3 / 4(\log _2 x)^{2}+(\log _2 x)-5 / 4}=\sqrt{2}$ has

(a) at least one real solution

(b) exactly one irrational solution

(d) all of the above.

(IIT 1989)

ANSWERS

1. (i) (c)	$(i i)(d)(iii)(b)$	2. $(b)$	3. (i)(b) (ii)(b)	4. (c)	5. $(d)$	6. (c)	7. $(d)$	8. (d)	9. (a)
10. $(c)$	11. $(c)$	12. $(d)$	13. (i)(b)(ii)(c)	14. (a)	15. (c)	16. $(b)$	17. (a)	18. $(d)$	19. $(c)$
20. $(i)(c)$	(ii) $(b)$	21. $(d)$	22. (a)	23. $(b)$	24. (d)	25. $(d)$	26. $(c)$	27. $(d)$	28. $(c)$
29. (c)	30. (a)	31. $(d)$	32. (b)	33. $(d)$	34. (b)	35. $(b)$	36. (c)	37. $(d)$	38. (a)
39. $(d)$	40. $(d)$	41. $(b)$	42. (d)	43. $(b)$	44. $(c)$	45. $(d)$	46. $(c)$	47. $(d)$	48. (c)
49. $(d)$	50. (c)

HINTS AND SOLUTIONS

(i) $\log _{\pi}(\log _2(\log _7 x))=0$

$\Rightarrow(\log _2(\log _7 x)=\pi^{0}=1 \quad.$ [Using $\log _{a} m=x \Rightarrow m=a^{x}$ ]

$\Rightarrow \log _7 x=2^{1}=2 \Rightarrow x=7^{2}$.

~~ 2. Hint. $x=\log _{b} a \Rightarrow x=\frac{\log _{e} a}{\log _{e} b}, y=\log _{c} b \Rightarrow y=\frac{\log _{e} b}{\log _{e} c}$ $z=\log _{a} c \Rightarrow z=\frac{\log _{e} c}{\log _{e} a}$.

~~ 3. (i) $7^{2 \log _7 5}=7^{\log _7 5^{2}}=5^{2} \quad[\stackrel{.\text{ Using } a^{\log _{a} x}=x]}{ }.$

(ii) $7^{\log _7(x^{2}-4 x+5)}=x-1$

$\Rightarrow x^{2}-4 x+5=x-1$. Now, solve.

~~ 4. Hint. $\frac{1}{\log _3 12}+\frac{1}{\log _4 12}=\log _{12} 3+\log _{12} 4$

$ [\text{ Using } \log _{a} x=\frac{1}{\log _{x} a}] $

~~ 5. Hint. Given exp. $=\log _{10}{10^{4}(\frac{a+10 b+10^{2} c}{a+10 b+10^{2} c})}$

~~ 6. Given exp. $=\log 6+2 \log 5+\log 4-\log 3-\log 2$

$ \begin{aligned} & =\log 6+\log (5)^{2}+\log 4-\log 3-\log 2 \\ & =\log 6+\log 25+\log 4-(\log 3+\log 2) \\ & =\log (6 \times 25 \times 4)-\log (3 \times 2) \\ & =\log (\frac{6 \times 25 \times 4}{3 \times 2})=\log _{10} 100=\log _{10} 10^{2} \\ & =2 \log _{10} 10=2 \times 1=\mathbf{2} \end{aligned} $

~~ 7. $\log \frac{a^{2}}{b c}+\log \frac{b^{2}}{c a}+\log \frac{c^{2}}{a b}=\log (\frac{a^{2}}{b c} \times \frac{b^{2}}{a c} \times \frac{c^{2}}{a b})$

$ =\log (\frac{a^{2} b^{2} c^{2}}{a^{2} b^{2} c^{2}})=\log 1=\mathbf{0} . $

~~ 8. Given, $\log _{r} 6=m$ and $\log _{r} 3=n$

Since, $\log _{r} 6=\log _{r}(2 \times 3)=\log _{r} 2+\log _{r} 3$

$\Rightarrow \log _{r} 2+\log _{r} 3=m$

$\Rightarrow \log _{r} 2+n=m \Rightarrow \log _{r} 2=m-n$

Now, $\log _{r}(r / 2)=\log _{r} r-\log _{r} 2=1-(m-n)=\mathbf{1}-\boldsymbol{m}+\boldsymbol{n}$.

~~ 9. $25^{[(-\frac{1}{4} \log _5 25)]}=5^{.[2(-1 / 4) \log _5 25)]}$

$ =5^{(-1 / 2 \log _5 25)}=5^{\log _5(25)^{-1 / 2}}=25^{-1 / 2}=\frac{1}{5} $

$(\because a^{\log _{a} x}=x)$

~~ 10. Given, $\log _{10} x-\log _{10} \sqrt{x}=\frac{2}{\log _{10} x}$

$\Rightarrow \log _{10}(\frac{x}{\sqrt{x}})=\frac{2}{\log _{10} x}$

$\Rightarrow \log _{10} \sqrt{x}=\frac{2}{\log _{10} x} \Rightarrow \frac{1}{2} \log _{10} x=\frac{2}{\log _{10} x}$

$\Rightarrow(\log _{10} x)^{2}=4 \Rightarrow \log _{10} x= \pm 2$

If $\log _{10} x=+2$ then $x=10^{2}=\mathbf{1 0 0}$

If $\log _{10} x=-2$ then $x=10^{-2}=\mathbf{1} / \mathbf{1 0 0}$.

~~ 11. Hint. Given, $\log _4(2 \times 4 \times x \times 16)=6 \Rightarrow \log _4(128 x)=6$

$\Rightarrow 128 x=4^{6}$

[Using $\log _{a} x=n \Rightarrow x=a^{n}$ ]

~~ 12. Hint. $2^{x} \cdot 3^{x+4}=7^{x} \Rightarrow \log _{e}(2^{x} \cdot 3^{x+4})=\log _{e} 7^{x}$

Taking $\log$ to the same base on both sides

$\Rightarrow x \log 2+(x+4) \log 3=x \log 7$

$\Rightarrow x(\log 7-\log 2-\log 3)=4 \log 3$

~~ 13. (i) Given, $1000 !=n$. Now, $\frac{1}{\log _2 n}+\frac{1}{\log _3 n}+\ldots+\frac{1}{\log _{1000} n}$ $=\log _{n} 2+\log _{n} 3+\ldots+\log _{n} 1000 \quad[.$ Using $.\frac{1}{\log _{b} a}=\log _{a} b]$ $=\log _{n}(2 \times 3 \times 4 \times \ldots \times 1000)=\log _{n}(1000 !)=\log _{n} n=1$.

~~ 14. Hint. $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (suppose)

$ \begin{aligned} \Rightarrow & \log _{e} x=k(b-c) \Rightarrow x=e^{k(b-c)} \\ & \log _{e} y=k(c-a) \Rightarrow y=e^{k(c-a)} \\ & \log _{e} z=k(a-b) \Rightarrow z=e^{k(a-b)} \\ \therefore & x^{b+c} \cdot y^{c+a} \cdot z^{a+b} \\ & =e^{k(b-c)(b+c)} \cdot e^{k(c-a)(c+a)} \cdot e^{k(a-b)(a+b)} \end{aligned} $

Now, complete.

~~ 15. Hint. $\log _{x} 484-\log _{x} 4+\log _{x} 14641-\log _{x} 1331=3$

$\Rightarrow \log _{x}(2^{2} \times 11^{2})-\log _{x}(2^{2})+\log _{x}(11^{4})-\log _{x}(11^{3})=3$

$\Rightarrow 2 \log _{x} 2+2 \log _{x} 11-2 \log _{x} 2+4 \log _{x} 11-3 \log _{x} 11=3$

$\Rightarrow 3 \log _{x} 11=3 \Rightarrow \log _{x} 11=1 \Rightarrow x^{1}=11 \Rightarrow x=11$.

~~ 16. Hint. $\log _{\sqrt{3}} 300=\log _{\sqrt{3}}[(\sqrt{3})^{2} \cdot 10^{2}]$

$=2 \log _{\sqrt{3}} \sqrt{3}+2 \log _{\sqrt{3}} 10=2+2 \log _{\sqrt{3}}(2 \times 5)$

~~ 17. $\log _{49}(28)=\log _{7^{2}}(7 \times 2^{2})=\log _{7^{2}} 7+\log _{7^{2}} 2^{2}$

$=\frac{1}{2} \log _7 7+\frac{2}{2} \log _7 2$

$[.$ Using $.\log _{a^{n}} x=\frac{1}{n} \log _{a} x, \log _{a^{n}}(x^{m})=\frac{m}{n} \log _{a} x]$

$=\frac{1}{2}+\lambda=\frac{1}{2}(2 \lambda+1)$.

~~ 18. Given, $\log _2(3 x-2)=\log _{\frac{1}{2}} x \Rightarrow \log _2(3 x-2)=\log _{2^{-1}} x$ $\Rightarrow \log _2(3 x-2)=\frac{1}{-1} \log _2 x \quad[.$ Using $.\log _{a^{n}} x=\frac{1}{n} \log _{a} x]$ $\Rightarrow \log _2(3 x-2)=(-1) \log _2 x=\log _2 x^{-1}$

[Using $n \log _{a} x=\log _{a} x^{n}$ ]

$\Rightarrow(3 x-2)=x^{-1}=\frac{1}{x} \Rightarrow 3 x^{2}-2 x-1=0$

$\Rightarrow(3 x+1)(x-1)=0 \Rightarrow x=-\frac{1}{3}$ or 1

$\Rightarrow x=1$, since $\log _2(3 x-2)$ is not defined when $x=-\frac{1}{3}$.

~~ 19. Since $\log _2 3+\log _2 12=\log _2(3 \times 12)=\log _2 36=\log _2 6^{2}$ $=2 \log _2 6$, therefore,

$\log _2 3, \log _2 6$ and $\log _2 12$ in A.P.

$\Rightarrow \frac{1}{\log _2 3}, \frac{1}{\log _2 6}, \frac{1}{\log _2 12}$ and in H.P. $\Rightarrow \log _3 2, \log _6 2, \log _{12} 2$ and in H.P.

$[.$ Using $.\log _{a} x=\frac{1}{\log _{x} a}]$

~~ 20. (i) $\frac{\log _3 5 \times \log _{25} 27 \times \log _{49} 7}{\log _{81} 3}=\frac{\log _3 5 \times \log _{5^{2}} 3^{3} \times \log _{7^{2}} 7}{\log _{3^{4}} 3}$

$=\frac{\log _3 5 \times \frac{3}{2} \log _5 3 \times \frac{1}{2} \log _7 7}{\frac{1}{4} \log _3 3}$

$[.$ Using $.\log _{a^{n}} x^{m}=\frac{m}{n} \log _{a} x, \log _{a^{n}} x=\frac{1}{n} \log _{a} x]$

$=3(\log _3 5 \times \log _5 3)=3 \times 1=3$ [Using $\log _{b} a \times \log _{a} b=1$ ]

(ii) Hint. Since $4 \cdot \sqrt[3]{4^{2}}=2^{2} \cdot(2^{4})^{\frac{1}{3}}=2^{2+\frac{4}{3}}=2^{\frac{10}{3}}$ and $1024=2^{10}$, therefore,

$ \begin{aligned} & \log _{4 \cdot \sqrt[3]{4^{2}}}(\frac{1}{1024})=\log _{2^{10 / 3}}(2^{-10})=\frac{-10}{\frac{10}{3}} \log _2 2 \\ & =-3 \times 1=-3 . \quad[\text{ Using } \log _{a^{n}} x^{m}=\frac{m}{n} \log _{a} x] . \end{aligned} $

~~ 21. $2^{\log _{10} 3 \sqrt{3}}=3^{k \log _{10} 2} \Rightarrow 2^{\log _{10}(3^{3 / 2})}=3^{k \log _{10} 2}$

$\Rightarrow 2^{\log _2(3^{3 / 2}) \cdot \log _{10} 2}=3^{k \log _{10} 2} \quad$ [Using $.\log _{a} x=\log _{b} x \cdot \log _{a} b]$

$\Rightarrow[2^{\log _2(3^{3 / 2})}]^{\log _{10} 2}=(3^{k})^{\log _{10} 2} \Rightarrow 2^{\log _2 3^{3 / 2}}=3^{k}$

$\Rightarrow 3^{3 / 2}=3^{k} \Rightarrow k=\frac{\mathbf{3}}{\mathbf{2}}$

$(\because a^{\log _{a} x}=x)$

~~ 22. $\frac{1}{\log _{a} b c+1}+\frac{1}{\log _{b} a c+1}+\frac{1}{\log _{c} a b+1}$

$ \begin{aligned} & =\frac{1}{\log _{a} b c+\log _{a} a}+\frac{1}{\log _{b} a c+\log _{b} b}+\frac{1}{\log _{c} a b+\log _{c} c} \\ & =\frac{1}{\log _{a}(a b c)}+\frac{1}{\log _{b}(a b c)}+\frac{1}{\log _{c}(a b c)} \\ & =\log _{a b c}(a)+\log _{a b c}(b)+\log _{a b c}(c)=\log _{a b c} a b c=1 \end{aligned} $

$ (\text{ Using } \log _{a} b=\frac{1}{\log _{b} a}) $

~~ 23. $(\log _{1 / 2} 2)(\log _{1 / 3} 3)(\log _{1 / 4} 4) \ldots \ldots \ldots(\log _{\frac{1}{1000}} 1000)$

$ \begin{aligned} & =(\frac{\log 2}{\log \frac{1}{2}})(\frac{\log 3}{\log \frac{1}{3}})(\frac{\log 4}{\log \frac{1}{4}}) \cdots \ldots \ldots .(\frac{\log 1000}{\log \frac{1}{1000}}) \\ & =(\frac{\log 2}{-\log 2})(\frac{\log 3}{-\log 3})(\frac{\log 4}{-\log 4}) \cdots \ldots \ldots . .(\frac{\log 1000}{-\log 1000}) \end{aligned} $

$(\because \log \frac{1}{2}=\log 1-\log 2=0-\log 2=-\log 2.$ and similarly for others)

$=(-1) \times(-1) \times(-1) \times \ldots \ldots \ldots \times(-1)=-1$

( $\because$ Number of terms is odd )

~~ 24. Given exp. $=\log _{10}(10^{1 / 2} 10^{1 / 4} 10^{1 / 8}.$ to $\infty$ )

$ \begin{aligned} & =\log _{10} 10^{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \ldots \ldots . \text{ to } \infty)} \\ & =(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots . . \text{ to } \infty) \cdot \log _{10} 10 \\ & {[\text{ Using } \log _{a} x^{n}=n \log _{a} x]} \\ & =\frac{1 / 2}{(1-1 / 2)} \times 1=1 \\ & (\text{ Using sum of GP of infinite terms }=\frac{a}{1-r}) \end{aligned} $

~~ 25. $\log _{a} m=x \Rightarrow a^{x}=m$

$\log _{1 / a} 1 / m=y \Rightarrow(1 / a)^{y}=1 / m \Rightarrow m=a^{y} \Rightarrow a^{y}=a^{x} \Rightarrow y=x$

~~ 26. $5^{\log x}-3^{\log x-1}=3^{\log x+1}-5^{\log x-1}$

$\Rightarrow 5^{\log x}-3^{\log x} \times 3^{-1}=3^{\log x} \cdot 3-5^{\log x} \cdot 5^{-1}$

$\Rightarrow 5^{\log x}-\frac{1}{3} \times 3^{\log x}=3 \times 3^{\log x}-\frac{1}{5} \times 5^{\log x}$

$\Rightarrow(3+\frac{1}{3}) 3^{\log x}=(1+\frac{1}{5}) 5^{\log x} \Rightarrow \frac{10}{3} \times 3^{\log x}=\frac{6}{5} \times 5^{\log x}$

$\Rightarrow \frac{3^{\log x}}{5^{\log x}}=\frac{6}{5} \times \frac{3}{10}=\frac{9}{25} \Rightarrow(\frac{3}{5})^{\log x}=(\frac{3}{5})^{2}$

$\Rightarrow \log _{10} x=2 \Rightarrow x=10^{2}=\mathbf{1 0 0}$

~~ 27. Let $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k$

$\Rightarrow \log a=k(b-c), \log b=k(c-a), \log c=k(a-b)$

Now let $a^{a} b^{b} c^{c}=p$. Then,

$\log p=\log a^{a}+\log b^{b}+\log c^{c}=a \log a+b \log b+c \log c$

$ \begin{aligned} & =a \times k(b-c)+b \times k(c-a)+c \times k(a-b) \\ & =k(a b-a c+b c-b a+c a-c b)=0 \end{aligned} $

$\Rightarrow \log p=\log 1$

(Putting $\log 1$ for 0 )

$\Rightarrow p=1 \Rightarrow a^{a} b^{b} c^{c}=\mathbf{1}$.

~~ 28. Using the formula $\log _{a^{n}} x^{m}=\frac{m}{n} \log _{a} x$, we have

$\log _{\sqrt{b}} a \log _{\sqrt[3]{c}} b \log _{\sqrt[4]{a}} c=\log _{b^{1 / 2}} a \log _{c^{1 / 3}} b \log _{a^{1 / 4}} c$

$ \begin{aligned} & =2 \log _{b} a \times 3 \log _{c} b \times 4 \log _{a} c \\ & =24 \frac{\log a}{\log b} \times \frac{\log b}{\log c} \times \frac{\log c}{\log a}=\mathbf{2 4 .} \end{aligned} $

~~ 29. $\log _3(3+x)+\log _3(8-x)-\log _3(9 x-8)=2-\log _3 9$

$\Rightarrow \log _3(3+x)+\log _3(8-x)-\log _3(9 x-8)+\log _3 9=2$

$\Rightarrow \log _3[\frac{(3+x)(8-x)(9)}{(9 x-8)}]=2$

$\Rightarrow \frac{9(24+8 x-3 x-x^{2})}{(9 x-8)}=3^{2}=9$

$\Rightarrow-x^{2}+5 x+24=9 x-8 \Rightarrow x^{2}+4 x-32=0$

$\Rightarrow(x+8)(x-4)=0 \Rightarrow x=-8,4$.

Taking the positive value $x=4$.

~~ 30. $x=\log _{a} b c \Rightarrow a^{x}=b c \Rightarrow a^{x+1}=a b c \Rightarrow a=(a b c)^{1 /(x+1)}$

Similarly, $b=(a b c)^{1 /(y+1)}, c=(a b c)^{1 /(z+1)}$

$ \begin{aligned} \therefore & a b c=(a b c)^{\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}} \\ \Rightarrow & \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1 \Rightarrow(y+1)(z+1)+(x+1) \\ & (z+1)+(x+1)(y+1)=(x+1)(y+1)(z+1) \\ \Rightarrow & y z+y+z+1+x z+x+z+1+x y+y+x+1=x y z+x y \\ & +y z+z x+x+y+z+1 \\ \Rightarrow & \boldsymbol{x}+\boldsymbol{y}+z+\mathbf{2}=\boldsymbol{x y z} . \end{aligned} $

~~ 31. $\alpha=\frac{1}{\log _{a} x}, \beta=\frac{1}{\log _{b} x}, \gamma=\frac{1}{\log _{c} x}$

$\Rightarrow \alpha=\log _{x} a, \beta=\log _{x} b, \gamma=\log _{x} c$

$\Rightarrow \alpha+\beta+\gamma=\log _{x} a+\log _{x} b+\log _{x} c=\log _{x}(a b c)$

$\Rightarrow \frac{1}{\alpha+\beta+\gamma}=\frac{1}{\log _{x}(a b c)}=\log _{a b c} x$.

~~ 32. $y=\frac{1}{a^{1-\log _{a} x}}=a^{-(1-\log _{a} x)}$

$\Rightarrow \log _{a} y=\frac{1}{1-\log _{a} x}$ and $\log _{a} z=\frac{1}{1-\log _{a} y}$

$\therefore \log _{a} z=\frac{1}{1-(\frac{1}{1-\log _{a} x})}=\frac{1-\log _{a} x}{-\log _{a} x}$

$\Rightarrow-\log _{a} z=-1+\frac{1}{\log _{a} x} \Rightarrow \frac{1}{\log _{a} x}=1-\log _{a} z$

$\Rightarrow \log _{a} x=\frac{1}{1-\log _{a} z}$

$\Rightarrow x=a^{\frac{1}{1-\log _{a} z}}=a^{k} \Rightarrow k=\frac{1}{1-\log _{a} z}$.

~~ 33. Since $\log _{a x} a=\frac{1}{\log _{a} a x}=\frac{1}{\log _{a} a+\log _{a} x}=\frac{1}{1+\log _{a} x}$ and

$ \begin{aligned} \log _{a^{2} x} a & =\frac{1}{\log _{a} a^{2} x}=\frac{1}{\log _{a} a^{2}+\log _{a x} x}=\frac{1}{2 \log _{a} a+\log _{a x} x} \\ & =\frac{1}{2+\log _{a} x} \end{aligned} $

Given, $2 \log _{x} a+\log _{a x} a+3 \log _{a^{2} x} a=0$

$ \Rightarrow \frac{2}{\log _{a} x}+\frac{1}{1+\log _{a} x}+\frac{3}{2+\log _{a} x}=0 $

Now let $\log _{a} x=t$, then $\frac{2}{t}+\frac{1}{1+t}+\frac{3}{2+t}=0$

$\Rightarrow 2(1+t)(2+t)+t(2+t)+3 t(1+t)=0$

$\Rightarrow 2(2+2 t+t+t^{2})+2 t+t^{2}+3 t+3 t^{2}=0$

$\Rightarrow 4+6 t+2 t^{2}+2 t+t^{2}+3 t+3 t^{2}=0$

$\Rightarrow 6 t^{2}+11 t+4=0$

$\Rightarrow(3 t+4)(2 t+1)=0 \Rightarrow t=-1 / 2,-4 / 3$

When $t=-1 / 2, \log _{a} x=-\frac{1}{2} \Rightarrow x=a^{-1 / 2}$

When $t=-\frac{4}{3}, \log _{a} x=-\frac{4}{3} \Rightarrow \boldsymbol{x}=\boldsymbol{a}^{-4 / 3}$

~~ 34. Given, $\log _2(5.2^{x}+1), \log _4(2^{1-x}+1), 1$ are in A.P.

$\Rightarrow \log _2(5.2^{x}+1)+1=2 \log _4(2^{1-x}+1)$

$\Rightarrow \log _2(5.2^{x}+1)+\log _2 2=2 \log _{2^{2}}(2^{1-x}+1)$

$\Rightarrow \log _2(5.2^{x}+1) .2=2 \times \frac{1}{2} \log _2(2^{1-x}+1)$

$\Rightarrow \log _2(10.2^{x}+2)=\log _2(2^{1-x}+1)$

$ (\because \log _{a^{n}} x=\frac{1}{n} \log _{a} x) $

$\Rightarrow 10.2^{x}+2=2^{1-x}+1 \Rightarrow 10.2^{x}+2=\frac{2}{2^{x}}+1$

Let $2^{x}=a$, then

$a+2=\frac{2}{a}+1 \Rightarrow 10 a+1=\frac{2}{a} \Rightarrow 10 a^{2}+a-2=0$

$\Rightarrow(5 a-2)(2 a+1)=0 \Rightarrow a=\frac{2}{5} \Rightarrow 2^{x}=\frac{2}{5}$

$(\because 2^{x}>0.$, reject $.a=-\frac{1}{2})$

$\Rightarrow \log 2^{x}=\log \frac{2}{5}$

$\Rightarrow x \log _2 2=\log _2 2-\log _2 5 \Rightarrow \boldsymbol{x}=\mathbf{1}-\log _2 \mathbf{5}$.

~~ 35. $\frac{1}{3} \log _3 M+3 \log _3 N=1+\log _{0.008} 5$

$\Rightarrow \log _3 M^{1 / 3}+\log _3 N^{3}=1+\log _{0.008} 5$

$\Rightarrow \log _3 M^{1 / 3} N^{3}=1+\log _{0.008} 5$

$\Rightarrow M^{1 / 3} N^{3}=3^{(1+\log _{0.008} 5)}$

$\Rightarrow M^{1 / 3} N^{3}=3^{1} \cdot 3^{\log _{0.008} 5}$

$\Rightarrow N^{9}=\frac{27}{M}(3^{3 \log 0.008^{5}})$

$\Rightarrow N^{9}=\frac{27}{M}(3^{\log _{(0.2)^{3}}(5^{3})})$

$\Rightarrow N^{9}=\frac{27}{M}(3^{\log _{0.2} 5}) \quad[\because \log _{a^{n}} x^{m}=\frac{m}{n} \log _{a} x]$

$\Rightarrow N^{9}=\frac{27}{M}(3^{\log _{1 / 5} 5})=\frac{1}{M}(27)(3^{-1})=\frac{\mathbf{9}}{\boldsymbol{M}}$.

~~ 36. Let each ratio $=k$ and base $=e$

$\Rightarrow \log _{e} x=k(a^{2}+a b+b^{2})$

$\Rightarrow(a-b) \log _{e} x=k(a-b)(a^{2}+a b+b^{2})$

$\Rightarrow \log _{e} x^{a-b}=k(a^{3}-b^{3}) \Rightarrow x^{a-b}=e^{k(a^{3}-b^{3})}$

Similarly, $y^{b-c}=e^{k(b^{3}-c^{3})}, z^{c-a}=e^{k(c^{3}-a^{3})}$

$\therefore x^{a-b} \cdot y^{b-c} \cdot z^{c-a}=e^{k(a^{3}-b^{3})} \cdot e^{k(b^{3}-c^{3})} \cdot e^{k(c^{3}-a^{3})}$

$ =e^{k[a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}]}=e^{0}=\mathbf{1} \text{. } $

~~ 37. $\log _{y} x \cdot \log _{z} x-\log _{x} x=\frac{\log x}{\log y} \cdot \frac{\log x}{\log z}-1=\frac{(\log x)^{2}}{\log y \cdot \log z}-1$

Similarly, $\log _{x} y \cdot \log _{z} y-\log _{y} y=\frac{(\log y)^{2}}{\log x \cdot \log z}-1$ and

$\log _{x} z \cdot \log _{y} z-\log _{z} z=\frac{(\log z)^{2}}{\log x \cdot \log y}-1$

$ \begin{aligned} & \therefore \text{ LHS }=\frac{(\log x)^{2}}{\log y \cdot \log z}-1+\frac{(\log y)^{2}}{\log z \cdot \log x}-1+\frac{(\log z)^{2}}{\log x \cdot \log y}-1 \\ &=\frac{(\log x)^{3}+(\log y)^{3}+(\log z)^{3}-3 \log x \cdot \log y \cdot \log z}{\log x \cdot \log y \cdot \log z}=0 \end{aligned} $

(given)

$\Rightarrow(\log x)^{3}+(\log y)^{3}+(\log z)^{3}-3 \log x \cdot \log y \cdot \log z=0$

$\Rightarrow \log x+\log y+\log z=0$

(if $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ )

$\Rightarrow \log x y z=0 \Rightarrow x y z=1$.

~~ 38. Let $h$ be the first term and $k$ be the common ratio of a GP, then

$ a=h k^{p-1}, \quad b=h k^{q-1}, \quad c=h k^{r-1} $

$\therefore(q-r) \log a+(r-p) \log b+(p-q) \log c$

$ \begin{aligned} & =\log [h k^{p-1}]^{q-r}+\log [h k^{q-1}]^{r-p}+\log [h k^{r-1}]^{p-q} \\ & =\log (h^{q-r+r-p+p-q})(k^{p-1})^{q-r}(k^{q-1})^{r-p}(k^{r-1})^{p-q} \\ & =\log (h^{\mathbf{o}} k^{0})=\log 1=\mathbf{0} \end{aligned} $

~~ 39. $1, \log _9(3^{1-x}+2), \log _3(4.3^{x}-1)$ are in A.P.

$\Rightarrow \log _3 3, \log _3(3^{1-x}+2)^{1 / 2}, \log _3(4.3^{x}-1)$ are in A.P.

$\Rightarrow 3,(3^{1-x}+2)^{1 / 2},(4.3^{x}-1)$ are in G.P.

Since $\log _9(3^{1-x}+2)=\log _{3^{2}}(3^{1-x}+2)=\frac{1}{2} \log _3(3^{1-x}+2)$,

using $.\log _{a^{n}} x=\frac{1}{n} \log _{a} x=\log _3(3^{1-x}+2)^{\frac{1}{2}}]$

$\Rightarrow[[3^{1-x}+2]^{1 / 2}]^{2}=3 \cdot(4 \cdot 3^{x}-1)$

$\Rightarrow 3^{1-x}+2=4.3^{x+1}-3$

$\Rightarrow 4.3^{x+1}-3^{1-x}=5 \Rightarrow 12.3^{x}-\frac{3}{3^{x}}=5$

Let $3^{x}=y$, then $12 y-\frac{3}{y}=5 \Rightarrow 12 y^{2}-5 y-3=0$

$\Rightarrow(3 y+1)(4 y-3)=0 \Rightarrow y=-\frac{1}{3}, \frac{3}{4}$

$\therefore$ Rejecting the negative value, we have $3^{x}=\frac{3}{4}$

$\Rightarrow x=\log _3 \frac{3}{4}$.

~~ 40. $S=\log m+\log \frac{m^{2}}{n}+\log \frac{m^{3}}{n^{2}}+$ .$n$ terms

$=\log [m \cdot \frac{m^{2}}{n} \cdot \frac{m^{3}}{n^{2}} \ldots \ldots \ldots \frac{m^{n}}{n^{n-1}}]=\log [\frac{m^{(1+2+3+4+\ldots \ldots \ldots+n)}}{n^{(1+2+3+\ldots \ldots . . .(n-1)}}]$

$=\log [\frac{m^{\frac{n(n+1)}{2}}}{n}]=\log [\frac{\boldsymbol{m}^{\boldsymbol{n}+\mathbf{1}}}{\boldsymbol{n}^{\boldsymbol{n}-\mathbf{1}}}]^{\boldsymbol{n} / \mathbf{2}}$.

~~ 41. $\log _{2 x} \sqrt{x}+\log _{2 \sqrt{x}} x=0$

Let $\log _2 x=t$. Then,

$ \log _{2 x} \sqrt{x}=\frac{\log _2 \sqrt{x}}{\log _2 2 x}=\frac{\frac{1}{2} \log _2 x}{\log _2 2+\log _2 x}=\frac{t / 2}{1+t} $

$ \log _{2 \sqrt{x}} x=\frac{\log _2 x}{\log _2 2 \sqrt{x}}=\frac{\log _2 x}{\log _2 2+\frac{1}{2} \log _2 x}=\frac{t}{1+t / 2} $

$\therefore \quad$ Substituting in $(i)$, we get

$ \frac{t / 2}{1+t}+\frac{t}{1+t / 2}=0 \Rightarrow \frac{t}{2+2 t}+\frac{2 t}{2+t}=0 $

$\Rightarrow t(2+t)+2 t(2+2 t)=0 \Rightarrow 2 t+t^{2}+4 t+4 t^{2}=0$

$\Rightarrow 5 t^{2}+6 t=0 \Rightarrow t(5 t+6)=0 \Rightarrow t=0$ or $-\frac{6}{5}$

$\Rightarrow \log _2 x=0 \Rightarrow x=2^{0}=1$ and $\log _2 x=-\frac{6}{5} \Rightarrow x=2^{-6 / 5}$

$\therefore x=1$ or $2^{-6 / 5}$

~~ 42. Taking $\log$ of both the sides to base 3 , we have,

$[(\log _3 x)^{2}-\frac{9}{2} \log _3 x+5] \log _3 x=\log _3 3^{3 / 2}=\frac{3}{2}(\because \log _3 3=1)$

$\Rightarrow 2(\log _3 x)^{3}-9(\log _3 x)^{2}+10 \log _3 x-3=0$

$\Rightarrow 2 y^{3}-9 y^{2}+10 y-3=0$

(Take $\log _3 x=y$ )

$\Rightarrow(y-1)(y-3)(2 y-1)=0$

(Factorising)

$\Rightarrow(\log _3 x-1)(\log _3 x-3)(2 \log _3 x-1)=0$

$\Rightarrow \log _3 x=1, \log _3 x=3,2 \log _3 x=1 \Rightarrow x=3^{1}, x=3^{3}, x^{2}=3^{1}$

$\Rightarrow x=(3,27, \sqrt{3})$

$\therefore$ There are three solutions.

~~ 43. $\log _{100}|x+y|=\frac{1}{2} \Rightarrow|x+y|=100^{\frac{1}{2}}$

$\Rightarrow|x+y|=10$ as ( -10 is inadmissible)

$\log _{10} y-\log _{10}|x|=\log _{100} 4$

$\Rightarrow \log _{10} \frac{y}{|x|}=\log _{10^{2}} 2^{2}=\log _{10} 2$

$[.$ Using $\log _{a^{n}}(x^{m})=\frac{m}{n} \log _{a} x$

$\Rightarrow \frac{y}{|x|}=2 \Rightarrow y=2|x|$

Substituting the value of $y$ from (ii) in (i), we get

$ |x+2| x||=10 $

If $x>0$, then $3 x=10 \Rightarrow x=\frac{10}{3}$

If $x<0$, then $x=10$.

$\therefore$ If $x=\frac{10}{3}$, then $y=\frac{20}{3}$ and if $x=10, y=20$.

~~ 44. Given, $\log (a+c)+\log (a-2 b+c)=2 \log (a-c)$

$\Rightarrow \log (a+c)(a-2 b+c)=\log (a-c)^{2}$

$\Rightarrow(a+c)(a-2 b+c)=(a-c)^{2}$

$\Rightarrow \not \not^{2}+c a-2 b a-2 b c+a c+\not \not^{2}=\not \alpha^{2}-2 a c+\not \not^{2}$

$\Rightarrow 4 a c=2 b a+2 b c \Rightarrow 2 a c=b(a+c)$

$\therefore b=\frac{2 a c}{a+c} \Rightarrow a, b, c$ are in H.P.

~~ 45. $5^{3 x^{2} \log _{10} 2}=2^{(x+\frac{1}{2}) \log _{10} 25}$

$\Rightarrow 5^{3 x^{2} \log _{10} 2}=2^{(\frac{2 x+1}{2}) \times 2 \log _{10} 5}=2^{(2 x+1) \log _{10} 5}$

$\Rightarrow 5^{3 x^{2} \log _{10} 2}=2^{(2 x+1) \log _2 5 \cdot \log _{10} 2}$

$\Rightarrow 5^{3 x^{2} \log _{10} 2}=[2^{\log _2 5^{(2 x+1)}}]^{\log _{10} 2}$

(using $\log _{a} x=\log _{b} x \cdot \log _{a} b$ )

$\Rightarrow(5^{3 x^{2}})^{\log _{10} 2}=(5^{2 x+1})^{\log _{10} 2} \quad[.$ Using $.a^{\log _{a} x}=x]$

$\Rightarrow 3 x^{2}=2 x+1 \Rightarrow 3 x^{2}-2 x-1=0$

$\Rightarrow(x-1)(3 x+1)=0$

$\Rightarrow x=1,-\frac{1}{3}$.

~~ 46. Let us assume $\log _2 7$ be a rational number. Then,

$\log _2 7=\frac{p}{q}$, where $p, q \in I$ and $q \neq 0$

$\Rightarrow 2^{p / q}=7 \Rightarrow 2^{p}=7^{q}$

This is not true as 2 is even and 7 is odd.

$\therefore$ Hence our assumption that $\log _2 7$ is a rational number is wrong.

$\therefore \log _2 7$ is an irrational number.

~~ 47. $x, y, z$ are in G.P. $\Rightarrow y^{2}=x z$

$(\log x-\log 2 y),(\log 2 y-\log 3 z)$ and $(\log 3 z-\log x)$ are in A.P. $\Rightarrow 2(\log 2 y-\log 3 z)=(\log x-\log 2 y)+(\log 3 z-\log x)$

$\Rightarrow 3 \log 2 y=3 \log 3 z \Rightarrow \log 2 y=\log 3 z \Rightarrow \boldsymbol{y}=\frac{\mathbf{3}}{\mathbf{2}} \boldsymbol{z}$.

$\therefore$ Putting the value of $y$ in $(i)$, we have

$(\frac{3}{2} z)^{2}=x z \Rightarrow x=\frac{9}{4} z$.

Now, by the cosine rule of triangles,

$\cos A=\frac{y^{2}+z^{2}-x^{2}}{2 y z}$,

where $x$ is the length of the

side opposite $\angle A$.

$ \begin{aligned} \Rightarrow \cos A & =\frac{(\frac{3}{2} z)^{2}+z^{2}-(\frac{9}{4} z)^{2}}{2 \times \frac{3}{2} z \times z}=\frac{\frac{9}{4} z^{2}+z^{2}-\frac{81}{16} z^{2}}{3 z^{2}} \\ & =\frac{\frac{9}{4}+1-\frac{81}{16}}{3}=\frac{1}{3} \times[\frac{36+16-81}{16}]=-\frac{29}{48}<0 \end{aligned} $

$\because \cos A$ is less than 0 , i.e, negative, $\angle A$ is obtused and the triangle is obtuse angled.

~~ 48. In a right angled triangle with $a, b$ as sides and $c$ as hypotenuse,

$ c^{2}=a^{2}+b^{2} $

(Pythagoras’ Theorem)

Now, given expression $=\frac{\log _{c+b} a+\log _{c-b} a}{2 \times \log _{c+b} a \times \log _{c-b} a}$

$=\frac{1}{2}[\frac{1}{\log _{c-b} a}+\frac{1}{\log _{c+b} a}]$ $=\frac{1}{2}[\log _{a}(c-b)+\log _{a}(c+b)]=\frac{1}{2}[\log _{a}[(c-b)(c+b)].$

$=\frac{1}{2}[\log _{a}(c^{2}-b^{2})]=\frac{1}{2} \log _{a} a^{2}=\log _{a} a=\mathbf{1}$.

~~ 49. Let

$\log _{\frac{3}{2}}[\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}}}} \ldots.$

$\Rightarrow p=\sqrt{4-\frac{1}{3 \sqrt{2}} p} \Rightarrow p^{2}=4-\frac{1}{3 \sqrt{2}} p \Rightarrow p^{2}+\frac{1}{3 \sqrt{2}} p-4=0$

$\Rightarrow p=\frac{-\frac{1}{3 \sqrt{2}} \pm \sqrt{\frac{1}{18}+16}}{2}=\frac{-\frac{1}{3 \sqrt{2}} \pm \frac{17}{3 \sqrt{2}}}{2}$

(Applying the formula for roots of Q.E.)

$\Rightarrow p=\frac{16}{3 \times 2 \sqrt{2}}$ or $\frac{-18}{3 \times 2 \sqrt{2}}=\frac{8}{3 \sqrt{2}}$ or $-\frac{3}{\sqrt{2}}$

Neglecting $p=\frac{-3}{\sqrt{2}}$ as $p \geq 0$, we have $p=\frac{8}{3 \sqrt{2}}$

$\therefore A=6+\log _{\frac{3}{2}}(\frac{1}{3 \sqrt{2}} \times \frac{8}{3 \sqrt{2}})=6+\log _{\frac{3}{2}}(\frac{4}{9})$

$ =6+\log _{\frac{3}{2}}(\frac{3}{2})^{-2}=6-2 \log _{\frac{3}{2}} 3 / 2=6-2=4 . $

~~ 50. Given, $x^{3 / 4(\log _2 x)^{2}}+\log _2 x-5 / 4=\sqrt{2}$

Taking $\log$ to the base 2 of both the sides, we have

$ \begin{aligned} {[\frac{3}{4}(\log _2 x)^{2}+(\log _2 x)-5 / 4] } & \log _2 x=\log _2 \sqrt{2} \\ & =\log _2 2^{1 / 2}=\frac{1}{2} \log _2 2=\frac{1}{2} \end{aligned} $

Let us assume $\log _2 x=a$. Then,

$(\frac{3}{4} a^{2}+a-\frac{5}{4}) a=\frac{1}{2} \Rightarrow 3 a^{3}+4 a^{2}-5 a=2$

$\Rightarrow 3 a^{3}+4 a^{2}-5 a-2=0$.

Using hit and trial method check for $a=1$.

$f(a)=3 a^{3}+4 a^{2}-5 a-2 \Rightarrow f(1)=3.1^{3}+4.1^{2}-5.1-2=0$

$\therefore(a-1)$ is a factor of $3 a^{3}+4 a^{2}-5 a-2$

$\therefore$ Now by dividing $3 a^{3}+4 a^{2}-5 a-2$ by $(a-1)$, we get

$3 a^{3}+4 a^{2}-5 a-2=(a-1)(3 a+1)(a+2)=0$

$\Rightarrow a=1$ or $a=-\frac{1}{3}$ or $a=-2$

$\Rightarrow \log _2 x=1$ or $\log _2 x=-\frac{1}{3}$ or $\log _2 x=-2$

$\Rightarrow x=2^{1}=2$ or $x=2^{-1 / 3}$ or $x=2^{-2}=\frac{1}{4}$

$\therefore$ The given equation has exactly three real solutions, wherein $x=2^{-1 / 3}$ is irrational.

SELF ASSESSMENT SHEET

If $\log _2[\log _7(x^{2}-x+37)]=1$, then what could be the value of $x$ ?

(a) 3

(b) 5

(d) None of these

(CAT 1997)

~~ 2. $\log _{\sqrt{3}} \sqrt{3 \sqrt{3 \sqrt{3 \sqrt{3}}}}=$

(a) $\frac{31}{32}$

(b) $\frac{15}{16}$

(d) $\frac{15}{8}$

~~ 3. Find the sum of ’ $n$ ’ terms of the series: $\log _2(\frac{x}{y})+\log _4(\frac{x}{y})^{2}+\log _8(\frac{x}{y})^{3}+\log _{16}(\frac{x}{y})^{4}+$

(a) $\log _2(\frac{x}{y})^{4 n}$

(b) $n \log _2(\frac{x}{y})$

(d) $\frac{1}{2} \log _2(\frac{x}{y})^{n}$

~~ 4. If $\log _{x} a, a^{x / 2}$ and $\log _{b} x$ are in $G P$, then $x$ is equal to :

(a) $\log _{a}(\log _{b} a)$

(b) $\log _{a}(\log _{e} a)-\log _{a}(\log _{e} b)$

(d) both (a) and (b)

~~ 5. If $\frac{\log x}{l+m-2 n}=\frac{\log y}{m+n-2 l}=\frac{\log z}{n+l-2 m}$, then $x y z$ is equal to : (a) 0

(b) 1

(d) 2

~~ 6. If $a=\log _{12} m$ and $b=\log _{18} m$, then $\frac{a-2 b}{b-2 a}$ equals

(a) $\log _3 2$

(b) $\log _2 3$

(d) 1

~~ 7. If $x=\log _{2 a} a, y=\log _{3 a} 2 a, z=\log _{4 a} 3 a$, then $x y z-2 y z$ equals

(a) $a^{3}$

(b) 1

(d) -1

~~ 8. The sum of $n$ terms of the series $\sum _{x=1}^{n} \log \frac{2^{x}}{3^{x-1}}$ is :

(a) $\log (\frac{3^{n-1}}{2^{n+1}})^{n / 2}$

(b) $\log (\frac{2^{n-1}}{3^{n+1}})^{n / 2}$

(d) $\log (\frac{2^{n+1}}{3^{n-1}})^{n / 2}$

(UPSEE 2011)

~~ 9. The number of meaningful solutions of

$ \log _4(x-1)=\log _2(x-3) \text{ is } $

(a) zero

(b) 1

(d) 3

(IIT 2001)

~~ 10. The value of $[0.16^{\log 2.5(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . .+\infty)}]^{1 / 2}$ is:

(a) -1

(b) 0

(d) None of these

SELF ASSESSMENT SHEET

~~ 2. (d)

~~ 3. (b)

~~ 4. (b)

~~ 5. (b)

~~ 6. (a)

~~ 7. $(d)$

~~ 8. $($ d)

~~ 9. (b)

~~ 10. $(d)$

HINTS AND SOLUTIONS

$\log _2[\log _7(x^{2}-x+37)]=1$

$\Rightarrow \log _7(x^{2}-x+37)=2^{1}=2$

$\Rightarrow x^{2}-x+37=7^{2}=49 \Rightarrow x^{2}-x-12=0$

Now solve for $x$.

~~ 2. Given expression $=\log _{\sqrt{3}} 3^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}$

~~ 3. Given series

$ =\log _{3^{1 / 2}} 3^{15 / 16}=\frac{15}{16} \times 2 \log _3 3=\frac{\mathbf{1 5}}{\mathbf{8}} . $

$=\log _2(\frac{x}{y})+\log _{2^{2}}(\frac{x}{y})^{2}+\log _{2^{3}}(\frac{x}{y})^{3}+\log _{2^{4}}(\frac{x}{y})^{4}+\ldots \ldots$.

$=\log _2(\frac{x}{y})+\log _2(\frac{x}{y})+\log _2(\frac{x}{y})+\log _2(\frac{x}{y})+\ldots . n$ terms

$=\log _2(\frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y} \ldots \ldots . n.$ terms $)$

$=\log _2(\frac{x}{y})^{n}=n \log _2(\frac{\boldsymbol{x}}{\boldsymbol{y}})$.

~~ 4. $\log _{x} a, a^{x / 2}, \log _{b} x$ are in GP $\Rightarrow[a^{x / 2}]^{2}=\log _{x} a \cdot \log _{b} x$

$\Rightarrow a^{x}=\frac{\log a}{\log x} \cdot \frac{\log x}{\log b}$

$\Rightarrow a^{x}=\frac{\log a}{\log b}=\log _{b} a$

$\Rightarrow x=\log _{a}(\log _{b} a)$

$=\log a(\frac{\log _{e} a}{\log _{e} b})=\log a(\log _{e} a)-\log a(\log _{e} b)$

~~ 5. Let $\frac{\log x}{l+m-2 n}=\frac{\log y}{m+n-2 l}=\frac{\log z}{n+l-2 m}=k$. Then

$\log x=k(l+m-2 n), \log y=k(m+n-2 l) ; \log z=k(n+l-2 m)$

$\Rightarrow \log x+\log y+\log z=k(l+m-2 n)+k(m+n-2 l)$

$+k(n+l-2 m)$

$\Rightarrow \log (x y z)=0 \Rightarrow \log (x y z)=\log 1 \Rightarrow x y z=1$.

~~ 6. $\frac{a-2 b}{b-2 a}=\frac{\log _{12} m-2 \log _{18} m}{\log _{18} m-2 \log _{12} m}$

$=\frac{\frac{\log m}{\log 12}-2 \frac{\log m}{\log 18}}{\frac{\log m}{\log 18}-2 \frac{\log m}{\log 12}}=\frac{\log m \log 18-2 \log m \log 12}{\log m \log 12-2 \log m \log 18}$

$=\frac{\log 18-2 \log 12}{\log 12-2 \log 18}=\frac{\log (3^{2} \times 2)-2 \log (2^{2} \times 3)}{\log (2^{2} \times 3)-2 \log (3^{2} \times 2)}$

$=\frac{2 \log 3+\log 2-4 \log 2-2 \log 3}{2 \log 2+\log 3-4 \log 3-2 \log 2}=\frac{-3 \log 2}{-3 \log 3}=\frac{\log 2}{\log 3}=\log _3 2$.

~~ 7. $x=\log _{2 a} a=\frac{\log a}{\log 2 a}, \quad y=\log _{3 a} 2 a=\frac{\log 2 a}{\log 3 a}$

$z=\log _{4 a} 3 a=\frac{\log 3 a}{\log 4 a}$

$\therefore x y z-2 y z=\frac{\log a}{\log 2 a} \cdot \frac{\log 2 a}{\log 3 a} \cdot \frac{\log 3 a}{\log 4 a}-2 \frac{\log 2 a}{\log 3 a} \cdot \frac{\log 3 a}{\log 4 a}$

$=\frac{\log a}{\log 4 a}-2 \frac{\log 2 a}{\log 4 a}=\frac{\log a-2 \log 2 a}{\log 4 a}$

$=\frac{\log a-\log (2 a)^{2}}{\log 4 a}=\frac{\log a / 4 a^{2}}{\log 4 a}=\frac{\log (4 a)^{-1}}{\log (4 a)}=\frac{-1 \cdot \log 4 a}{\log 4 a}=-\mathbf{1}$.

~~ 8. $\sum _{x=1}^{n} \log \frac{2^{x}}{3^{x-1}}=\log (\frac{2^{1}}{3^{0}})+\log (\frac{2^{2}}{3^{1}})+\log (\frac{2^{3}}{3^{2}})+\ldots+\log (\frac{2^{n}}{3^{n-1}})$

$=\log (\frac{2^{1}}{3^{0}} \cdot \frac{2^{2}}{3^{1}} \cdot \frac{2^{3}}{3^{2}} \cdot \ldots \ldots . \frac{2^{n}}{3^{n-1}})$

$=\log (\frac{2^{1+2+3+\ldots \ldots .+n}}{3^{1+2+3+\ldots \ldots .+(n-1)}})=\log [\frac{2^{\frac{n(n+1)}{2}}}{3^{\frac{n(n-1)}{2}}}]=\log [\frac{\mathbf{2}^{\boldsymbol{n}+\mathbf{1}}}{\mathbf{3}^{\boldsymbol{n}-\mathbf{1}}}]^{\boldsymbol{n} / \mathbf{2}}$

~~ 9. $\log _4(x-1)=\log _2(x-3) \Rightarrow \log _{2^{2}}(x-1)=\log _2(x-3)$

$\Rightarrow \frac{1}{2} \log _2(x-1)=\log _2(x-3) \Rightarrow \log _2(x-1)=2 \log _2(x-3)$

$[.$ Using $.\log _{a^{m}}(b^{n})=\frac{n}{m} \log _{a} b]$

$\Rightarrow \log _2(x-1)=\log _2(x-3)^{2}$

$\Rightarrow(x-1)=(x-3)^{2} \Rightarrow x-1=x^{2}-6 x+9$

$\Rightarrow x^{2}-7 x+10=0 \Rightarrow(x-2)(x-5)=0 \Rightarrow x=2$ or 5

Neglecting $x=2$ as $\log _2(x-3)$ is defined when $x>2$.

$\Rightarrow$ There is only one meaningful solution of the given equation.

~~ 10.

$.[(0.16)^{\log _{2.5}[\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \ldots \infty.}]]^{1 / 2}=[(0.4)^{2 \log _{2.5}[\frac{1 / 3}{1-1 / 3}]}]^{1 / 2}$

$ \begin{matrix} {[\because \text{ Sum of infinite GP }=\frac{\text{ First term }}{1-\text{ common ratio }}]} \\ =[(0.4)^{\log _{5 / 2}(\frac{1}{2})}]=[(0.4)^{\log _{5 / 2}(\frac{1}{2})}]=[0.4^{\log _{(2 / 5)^{-1}(2)^{-1}}}] \\ =0.4^{\log _{2 / 5} 2}=0.4^{\log _{0.4} 2}=2 . \quad[\because \log _{a} m^{b^{n}}=\frac{n}{m} \log _{a} b] \\ {[\text{ Using } a^{\log _{a} x}=x]} \end{matrix} $

Polynomials

KEY FACTS

A function $f(x)$ of the form $f(x)=a_0+a_1 x+a_2 x^{2}+$ $+a_n x^{n}$ where $a_0, a_1, a_2, \ldots \ldots . ., a_n$ are real numbers, $a_n \neq 0$ and $n$ is a non negative integer is called a polynomial in $x$. The real numbers $a_0, a_1, a_2, \ldots \ldots \ldots ., a_n$ are called coefficients of the polynomial.

Ex. (a) $6 x^{2}-8 x+5$ is a polynomial with integral coefficients.

(b) $\frac{9}{5} x^{3}+\frac{4}{7} x^{2}-8$ is a polynomial with rational coefficients.

2. Types of Polynomials

Monomial : A polynomial having only one term as $9, \sqrt{2} x, \frac{1}{4} x^{2}$, etc.
Binomial : A polynomial having only two terms as $4 x-5,6 x^{2}+8 x$, etc.
Trinomial : A polynomial having only three terms as $4 x^{2}-\sqrt{2} x+\frac{1}{3}$

3. Degree of a Polynomial :

The degree of a polynomial in one variable is the highest exponent of the variable in that polynomial. Degree of $9 x^{7}-6 x^{5}+4 x^{3}+8$ is 7 .
The degree of a polynomial in more than one variable is the highest sum of the powers of the variables. Degree of $4 x^{5}-6 x^{2} y^{4}+8-3 x y^{6}$ is $\mathbf{1}+\mathbf{6}=\mathbf{7}$.
A polynomial is said to be linear, quadratic, cubic or biquadratic if its degree is 1, 2, 3 or 4 respectively.
A constant is a polynomial of degree 0 .

4. Division of a Polynomial by Another Polynomial.

If $f(x)$ and $g(x)$ are two polynomials, $g(x) \neq 0$, such that $f(x)=g(x) . q(x)+r(x)$ where degree of $r(x)<$ degree of $f(x)$, then $f(x)$ is divided by $g(x)$, and it gives $\boldsymbol{q}(\boldsymbol{x})$ as quotient and $\boldsymbol{r}(\boldsymbol{x})$ as remainder.

Note: If $r(x)=0$, then the divisor $g(x)$ is a factor of $f(x)$.

Remainder Theorem : If $\boldsymbol{f}(\boldsymbol{x})$ be any polynomial of degree $\geq 1$, and $a$ be any number, then if $f(x)$ is divided by $(x-a)$, the remainder is $\boldsymbol{f}(\boldsymbol{a})$.

Ex. (a) The remainder when $f(x)=(5 x^{2}-4 x-1)$ is divided by $(x-1)$ is $f(1)=5.1^{2}-4.1-1=\mathbf{0}$.

(b) The remainder when $f(x)=x^{4}+2 x^{3}-3$ is divided by $(x+2)$ is $f(-2)=(-2)^{4}+2 \cdot(-2)^{3}-3=16-16-3=-\mathbf{3}$.

Factor Theorem : Let $f(x)$ be a polynomial of degree $n>0$. If $f(a)=0$, for any real number $a$, then $(x-a)$ is a factor of $f(x)$.

Conversely, if $(x-a)$ is a factor of $f(x)$, then $f(a)=0$.

Ex. $f(x)=x^{3}-6 x^{2}+11 x-6$ is exactly divisible by $(x-1)$ as $f(1)=1^{3}-6.1^{2}+11.1-6=\mathbf{0}$.

7. Some useful Identities :

$(a+b)^{2}=a^{2}+2 a b+b^{2}$
$(a-b)^{2}=a^{2}-2 a b+b^{2}=(a+b)^{2}-4 a b$
$(a+b)(a-b)=a^{2}-b^{2}$
$(a+b)^{2}+(a-b)^{2}=2(a^{2}+b^{2})$
$(a+b)^{2}-(a-b)^{2}=4 a b$
$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
$(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
$a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-a b)$
$a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+a b)$
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$(a^{3}+b^{3}+c^{3})-3 a b c=(a+b+c)(a^{2}+b^{2}+c^{2}-a b-c a-b c)$
$a+b+c=0 \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c$
$(a^{n}-b^{n})$ is divisible by $(a-b)$ for all values of $n$.
$(a^{n}-b^{n})$ is divisible by $(a+b)$ for only even values of $n$.
$(a^{n}+b^{n})$ is never divisible by $(a-b) \quad(a^{n}+b^{n})$ is divisible by $(a+b)$ only when $n$ is odd.

Note : When a polynomial $f(x)$ is divided by $(x-a)$ and $(x-b)$, the respective remainders are $A$ and $B$. Then, if the same polynomial is divided by $(x-a)(x-b)$, then the remainder will be :

$ \frac{A-B}{a-b} x+\frac{B a-A b}{a-b} $

Homogeneous Expressions : If all the terms of an algebraic expression are of the same degree, then such an expression is called homogeneous expression.

Homogeneous expressions in $(x, y)$ of - Homogeneous expression in $(x, y, z)$ of

Degree $1 \to p x+q y$ Degree $1 \to p x+q y+r z$

Degree $2 \to p x^{2}+r x y+q y^{2}$

Degree $2 \to p x^{2}+q y^{2}+r z^{2}+s x y+t y z+u z x$

Degree $3 \to p x^{3}+r x^{2} y+s x y^{2}+q y^{3}$

Degree $3 \to a x^{3}+b y^{3}+c z^{3}+d x^{2} y+e x y^{2}+f y^{2} z+g y z^{2}+h z^{2} x+k z x^{2}$.

Symmetric Expression : An algebraic expression $f(x, y)$ in two variables $x, y$ is called a symmetric expression if $f(x, y)=f(y, x)$.

An algebraic expression $f(x, y, z)$ is said to be a cyclic expression, if $f(x, y, z)=f(y, z, x)=f(z, x, y)$

e.g. $f(a, b, c)=a(b-c)+b(c-a)+c(a-b)$

$\sum$ (Sigma) is used for the sum of the terms of a cyclic expression.

Ex. $\sum _{x, y, z} x^{3}(y-z)=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$

$\pi(\mathbf{P i})$ is used for the product of the terms of a cyclic expression.

Ex. $\underset{a, b, c}{\pi}(a-b)=(a-b)(b-c)(c-a)$

11. Horner’s Method of Synthetic Division for Factorization

Ex. Divide $3 x^{3}-2 x^{2}-19 x+22$ by $(x-2)$

Sol.

$\therefore f(x)=(x-2)(3 x^{2}+4 x-11)$

Step 1: Write the coefficients of the descending powers of $x$ in the first horizontal row.

Step 2: The multiplier is obtained by putting the divisor $(x-2)=0 \Rightarrow x=2$.

Step 3: Now below the $1^{\text{st }}$ coefficient, i.e., 3 in the first horizontal row, put 0 and add $3+0$, i.e., 3 .

Now $3 \times$ multiplier $=3 \times 2=6=2$ nd element of 2 nd horizontal row. $-2+6=4$

Now $4 \times$ multiplier $=4 \times 2=8=3$ rd element of 2 nd horizontal row. $-19+8=-11$

For the last element again $-11 \times 2=22$ and $22+(-22)=0$.

The first three figures in the third row stand for the coefficients of descending powers of $x$ of quotient and the last entry is for the remainder.

SOLVED EXAMPLES

~~ Ex. 1 . For what value of $p$ is the coefficient of $x^{2}$ in the product $(2 x-1)(x-k)(p x+1)$ equal to 0 and the constant term equal to 2 ?

(CDS 2005)

Sol. $(2 x-1)(x-k)(p x+1)=(2 x-1)(p x^{2}+x-k p x-k)$

$ \begin{aligned} & =2 p x^{3}+2 x^{2}-2 k p x^{2}-2 k x-p x^{2}-x+k p x+k \\ & =2 p x^{3}+x^{2}[2-2 k p-p]-x[2 k+1-k p]+k \end{aligned} $

Here constant term $=k=\mathbf{2}$.

Coefficient of $x^{2}=2-2 k p-p=2-4 p-p=2-5 p$

Given, $2-5 p=0 \Rightarrow p=\frac{\mathbf{2}}{\mathbf{5}}$.

~~ Ex. 2 . For what value of $m$ will the expression $3 x^{3}+m x^{2}+4 x-4 m$ be divisible by $x+2$ ?

(CDS 2005)

Sol. $f(x)=3 x^{3}+m x^{2}+4 x-4 m$

$f(x)$ is divisible by $(x+2)$ if $f(-2)=0$

Now $f(-2)=3(-2)^{3}+m(-2)^{2}+4(-2)-4 m=-24+4 m-8-4 m=-32 \neq 0$

$\therefore$ No such value of $m$ exists for which $(x+2)$ is a factor of the given expression.

~~ Ex. 3 . If $x^{5}-9 x^{2}+12 x-14$ is divisible by $(x-3)$, what is the remainder ?

(CDS 2011)

Sol. Let $f(x)=x^{5}-9 x^{2}+12 x-14$

$f(x)$ is divisible by $(x-3)$ so remainder $=f(3)$.

$\therefore f(3)=(3)^{5}-9(3)^{2}+12(3)-14=243-81+36-14=\mathbf{1 8 4}$.

~~ Ex. 4 . If the expressions $(p x^{3}+3 x^{2}-3)$ and $(2 x^{3}-5 x+p)$ when divided by $(x-4)$ leave the same remainder, then what is the value of $p$ ?

Sol. Let $f(x)=p x^{3}+3 x^{2}-3$

$ g(x)=2 x^{3}-5 x+p $

When divisible by $x-4$, the remainders for the given expressions are $f(4)$ and $g(4)$ respectively.

$ \begin{aligned} & f(4)=p(4)^{3}+3(4)^{2}-3=64 p+48-3=64 p+45 \\ & g(4)=2(4)^{3}-5(4)+p=128-20+p=108+p . \end{aligned} $

Given, $f(4)=g(4) \Rightarrow 64 p+45=108+p \Rightarrow 63 p=63 \Rightarrow p=\mathbf{1}$.

~~ Ex. 5 . What is/are the factors of $(x^{29}-x^{24}+x^{13}-1)$ ?

(a) $(x-1)$ only

(b) $(x+1)$ only

(d) Neither $(x-1)$ nor $(x+1)$

(CDS 2008)

Sol. For $(x-1)$ to be a factor of the given expression, the value of expression at $x=1$ is

$(1)^{29}-(1)^{24}+(1)^{13}-1=1-1+1-1=0$

$\therefore(x-1)$ is a factor of $x^{29}-x^{24}+x^{13}-1$

Similarly for $(x+1)$ to be the factor, the value of expression at $x=-1$ is

$(-1)^{29}-(-1)^{24}+(-1)^{13}-1=-1-1-1-1=-4 \neq 0$

$\therefore(x+1)$ is not a factor of $x^{29}-x^{24}+x^{13}-1$.

Hence, (a) is the correct option.

~~ Ex. 6 . Which one of the following is one of the factors of $x^{2}(y-z)+y^{2}(z-x)-z(x y-y z-z x)$ ?

(a) $(x-y)$

(b) $(x+y-z)$

(d) $(x+y+z)$

(CDS 2007)

Sol. $x^{2}(y-z)+y^{2}(z-x)-z(x y-y z-z x)$

$ \begin{aligned} & =x^{2} y-x^{2} z+y^{2} z-y^{2} x-z x y+y z^{2}+z^{2} x \\ & =x y(x-y-z)+z^{2}(x+y)-z(x^{2}-y^{2}) \\ & =x y(x-y-z)-z(x+y)(x-y-z)=(x-y-z)(x y-y z-z x) \end{aligned} $

Hence, $(c)$ is the correct option.

~~ Ex. 7 . Without actual division show that $2 x^{4}-6 x^{3}+3 x^{2}+3 x-2$ is exactly divisible by $x^{2}-3 x+2$.

Sol. Let $f(x)=2 x^{4}-6 x^{3}+3 x^{2}+3 x-2$ and $g(x)=x^{2}-3 x+2=x^{2}-2 x-x+2=x(x-2)-1(x-2)=(x-2)(x-1)$

For $f(x)$ to be exactly divisible by $g(x),(x-1)$ and $(x-2)$ should be the factors of $f(x)$, i.e.,

$ f(1)=0 \text{ and } f(2)=0 \text{. } $

Now,

$ \begin{aligned} & f(1)=2 \cdot(1)^{4}-6 \cdot(1)^{3}+3 \cdot(1)^{2}+3 \cdot 1-2=2-6+3+3-2=0 \\ & f(2)=2 \cdot(2)^{4}-6(2)^{3}+3(2)^{2}+3 \cdot 2-2=32-48+12+6-2=0 . \end{aligned} $

$\therefore(x-1)$ and $(x-2)$ are factors of $f(x) \Rightarrow f(x)$ is exactly divisible by $g(x)$.

~~ Ex. 8 . If $a+b+c=0$, then what is the value of $a^{4}+b^{4}+c^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}$ ?

(CDS 2005)

Sol. Given, $a+b+c=0$.

Now, $a^{4}+b^{4}+c^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}=(a^{2}+b^{2}+c^{2})^{2}-4 a^{2} b^{2}-4 b^{2} c^{2}-4 c^{2} a^{2}$

$ \begin{aligned} & =[(a+b+c)^{2}-2 a b-2 b c-2 c a]^{2}-4 a^{2} b^{2}-4 b^{2} c^{2}-4 c^{2} a^{2} \\ & =[0^{2}-2 a b-2 b c-2 c a]^{2}-4 a^{2} b^{2}-4 b^{2} c^{2}-4 c^{2} a^{2} \\ & =4 a^{2} b^{2}+4 b^{2} c^{2}+4 c^{2} a^{2}+8 a b^{2} c+8 a b c^{2}+8 a^{2} b c-4 a^{2} b^{2}-4 b^{2} c^{2}-4 c^{2} a^{2} \\ & =8 a b^{2} c+8 a b c^{2}+8 a^{2} b c=8 a b c(b+c+a)=8 a b c .0=\mathbf{0} . \end{aligned} $

~~ Ex. 9 . If $x=\frac{a-b}{a+b}, y=\frac{b-c}{b+c}, z=\frac{c-a}{c+a}$, then what is the value of $\frac{1+x}{1-x} \cdot \frac{1+y}{1-y} \cdot \frac{1+z}{1-z}$ ?

(CDS 2006)

Sol. $x=\frac{a-b}{a+b} \Rightarrow \frac{1}{x}=\frac{a+b}{a-b}$

$\Rightarrow \frac{1+x}{1-x}=\frac{a+b+a-b}{a+b-a+b}=\frac{2 a}{2 b} \Rightarrow \frac{1+x}{1-x}=\frac{a}{b}$

(Applying componendo and dividendo)

Similarly, $\frac{1+y}{1-y}=\frac{b}{c}, \frac{1+z}{1-z}=\frac{c}{a} \therefore \frac{1+x}{1-x} \cdot \frac{1+y}{1-y} \cdot \frac{1+z}{1-z}=\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}=\mathbf{1}$.

~~ Ex. 10 . If $x+y+z=0$, then what is $[\frac{(y-z-x)}{2}]^{3}+[\frac{(z-x-y)}{2}]^{3}+[\frac{(x-y-z)}{2}]^{3}$ equal to ?

Sol. $(\frac{y-z-x}{2})^{3}+(\frac{z-x-y}{2})^{3}+(\frac{x-y-z}{2})^{3}$

$ =(\frac{y-(z+x)}{2})^{3}+(\frac{z-(x+y)}{2})^{3}+(\frac{x-(y+z)}{2})^{3} $

$ =(\frac{y-(-y)}{2})^{3}+(\frac{z-(-z)}{2})^{3}+(\frac{x-(-x)}{2})^{3} $

$(\because x+y+z=0)$

$=(\frac{2 y}{2})^{3}+(\frac{2 z}{2})^{3}+(\frac{2 x}{2})^{3}=y^{3}+z^{3}+x^{3}=\mathbf{3 x y z} . \quad(\because a^{3}+b^{3}+c^{3}=3 a b c.$, if $.a+b+c=0)$

~~ Ex. 11 . If $x^{2}-4 x+1=0$, then what is the value of $x^{3}+\frac{1}{x^{3}}$ ?

Sol. $x^{2}-4 x+1=0$

$ \begin{aligned} & \Rightarrow x=\frac{4 \pm \sqrt{16-4 \times 1 \times 1}}{2 \times 1}=\frac{4 \pm 2 \sqrt{3}}{2}=2 \pm \sqrt{3} \\ & \therefore \quad x^{3}+\frac{1}{x^{3}}=(2+\sqrt{3})^{3}+\frac{1}{(2+\sqrt{3})^{3}}=(2+\sqrt{3})^{3}+[\frac{(2-\sqrt{3}) \times 1}{(2+\sqrt{3})(2-\sqrt{3})}]^{3}=(2+\sqrt{3})^{3}+(2-\sqrt{3})^{3} \\ & =2^{3}+(\sqrt{3})^{3}+3 \times 2 \times \sqrt{3}(2+\sqrt{3})+2^{3}-(\sqrt{3})^{3}-3 \times 2 \times \sqrt{3}(2-\sqrt{3}) \\ & =8+18+8+18=\mathbf{5 2} \text{. Similarly for } x=2-\sqrt{3}, x^{3}+\frac{1}{x^{3}}=\mathbf{5 2} \text{. } \end{aligned} $

~~ Ex. 12 . If $\frac{1}{y+z}+\frac{1}{z+x}=\frac{2}{x+y}$, then what is $(x^{2}+y^{2})$ equal to ?

Sol. $\frac{1}{y+z}+\frac{1}{z+x}=\frac{2}{x+y}$

$\Rightarrow \frac{1}{y+z}-\frac{1}{x+y}=\frac{1}{x+y}-\frac{1}{z+x} \Rightarrow \frac{(x+y)-(y+z)}{(y+z)(x+y)}=\frac{(z+x)-(x+y)}{(x+y)(z+x)}$

$\Rightarrow \frac{x-z}{y+z}=\frac{z-y}{z+x} \Rightarrow(x-z)(x+z)=(z-y)(z+y) \quad \Rightarrow x^{2}-z^{2}=z^{2}-y^{2} \Rightarrow x^{2}+y^{2}=\mathbf{2} z^{2}$.

~~ Ex. 13 . If the sum and difference of two expressions are $5 a^{2}-a-4$ and $a^{2}+9 a-10$ respectively, then what is their LCM ?

Sol. Let $P$ and $Q$ be the two expressions. Then,

$ \begin{aligned} & P+Q=5 a^{2}-a-4 \\ & P-Q=a^{2}+9 a-10 \end{aligned} $

Adding $(i)$ and (ii)

$\Rightarrow 2 P=6 a^{2}+8 a-14 \Rightarrow P=3 a^{2}+4 a-7=(a-1)(3 a+7)$

From $(i), Q=(5 a^{2}-a-4)-(3 a^{2}+4 a-7)=2 a^{2}-5 a+3=(a-1)(2 a-3)$

$\therefore$ LCM of $P$ and $Q=(a-1)(2 a-3)(3 a+7)$.

~~ Ex. 14 . Without actual division, show that $(x-1)^{2 n}-x^{2 n}+2 x-1$ is divisible by $2 x^{3}-3 x^{2}+x$.

Sol. Let $f(x)=(x-1)^{2 n}-x^{2 x}+2 x-1$

$ g(x)=2 x^{3}-3 x^{2}+x=x(2 x^{2}-3 x+1) $

Now $g(x)=0 \Rightarrow x[2 x^{2}-3 x+1]=0 \Rightarrow x[2 x^{2}-2 x-x+1]=0$

$ \Rightarrow[2 x(x-1)-1(x-1)]=0 \Rightarrow(2 x-1)(x-1)=0 \Rightarrow x=\frac{1}{2}, 1 $

$\therefore$ For $f(x)$ to be exactly divisible by $g(x), f(\frac{1}{2})=f(1)$ should be all zero.

$ \begin{aligned} f(\frac{1}{2}) & =(\frac{1}{2}-1)^{2 n}-(\frac{1}{2})^{2 n}+2 \times \frac{1}{2}-1=(-\frac{1}{2})^{2 n}-(\frac{1}{2})^{2 n}+1-1=(\frac{1}{2})^{2 n}-(\frac{1}{2})^{2 n}+1-1=0 \\ f(1) & =(1-1)^{2 n}-1^{2 n}+2 \times 1-1=0-1+2-1=0 . \end{aligned} $

$\therefore[(x-1)^{2 n}-x^{2 n}+2 x-1]$ is completely divisible by $2 x^{3}-3 x^{2}+x$.

~~ Ex. 15 . If the HCF of $(x^{2}+x-12)$ and $(2 x^{2}-k x-9)$ is $(x-k)$, then what is the value of $k$ ?

(CDS 2008)

Sol. Since $(x-k)$ is the HCF of $(x^{2}+x+12)$ and $(2 x^{2}-k x-9)$ $(x-k)$ will be a factor of $2 x^{2}-k x-9$

$\therefore 2 . k^{2}-k \cdot k-9=0 \Rightarrow k^{2}-9=0 \Rightarrow k= \pm 3$

Also, the factors of $(x^{2}+x-12)=(x+4)(x-3) \therefore \boldsymbol{k}=\mathbf{3}$.

PRACTICE SHEET

LEVEL-1

When $x^{13}+1$ is divided by $x-1$, the remainder is :

(a) 1

(b) -1

(d) 2

~~ 2. If $x^{3}+5 x^{2}+10 k$ leaves remainder $-2 x$ when divided by $x^{2}+2$, then what is the value of $k$ ?

(a) -2

(b) -1

(d) 2

(CDS 2012)

~~ 3. $x^{4}+x y^{3}+x^{3} y+x z^{3}+y^{4}+y z^{3}$ is divisible by:

(a) $(x-y)$ only

(b) $(x^{3}+y^{3}+z^{3})$ only

(d) None of the above

(CDS 2012)

~~ 4. For what value of $k$, will the expression $(3 x^{3}-k x^{2}+4 x+16)$ be divisible by $(x-k / 2)$ ?

(a) 4

(b) -4

(d) 0

(CDS 2007)

~~ 5. When $(x^{3}-2 x^{2}+p x-q)$ is divided by $(x^{2}-2 x-3)$, the remainder is $(x-6)$, What are the values of $p$ and $q$ respectively?

(a) $-2,-6$

(b) $2,-6$

(d) 2,6

(CDS 2009)

~~ 6. If $\frac{x^{3}+a x^{2}+b x+4}{x^{2}+x-2}$ is a polynomial of degree 1 in $x$, then what are the values of $a, b$ respectively?

(a) $-1,-4$

(b) $-1,4$

(d) 3,4

(CDS 2005)

~~ 7. When $a+b+c+3 a^{1 / 3} b^{2 / 3}+3 a^{2 / 3} b^{1 / 3}$ is divided by $a^{1 / 3}+b^{1 / 3}+c^{1 / 3}$, what is the remainder?

(a) $3 a$

(b) $3 b$

(d) $c^{2 / 3}$

(CDS 2005)

~~ 8. If the polynomials $a x^{3}+4 x^{2}+3 x-4$ and $x^{3}-4 x+a$ leave the same remainder when divided by $(x-3)$, the value of $a$ is :

(a) 2

(b) $-3 / 2$

(d) 4

~~ 9. Let $R_1$ and $R_2$ be the remainders when the polynomials $x^{3}+2 x^{2}-5 a x-7$ and $x^{2}+a x^{2}-12 x+6$ are divided by $(x+1)$ and $(x-2)$ respectively. If $2 R_1+R_2=6$, the value of $a$ is :

(a) -2

(b) 1

(d) 2

~~ 10. If both $(x-2)$ and $(x-1 / 2)$ are factors of $p x^{2}+5 x+r$, then:

(a) $p=2 r$

(b) $p+r=0$

(d) $p \times r=1$

~~ 11. If the expression $a x^{2}+b x+c$ is equal to 4 , when $x=0$, leaves a remainder 4 when divided by $x+1$ and leaves a remainder 6 when divided by $x+2$, then the values of $a, b$ and $c$ are respectively,

(a) $1,1,4$

(b) $2,2,4$

(d) $4,4,4$

~~ 12. $f(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b$ is a polynomial such that when it is divided by $(x-1)$ and $(x+1)$, the remainders are respectively 5 and 19 . Determine the remainder when $f(x)$ is divided by $(x-2)$.

(a) 6

(b) 10

(d) 8

~~ 13. If $(x^{2}-1)$ is a factor of $a x^{4}+b x^{3}+c x^{2}+d x+e$, then :

(a) $a+c+e=0$

(b) ace $=1$

(d) Both (a) and (c)

~~ 14. What is $\frac{x^{2}-3 x+2}{x^{2}-5 x+6} \div \frac{x^{2}-5 x+4}{x^{2}-7 x+12}$ equal to ?

(a) $\frac{x+3}{x-3}$

(b) 1

(d) 2

(CDS 2011)

~~ 15. If the expression $(p x^{3}+x^{2}-2 x-q)$ is divisible by $(x-1)$ and $(x+1)$, then the values of $p$ and $q$ respectively are ?

(a) $2,-1$

(b) $-2,1$

(d) 2,1

(CDS 2010)

LEVEL-2

~~ 16. When $x^{40}+2$ is divided by $x^{4}+1$, what is the remainder?

(a) 1

(b) 2

(d) 4

(CDS 2009)

~~ 17. If the remainder of the polynomial $a_0+a_1 x+a_2 x^{2}+\ldots \ldots$. $+a_n x^{n}$ when divided by $(x-1)$ is 1 , then which one of the following is correct?

(a) $a_0+a_2+\ldots \ldots=a_1+a_3+\ldots \ldots$

(b) $a_0+a_2+\ldots \ldots=1+a_1+a_3+\ldots \ldots$

(d) $1-a_0-a_2-\ldots \ldots=a_1+a_3+\ldots \ldots$

(CDS 2009)

~~ 18. The remainder when $1+x+x^{2}+x^{3}+\ldots \ldots . .+x^{1007}$ is divided by $(x-1)$ is

(a) 1006

(b) 1008

(d) 0

~~ 19. A cubic polynomial $f(x)$ is such that $f(1)=1, f(2)=2$, $f(3)=3$ and $f(4)=5$, then $f(6)$ equals :

(a) 7

(b) 6

(d) 13

~~ 20. If the polynomial $x^{6}+p x^{5}+q x^{4}-x^{2}-x-3$ is divisible by $x^{4}-1$, then the value of $p^{2}+q^{2}$ is :

(a) 1

(b) 9

(d) 13

(CDS 2001)

~~ 21. The factors of $x^{8}+x^{4}+1$ are :

(a) $(x^{4}+1-x^{2})(x^{2}+1+x)(x^{2}+1-x)$

(b) $(x^{4}+1-x^{2})(x^{2}-1+x)(x^{2}+1+x)$

(d) $(x^{4}-1+x^{2})(x^{2}+1-x)(x^{2}+1+x)$

(CDS 1999)

~~ 22. If the polynomial $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$ is divided by $(x^{2}+1)$, then the remainder is :

(a) 1

(b) $x^{2}+4$

(d) $x$

~~ 23. If $(x+k)$ is a common factor of $x^{2}+p x+q$ and $x^{2}+l x+m$, then the value of $k$ is

(a) $\frac{p+q}{l+m}$

(b) $\frac{p-l}{q-m}$

(d) $\frac{q-m}{p-l}$

~~ 24. If $(x-1)$ is a factor of $A x^{3}+B x^{2}-36 x+22$ and $2^{B}=64^{A}$, find $A$ and $B$ ?

(a) $A=4, B=16$

(b) $A=6, B=24$

(d) $A=8, B=16$

~~ 25. When a polynomial $f(x)$ is divided by $(x-3)$ and $(x+6)$, the respective remainders are 7 and 22 . What is the remainder when $f(x)$ is divided by $(x-3)(x+6)$ ?

(a) $\frac{-5}{3} x+12$

(b) $-\frac{7}{3} x+14$

(d) $-\frac{7}{3} x+12$

~~ 26. If $p(x)$ is a common multiple of degree 6 of the polynomials $f(x)=x^{3}+x^{2}-x-1$ and $g(x)=x^{3}-x^{2}+x-1$, then which one of the following is correct ?

(a) $p(x)=(x-1)^{2}(x+1)^{2}(x^{2}+1)$

(b) $p(x)=(x-1)(x+1)(x^{2}+1)^{2}$

(d) $p(x)=(x-1)^{2}(x^{4}+1)$

(CDS 2012)

~~ 27. Which one of the following is divisible by $(1+a+a^{5})$ and $(1+a^{4}+a^{5})$ individually?

(a) $(a^{2}+a+1)(a^{3}+a^{2}+1)(a^{3}+a+1)$

(b) $(a^{4}-a+1)(a^{3}+a^{2}+1)(a^{3}+a-1)$

(d) $(a^{2}+a+1)(a^{3}-a^{2}+1)(a^{3}-a+1)$

(CDS 2005)

~~ 28. Consider the following statements :

~~ 29. $a^{n}+b^{n}$ is divisible by $a+b$ if $n=2 k+1$, where $k$ is a positive integer.

~~ 30. $a^{n}-b^{n}$ is divisible by $a-b$ if $n=2 k$, where $k$ is a positive integer. Which of the statements given above is/are correct?

(a) 1 only

(b) 2 only

(d) Neither 1 nor 2

(CDS 2005)

~~ 29. If $(x-2)$ is a common factor of the expressions $x^{2}+a x+b$ and $x^{2}+c x+d$, then $\frac{b-d}{c-a}$ is equal to

(a) -2

(b) -1

(d) 2

(EAMCET 2004)

~~ 30. Let $a \neq 0$ and $p(x)$ be a polynomial of degree greater than 2 . If $p(x)$ leaves remainders $a$ and $-a$, when divided respectively by $(x+a)$ and $(x-a)$, then the remainder when $p(x)$ is divided by $(x^{2}-a^{2})$ is

(a) $-2 x$

(b) $-x$

(d) $2 a$

(EAMCET 2003)

~~ 31. If $9 x^{2}+3 p x+6 q$ when divided by $(3 x+1)$ leaves a remainder $(-\frac{3}{4})$ and $q x^{2}+4 p x+7$ is exactly divisible by $(x+1)$, then the values of $p$ and $q$ respectively will be :

(a) $0, \frac{7}{4}$

(b) $-\frac{7}{4}, 0$

(d) $\frac{7}{4}, 0$

~~ 32. What should be subtracted from $27 x^{3}-9 x^{2}-6 x-5$ to make it exactly divisible by $(3 x-1)$

(a) -5

(b) -7

(d) 7

(CDS 2009)

~~ 33. The values of $a, b$ and $c$ respectively for the expression $f(x)=x^{3}+a x^{2}+b x+c$, if $f(1)=f(2)=0$ and $f(4)=f(0)$ are :

(a) $9,20,12$

(b) $-9,-20,12$

(d) $-9,-20,-12$

~~ 34. The remainder, when $x^{200}$ is divided by $x^{2}-3 x+2$ is

(a) $(2^{200}-1) x+(-2^{200}+2)$

(b) $(2^{200}+1) x+(-2^{200}-2)$

(d) $2^{100}$

~~ 35. (i) For $a \neq b$, if $x+k$ is the HCF of $x^{2}+a x+b$ and $x^{2}+b x$ $+a$, then the value of $a+b$ is equal to

(a) -2

(b) -1

(d) 2

(Type Raj PET 2004, Kerala PET 2004)

(ii) If $(x+k)$ is the HCF of $a x^{2}+a x+b$ and $x^{2}+c x+d$, then what is the value of $k$ ?

(a) $\frac{b+d}{a+c}$

(b) $\frac{a+b}{c+d}$

(d) None of these

(CDS 2008)

LEVEL-3

~~ 36. The value of $\frac{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}{(a-b)(b-c)(c-a)}$ is :

(a) 1

(b) 3

(d) Zero

(CDS 2000)

~~ 37. If $a^{2}=b y+c z, b^{2}=c z+a x, c^{2}=a x+b y$, then the value of $\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{c+z}$ will be:

(a) $a+b+c$

(b) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

(d) 0

(CDS 2001)

~~ 38. If $x+y+z=0$, then $x(y-z)^{3}+y(z-x)^{3}+z(x-y)^{3}$ equals

(a) 0

(b) $y+z$

(d) $(z+x)^{2}$

~~ 39. If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{(a+b+c)}$, where $a+b+c \neq 0, a b c \neq 0$, what is the value of $(a+b)(b+c)(c+a)$ ?

(a) 0

(b) 1

(d) 2

(CDS 2005)

~~ 40. If $\frac{x}{(b-c)(b+c-2 a)}=\frac{y}{(c-a)(c+a-2 b)}$

$=\frac{z}{(a-b)(a+b-2 c)}$, what is the value of $x+y+z$ ?

(a) $(a+b+c)$

(b) $a^{2}+b^{2}+c^{2}$

(d) 1

(CDS 2005)

~~ 41. If $a+b+c=0$, then find the value of :

$\frac{a^{2}}{a^{2}-b c}+\frac{b^{2}}{b^{2}-c a}+\frac{c^{2}}{c^{2}-a b}$

(a) 4

(b) 2

(d) 0

(MAT 2005)

~~ 42. The value of $\frac{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}{(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3}+(z^{2}-x^{2})^{3}}$ is :

(a) 1

(b) $[2(x+y+z)]^{-1}$

(d) 0

(CDS 2004)

~~ 43. If $a+b+c=0$, then $a^{2}+a b+b^{2}$ is equal to :

(a) $b^{2}-b c+c^{2}$

(b) $c^{2}-a b$

(d) 0

(CDS 2004)

~~ 44. If $p q r=1$, the value of $\frac{1}{[1+p+q^{-1}]}+\frac{1}{[1+q+r^{-1}]}$

$+\frac{1}{[1+r+p^{-1}]}$ will be equal to :

(a) 1

(b) 0

(d) -2

(CDS 2004)

~~ 45. If $a=\frac{x y}{x+y}, b=\frac{x z}{x+z}$ and $c=\frac{y z}{y+z}$, where $a, b$ and $c$ are non-zero, then what is $x$ equal to ?

(a) $\frac{2 a b c}{a c+b c-a b}$

(b) $\frac{2 a b c}{a b-a c+b c}$

(d) $\frac{2 a b c}{a b+a c-b c}$

~~ 46. If $a+b+c=0$, then what is the value of

(CDS 2006) $\frac{a^{2}+b^{2}+c^{2}}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$

(a) 1

(b) 3

(d) 0

(CDS 2006)

~~ 47. If $x+y+z=2 s$, then what is

$(s-x)^{3}+(s-y)^{3}+3(s-x)(s-y) z$ equal to :

(a) $z^{3}$

(b) $-z^{3}$

(d) $y^{3}$

(CDS 2007)

~~ 48. If $x+\frac{1}{x}=p$, then $x^{6}+\frac{1}{x^{6}}$ equals to :

(a) $p^{6}+6 p$

(b) $p^{6}-6 p$

(d) $p^{6}-6 p^{4}+9 p^{2}-2$

(CDS 2007)

~~ 49. If $x+y+z=0$, then what is the value of :

$\frac{1}{x^{2}+y^{2}-z^{2}}+\frac{1}{y^{2}+z^{2}-x^{2}}+\frac{1}{z^{2}+x^{2}-y^{2}}$ ?

(a) $\frac{1}{x^{2}+y^{2}+z^{2}}$

(b) 1

(d) 0

(CDS 2010)

~~ 50. If $x+y+z=0$, then $\frac{x^{2}}{2 x^{2}+y z}+\frac{y^{2}}{2 y^{2}+z x}+\frac{z^{2}}{2 z^{2}+x y}=$

(a) 4

(b) 2

(d) 1 ~~ 51. If $(b+c-a) x=(c+a-b) y=(a+b-c) z=2$, then $(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})(\frac{1}{x}+\frac{1}{y})$ is equals:

(a) $a^{2} b^{2} c^{2}$

(b) $a b c$

(d) $(a b c)^{2}$

~~ 52. If $a^{x}=(x+y+z)^{y}, a^{y}=(x+y+z)^{z}, a^{z}=(x+y+z)^{x}$, then:

(a) $3(x+y+z)=a$

(b) $2 a=x+y+z$

(d) $x=y=z=a / 3$

~~ 53. If $x+\frac{1}{x}=a$, then what is the value of $x^{3}+x^{2}+\frac{1}{x^{3}}+\frac{1}{x^{2}}$ ?

(a) $a^{3}+a^{2}$

(b) $a^{3}+a^{2}-5 a$

(d) $a^{3}+a^{2}-4 a-2$

(CDS 2012)

~~ 54. If $x^{1 / 3}+y^{1 / 3}+z^{1 / 3}=0$, then what is $(x+y+z)^{3}$ equal to ?

(a) 1

(b) 3

(d) $27 x y z$

~~ 55. $\frac{2 a}{a+b}+\frac{2 b}{b+c}+\frac{2 c}{c+a}+\frac{(b-c)(c-a)(a-b)}{(b+c)(c+a)(a+b)}$ equals

(a) 0

(b) -1

(d) 2

ANSWERS

1. $(d)$	2. (c)	3. $(b)$	4. $(b)$	5. $(c)$	6. $(a)$	7. $(c)$	8. $(c)$	9. $(d)$	10. (c)
11. (a)	12. $(b)$	13. $(d)$	14. (b)	15. $(d)$	16. $(c)$	17. $(d)$	18. $(b)$	19. $(b)$	20. $(c)$
21. (a)	22. (c)	23. $(d)$	24. (c)	25. (a)	26. (a)	27. (b)	28. (c)	29. (d)	30. $(b)$
31. $(d)$	32. $(b)$	33. $(c)$	34. (a)	35. $(i)$	$(b)(i i)(d)$	36. $(b)$	37. (c)	38. (a)	39. (a)
40. (c)	41. $(b)$	42. (c)	43. $(b)$	44. (a)	45. (a)	46. (c)	47. (a)	48. $(d)$	49. $(d)$
50. $(d)$	51. $(b)$	52. $(d)$	53. $(c)$	54. $(d)$	55. $(c)$

HINTS AND SOLUTIONS

Remainder when $x^{13}+1$ is divided by $(x-1)=1^{13}+1=\mathbf{2}$.

$ \frac{-5 x^{2}+10}{-2 x+10 k-10}=\text{ Remainder } $

Given, $-2 x+10 k-10=-2 x$

$\Rightarrow 10 k=10 \Rightarrow k=1$.

~~ 3. Hint. $x^{4}+x y^{3}+x^{3} y+x z^{3}+y^{4}+y z^{3}$

$ \begin{aligned} & =(x^{4}+x y^{3}+x z^{3})+(x^{3} y+y^{4}+y z^{3}) \\ & =x(x^{3}+y^{3}+z^{3})+y(x^{3}+y^{3}+z^{3}) \\ & =(x+y)(x^{3}+y^{3}+z^{3}) \end{aligned} $

~~ 4. Let $f(x)=3 x^{3}-k x^{2}+4 x+16$. Then, $f(x)$ will be divisible by $(x-k / 2)$ if $f(k / 2)=0$

$\Rightarrow 3 .(k / 2)^{3}-k .(k / 2)^{2}+4(k / 2)+16=0$

$\Rightarrow \frac{3 k^{3}}{8}-\frac{k^{3}}{4}+\frac{4 k}{2}+16=0$

$\Rightarrow \frac{3 k^{3}-2 k^{3}+16 k+128}{8}=0$

$\Rightarrow k^{3}+16 k+128=0 \Rightarrow(k+4)(k^{2}-4 k+32)=0$

$\Rightarrow k+4=0 \Rightarrow k=-4$.

~~ 5. alt text

Given, $(p+3) x-q=x-6$

$\Rightarrow p+3=1$ and $q=6$

$\Rightarrow p=-\mathbf{2}, q=\mathbf{6}$

~~ 6. alt text

As the given polynomial is of degree 1 , the degree of the remainder should be less than 1, i.e., 0 , i.e., the remainder has only a constant term.

$\Rightarrow b-a+3=0$ and $2 a+2=0 \Rightarrow a=-1$

$\therefore b-(-1)+3=0 \Rightarrow b=-4$.

$\therefore a=-1, b=-4$.

~~ 7. Let $a^{1 / 3}=x, b^{1 / 3}=y, c^{1 / 3}=z$. Then,

$a+b+c+3 a^{1 / 3} b^{2 / 3}+3 a^{2 / 3} b^{1 / 3}=x^{3}+y^{3}+z^{3}+3 x y^{2}+3 x^{2} y$ and $a^{1 / 3}+b^{1 / 3}+c^{1 / 3}=x+y+z$. Now $x^{3}+y^{3}+z^{3}+3 x y^{2}+3 x^{2} y$

$ \begin{aligned} & =x^{3}+3 x y^{2}+3 x^{2} y+y^{3}+z^{3} \\ & =(x+y)^{3}+z^{3} \\ & =[x+y+z][(x+y)^{2}-(x+y) z+z^{2}] . \end{aligned} $

$\therefore$ Given expression is completely divisible by $(x+y+z)$, i.e., by $a^{1 / 3}+b^{1 / 3}+c^{1 / 3}$.

~~ 8. Let $f(x)=a x^{3}+4 x^{2}+3 x-4$

$ g(x)=x^{3}-4 x+a . $

Remainders when $f(x)$ and $g(x)$ are divided by $(x-3)$ are $f(3)$ and $g(3)$ respectively. Now,

$ \begin{aligned} f(3) & =a \cdot(3)^{3}+4 \cdot(3)^{2}+3 \cdot 3-4 \\ & =27 a+36+9-4=27 a+41 \\ g(3) & =(3)^{3}-4 \cdot(3)+a=27-12+a=15+a \end{aligned} $

Given, $f(3)=g(3)$

$\therefore 27 a+41=15+a \Rightarrow 26 a=-26 \Rightarrow a=-\mathbf{1}$.

~~ 9. Let $f(x)=x^{3}+2 x^{2}-5 a x-7$

$ \begin{aligned} \therefore \quad R_1 & =f(-1)=(-1)^{3}+2 \cdot(-1)^{2}-5 \cdot a \cdot(-1)-7 \\ & =-1+2+5 a-7=5 a-6 \end{aligned} $

$ \begin{aligned} g(x) & =x^{3}+a x^{2}-12 x+6 \\ R_2 & =g(2)=(2)^{3}+a \cdot(2)^{2}-12 \cdot(2)+6 \\ & =8+4 a-24+6=4 a-10 \end{aligned} $

Given, $2 R_1+R_2=6 \Rightarrow 2(5 a-6)+(4 a-10)=6$

$\Rightarrow \quad 10 a-12+4 a-10=6$

$\Rightarrow \quad 14 a-22=6 \Rightarrow 14 a=28 \Rightarrow a=\mathbf{2}$.

~~ 10. Let $f(x)=p x^{2}+5 x+r$

Since, $(x-2)$ and $(x-\frac{1}{2})$ are the factors of $f(x)$, therefore,

$f(2)=0$ and $f(\frac{1}{2})=0$.

$\therefore f(2)=p \times(2)^{2}+5 \times 2+r=4 p+10+r=0$

$f(\frac{1}{2})=p \times(\frac{1}{2})^{2}+5 \times \frac{1}{2}+r=\frac{p}{4}+\frac{5}{2}+r=p+10+4 r=0$

(i) and (ii) $\Rightarrow 4 p+10+r=p+10+4 r \Rightarrow 3 p=3 r \Rightarrow \boldsymbol{p}=\boldsymbol{r}$.

~~ 11. Given exp. $f(x)=a x^{2}+b x+c$

$\therefore$ When $x=0, a .0+b .0+c=4 \Rightarrow \boldsymbol{c}=\mathbf{4}$.

The remainders when $f(x)$ is divided by $(x+1)$ and $(x+2)$ respectively are $f(-1)$ and $f(-2)$.

$ \begin{matrix} \therefore & f(-1)=a \cdot(-1)^{2}+b \cdot(-1)+c=4 \\ \Rightarrow & a-b+c=4 \Rightarrow a-b+4=4 \Rightarrow a-b=0 \\ & f(-2)=a \cdot(-2)^{2}+b(-2)+c=6 \\ \Rightarrow & 4 a-2 b+4=6 \Rightarrow 4 a-2 b=2 \end{matrix} $

Solving (i) and (ii) simultaneously we get, $\boldsymbol{a}=\mathbf{1}, \boldsymbol{b}=\mathbf{1}$.

~~ 12. When $f(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b$ is divided by $(x-1)$ and $(x+1)$, the remainders are 5 and 19 respectively.

i.e., $f(1)=5$ and $f(-1)=19$

$\Rightarrow 1-2+3-a+b=5$ and $1+2+3+a+b=19$

$\Rightarrow-a+b=3$ and $a+b=13$

Adding the two equations, we get $2 b=16 \Rightarrow b=8 \Rightarrow a=5$

$\therefore f(x)=x^{4}-2 x^{3}+3 x^{2}-a x+b$

$ =x^{4}-2 x^{3}+3 x^{2}-5 x+8 $

$\therefore$ Remainder, when $f(x)$ is divided by $(x-2)$ is equal to $f(2)$

$\therefore f(2)=2^{4}-2.2^{3}+3.2^{2}-5.2+8$

$ =16-16+12-10+8=\mathbf{1 0} \text{. } $

~~ 13. Let $f(x)=a x^{4}+b x^{4}+c x^{2}+d x+e$ be the given polynomial.

Then, $(x^{2}-1)$ is a factor of $f(x)$.

$\Rightarrow(x-1)(x+1)$ is a factor of $f(x)$

$\Rightarrow(x-1)$ and $(x+1)$ are factors of $f(x)$

$\Rightarrow f(1)=0$ and $f(-1)=0$

$\Rightarrow a+b+c+d+e=0$ and $a-b+c-d+e=0$.

Adding and subtracting the two equations, we get

$ 2(a+c+e)=0 \text{ and } 2(b+d)=0 $

$\Rightarrow a+c+e=0$ and $b+d=0$.

~~ 14. $\frac{x^{2}-3 x+2}{x^{2}-5 x+6} \div \frac{x^{2}-5 x+4}{x^{2}-7 x+12}$

$ \begin{aligned} & =\frac{(x-1)(x-2)}{(x-3)(x-2)} \div \frac{(x-4)(x-1)}{(x-3)(x-4)} \\ & =\frac{(x-1)}{(x-3)} \div \frac{(x-1)}{(x-3)}=\frac{(x-1)}{(x-3)} \times \frac{(x-3)}{(x-1)}=\mathbf{1} . \end{aligned} $

~~ 15. $p x^{3}+x^{2}-2 x-q$ is divisible by $(x-1)$ and $(x+1)$

$\Rightarrow p(1)^{3}+(1)^{2}-2(1)-q=0 \Rightarrow p-q=1$

and $p(-1)^{3}+(-1)^{2}-2(-1)-q=0 \Rightarrow p+q=3$

Solving $(i)$ and $(i i) p=2, q=1$.

~~ 16. Put $x^{4}=-1$ in $f(x)=x^{40}+2$

Remainder $=(x^{4})^{10}+2=(-1)^{10}+2=\mathbf{3}$.

~~ 17. Let $f(x)=a_0+a_1 x+a_2 x^{2}+\ldots \ldots \ldots .+a_n x^{n}$

Given, $f(1)=1$

$\Rightarrow a_0+a_1+a_2+\ldots \ldots . .+a_n=1$

$\Rightarrow 1-a_0-a_2-\ldots \ldots .=a_1+a_3+\ldots \ldots \ldots$.

~~ 18. Required remainder $=f(1)$

$ \begin{aligned} & =1+1+1+1+1 \ldots \ldots \ldots+1 \text{ (1008 times }) \\ & =1008 \times 1=\mathbf{1 0 0 8} \end{aligned} $

~~ 19. Let the cubic polynomial be :

$ f(x)=a x^{3}+b x^{2}+c x+d . $

Given, $f(1)=1 \Rightarrow a+b+c+d=1$

$ f(2)=2 \Rightarrow 8 a+4 b+2 c+d=2 $

$ \begin{aligned} & f(3)=4 \Rightarrow 27 a+9 b+3 c+d=3 \\ & f(4)=5 \Rightarrow 125 a+25 b+5 c+d=5 \end{aligned} $

(ii) - (i) $\Rightarrow 7 a+3 b+c=1$

(iii) - (ii) $\Rightarrow 19 a+5 b+c=1$

(iv) - (iii) $\Rightarrow 98 a+16 b+2 c=2$

(vi) $-(v) \Rightarrow 12 a+2 b=0 \Rightarrow 6 a+b=0$

(vii) $-2(v i) \Rightarrow 60 a+6 b=0 \Rightarrow 10 a+b=0$ Putting values of $a, b, c, d$ in $f(x)=a x^{3}+b x^{2}+c x+d$, we get the polynomial $f(x)=x \Rightarrow f(6)=\mathbf{6}$.

~~ 20. $f(x)=x^{6}+p x^{5}+q x^{4}-x^{2}-x-3$

$ =x^{4} \cdot x^{2}+p \cdot x^{4} x+q \cdot x^{4}-x^{2}-x-3 $

As $(x^{4}-1)$ is a factor of $f(x)$, so putting $x^{4}=1$, we get

$x^{2}+p x+q-x^{2}-x-3=0$

$\Rightarrow(p-1) x+(q-3)=0 \Rightarrow p-1=0$ and $q-3=0$

$\Rightarrow p=1$ and $q=3$.

$\therefore p^{2}+q^{2}=1+9=\mathbf{1 0}$.

~~ 21. $x^{8}+x^{4}+1=x^{8}+2 x^{4}+1-x^{4}$ (Adding and subtracting $.x^{4})$

$=(x^{4}+1)^{2}-(x^{2})^{2}=(x^{4}+1+x^{2})(x^{4}+1-x^{2}) \\ $

$ =[(x^{4}+2 x^{2}+1)-x^{2}](x^ 4 + 1-x^2) \\ $

$ =[(x^{2}+1)^{2}-(x)^{2}] (x^4 + 1 - x^2) \\ $

$ =(x^{2}+1+x)(x^{2}+1-x)(x^{4}+1-x^{2}) $

~~ 22. $f(x)=x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$

Putting $x^{2}=-1$, we get

$ \begin{aligned} f(x) & =(x^{2})^{9} \cdot x+(x^{2})^{8} \cdot x+(x^{2})^{6} \cdot x+(x^{2})^{5} \cdot x+(x^{2})^{2} \cdot x+x^{2} \cdot x \\ & =(-1)^{9} x+(-1)^{8} \cdot x+(-1)^{6} \cdot x+(-1)^{5} \cdot x+(-1)^{2} \cdot x+(-1) \cdot x \\ & =-x+x+x-x+x-x=-x . \end{aligned} $

~~ 23. Let $f(x)=x^{2}+p x+q$

$g(x)=x^{2}+l x+m$.

Since $(x+k)$ is a common factor of $f(x)$ and $g(x)$,

$ \begin{gathered} f(-k)=k^{2}-p k+q=0 \\ g(-k)=k^{2}-l k+m=0 \\ \Rightarrow k^{2}-p x+q=k^{2}-l k+m \\ \Rightarrow q-m=(p-l) k \Rightarrow k=\frac{q-m}{p-l} \end{gathered} $

~~ 24. Since $(x-1)$ is a factor of $A x^{3}+B x^{2}-36 x+22$

$\therefore A(1)^{3}+B(1)^{2}-36(1)+22=0$

$\Rightarrow A+B=14$

and $2^{B}=64^{A} \Rightarrow 2^{B}=(2^{6})^{A} \Rightarrow B=6 A$

$\therefore$ From (i) and (ii) $A=2, B=12$.

~~ 25. The function $f(x)$ is not known. Here,

$ \begin{matrix} a=3, & & b=-6 \\ A=7, & B=22 \quad & \\ A \text{ Refer to Key Fact } 8] \end{matrix} $

$\therefore$ Required remainder $=\frac{A-B}{a-b} x+\frac{B a-A b}{a-b}$

$ =\frac{7-22}{3-(-6)} x+\frac{22 \times 3-7 \times(-6)}{3-(-6)}=\frac{-5}{\mathbf{3}} x+\mathbf{1 2} . $

~~ 26. $f(x)=x^{3}+x^{2}-x-1$

$g(x)=x^{3}-x^{2}+x-1$

$f(x) \cdot g(x)=(x^{3}+x^{2}-x-1) \cdot(x^{3}-x^{2}+x-1)$

$ \begin{aligned} & =x^{6}-x^{\not b}+\chi^{A}-x^{\not b}+\chi^{\not}-x^{4}+x^{\not}-\chi^{\not} \\ & -x^{4}+\not x^{\not 2}-\not x^{\not 2}+\not x-\not x^{\not 2}+x^{2}-\not x+1 \\ & =x^{6}-x^{4}-x^{2}+1 \\ & =x^{4}(x^{2}-1)-(x^{2}-1)=(x^{2}-1)(x^{4}-1) \\ & =(x-1)(x+1)[(x^{2})^{2}-1] \\ & =(x-1)(x+1)[(x^{2}-1)(x^{2}+1)] \end{aligned} $

$ \therefore p(x)=x^{6}-x^{4}-x^{2}+1 $

$ \begin{aligned} & =(x-1)(x+1)[(x-1)(x+1)(x^{2}+1)] \\ & =(x-1)^{2}(x+1)^{2}(x^{2}+1) . \end{aligned} $

~~ 27. The given expression has to be divided by $(1+a+a^{5})$ and $(1+a^{4}+a^{5})$ individually, so highest power of $a$ is $5+5=10$, followed by $5+4=9$, and both are positive.

In options $(a)$ and $(d)$ the highest power of $a$ is 8 , hence these options are not acceptable.

The only choices are (b) and (c), but in option (c) $a^{10}$ is positive but $a^{9}$ is negative and in (b) both $a^{10}$ and $a^{9}$ are positive. Hence $(b)$ is the correct option.

~~ 28. Statement (1) is correct as for $k=1, n=2 \times 1+1=3$.

$\therefore a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-a b)$ which is divisible by $(a+b)$, statement (2) is also correct as for $k=1, n=2$,

$\therefore a^{2}-b^{2}=(a-b)(a+b)$ which is divisible by $(a-b)$.

~~ 29. $(x-2)$ is a common factor of $(x^{2}+a x+b)$ and $(x^{2}+c x+d)$

$\Rightarrow 4+2 a+b=0$

and $4+2 c+d=0$

$\therefore 2 a+b=2 c+d \Rightarrow b-d=2(c-a) \Rightarrow \frac{b-d}{c-a}=\mathbf{2}$.

~~ 30. Let $r x+t$ be the remainder, $q(x)$ be the quotient when $p(x)$ is divided by $x^{2}-a^{2}$.

$\therefore p(x)=(x^{2}-a^{2}) \cdot q x+r x+t$

Given, $p(x)$ leaves remainders $a$ and $-a$ respectively when divided by $(x+a)$ and $(x-a)$.

$\therefore p(-a)=a$ and $p(a)=-a$

Putting $x=-a$ in (i), we get

$p(-a)=0 . q(-a)+(-r a+t)$

$\Rightarrow a=-r a+t$

Putting $x=a$, in $(i)$, we get

$p(a)=0 . q(a)+(r a+t)$

$\Rightarrow-a=r a+t$

$\therefore$ Adding (ii) and (iii), we get $2 t=0 \Rightarrow t=0 \Rightarrow r=-1$

$\therefore$ Required remainder $=r x+t=-\boldsymbol{x}$.

~~ 31. Given, $(9 x^{2}+3 p x+6 q)$, when divided by $(3 x+1)$ leaves a remainder $-\frac{3}{4}$

$\therefore f(x)=9 x^{2}+3 p x+6 q-(-\frac{3}{4})=(9 x^{2}+3 p x+6 q+\frac{3}{4})$ is exactly divisible by $(3 x+1)$

$ \begin{aligned} \therefore f(-\frac{1}{3})=0 & \Rightarrow 9(-\frac{1}{3})^{2}+3 p \cdot(-\frac{1}{3})+6 q+\frac{3}{4}=0 \\ & \Rightarrow 6 q-p+\frac{7}{4}=0 \\ & \Rightarrow 24 q-4 p+7=0 \end{aligned} $

Now, the expression $g(x)=q x^{2}+4 p x+7$ is exactly divisible by $x+1$

$\Rightarrow g(-1)=0 \Rightarrow q-4 p+7=0$

Solving equations $(i)$ and (ii), we get $q=\mathbf{0}, p=\frac{7}{\mathbf{4}}$.

~~ 32. To make $f(x)=27 x^{3}-9 x^{2}-6 x-5$ exactly divisible by $(3 x-1)$, the remainder obtained on division should be subtracted.

Remainder $=f(+\frac{1}{3})=27 \times(\frac{1}{3})^{3}-9 \times(\frac{1}{3})^{2}-6 \times \frac{1}{3}-5$

$ =1-1-2-5=-7 . $

$ (\because 3 x-1=0 \Rightarrow x=\frac{1}{3}) $

~~ 33. Given, $f(x)=x^{3}+l x^{2}+m x+n$.

$f(1)=f(2)=0 \Rightarrow(x-1)$ and $(x-2)$ are factors of $f(x)$.

Since, $f(x)$ is polynomial of degree 3 , it shall have three linear factors. So, let the third factor be $(x-k)$.

Then, $f(x)=(x-1)(x-2)(x-k)$

$\Rightarrow f(x)=x^{3}+l x^{2}+m x+n=(x-1)(x-2)(x-k)$

Given, $f(4)=f(0)$

$\Rightarrow(4-1)(4-2)(4-k)=(-1)(-2)(-k)$

$\Rightarrow 24-6 k=-2 k \Rightarrow 4 k=24 \Rightarrow k=6$

$\therefore f(x)=(x-1)(x-2)(x-6)=(x^{2}-3 x+2)(x-6)$

$=x^{3}-9 x^{2}+20 x-12$

$\therefore x^{3}+l x^{2}+m x+n=x^{3}-9 x^{2}+20 x-12$

$\Rightarrow l=-9, m=20, n=-12$.

~~ 34. Let $x^{200}=(x^{2}-3 x+2) \cdot Q(x)+l x+m$

where, $Q(x)=$ quotient and $(l x+m)$ is the remainder

Now $(x^{2}-3 x+2)=0 \Rightarrow(x-1)(x-2)=0 \Rightarrow x=1,2$.

Substituting $x=1$ in $(i)$, we have,

$1^{200}=0 . Q \cdot(x)+l+m$

Similarly, for $x=2$,

$2^{200}=0 . Q(x)+2 l+m$

$\therefore l+m=1,2 l+m=2^{200}$

Solving we get, $l=2^{200}-1$ and $m=2-2^{200}$

Hence remainder $=l x+m=(2^{200}-1) x+(-2^{200}+2)$.

~~ 35. (i) Since $x+k$ is the HCF of the given expressions, therefore, $x=-k$ will make each expression zero.

$k^{2}-a k+b=0$

$k^{2}-b k+a=0$

Solving (i) and (ii) by the rule of cross multiplication,

$ \frac{k^{2}}{-a^{2}+b^{2}}=\frac{k}{b-a}=\frac{1}{-b+a} $

From last two relations, $k=\frac{b-a}{-(b-a)}=-1$

$ \begin{aligned} & \therefore \frac{k^{2}}{-a^{2}+b^{2}}=\frac{1}{-b+a} \Rightarrow \frac{(-1)^{2}}{-a^{2}+b^{2}}=\frac{1}{-b+a} \\ & \Rightarrow \frac{1}{(b-a)(b+a)}=\frac{-1}{(b-a)} \Rightarrow a+b=-1 . \end{aligned} $

(ii) Hint. $a k^{2}-a k+b=0$

$k^{2}-c k+d=0$

Solving by the rule of cross multiplication,

$\frac{k^{2}}{-a d+b c}=\frac{k}{b-a d}=\frac{1}{-a c+a}$

$\Rightarrow k=\frac{b-a d}{a(1-c)}, \frac{b c-a d}{b-a d}$.

~~ 36. Since $a+b+c=0 \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c$.

So, as $(a-b)+(b-c)+(c-a)=0$

$\Rightarrow(a-b)^{3}+(b-c)^{3}+(c-a)^{3}=3(a-b)(b-c)(c-a)$

$\therefore$ Given expression $=\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=\mathbf{3}$.

~~ 37. $a^{2}=b y+c z \Rightarrow a^{2}+a x=a x+b y+c z$

$\Rightarrow a(a+x)=a x+b y+c z$

Similarly, $b^{2}=c z+a x \Rightarrow b(b+y)=a x+b y+c z$

and $c^{2}=a x+b y \Rightarrow c(c+z)=a x+b y+c z$

Hence, $\frac{x}{a+x}+\frac{y}{b+y}+\frac{c}{c+z}$

$ \begin{aligned} & =\frac{a x}{a(a+x)}+\frac{b y}{b(b+y)}+\frac{c z}{c(c+z)} \\ & =\frac{x \cdot a}{a x+b y+c z}+\frac{y \cdot b}{a x+b y+c z}+\frac{z \cdot c}{a x+b y+c z} \\ & =\frac{a x+b y+c z}{a x+b y+c z}=\mathbf{1} . \end{aligned} $

~~ 38. Now, $x+y+z=0$

$ \begin{aligned} & \Rightarrow x=-y-z, y=-x-z, z=-x-y \\ & \therefore x(y-z)^{3}+y(z-x)^{3}+z(x-y)^{3} \\ & =(-y-z)(y-z)^{3}+(-z-x)(z-x)^{3}+(-x-y)(x-y)^{3} \\ & =-(y+z)(y-z)^{3}-(z+x)(z-x)^{3}-(x+y)(x-y)^{3} \\ & =-[(y^{2}-z^{2})(y-z)^{2}+(z^{2}-x^{2})(z-x)^{2}+(x^{2}-y^{2})(x-y)^{2}] \\ & =-[(y^{2}-z^{2})(y^{2}-2 y z+z^{2})+(z^{2}-x^{2})(z^{2}-2 x z+x^{2}). \\ & .+(x^{2}-y^{2})(x^{2}-2 x y+y^{2})] \\ & =-[(y^{A}-z^{A})-2 y z(y^{2}-z^{2})+(z^{A}-x^{A}). \\ & .-2 x z(z^{2}-x^{2})+(\chi^{4}-y^{4})-2 x y(x^{2}-y^{2})] \end{aligned} $

~~ 39. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$

$\Rightarrow(a+b+c)[\frac{b c+a c+a b}{a b c}]=1$

$\Rightarrow(a+b+c)(b c+a c+a b)=a b c$

$\Rightarrow a b c+a^{2} c+a^{2} b+b^{2} c+a b c+a b^{2}+b c^{2}+a c^{2}+a b c=a b c$

$\Rightarrow a^{2}(c+b)+b c(c+b)+a b(c+b)+a c(c+b)=0$

$\Rightarrow(b+c)(a^{2}+b c+a b+a c)=0$

$\Rightarrow(b+c)(a^{2}+a b+b c+a c)=0$

$\Rightarrow(b+c)[a(a+b)+c(a+b)]=0$

$\Rightarrow(b+c)(a+b)(c+a)=0$.

~~ 40. Let $\frac{x}{(b-c)(b+c-2 a)}=\frac{y}{(c-a)(c+a-2 b)}$

$ =\frac{z}{(a-b)(a+b-2 c)}=k . $

Then, $x=k(b-c)(b+c-2 a)$

$ \begin{aligned} & y=k(c-a)(c+a-2 b) \\ & z=k(a-b)(a+b-2 c) \\ & \therefore x+y+z=k(b-c)(b+c-2 a)+k(c-a)(c+a-2 b) \\ & +k(a-b)(a+b-2 c) \\ & =k(b^{2}-c^{2}-2 a b+2 c a)+k(c^{2}-a^{2}-2 b c+2 a b) \\ & +k(a^{2}-b^{2}-2 c a+2 b c) \\ & =k(b^{2}-c^{2}-2 a b+2 c a+c^{2}-a^{2}-2 b c+2 a b+a^{2}-b^{2}. \\ & -2 c a+2 b c) \\ & =k \times 0=\mathbf{0} \text{. } \end{aligned} $

~~ 41. $a+b+c=0$

$\Rightarrow a^{2}=(b+c)^{2}$ or $a=-b-c$

$\therefore$ Given expression $=\frac{a^{2}}{a^{2}-b c}+\frac{b^{2}}{b^{2}-c a}+\frac{c^{2}}{c^{2}-a b}$

$ \begin{aligned} & =\frac{(b+c)^{2}}{(b+c)^{2}-b c}+\frac{b^{2}}{b^{2}+c(b+c)}+\frac{c^{2}}{c^{2}+b(b+c)} \\ & =\frac{(b+c)^{2}}{b^{2}+c^{2}+b c}+\frac{b^{2}}{b^{2}+c^{2}+b c}+\frac{c^{2}}{c^{2}+b^{2}+b c} \\ & =\frac{b^{2}+c^{2}+2 b c+b^{2}+c^{2}}{b^{2}+c^{2}+b c}=\frac{2(b^{2}+c^{2}+b c)}{(b^{2}+c^{2}+b c)}=\mathbf{2} . \end{aligned} $

~~ 42. $\frac{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}{(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3}+(z^{2}-x^{2})^{3}}$

$ =\frac{3(x-y)(y-z)(z-x)}{3(x^{2}-y^{2})(y^{2}-z^{2})(z^{2}-x^{2})} $

$ \begin{matrix} { \begin{bmatrix} \because a+b+c=0 \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c \\ \text{ Here }(x-y)+(y-z)+(z-x)=0 \\ \text{ and }(x^{2}-y^{2})+(y^{2}-z^{2})+(z^{2}-x^{2})=0 \end{bmatrix} } \\ =\frac{3(x-y)(y-z)(z-x)}{3(x+y)(x-y)(y+z)(y-z)(z+x)(z-x)} \\ =\frac{1}{(x+y)(y+z)(z+x)}=[(x+y)(y+z)(z+x)]^{-1} \end{matrix} $

~~ 43. If $a+b+c=0$, then $a^{3}+b^{3}+c^{3}-3 a b c=0$

$ \begin{aligned} & \Rightarrow(a+b)(a^{2}-a b+b^{2})+c^{3}=3 a b c \\ & \Rightarrow(-c)(a^{2}-a b+b^{2})+c^{3}=3 a b c \quad[\because(a+b)=-c] \\ & \Rightarrow a^{2}-a b+b^{2}-c^{2}=-3 a b \\ & \Rightarrow a^{2}-a b+b^{2}+2 a b-c^{2}=-3 a b+2 a b \end{aligned} $

$ =a^{2}+a b+b^{2}=\boldsymbol{c}^{2}-\boldsymbol{a} \boldsymbol{b} . $

~~ 44. $\frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}$

$=\frac{1}{1+p+\frac{1}{q}}+\frac{1}{1+q+\frac{1}{r}}+\frac{1}{1+r+\frac{1}{p}}$ $=\frac{q}{q+p q+1}+\frac{r}{r+q r+1}+\frac{p}{p+p r+1}$

$=\frac{q}{q+\frac{1}{r}+1}+\frac{r}{r+\frac{1}{p}+1}+\frac{p}{p+p r+1} \quad[\because p q r=1]$

$=\frac{q r}{q r+1+r}+\frac{p r}{p r+1+p}+\frac{p}{p+p r+1}$

$=\frac{q r}{\frac{1}{p}+1+r}+\frac{p r}{p r+1+p}+\frac{p}{p+p r+1}$

$=\frac{p q r}{1+p+p r}+\frac{p r}{p r+1+p}+\frac{p}{p+p r+1}$

$=\frac{p q r+p r+p}{1+p+p r}=\frac{1+p r+p}{1+p+p r}=\mathbf{1}$.

~~ 45. Given, $c=\frac{y z}{y+z} \Rightarrow c y+c z=y z \Rightarrow y z-c z=c y \Rightarrow z(y-c)=c y$

$\Rightarrow z=\frac{c y}{y-c}$

Also $b=\frac{x z}{x+z} \Rightarrow z=\frac{b x}{x-b}$

$\therefore \frac{c y}{y-c}=\frac{b x}{x-b} \Rightarrow c y x-c y b=b x y-b x c$

$\Rightarrow c y x-c y b-b x y=-b x c$

$\Rightarrow-y(b x+b c-c x)=-b x c$

$\Rightarrow y=\frac{b x c}{b x+b c-c x}$

Now, $a=\frac{x y}{x+y} \Rightarrow y=\frac{a x}{x-a}$.

$\therefore \frac{b x c}{b x+b c-c x}=\frac{a x}{x-a}$

$\Rightarrow a b x^{2}+a b c x-a c x^{2}=b x^{2} c-a b c x$

$\Rightarrow 2 a b c x=x^{2}(b c+a c-a b) \Rightarrow x=\frac{2 a b c}{(b c+a c-a b)}$

~~ 46. $a+b+c=0 \Rightarrow(a+b+c)^{2}=0$

$\Rightarrow a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=0$

$\Rightarrow a^{2}+b^{2}+c^{2}=-(2 a b+2 b c+2 c a)$

Now, $\frac{a^{2}+b^{2}+c^{2}}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$

$=\frac{a^{2}+b^{2}+c^{2}}{a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 b c+c^{2}+a^{2}-2 c a}$

$ \begin{aligned} & =\frac{a^{2}+b^{2}+c^{2}}{2(a^{2}+b^{2}+c^{2})-(2 a b+2 b c+2 c a)} \\ & =\frac{a^{2}+b^{2}+c^{2}}{2(a^{2}+b^{2}+c^{2})+(a^{2}+b^{2}+c^{2})} \\ & =\frac{a^{2}+b^{2}+c^{2}}{3(a^{2}+b^{2}+c^{2})}=\frac{\mathbf{1}}{\mathbf{3}} . \end{aligned} $

~~ 47. $x+y+z=2 s$

Also, $(s-x)+(s-y)+(-z)=2 s-(x+y+z)$

$ =2 s-2 s=0 \text{. } $

$\Rightarrow(s-x)^{3}+(s-y)^{3}+(-z)^{3}-3(s-x)(s-y)(-z)=0$

$ \begin{bmatrix} \because a+b+c=0 \\ \Rightarrow a^{3}+b^{3}+c^{3}-3 a b c=0 \end{bmatrix} $

$ \Rightarrow(s-x)^{3}+(s-y)^{3}+3(s-x)(s-y)(z)=z^{3} $

~~ 48. Given, $x+\frac{1}{x}=p$

$ \Rightarrow(x+\frac{1}{x})^{2}=p^{2} $

$\Rightarrow x^{2}+\frac{1}{x^{2}}+2=p^{2} \quad \Rightarrow x^{2}+\frac{1}{x^{2}}=p^{2}-2$

$\Rightarrow(x^{2}+\frac{1}{x^{2}})^{3}=(p^{2}-2)^{3}$

$\Rightarrow x^{6}+\frac{1}{x^{6}}+3(x^{2}+\frac{1}{x^{2}})=p^{6}-8+6 p^{2}(p^{2}-2)$

$\Rightarrow x^{6}+\frac{1}{x^{6}}+3(p^{2}-2)=p^{6}-8+6 p^{2}(p^{2}-2)$

$\Rightarrow x^{6}+\frac{1}{x^{6}}=\boldsymbol{p}^{6}-6 \boldsymbol{p}^{4}-\mathbf{9} \boldsymbol{p}^{2}-\mathbf{2}$

~~ 49. Given, $x+y+z=0 \Rightarrow x+y=-z$

$\Rightarrow x^{2}+y^{2}+2 x y=z^{2} \Rightarrow x^{2}+y^{2}=z^{2}-2 x y$

$\therefore \frac{1}{x^{2}+y^{2}-z^{2}}=\frac{1}{z^{2}-2 x y-z^{2}}=\frac{1}{-2 x y}=-\frac{1}{2 x y}$

Similarly, $\frac{1}{y^{2}+z^{2}-x^{2}}=-\frac{1}{2 y z}$ and $\frac{1}{z^{2}+x^{2}-y^{2}}=-\frac{1}{2 z x}$

$ \begin{aligned} & \therefore \frac{1}{x^{2}+y^{2}-z^{2}}+\frac{1}{y^{2}+z^{2}-x^{2}}+\frac{1}{z^{2}+x^{2}-y^{2}} \\ & =-\frac{1}{2 x y}-\frac{1}{2 y z}-\frac{1}{2 z x}=-\frac{1}{2}[\frac{z+x+y}{x y z}]=\mathbf{0} \end{aligned} $

$[\because x+y+z=0]$

~~ 50. $x+y+z=0 \Rightarrow x=-y-z$

$ y=-x-z $

$ z=-x-y $

Now, $\frac{x^{2}}{2 x^{2}+y z}+\frac{y^{2}}{2 y^{2}+z x}+\frac{z^{2}}{2 z^{2}+x y}$

$=\frac{x^{2}}{x^{2}+x \cdot x+y z}+\frac{y^{2}}{y^{2}+y \cdot y+z x}+\frac{z^{2}}{z^{2}+z \cdot z+x y}$

$ \begin{aligned} & =\frac{x^{2}}{x^{2}+x(-y-z)+y z}+\frac{y^{2}}{y^{2}+y(-x-z)+z x} \\ & +\frac{z^{2}}{z^{2}+z(-x-y)+x y} \\ & =\frac{x^{2}}{x^{2}-x y-x z+y z}+\frac{y^{2}}{y^{2}-y x-z y+z x} \\ & =\frac{x^{2}}{x(x-y)-z(x-y)}+\frac{z^{2}}{y(y-x)-z x-z y+x y} \\ & =\frac{y^{2}-\frac{(y-x)}{z(z-x)-y(z-x)}}{(x-y)(x-z)}+\frac{z^{2}}{(y-x)(y-z)}+\frac{z^{2}}{(z-x)(z-y)} \\ & =-[\frac{x^{2}}{x^{2}}+\frac{y^{2}}{(x-y)(z-x)}-\frac{z^{2}}{(x-y)(y-z)}-\frac{(z-x)(y-z)}{(x-y)(y-z)(z-x)}. \\ & =-[\frac{x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)}{(x-z)(z-x)}] \end{aligned} $

Factorising the numerator.

~~ 51. Given, $(b+c-a) x=(c+a-b) y=(a+b-c) z=2$

$ \begin{aligned} & \Rightarrow x=\frac{2}{(b+c-a)} ; y=\frac{2}{(c+a-b)} ; z=\frac{2}{(a+b-c)} \\ & \therefore \frac{1}{x}=\frac{b+c-a}{2} ; \frac{1}{y}=\frac{c+a-b}{2} ; \frac{1}{z}=\frac{a+b-c}{2} \\ & \therefore(\frac{1}{y}+\frac{1}{z})=(\frac{c+a-b}{2}+\frac{a+b-c}{2})=\frac{2 a}{2}=a \\ & (\frac{1}{z}+\frac{1}{x})=(\frac{a+b-c}{2}+\frac{b+c-a}{2})=\frac{2 b}{2}=b \\ & (\frac{1}{x}+\frac{1}{y})=(\frac{b+c-a}{2}+\frac{a+c-b}{2})=\frac{2 c}{2}=c \\ & \therefore(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})(\frac{1}{x}+\frac{1}{y})=a \cdot b \cdot c=\boldsymbol{a} b c . \end{aligned} $

~~ 52. $a^{x} \cdot a^{y} \cdot a^{z}=(x+y+z)^{x+y+z}$

$\Rightarrow a^{x+y+z}=(x+y+z)^{x+y+z}$

$\Rightarrow a=(x+y+z)$

Now, $(x+y+z)^{y}=a^{x}$ (given)

$\Rightarrow(x+y+z)^{y}=(x+y+z)^{x} \Rightarrow y=x$

Similarly, $y=z$ and $z=x$.

$\therefore x=y=z=\frac{x+y+z}{3}=\frac{\boldsymbol{a}}{\mathbf{3}}$.

~~ 53. Given, $x+\frac{1}{x}=a$.

Now, $x^{3}+x^{2}+\frac{1}{x^{3}}+\frac{1}{x^{2}}=(x^{3}+\frac{1}{x^{3}})+(x^{2}+\frac{1}{x^{2}})$

$=(x+\frac{1}{x})^{3}-3(x+\frac{1}{x})+(x+\frac{1}{x})^{2}-2$

$=a^{3}-3 a+a^{2}-2=\boldsymbol{a}^{3}+\boldsymbol{a}^{\mathbf{2}}-\mathbf{3} \boldsymbol{a}-\mathbf{2}$.

~~ 54. If $x^{1 / 3}+y^{1 / 3}+z^{1 / 3}=0$, then

$(x^{1 / 3})^{3}+(y^{1 / 3})^{3}+(z^{1 / 3})^{3}=3 x^{1 / 3} y^{1 / 3} z^{1 / 3}$

$\Rightarrow x+y+z=3 x^{1 / 3} y^{1 / 3} z^{1 / 3}$

$ \begin{bmatrix} \because a+b+c=0 \\ \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c \end{bmatrix} $

Now taking the cube of both the sides, we have $(x+y+z)^{3}=(3 x^{1 / 3} y^{1 / 3} z^{1 / 3})^{3}=\mathbf{2 7} \boldsymbol{x y z}$.

~~ 55. Given expression

$ \begin{matrix} =\frac{2 a(b+c)(c+a)+2 b(a+b)(c+a)+2 c(a+b)(b+c)}{(a+b)(b+c)(c+a)} \\ +\frac{(b-c)(c-a)(a-b)}{(b+c)(c+a)(a+b)} \end{matrix} $

$2 a(b c+c^{2}+a b+a c)+2 b(a c+b c+a^{2}+a b)$

$=\frac{+2 c(a b+b^{2}+a c+b c)+(b c-c^{2}-a b+a c)(a-b)}{(b+c)(c+a)(a+b)}$

$2 a b c+2 a c^{2}+2 a^{2} b+2 a^{2} c+2 a b c+2 b^{2} c+2 b a^{2}$

$+2 a b^{2}+2 a b c+2 c b^{2}+2 a c^{2}+2 b c^{2}+a b c-a c^{2}$

$=\frac{-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c}{(b+c)(c+a)(a+b)}$

$=\frac{6 a b c+3 a c^{2}+3 a^{2} b+3 a^{2} c+3 b^{2} c+3 a b^{2}+3 b c^{2}}{(b+c)(c+a)(a+b)}$

$=\frac{3[2 a b c+a c^{2}+a^{2} b+a^{2} c+b^{2} c+a b^{2}+b c^{2}]}{(b+c)(c+a)(a+b)}$

$=\frac{3(b+c)(c+a)(a+b)}{(b+c)(c+a)(a+b)}=\mathbf{3}$.

SELF ASSESSMENT SHEET

$(x^{n}-a^{n})$ is divisible by $(x-a)$

(a) for all values of $n$

(b) for even values of $n$

(d) only for prime values of $n$

~~ 2. If $(x+1)$ is a factor of $x^{4}+9 x^{3}+7 x^{2}+9 a x+5 a^{2}$, then :

(a) $a=137$

(b) $5 a^{2}-9 a-1=0$

(d) $a=\sqrt{131}$

(CDS 2004)

~~ 3. When $x^{3}+2 x^{2}+4 x+b$ is divided by $(x+1)$, the quotient is $x^{2}+a x+3$ and the remainder is $-3+2 b$. What are the values of $a$ and $b$ respectively?

(a) 1,0

(b) $-1,0$

(d) $-1,-1$

~~ 4. If $x^{3}+p x+q$ and $x^{3}+q x+p$ have a common factor, then which of the following is correct?

(a) $p+q=0$

(b) $p+q-1=0$

(d) $p-q+1=0$

(CDS 2005)

~~ 5. $(2 x-3 y)^{3}+(3 y-4 z)^{3}+(4 z-2 x)^{3}$ can be factorised into which one of the following ?

(a) $(2 x+3 y+4 z)(2 x-3 y-4 z)$

(b) $(2 x+3 y-4 z)(2 x-3 y+4 z)$

(d) $6(2 x-3 y)(3 y-4 z)(2 z-x)$

(CDS 2005)

~~ 6. If $a x^{3}+b x^{2}+x-6$ has $(x+2)$ as a factor and leaves a remainder 4 , when divided by $(x-2)$, the value of $a$ and $b$ respectively are :

(a) $1,-2$

(b) 2,1

(d) $1,-1$

~~ 7. If $4 x^{2}-6 x+m$ is divisible by $x-3$, which one of the following is the greatest divisor of $m$ ?

(a) 9

(b) 12

(d) 36

(CDS 2006)

~~ 8. If $y=x+\frac{1}{x}$, then $x^{4}+x^{3}-4 x^{2}+x+1=0$ can be reduced to which one of the following ? $(x \neq 0)$

(a) $y^{2}+y-2=0$

(b) $y^{2}+y-4=0$

(d) $y^{2}+y+6=0$

(CDS 2006)

~~ 9. If $x^{2}=y+z, y^{2}=z+x, z^{2}=x+y$, then what is the value of: $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ ?

(a) 1

(b) 0

(d) 2

(CDS 2007)

~~ 10. $\frac{(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$ on simplification is equal to :

(a) 1

(b) $(a-b)(b-c)(c-a)$

(d) 0

ANSWERS

~~ 2. (b)

~~ 3. (a)

~~ 4. $(c)$

~~ 5. $(d)$

~~ 6. (c)

~~ 7. $(c)$

~~ 8. (c)

~~ 9. (a)

~~ 10. (c)

HINTS AND SOLUTIONS

~~ 2. Let $f(x)=x^{4}+9 x^{3}+7 x^{2}+9 a x+5 a^{2}$.

If $(x+1)$ is a factor of $f(x)$, then $\begin{aligned} f(-1)=0 & \Rightarrow(-1)^{4}+9(-1)^{3}+7(-1)^{2}+9 a(-1)+5 a^{2}=0 \\ & \Rightarrow 1-9+7-9 a+5 a^{2}=0 \Rightarrow \mathbf{5} \boldsymbol{a}^{\mathbf{2}}-\mathbf{9} \boldsymbol{a}-\mathbf{1}=\mathbf{0} .\end{aligned}$

~~ 3. Let $f(x)=x^{3}+2 x^{2}+4 x+b$.

When divided by $(x+1)$, the remainder $=f(-1)$

Given, remainder $=-3+2 b$

$\therefore-3+2 b=f(-1)=(-1)^{3}+2(-1)^{2}+4(-1)+b$

$\Rightarrow-3+2 b=-1+2-4+b$

$\Rightarrow-3+2 b=-3+b$.

This is only possible when $\boldsymbol{b}=\mathbf{0}$.

$\therefore f(x)=x^{3}+2 x^{2}+4 x$.

Now dividing $f(x)$ by $(x+1)$, we see that

alt text

$\therefore$ Quotient $=x^{2}+a x+3=x^{2}+x+3 \Rightarrow a=\mathbf{1}$. ~~ 4. Let the common factor be $x-\alpha$, then $x=\alpha$ will make the given expressions zero, i.e.,

$ \begin{aligned} & \alpha^{2}+p \alpha+q=0 \\ & \alpha^{2}+q \alpha+p=0 \end{aligned} $

Solving by the rule of cross-multiplication, we have

$ \frac{\alpha^{2}}{p^{2}-q^{2}}=\frac{\alpha}{q-p}=\frac{1}{q-p} $

From, last two relation, $\alpha=1$.

$\Rightarrow \alpha^{2}=\frac{p^{2}-q^{2}}{q-p} \Rightarrow 1=-(p+q) \Rightarrow \boldsymbol{p}+\boldsymbol{q}+\mathbf{1}=\mathbf{0}$.

~~ 5. Since $(2 x-3 y)+(3 y-4 z)+(4 z-2 x)=0$, therefore,

$(2 x-3 y)^{3}+(3 y-4 z)^{3}+(4 z-2 x)^{3}$

~~ 6. Let $f(x)=a x^{3}+b x^{2}+x-6$

$(x+2)$ is a factor of $f(x) \Rightarrow f(-2)=0$

$\therefore f(-2)=-8 a+4 b-2-6=0$

$\Rightarrow-8 a+4 b-8=0 \Rightarrow-2 a+b=2$

$(x-2)$ leaves a remainder 4 , when dividing $f(x) \Rightarrow f(2)=4$

$\therefore f(2)=8 a+4 b+2-6=4 \Rightarrow 8 a+4 b-8=0$

$\Rightarrow 2 a+b=2$

$\therefore$ From $(i)$ and $(i i) b=\mathbf{2}, a=\mathbf{0}$.

~~ 7. If $f(x)=4 x^{2}-6 x+m$ is divisible by $(x-3)$, then $f(3)=0$ $\Rightarrow 4(3)^{2}-6.3+m=0 \Rightarrow 36-18+m=0 \Rightarrow m=-18$.

The greatest divisor of $m=\mathbf{1 8}$.

~~ 8. $x^{4}+x^{3}-4 x^{2}+x+1=0$

$\Rightarrow x^{2}+x-4+\frac{1}{x}+\frac{1}{x^{2}}=0$

$\Rightarrow x^{2}+\frac{1}{x^{2}}+x+\frac{1}{x}-4=0$

$\Rightarrow(x+\frac{1}{x})^{2}-2+(x+\frac{1}{x})-4=0$

$\Rightarrow y^{2}+y-6=0$ ~~ 9. $x^{2}=y+z \Rightarrow x^{2}+x=x+y+z$ $\Rightarrow x(x+1)=x+y+z \Rightarrow \frac{x}{x+y+z}=\frac{1}{x+1}$

Similarly, $\frac{1}{y+1}=\frac{y}{x+y+z}$ and $\frac{1}{z+1}=\frac{z}{x+y+z}$

$ \begin{aligned} \therefore \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}= & \frac{x}{x+y+z}+\frac{y}{x+y+z} \\ & +\frac{z}{x+y+z}=\frac{x+y+z}{x+y+z}=\mathbf{1} . \end{aligned} $

~~ 10. Use the identity, if $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$.

Quadratic Equations

KEY FACTS

An equation in which the highest power of the variable is 2 is called a quadratic equation. $a x^{2}+b x+c=0$, where $a, b, c$ are constants is a general quadratic equation and $a \neq 0$, and $a, b, c \in R$.

2. Solving a Quadratic Equation : To find the roots of a quadratic equation is called solving a quadratic equation.

(a) Method I : Factorising the quadratic equation into linear factors.

The quadratic expression $a x^{2}+b x+c=0$ can be expressed as a product of two linear factors as the degree of the algebraic expression here is $\mathbf{2}$.

Let $a x^{2}+b x+c=(m x+n)(e x+f)$, where $m \neq 0, e \neq 0$.

Then, $a x^{2}+b x+c=0 \Rightarrow(m x+n)(e x+f)=0$

$\Rightarrow(m x+n)=0$ or $(e x+f)=0$

$\Rightarrow x=-\frac{n}{m}$ or $x=-\frac{f}{e}$

$\therefore$ The two roots of $a x^{2}+b x+c=0$ are $\frac{-n}{m}$ and $\frac{-f}{e}$.

(b) Method II : Using the formula.

$a x^{2}+b x+c=0$ $(a \neq 0)$

$\Rightarrow a x^{2}+b x=-c$ (Transposing the constant term)

$\Rightarrow x^{2}+\frac{b}{a} x=-\frac{c}{a}$ (Dividing by the coefficient of $x^{2}$ )

$\Rightarrow x^{2}+\frac{b}{a} x+\frac{b^{2}}{4 a^{2}}=\frac{b^{2}}{4 a^{2}}-\frac{c}{a} \quad(.$ Adding $(\frac{b}{2 a})^{2}$ on both the sides to make LHS a perfect square $)$

$\Rightarrow(x+\frac{b}{2 a})^{2}=\frac{b^{2}-4 a c}{4 a^{2}} \Rightarrow x+\frac{b}{2 a}=\frac{ \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\Rightarrow x=-\frac{b}{2 a} \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

The formula $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Hence, the roots of the equation $a x^{2}+b x+c=0$ are $\frac{-\boldsymbol{b}+\sqrt{\boldsymbol{b}^{\mathbf{2}}-\mathbf{4 a c}}}{\mathbf{2} \boldsymbol{a}}$ and $\frac{-\boldsymbol{b}-\sqrt{\boldsymbol{b}^{\mathbf{2}}-\mathbf{4 a c}}}{\mathbf{2} \boldsymbol{a}}$

These two values are called the roots of the equation and are also called the zeros of the function defined by $f(x)=a x^{2}+b x+c$.

3. Equations Reducible to Quadratic Equations

Type I : $a x^{2 n}+b x^{n}+c=0$

In this type of an equation, we put $x^{n}=y$. So, $a x^{2 n}+b x^{n}+c=0$ reduces to $a y^{2}+$ by $+c=0$.

Now solve for $y$ and hence for $x$.

Example. Solve : $4^{1+x}+4^{1-x}=10$ for $x$.

Sol. $4^{1+x}+4^{1-x}=10 \quad \Rightarrow 4^{1} .4^{x}+4^{1} .4^{-x}=10 \quad \Rightarrow 4.4^{x}+\frac{4}{4^{x}}=10$

$\Rightarrow 4.4^{2 x}+4=10 \times 4^{x} \Rightarrow 4.4^{2 x}-10.4^{x}+4=0$.

Let $4^{x}=y$. Then, the given equation becomes $4 y^{2}-10 y+4=0$

$ \begin{matrix} \Rightarrow 2 y^{2}-5 y+2=0 & \Rightarrow(y-2)(2 y-1)=0 & \Rightarrow y=2 \text{ or } \frac{1}{2} \\ \Rightarrow 4^{x}=2 \text{ or, } 4^{x}=+\frac{1}{2} \quad \Rightarrow 2^{2 x}=2^{1} \text{ or } 2^{2 x}=2^{-1} & \Rightarrow 2 x=1 \text{ or } 2 x=-1 \Rightarrow x=\frac{\mathbf{1}}{\mathbf{2}} \text{ or } \frac{-\mathbf{1}}{\mathbf{2}} . \end{matrix} $

Type II : $a z+\frac{b}{z}=c$, where $a, b, c$ are constants.

Example. If $\sqrt{\frac{2 x^{2}+x+2}{x^{2}+3 x+1}}+2 \sqrt{\frac{x^{2}+3 x+1}{2 x^{2}+x+2}}-3=0$, find $x$.

Sol. Let $\sqrt{\frac{2 x^{2}+x+2}{x^{2}+3 x+1}}=y$. Then, $y+2 \times \frac{1}{y}-3=0$

$ \begin{aligned} & \Rightarrow y^{2}-3 y+2=0 \Rightarrow(y-1)(y-2)=0 \Rightarrow y=1 \text{ or } 2 \\ & \therefore \quad y=1 \Rightarrow \sqrt{\frac{2 x^{2}+x+2}{x^{2}+3 x+1}}=1 \Rightarrow \frac{2 x^{2}+x+2}{x^{2}+3 x+1}=1 \Rightarrow 2 x^{2}+x+2=x^{2}+3 x+1 \\ & \Rightarrow \quad x^{2}-2 x+1=0 \quad \Rightarrow(x-1)^{2}=0 \Rightarrow x=1 \\ & y=2 \Rightarrow \sqrt{\frac{2 x^{2}+x+2}{x^{2}+3 x+1}}=2 \Rightarrow \frac{2 x^{2}+x+2}{x^{2}+3 x+1}=4 \\ & \Rightarrow 2 x^{2}+x+2=4 x^{2}+12 x+4 \Rightarrow 2 x^{2}+11 x+2=0 \\ & \therefore x=\frac{-11 \pm \sqrt{121-4 \times 2 \times 2}}{2 \times 2}=\frac{-11 \pm \sqrt{121-16}}{4}=\frac{-11 \pm \sqrt{105}}{4} . \quad \therefore x=1, \frac{-\mathbf{1 1} \pm \sqrt{\mathbf{1 0 5}}}{4} . \end{aligned} $

Type III : Equations of type $(x+a)(x+b)(x+c)(x+d)+k=0$, where the sum of two of the quantities $a, b, c, d$ is equal to the sum of the other two.

Example. $(x+1)(x+2)(x+3)(x+4)+1=0$

Sol. $[(x+1)(x+4)][(x+2)(x+3)]+1=0$

$\Rightarrow(x^{2}+5 x+4)(x^{2}+5 x+6)+1=0$

Let $x^{2}+5 x=y$. Then, $(y+4)(y+6)+1=0$

$\Rightarrow y^{2}+10 y+24+1=0 \Rightarrow y^{2}+10 y+25=0 \Rightarrow(y+5)^{2}=0 \Rightarrow y=-5$

$\therefore x^{2}+5 x=-5 \Rightarrow x^{2}+5 x+5=0 . \Rightarrow x=\frac{-5 \pm \sqrt{25-20}}{2}=\frac{-\mathbf{5} \pm \sqrt{\mathbf{5}}}{\mathbf{2}}$.

4. Important Properties of Inequalities

An inequality will still hold after each side has been increased, diminished, multiplied or divided by the same positive quantity.
In an inequality any term may be transposed from one side to the other if its sign is changed.
Both the sides of an inequality can be multiplied or divided by the same negative number by reversing the sign of inequality.
Nature of Roots. A quadratic equation $a x^{2}+b x+c=0, a \neq 0$, has two roots which by the quadratic formula are as under :

$ \frac{-b+\sqrt{b^{2}-4 a c}}{2 a} \text{ and } \frac{-b-\sqrt{b^{2}-4 a c}}{2 a} $

The expression $\boldsymbol{b}^{2}-\mathbf{4} \boldsymbol{a} \boldsymbol{c}$ is called the discriminant.

Examining the nature of the roots means to see what type of roots the equation has, that is, whether they are real or imaginary, real or irrational, equal or unequal. The nature of the roots depends entirely on the value of the discriminant $D=b^{2}-4 a c$

Thus, if $a, b, c$ are rational, then

I. If $\boldsymbol{D}=\boldsymbol{b}^{2}-\mathbf{4} \boldsymbol{a} \boldsymbol{c}>\mathbf{0}$ (i.e, positive), the roots are real and unequal.

Also,

(a) If $\boldsymbol{D}=\boldsymbol{b}^{2}-4 \boldsymbol{4} \boldsymbol{c}$ is a perfect square, the roots are rational.

(b) If $\boldsymbol{D}=\boldsymbol{b}^{2}-4 \boldsymbol{a c}$ is not a perfect square, the roots are irrational.

(c) If $\boldsymbol{D}=\boldsymbol{b}^{2}-4 \boldsymbol{4} \boldsymbol{c}=\mathbf{0}$, the roots are equal, each being equal to $\frac{-\boldsymbol{b}}{2 \boldsymbol{a}}$.

So, $a x^{2}+b x+c=0$ is a perfect square if $D=0$.

II. If $\boldsymbol{D}=\boldsymbol{b}^{\mathbf{2}}-\mathbf{4} \boldsymbol{a} \boldsymbol{c}<\mathbf{0}$ (i.e., negative), the roots are imaginary (complex).

~~ Example. Examine the nature of the roots of the equations: (i) $2 x^{2}+2 x+3=0$ (ii) $2 x^{2}-7 x+3=0$ (iii) $x^{2}-5 x-2=0$ (iv) $4 x^{2}-4 x+1=0$.

Sol. (i) $2 x^{2}+2 x+3=0$

(Here, $a=2, b=2, c=3$ )

$\therefore D=b^{2}-4 a c=(2)^{2}-4 \times 2 \times 3=4-24=\mathbf{- 2 0}<\mathbf{0}$

Hence, roots are imaginary.

(ii) $2 x^{2}-7 x+3=0 \quad$ (Here, $a=2, b=-7, c=3$ )

$\therefore D=b^{2}-4 a c=49-24=\mathbf{2 5}>\mathbf{0}$ and a perfect square

Hence, roots are real and rational.

(iii) $x^{2}-5 x-2=0$.

(Here, $a=1, b=-5, c=-2$ )

$\therefore D=b^{2}-4 a c=25+8=\mathbf{3 3}>\mathbf{0}$ and not a perfect square

Hence, roots are real and irrational. (iv) $4 x^{2}-4 x+1=0$ (Here, $a=4, b=-4, c=1$ ) $\therefore D=b^{2}-4 a c=16-16=\mathbf{0}$.

Hence, roots are real and equal.

6. Sum and Product of Roots:

If the two roots of the quadratic equation $a x^{2}+b x+c=0$ obtained by the quadratic formula be denoted by $\alpha$ and $\beta$, then we have

$ \begin{gathered} \alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}, \beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a} \\ \therefore \text{ Sum of roots }=\alpha+\beta=-\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}+\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}=\frac{-2 b}{2 a}=\frac{-\boldsymbol{b}}{\boldsymbol{a}} \\ \text{ Product of roots }=\alpha \beta=(\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}) \times(\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}) \\ \qquad=\frac{(-b)^{2}-(\sqrt{b^{2}-4 a c})^{2}}{4 a^{2}}=\frac{b^{2}-(b^{2}-4 a c)}{4 a^{2}}=\frac{4 a c}{4 a^{2}}=\frac{\boldsymbol{c}}{\boldsymbol{a}} . \end{gathered} $

Thus, $\quad$ Sum of roots $=\frac{- \text{ Coeff. of } x}{\text{ Coeff. of } x^{2}} ;$ Product of roots $=\frac{\text{ Constant term }}{\text{ Coeff. of } x^{2}}$

Thus, if $\alpha, \beta$ be the roots of the equation $6 x^{2}-5 x+7=0$, then $\alpha+\beta=-(-5 / 6)=5 / 6, \alpha \beta=7 / 6$.

7. Values of the Symmetric Functions of the Roots

If $\alpha, \beta$ be the roots of a given quadratic equation and we wish to find the value of a symmetric function of $\alpha$ and $\beta$, we can do so by proceeding as follows :

Method:
Step I. Write the values of $\alpha+\beta$ and $\alpha \beta$ from the given equation.
Step II. Express the given function in terms of $\alpha+\beta$ and $\alpha \beta$.
Step III. Substitute the values of $\alpha+\beta$ and $\alpha \beta$ from step 1.

Caution:Do not find the values of $\alpha$ and $\beta$ separately.

The following algebraic relations can be very useful :

$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$
$\alpha^{2}-\beta^{2}=(\alpha+\beta)(\alpha-\beta)=(\alpha+\beta) \sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$
$\alpha^{3}+\beta^{3}=(\alpha+\beta)(\alpha^{2}-\alpha \beta+\beta^{2})=(\alpha+\beta)[(\alpha+\beta)^{2}-3 \alpha \beta]=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$\alpha^{3}-\beta^{3}=(\alpha-\beta)(\alpha^{2}+\alpha \beta+\beta^{2})=(\alpha-\beta)[(\alpha-\beta)^{2}+3 \alpha \beta]=(\alpha-\beta)^{3}+3 \alpha \beta(\alpha-\beta)$
$\alpha^{4}-\beta^{4}=(\alpha^{2}+\beta^{2})^{2}-2 \alpha^{2} \beta^{2}=[(\alpha+\beta)^{2}-2 \alpha \beta]^{2}-2(\alpha \beta)^{2}$
$\alpha^{4}-\beta^{4}=(\alpha^{2}-\beta^{2})(\alpha^{2}+\beta^{2})=(\alpha-\beta)(\alpha+\beta)(\alpha^{2}+\beta^{2})=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}(\alpha+\beta)[(\alpha+\beta)^{2}-2 \alpha \beta]$
Formation of Equations with given Roots :

Suppose we have to form the equation whose roots are $\alpha$ and $\beta$. Then, as $x=\alpha, x=\beta$ are the roots of the equation, so $(x-\alpha)=0$ and $(x-\beta)=0$

$\therefore(x-\alpha)(x-\beta)=0$

$\Rightarrow x^{2}-(\alpha+\beta) x+\alpha \beta=0$

$\Rightarrow \boldsymbol{x}^{2}$ - (Sum of roots) $\boldsymbol{x}+$ Product of roots $=0$.

Thus, the equation whose roots are 5 and 7 is $x^{2}-(5+7) x+5 \times 7=0 \Rightarrow x^{2}-12 x+35=0$.

To find the condition when a relation between the two roots is given

Step I. Let one root be $\alpha$. Write the other root using the given relation.

Step II. Write the sum and product of the roots.

Step III. Eliminate a from the two relations obtained in Step II.

~~ Ex. Find the condition that one root of $a x^{2}+b x+c=0$ may be four times the other.

Sol. Let the roots be $\alpha$ and $4 \alpha$. Then,

$ \begin{matrix} \alpha+4 \alpha=5 \alpha=-\frac{b}{a} \\ \alpha .4 \alpha=4 \alpha^{2}=\frac{c}{a} \end{matrix} $

From (i) $\quad \alpha=-\frac{b}{5 a}$

$\therefore$ From (ii) 4. $(-\frac{b}{5 a})^{2}=\frac{c}{a} \Rightarrow \frac{4 b^{2}}{25 a^{2}}=\frac{c}{a} \Rightarrow \mathbf{4} \boldsymbol{b}^{2}=\mathbf{2 5} \boldsymbol{a} \boldsymbol{c}$.

Special Roots: For a quadratic equation $a x^{2}+b x+c=0, a \neq 0, a, b, c \in R$ whose roots are $\alpha$ and $\beta$.

(a) Reciprocal roots

If $\alpha=\frac{1}{\beta}$, then $\alpha \beta=1 \Rightarrow \alpha \beta=\frac{c}{a}=1 \Rightarrow \boldsymbol{c}=\boldsymbol{a}$

Thus the roots of a quadratic equation will be reciprocal of each other if coefficient of $x^{2}=$ constant term.

(b) Zero roots

Case I : When one root is zero, say $\alpha=0$.

Then, $\alpha \beta=0 \Rightarrow \frac{c}{a}=0, \boldsymbol{c}=\mathbf{0}$ as $a \neq 0$

Case II : When both roots are zero, i.e, $\alpha=0, \beta=0$

Then, $\alpha+\beta=0$ and $\alpha \beta=0$

$\Rightarrow-\frac{b}{a}=0$ and $\frac{c}{a}=0 \Rightarrow \boldsymbol{b}=\mathbf{0}$ and $\boldsymbol{c}=\mathbf{0}$ as $a \neq 0$.

Let $\alpha, \beta$ be the roots of $a x^{2}+b x+c=0$

Then, the equation whose roots are $\frac{1}{\alpha}, \frac{1}{\beta}$ is $c x^{2}+b x+a=0$

Now if one root of (ii) is zero then $a=0 \Rightarrow$ the corresponding root of $(i)$ is $\frac{1}{0}=\infty$.

If both the roots of (ii) are zero, then $a=0, b=0 \Rightarrow$ both the corresponding roots of ( $i$ ) are infinitely large.

Thus, For one root to be infinite, $\boldsymbol{a}=\mathbf{0}$;

For both roots to be infinite, $\boldsymbol{a}=\mathbf{0}, \boldsymbol{b}=\mathbf{0}$

Signs of the Roots

(a) Positive roots : Both the roots will be positive if $\alpha+\beta$ and $\alpha \beta$ are both positive, i.e, $-\frac{b}{a}$ and $\frac{c}{a}$ are positive. It will be so when $b$ and $a$ are of opposite signs and $c$ and $a$ are of the same sign.

(b) Negative roots : Both the roots will be negative if $\alpha+\beta$ is negative and $\alpha \beta$ is positive, i.e., $-\frac{b}{a}$ is negative and $\frac{c}{a}$ is positive, i.e., when $\boldsymbol{a}, \boldsymbol{b}$ and $\boldsymbol{c}$ all have the same sign.

(c) Roots of opposite signs. It will occur when $\alpha \beta$ is negative, i.e., $\boldsymbol{c}$ and $\boldsymbol{a}$ are of opposite signs.

(d) Roots equal in magnitude but opposite in sign. It will occur if $\alpha+\beta=0$, i.e., $-\frac{b}{a}=0$, i.e. $b=0$.

For real solutions the signs of $c$ and $a$ should be opposite.

Common Roots
To find the condition that two quadratic equations may have one common root.

Let the two quadratic equations be $a x^{2}+b x+c=0, a_1 x^{2}+b_1 x+c_1=0$ and let $\alpha$ be their common root. Then,

$ \begin{matrix} a \alpha^{2}+b \alpha+c=0 \\ a_1 \alpha^{2}+b_1 \alpha+c_1=0 \end{matrix} $

Solving them by the rule of cross-multiplication, we have, $\frac{\alpha^{2}}{b c_1-c b_1}=\frac{\alpha}{c a_1-a c_1}=\frac{1}{a b_1-a_1 b}$

$ \begin{aligned} & \alpha^{2}=\frac{b c_1-c b_1}{a b_1-a_1 b}, \alpha=\frac{c a_1-c a_1}{a b_1-a_1 b} \\ & \Rightarrow \quad \frac{b c_1-c b_1}{a b_1-a_1 b}=\frac{(c a_1-a c_1)^{2}}{(a b_1-a_1 b)^{2}} \Rightarrow(b c_1-c b_1)(a b_1-a_1 b)=(c a_1-a c_1)^{2} \end{aligned} $

To find the condition that the two quadratic equations may have both the roots common.

Let the common roots of the equations $a x^{2}+b x+c=0$ and $a_1 x^{2}+b_1 x+c_1=0$ be $\alpha$ and $\beta$. Then,

From first equation, $\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$

From second equation, $\alpha+\beta=-\frac{b_1}{a_1}, \alpha \beta=\frac{c_1}{a_1}$

So, $-\frac{b}{a}=-\frac{b_1}{a_1}$ and

$\frac{c}{a}=+ \frac{c_1}{a_1} \Rightarrow \frac{\boldsymbol a}{\boldsymbol a_1} = \frac{\boldsymbol b}{\boldsymbol b_1} = \frac{\boldsymbol c}{\boldsymbol c_1}$.

13. Transformed Equations

If $f(x)=a x^{2}+b x+c=0$ is a quadratic equation, then the equation whose roots are the :

(a) Reciprocals of the roots of $\boldsymbol{f}(\boldsymbol{x})=\mathbf{0}$ is $\boldsymbol{f}(\frac{\mathbf{1}}{\boldsymbol{x}})=\mathbf{0}$ i.e., $\frac{a}{x^{2}}+\frac{b}{x}+c=0 \Rightarrow \boldsymbol{c} \boldsymbol{x}^{2}+\boldsymbol{b x}+\boldsymbol{a}=\mathbf{0}$.

(b) Roots of $f(x)=0$, each increased by a constant $k$ is $f(x-k)=0$, i.e., $a(x-k)^{2}+b(x-k)+c=0$.

(d) Roots of $\boldsymbol{f}(\boldsymbol{x})=\mathbf{0}$ with signs changed is $f(-x)=0$, i.e. $a(-x)^{2}+b(-x)+c=0 \Rightarrow \boldsymbol{a} \boldsymbol{x}^{2}-\boldsymbol{b} \boldsymbol{x}+\boldsymbol{c}=\mathbf{0}$. (e) Roots of $\boldsymbol{f}(\boldsymbol{x})=\mathbf{0}$ with each multiplied by $\boldsymbol{k} \neq \boldsymbol{0}$ is $\boldsymbol{f}(\frac{\boldsymbol{x}}{\boldsymbol{k}})=\mathbf{0}$ i.e. $a(\frac{x}{k})^{2}+b(\frac{x}{k})+c=0$ i.e., $\boldsymbol{a} \boldsymbol{x}^{2}+\boldsymbol{k} \boldsymbol{b} \boldsymbol{x}$ $+\boldsymbol{k}^{2} \boldsymbol{c}=\mathbf{0}$

14. Relation between the roots of a cubic equation and its coefficients.

Let the cubic equation be $x^{3}+S_1 x^{2}+S_2 x+S_3=0$, where $S_1, S_2, S_3$ are the coefficients.

Let $\alpha, \beta, \gamma$ be the roots of the given cubic equation. Then,

$ S_1=-(\alpha+\beta+\gamma), \quad S_2=(\alpha \beta+\beta \gamma+\gamma \alpha), \quad S_3=-(\alpha \beta \gamma) $

Conversely, if the roots of a cubic equation are given as $\alpha_1, \beta_1, \gamma$, then its equation can be written as :

$ \begin{aligned} & \boldsymbol{x}^{3}-\boldsymbol{S}_1 \boldsymbol{x}^{2}+\boldsymbol{S}_2 \boldsymbol{x}-\boldsymbol{S}_3=\mathbf{0} \text{, where } \\ & S_1=(\alpha+\beta+\gamma), S_2=(\alpha \beta+\beta \gamma+\gamma \alpha) \text{ and } S_3=\alpha \beta \gamma \end{aligned} $

15. Relation between the roots of a bi-quadratic equation (degree 4) and its coefficients.

A bi-quadratic equation, whose roots are $\alpha, \beta, \gamma$ and $\delta$ is

$ x^{4}-S_1 x^{3}+S_2 x^{2}-S_2 x+S_4=0 $

where $\quad S_1=\alpha+\beta+\gamma+\delta, S_2=\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta, S_3=\alpha \beta \gamma+\alpha \beta \delta+\alpha \gamma \delta+\beta \gamma \delta_L, S_4=\alpha \beta \gamma \delta$.

SOLVED EXAMPLES

~~ Ex. 1 . What are the roots of the equation $(a+b+x)^{-1}=a^{-1}+b^{-1}+x^{-1}$ ?

(CDS 2007)

Sol. Given, $\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x} \Rightarrow \frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}$

$ \begin{aligned} & \Rightarrow \frac{x-(a+b+x)}{x(a+b+x)}=\frac{a+b}{a b} \Rightarrow \frac{-(a+b)}{x(a+b+x)}=\frac{a+b}{a b} \Rightarrow-a b=x^{2}+(a+b) x \\ & \Rightarrow x^{2}+(a+b) x+a b=0 \Rightarrow(x+a)(x+b)=0 \Rightarrow \boldsymbol{x}=-\boldsymbol{a},-\boldsymbol{b} . \end{aligned} $

~~ Ex. 2 . What is one of the roots of the equation $\sqrt{\frac{2 x}{3-x}}-\sqrt{\frac{3-x}{2 x}}=\frac{3}{2}$ ?

(a) 1

(b) 2

(d) 4

(CDS 2008)

Sol. Given equation is $\sqrt{\frac{2 x}{3-x}}-\sqrt{\frac{3-x}{2 x}}=\frac{3}{2}$

Let $\sqrt{\frac{2 x}{3-x}}=a$. Then, the given equation reduces to $a-\frac{1}{a}=\frac{3}{2}$

$\Rightarrow 2(a^{2}-1)=3 a \Rightarrow 2 a^{2}-3 a-2=0 \quad \Rightarrow \quad 2 a^{2}-4 a+a-2=0$

$\Rightarrow 2 a(a-2)+1(a-2)=0 \Rightarrow(2 a+1)(a-2)=0$

$\Rightarrow a-2=0 \Rightarrow a=2$ or $(2 a+1)=0 \Rightarrow a=-\frac{1}{2}$

$\therefore \sqrt{\frac{2x}{3-x}}=2 \Rightarrow 2x=4(3-x) \text{ or } \quad \sqrt{\frac{2x}{3-x}} = -\frac{1}{2} \Rightarrow \frac{2x}{3-x} $

$\Rightarrow 6 x=12 \Rightarrow x=2 \text { or } \quad \Rightarrow 8 x=3-x \Rightarrow 9 x=3 \Rightarrow x=\frac{1}{3} \\ $

Hence, according to the given options, $(b)$ is correct.

~~ Ex. 3 . If $3^{x}+27(3^{-x})=12$, then what is the value of $x$ ?

(CDS 2009)

Sol. Given, $3^{x}+27(3^{-x})=12$

Let $3^{x}=y$. Then, $y+\frac{27}{y}=12 \Rightarrow y^{2}-12 y+27=0$

$\Rightarrow y^{2}-9 y-3 y+27=0 \Rightarrow(y-3)(y-9)=0 \quad \Rightarrow y=3,9$

$\Rightarrow 3^{x}=3$ or $3^{x}=9 \Rightarrow x=1$ or 2 .

~~ Ex. 4 . What is the ratio of sum of squares of roots to the product of the roots of the equation $7 x^{2}+12 x+18=0$ ?

(CDS 2009)

Sol. Let $\alpha, \beta$ be the roots of the equation $7 x^{2}+12 x+18=0$.

$[.$ For a quadratic equation $a x^{2}+b x+c=0$, sum of roots $=-\frac{b}{a}$, product of roots $.=+\frac{c}{a}]$

$\therefore \alpha+\beta=-\frac{12}{7}$ and $\alpha \beta=\frac{18}{7}$

$\Rightarrow(\alpha+\beta)^{2}=(\frac{-12}{7})^{2} \Rightarrow \alpha^{2}+\beta^{2}+2 \alpha \beta=\frac{144}{49}$

$\Rightarrow \alpha^{2}+\beta^{2}=\frac{144}{49}-\frac{36}{7}=\frac{-108}{49}$

$\therefore$ Required ratio $=\alpha^{2}+\beta^{2}: \alpha \beta=\frac{\frac{-108}{49}}{18 / 7}=-\frac{6}{7}=-6: 7$.

~~ Ex. 5 . What is the value of a for which the equation $2 x^{2}+2 \sqrt{6} x+a=0$ has equal roots? (Kerala PET 2010)

Sol. The equation $2 x^{2}+2 \sqrt{6} x+a=0$ has equal roots if the discriminant $D=0$.

$ \begin{matrix} \therefore \text{ Here, } D=(2 \sqrt{6})^{2}-4 \times(2) \times(a)=0 & {[D=b^{2}-4 a c \text{ for } a x^{2}+b x+c=0]} \\ \Rightarrow 24-8 a & =0 \Rightarrow a=3 . \end{matrix} $

~~ Ex. 6 . Of the following quadratic equations, which is the one whose roots are 2 and -15 ?

(a) $x^{2}-2 x+15=0$

(b) $x^{2}+15 x-2=0$

(d) $x^{2}-30=0$.

(MAT)

Sol. Sum of roots $=2+(-15)=-13$; Product of roots $=2 \times(-15)=-30$.

$\therefore$ Required equation is $x^{2}-$ (sum of roots) $x+$ product of roots $=0$

$\Rightarrow$ Reqd. equation $=\boldsymbol{x}^{2}-\mathbf{1 3} \boldsymbol{x}-\mathbf{3 0}=\mathbf{0}$.

~~ Ex. 7 . If one of the roots of the equation $x^{2}+a x+3=0$ is 3 and one of the roots of the equation $x^{2}+a x+b=0$ is three times the other root, then what is the value of $b$ ?

(J&K CET 2005)

Sol. Let 3 and $\alpha$ be the roots of the equation $x^{2}+a x+3=0$

Then, sum of roots $=3+\alpha=-a \quad \ldots(i), \quad$ Product of roots $=3 \alpha=3$

From (ii) $\alpha=1 . \quad \therefore$ Substituting $\alpha=1$ in (i), we get $a=-4$.

$\therefore$ The second equation $x^{2}+a x+b=0$ becomes $x^{2}-4 x+b=0$.

Let $\beta$ and $3 \beta$ be the roots of this equation. Then, sum of roots $\beta+3 \beta=4 \Rightarrow 4 \beta=4 \Rightarrow \beta=1$

and Product of roots $=\beta \times 3 \beta=b \Rightarrow 3 \beta^{2}=b \Rightarrow \boldsymbol{b}=\mathbf{3}$.

~~ Ex. 8 . If $\alpha, \beta$ be the two roots of the equation $x^{2}+x+1=0$, then the equation whose roots are $\alpha / \beta$ and $\beta / \alpha$ is ?

(a) $x^{2}-x-1=0$

(b) $x^{2}-x+1=0$

(d) $x^{2}+x+1=0$

(UPSEE 2005)

Sol. Let $\alpha, \beta$ be the roots of the equations $x^{2}+x+1=0$. Then,

Sum of roots $=\alpha+\beta=-1$, Product of roots $=\alpha \beta=1$

Now the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

$ \begin{aligned} & x^{2}-(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}) x+(\frac{\alpha}{\beta} \times \frac{\beta}{\alpha})=0 . \\ & \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}=\frac{(-1)^{2}-2(1)}{1}=-1 \text{ and } \frac{\alpha}{\beta} \times \frac{\beta}{\alpha}=1 . \end{aligned} $

$\therefore$ Required equation $=\boldsymbol{x}^{2}+\boldsymbol{x}+\mathbf{1}=\mathbf{0}$.

~~ Ex. 9 . If the difference in the roots of the equation $x^{2}-p x+q=0$ is unity, then which one of the following is correct?

(a) $p^{2}+4 q=1$

(b) $p^{2}-4 q=1$

(d) $p^{2}-4 q=-1$.

(CDS 2005)

Sol. Given, $x^{2}-p x+q=0$

Let $\alpha, \beta$ be the roots of the given equation. Then,

$ \alpha+\beta=-\frac{(-p)}{1}=p \quad \ldots(i), \quad \alpha \beta=\frac{q}{1}=q $

Also, $\alpha-\beta=1$ (given)

$\therefore$ From (i) and (iii), $2 \alpha=p+1 \Rightarrow \alpha=\frac{p+1}{2}$

$\therefore$ From (i) and (iii), $2 \beta=p-1 \Rightarrow \beta=\frac{p-1}{2}$

Substituting these values of $\alpha$ and $\beta$ in (ii), we have $(\frac{p+1}{2})(\frac{p-1}{2})=q$

$\Rightarrow \frac{p^{2}-1}{4}=q \Rightarrow p^{2}-1=4 q \Rightarrow \boldsymbol{p}^{2}-4 q=\mathbf{1}$.

~~ Ex. 10 . If the roots of the equation $x^{2}+x+1=0$ are in the ratio of $m: n$, then which one of the following relation holds?

(a) $m+n+1=0$

(b) $\frac{m}{n}+\frac{n}{m}+1=0$

(d) $\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}+1=0$.

Sol. $x^{2}+x+1=0$

$x^{2}+x+1=0$ $\therefore$ Roots are $=\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \quad($ where $=i=\sqrt{-1})$

Given, $\frac{\frac{-1+\sqrt{3} i}{2}}{\frac{-1-\sqrt{3} i}{2}}=\frac{m}{n} \Rightarrow \frac{-1+\sqrt{3} i}{-1-\sqrt{3} i}=\frac{m}{n}$

$\Rightarrow \frac{m+n}{m-n}=\frac{(-1+\sqrt{3} i)+(-1-\sqrt{3} i)}{(-1+\sqrt{3} i)-(-1-\sqrt{3} i)}=\frac{-2}{2 \sqrt{3} i}$

(Applying componendo and dividendo)

$\Rightarrow \frac{m+n}{m-n}=\frac{-1}{\sqrt{3} i}=\frac{i^{2}}{\sqrt{3} i}=\frac{i}{\sqrt{3}}$

$(\because i^{2}=-1)$

$\Rightarrow(\frac{m+n}{m-n})^{2}=(\frac{i}{\sqrt{3}})^{2} \Rightarrow \frac{m^{2}+n^{2}+2 m n}{m^{2}+n^{2}-2 m n}=\frac{-1}{3} \Rightarrow \frac{(m^{2}+n^{2}+2 m n)+(m^{2}+n^{2}-2 m n)}{(m^{2}+n^{2}+2 m n)-(m^{2}+n^{2}-2 m n)}=\frac{-1+3}{-1-3}$

$\Rightarrow \frac{2(m^{2}+n^{2})}{2(2 m n)}=\frac{2}{-4} \Rightarrow \frac{m^{2}+n^{2}}{2 m n}=\frac{1}{-2} \Rightarrow \frac{m^{2}+n^{2}}{m n}=-1 \Rightarrow \frac{\boldsymbol{m}}{\boldsymbol{n}}+\frac{\boldsymbol{n}}{\boldsymbol{m}}+\mathbf{1}=\mathbf{0}$.

~~ Ex. 11 . If the roots of the equation $x^{2}-2 a x+a^{2}+a-3=0$ are real and less than 3 , then which one of the following is correct?

(a) $a<2$

(b) $2<a<3$

(d) $a>4$

(CDS 2012)

Sol. If the roots of the equation $x^{2}-2 a x+a^{2}-a-3=0$ are real and less than 3 , then $D \geq 0$ and $f(3)>0$.

$\Rightarrow 4 a^{2}-4(a^{2}+a-3) \geq 0$ and $(3)^{2}-2 a(3)+a^{2}+a-3>0$

$\Rightarrow a^{2}-a^{2}-a+3 \geq 0$ and $9-6 a+a^{2}+a-3>0$

$\Rightarrow-a+3 \geq 0$ and $a^{2}-5 a+6>0 \Rightarrow a-3 \leq 0$ and $(a-2)(a-3)>0$

$\Rightarrow a \leq 3$ and $a<2$ or $a>3 \Rightarrow \boldsymbol{a}<\mathbf{2}$.

~~ Ex. 12 . What are the number of solutions for real $x$, which satisfy the equation

$ 2 \log _2 \log _2 x+\log _{1 / 2} \log _2(2 \sqrt{2} x)=1 ? $

Sol. $2 \log _2 \log _2 x+\log _{1 / 2} \log _2(2 \sqrt{2} x)=1$

$\Rightarrow 2 \log _2 \log _2 x-\log _2 \log _2(2 \sqrt{2} x)=1$ $\Rightarrow \log _2(\log _2 x)^{2}-\log _2(\log _2(2 \sqrt{2} x))=\log _2 2$ $\Rightarrow \log _2 \frac{(\log _2 x)^{2}}{\log _2(2 \sqrt{2} x)}=\log _2 2 \Rightarrow \frac{(\log _2 x)^{2}}{\log _2(2 \sqrt{2} x)}=2 \Rightarrow(\log _2 x)^{2}=2 \log _2(2 \sqrt{2} x)$ $.\Rightarrow(\log _2 x)^{2}=2 \log _2(2^{3 / 2} x) \Rightarrow(\log _2 x)^{2}=2[\frac{3}{2} \log 2 x]=2[3 / 2{\log _2 x)+\log _2 x}]$ $\Rightarrow(\log _2 x)^{2}=3+2 \log _2 x \quad \Rightarrow(\log _2 x)^{2}-2 \log _2 x-3=0$ $\Rightarrow(\log _2 x-3)(\log _2 x+1)=0 \Rightarrow \log _2 x=3$ or $\log _2 x=-1 \Rightarrow x=2^{3}=8$ or $x=2^{-1}=\frac{1}{2}$

But for $x=\frac{1}{2}, \log _2 \log _2(\frac{1}{2})$ is undefined so $\boldsymbol{x}=\mathbf{8}$ is the only possible value of $x$.

~~ Ex. 13 . If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}+a x^{2}+b x+c=0$, then what is $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to ?

Sol. Given $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}+a x^{2}+b x+c=0$. Then,

$ \begin{aligned} & S_1=\alpha+\beta+\gamma=-\frac{\text{ coefficient of } x^{2}}{\text{ coefficient of } x^{3}}=-a ; \quad S_2=\alpha \beta+\beta \gamma+\alpha \gamma=\frac{\text{ coefficient of } x}{\text{ coefficient of } x^{3}}=b \\ & S_3=\alpha \beta \gamma=\frac{- \text{ constant term }}{\text{ coefficient of } x^{3}}=-c \\ & \therefore \alpha^{-1}+\beta^{-1}+\gamma^{-1}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta \gamma+\alpha \gamma+\alpha \beta}{\alpha \beta \gamma}=\frac{S_2}{S_3}=\frac{b}{-c}=-\frac{b}{c} . \end{aligned} $

~~ Ex. 14 . If $\alpha, \beta$ are the roots of the equation $9 x^{2}+6 x+1=0$, then write the equation with roots $\frac{1}{\alpha}, \frac{1}{\beta}$.

Sol. As $\alpha, \beta$ are the roots of the equation $9 x^{2}+6 x+1=0, \alpha+\beta=-\frac{6}{9}=-\frac{2}{3}, \alpha \beta=\frac{1}{9}$

$\therefore$ Required equation $=x^{2}-($ Sum of roots $) x+$ Product of roots $=0$

$ \begin{aligned} & \text{ i.e., } x^{2}-(\frac{1}{\alpha}+\frac{1}{\beta}) x+\frac{1}{\alpha} \cdot \frac{1}{\beta}=0 \text{, i.e., } x^{2}-(\frac{\beta+\alpha}{\gamma \beta}) x+\frac{1}{\alpha \beta}=0 \\ & \text{ i.e., } x^{2}-(\frac{-2 / 3}{1 / 9}) x+9=0 \text{, i.e., } \boldsymbol{x}^{2}+\mathbf{6 x + 9}=\mathbf{0} \end{aligned} $

Alternatively,

The equation whose roots $(\frac{1}{\alpha}, \frac{1}{\beta})$ are the reciprocals of the roots $(\alpha, \beta)$ of the equation $9 x^{2}+6 x+1=0$ can be obtained by replacing $x$ by $\frac{1}{x}$ in the given equation.

$\therefore$ Required equation is : $9(\frac{1}{x})^{2}+6(\frac{1}{x})+1=0 \Rightarrow \boldsymbol{x}^{2}+\mathbf{6 x}+\mathbf{9}=\mathbf{0}$.

~~ Ex. 15 . Find the values of $k$ for which the equations $x^{2}-k x-21=0$ and $x^{2}-3 k x+35=0$ will have a common root?

Sol. Let $\alpha$ be the common root of both the given equations. Then $\alpha$ satisfies both the equations. So,

$ \begin{aligned} \alpha^{2}-k \alpha-21 & =0 \\ \alpha^{2}-3 k \alpha+35 & =0 \end{aligned} $

Solving equations (i) and (ii) simultaneously, we get

$ \begin{gathered} \frac{\alpha^{2}}{-35 k-63 k}=\frac{\alpha}{-21-35}=\frac{1}{-3 k+k} \\ \Rightarrow \alpha^{2}=\frac{-98 k}{-2 k}=49 \text{ and } \alpha=\frac{-56}{-2 k}=\frac{28}{k} \\ \therefore 49=(\frac{28}{k})^{2} \Rightarrow k^{2}=\frac{28 \times 28}{49}=16 \Rightarrow k= \pm 4 . \end{gathered} $

$ \begin{bmatrix} a_1 x+b_1 y+c_1=0 ; a_2 x+b_2 y+c_2=0 \\ \frac{x}{b_1 c_2-b_2 c_1}=\frac{y}{c_1 a_2-c_2 a_1}=\frac{1}{a_1 b_2-a_2 b_1} \end{bmatrix} $

PRACTICE SHEET

What are the roots of the equation $\log _{10}(x^{2}-6 x+45)=2$ ?

(a) $9,-5$

(b) $-9,5$

(d) $-11,5$

(a) $\frac{5}{13}$

(b) $\frac{7}{13}$

(d) $\frac{11}{3}$

(CDS 2010)

~~ 2. What is one of the values of $x$ in the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$ ~~ 3. What are the roots of the equation $4^{x}-3.2^{x+2}+32=0$ ?

(a) 1,2

(b) 3,4

(d) 1,3

~~ 4. What are the roots of the quadratic equation $a^{2} b^{2} x^{2}-(a^{2}+b^{2}) x+1=0$ ?

(a) $\frac{1}{a^{2}}, \frac{1}{b^{2}}$

(b) $-\frac{1}{a^{2}},-\frac{1}{b^{2}}$

(d) $-\frac{1}{a^{2}}, \frac{1}{b^{2}}$

(CDS 2011)

~~ 5. If the roots of the equation $(c^{2}-a b) x^{2}-2(a^{2}-b c) x+$ $(b^{2}-a c)=0$ for $a \neq 0$ are real and equal, then the value of $a^{3}+b^{3}+c^{3}$ is :

(a) $a b c$

(b) $3 a b c$

(d) None of these

(MAT 2003)

~~ 6. If $\sin \theta$ and $\cos \theta$ are the roots of the equations $a x^{2}-b x+c=0$, then which of the following is correct?

(a) $a^{2}+b^{2}+2 a c=0$

(b) $a^{2}-b^{2}+2 a c=0$

(d) $a^{2}-b^{2}-2 a c=0$.

(CDS 2011)

~~ 7. The roots of the quadratic equation $x^{2}-2 \sqrt{3} x-22=0$ are:

(a) imaginary

(b) real, rational, equal

(d) real, irrational, unequal

(WBJEE 2010)

~~ 8. Which one of the following is the equation whose roots are respectively three times the roots of the equation $a x^{2}+b x+c=0$ ?

(a) $a x^{2}+3 b x+c=0$

(b) $a x^{2}+3 b x+9 c=0$

(d) $a x^{2}+b x+3 c=0$

(CDS 2007)

~~ 9. If $\alpha, \beta$ are the roots of the quadratic equation $a x^{2}+b x+c=0$, then $\alpha \beta^{2}+\alpha^{2} \beta+\alpha \beta$ equals :

(a) $\frac{b c}{-a^{2}}$

(b) 0

(d) $\frac{c(a-b)}{a^{2}}$

(AMU 2000)

~~ 10. For what value of $m$ the ratio of the roots of the equation $12 x^{2}-m x+5=0$ is $3: 2$ ?

(a) $5 \sqrt{10}$

(b) $10 \sqrt{5}$

(d) $15 \sqrt{5}$

(Rajasthan PET 2002)

~~ 11. If the roots of the equation $a x^{2}+b x+c=0$ are equal in magnitude but opposite in sign, then which one of the following is correct?

(a) $a=0$

(b) $b=0$

(d) $b=0, c \neq 0, a \neq 0$.

(CDS 2005)

~~ 12. If $2 x^{2}-7 x y+3 y^{2}=0$, then the value of $x: y$ is

(a) $3: 2$

(b) $2: 3$

(d) $5: 6$

(MAT 2003)

~~ 13. If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^{2}+b x+c=0$, such that $\beta=\alpha^{1 / 3}$, then

(a) $(a^{3} b)^{1 / 4}+(a c^{3})^{1 / 4}+a=0$ (b) $(a^{3} c)^{1 / 4}+(a c^{3})^{1 / 4}+b=0$

(d) $(b^{3} c)^{1 / 4}+(b c^{3})^{1 / 4}+a=0$

(Kerala PET 2003)

~~ 14. If the roots of the equation $a(b-c) x^{2}+b(c-a) x$ $+c(a-b)=0$ are equal, then $a, b, c$ are in :

(a) AP

(b) GP

(d) None of these

~~ 15. If an integer $P$ is chosen at random in the interval $0 \leq p \leq 5$, the probability that the roots of the equation $x^{2}+p x$ $+\frac{p}{4}+\frac{1}{2}=0$ are real is

(a) $\frac{2}{3}$

(b) $\frac{2}{5}$

(d) $\frac{4}{5}$

~~ 16. Two students $A$ and $B$ solve an equation of the form $x^{2}+p x+q=0 . A$ starts with a wrong value of $p$ and obtains the roots as 2 and $6 . B$ starts with a wrong value of $q$ and gets the roots as 2 and -9 . What are the correct roots of the equations?

(a) 3 and -4

(b) -3 and -4

(d) 3 and 4

(CDS 2012)

~~ 17. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-6 x+6=0$, what is $\alpha^{3}+\beta^{3}+\alpha^{2}+\beta^{2}+\alpha+\beta$ equal to ?

(a) 150

(b) 138

(d) 124

(CDS 2011)

~~ 18. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+q=0$, then $-\alpha^{-1}$ and $-\beta^{-1}$ are the roots of which one of the following equations?

(a) $q x^{2}-p x+1=0$

(b) $q^{2}+p x+1=0$

(d) $x^{2}-p x+q=0$

(CDS 2010)

~~ 19. The number of solution of $\log _4(x-1)=\log _2(x-3)$ is :

(a) 0

(b) 5

(d) 3

(AMU 2007)

~~ 20. The equation $e^{\sin x}-e^{-\sin x}-4=0$ has

(a) no real roots

(b) exactly one real root

(d) infinite number of real roots.

(AIEEE 2012)

~~ 21. If $5^{56}(\frac{1}{5})^{x}(\frac{1}{5})^{\sqrt{x}}>1$, then $x$ satisfies :

(a) $[0,49)$

(b) $(49,64]$

(d) $[49,64)$

(DCE 2007)

~~ 22. The sum of the roots of the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ is zero. What is the product of the roots of the equation?

(a) $-\frac{(a+b)}{2}$

(b) $\frac{(a+b)}{2}$

(d) $\frac{(a^{2}+b^{2})}{2}$

(CDS 2010)

~~ 23. For what value of $k$ will the roots of the equation $k x^{2}-5 x+6=0$ be in the ratio $2: 3$ ?

(a) 0

(b) 1

(d) 2

(CDS 2010)

~~ 24. The number of real solutions of the equation $2|x|^{2}-5|x|+2=0$ is :

(a) 0

(b) 4

(d) None of these

~~ 25. If $p, q, r$ are positive and are in A.P., the roots of quadratic equation $p x^{2}+q x+r=0$ are real for :

(a) $|\frac{r}{p}-7| \geq 4 \sqrt{3}$

(b) $|\frac{p}{r}-7| \geq 4 \sqrt{3}$

(d) no $p$ and $r$

~~ 26. The values of $x$ which satisfy the expression $(5+2 \sqrt{6})^{x^{2}+3}+(5-2 \sqrt{6})^{x^{2}-3}=10$ are :

(a) $\pm 2, \pm \sqrt{3}$

(b) $\pm \sqrt{2}, \pm 4$

(d) $2, \sqrt{2}, \sqrt{3}$

~~ 27. If $\alpha, \beta, \gamma$ are the roots of the equation $2 x^{3}-3 x^{2}+6 x+1$ $=0$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is equal to

(a) $\frac{-15}{4}$

(b) $\frac{-9}{4}$

(d) 4

(KCET 2005)

~~ 28. If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}+4 x+1=0$, then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$ is equal to

(a) 2

(b) 3

(d) 5

(UPSEE 2003)

~~ 29. If the roots of $x^{3}-12 x^{2}+12 x-28=0$ are in A.P, their common difference is

(a) $\pm 3$

(b) $\pm 2$

(d) None of these

(Rajasthan PET 2001)

~~ 30. The quadratic equation whose roots are three times the roots of $3 a x^{2}+3 b x+c=0$ is

(a) $a x^{2}+b x+3 c=0$

(b) $a x^{2}+3 b x+c=0$

(d) $9 a x^{2}+9 b x+c=0$.

(WBJEE 2009)

~~ 31. If $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$ and if $p x^{2}+q x+r=0$ has roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$, then $r$ equals:

(a) $a b c$

(b) $a+2 b$

(d) $a b+b c+c a$. ~~ 32. The equation whose roots are the negatives of the roots of the equation $x^{7}+3 x^{5}+x^{3}-x^{2}+7 x+2=0$ is :

(a) $x^{7}+3 x^{5}+x^{3}-x^{2}-7 x-2=0$

(b) $x^{7}+3 x^{5}+x^{3}-x^{2}+7 x-2=0$

(d) $x^{7}+3 x^{5}+x^{3}+x^{2}+7 x-2=0$

(EAMCET 2001)

~~ 33. Given that $\alpha, \gamma$ are the roots of the equation $A x^{2}-4 x+1=0$ and $\beta, \delta$ are the roots of the equation $B x^{2}-6 x+1=0$, then the values of $A$ and $B$ respectively such that $\alpha, \beta, \gamma$ and $\delta$ are in H.P are :

(a) $-5,9$

(b) $3 / 2,5$

(d) None of these

~~ 34. Let $\alpha, \beta$ be the roots of the equation $(x-a)(x-b)=c, c \neq 0$, then the roots of the equation $(x-\alpha)(x-\beta)+c=0$ are :

(a) $a, c$

(b) $b, c$

(d) $a+c, b+c$

~~ 35. If the roots of the equation $x^{3}-a x^{2}+b x-c=0$ are three consecutive integers, then what is the smallest possible value of $b$ ?

(a) $-\frac{1}{\sqrt{3}}$

(b) -1

(d) 1

(CAT)

~~ 36. If two equations $x^{2}+a^{2}=1-2 a x$ and $x^{2}+b^{2}=1-2 b x$ have only one common root, then

(a) $(a-b)=-1$

(b) $|a-b|=1$

(d) $|a-b|=2$

(DCE 2004)

~~ 37. If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-3 x+11=0$, then the equation whose roots are $(\alpha+\beta),(\beta+\gamma),(\gamma+\alpha)$ is :

(a) $x^{3}+3 x+11=0$

(b) $x^{3}-3 x+11=0$

(d) $x^{3}-3 x-11=0$ ~~ 38. If $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0$, and $\alpha+k, \beta+k$ are the roots of $p x^{2}+q x+r=0$, then $k=$

(a) $-\frac{1}{2}(a / b-p / q)$

(b) $(a / b-p / q)$

(d) $(a b-p q)$

~~ 39. Find the value of $\sqrt{6+\sqrt{6+\sqrt{6+\ldots . \infty}}}$

(a) -4

(b) 2

(d) 6

~~ 40. The roots of $(x-a)(x-a-1)+(x-a-1)(x-a-2)$ $+(x-a)(x-a-2)=0, a \in R$ are always :

(a) imaginary

(b) real and distinct

(d) rational and equal

ANSWERS

1. (c)	2. $(c)$	3. $(c)$	4. $(a)$	5. $(b)$	6. $(b)$	7. (c)	8. $(b)$	9. $(d)$	10. (a)
11. $(d)$	12. $(c)$	13. $(b)$	14. $(c)$	15. (a)	16. $(b)$	17. $(b)$	18. $(a)$	19. $(b)$	20. (a)
21. (a)	22. (c)	23. (b)	24. (b)	25. (b)	26. (c)	27. (a)	28. $(c)$	29. (a)	30. (c)
31. (c)	32. $(d)$	33. $(c)$	34. (c)	35. $(b)$	36. $(d)$	37. $(d)$	38. $(c)$	39. (c)	40. $(b)$

HINTS AND SOLUTIONS

Given, $\log _{10}(x^{2}-6 x+45)=2 \Rightarrow x^{2}-6 x+45=10^{2}=100$

$\Rightarrow x^{2}-6 x-55=0 \Rightarrow x^{2}-11 x+5 x-55=0$

$\Rightarrow x(x-11)+5(x-11)=0 \Rightarrow(x+5)(x-11)=0$

$\Rightarrow x=-5$ or 11 .

~~ 2. Let $\sqrt{\frac{x}{1-x}}=y$. Then, the given equation reduces to

$y+\frac{1}{y}=\frac{13}{6} \Rightarrow 6(y^{2}+1)=13 y$

$\Rightarrow 6 y^{2}-13 y+6=0 \Rightarrow 6 y^{2}-9 y-4 y+6=0$

$\Rightarrow 3 y(2 y-3)-2(2 y-3)=0$

$\Rightarrow(3 y-2)(2 y-3)=0 \Rightarrow y=\frac{2}{3}$ and $\frac{3}{2}$

when $y=\frac{2}{3}, \sqrt{\frac{x}{1-x}}=\frac{2}{3} \Rightarrow \frac{x}{1-x}=\frac{4}{9}$

$\Rightarrow 9 x=4-4 x \Rightarrow 13 x=4 \Rightarrow x=\frac{\mathbf{4}}{\mathbf{1 3}}$

when $y=\frac{3}{2}, \sqrt{\frac{x}{1-x}}=\frac{3}{2} \Rightarrow \frac{x}{1-x}=\frac{9}{4}$

$\Rightarrow 4 x=9-9 x \Rightarrow 13 x=9 \Rightarrow x=\frac{\mathbf{9}}{\mathbf{1 3}}$.

~~ 3. $4^{x}-3.2^{x+2}+32=0$

$\Rightarrow 2^{2 x}-3 \cdot 2^{2} \cdot 2^{x}+32=0 \Rightarrow 2^{2 x}-12 \cdot 2^{x}+32=0$

Let $2^{x}=a$. Then, $a^{2}-12 a+32=0$

$\Rightarrow(a-8)(a-4)=0 \Rightarrow a=8$ and 4

$\Rightarrow 2^{x}=8$ and $2^{x}=4 \Rightarrow \boldsymbol{x}=\mathbf{3}$ and $\boldsymbol{x}=\mathbf{2}$.

~~ 4. Let the roots of the equation $a^{2} b^{2} x^{2}-(a^{2}+b^{2}) x+1=0$ be $\alpha$ and $\beta$. Then,

$ \begin{aligned} \alpha+\beta & =\frac{a^{2}+b^{2}}{a^{2} b^{2}} \quad \ldots .(i), \quad \alpha \beta=\frac{1}{a^{2} b^{2}} \\ \alpha-\beta & =\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta} \\ & =\sqrt{(\frac{a^{2}+b^{2}}{a^{2} b^{2}})^{2}-\frac{4}{a^{2} b^{2}}} \\ & =\sqrt{\frac{a^{4}+b^{4}+2 a^{2} b^{2}-4 a^{2} b^{2}}{(a^{2} b^{2})^{2}}} \\ & =\sqrt{\frac{(a^{2}-b^{2})^{2}}{(a^{2} b^{2})^{2}}}=\frac{a^{2}-b^{2}}{a^{2} b^{2}} \end{aligned} $

$\therefore$ On solving (i) and (ii), we get $\alpha=\frac{\mathbf{1}}{\boldsymbol{b}^{2}}, \beta=\frac{\mathbf{1}}{\boldsymbol{a}^{2}}$.

~~ 5. Given that the roots are real and equal,

$D=0 \Rightarrow b^{2}-4 a c=0$ for $a x^{2}+b x+c=0$.

$\therefore[-2(a^{2}-b c)]^{2}-4(c^{2}-a b)(b^{2}-a c)=0$

$ \begin{aligned} & \Rightarrow 4[a^{4}+b^{2} c^{2}-2 a^{2} b c-c^{2} b^{2}+a c^{3}+a b^{3}-a^{2} b c)=0 \\ & \Rightarrow 4 a(a^{3}+b^{3}+c^{3}-3 a b c)=0 \\ & \Rightarrow \boldsymbol{a}^{3}+\boldsymbol{b}^{\mathbf{3}}+\boldsymbol{c}^{\mathbf{3}}=\mathbf{3} \boldsymbol{a} \boldsymbol{b} \boldsymbol{c} . \end{aligned} $

~~ 6. As $\sin \theta$ and $\cos \theta$ are the roots of the equation $a x^{2}-b x+c=0$.

$\therefore \sin \theta+\cos \theta=\frac{b}{a}$ and $\sin \theta \cos \theta=\frac{c}{a}$

$\Rightarrow(\sin \theta+\cos \theta)^{2}=\frac{b^{2}}{a^{2}}$

$\Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=\frac{b^{2}}{a^{2}}$

$\Rightarrow 1+\frac{2 c}{a}=\frac{b^{2}}{a^{2}} \Rightarrow \frac{2 c}{a}=\frac{b^{2}}{a^{2}}-1=\frac{b^{2}-a^{2}}{a^{2}}$

$\Rightarrow 2 a c=b^{2}-a^{2} \Rightarrow \boldsymbol{a}^{2}-\boldsymbol{b}^{2}+\mathbf{2 a c}=\mathbf{0}$.

~~ 7. Given equation is $x^{2}-2 \sqrt{3} x-22=0$.

Discriminant $=D=b^{2}-4 a c(.$ for $.a x^{2}+b x+c=0)$

$ =(-2 \sqrt{3})^{2}-4(-22)=12+88=100 $

As $D>0$ and is a perfect square, the roots are real, rational and unequal.

~~ 8. Let $\alpha, \beta$ be the roots of the equation $a x^{2}+b x+c=0$. Then,

$ \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a} . $

By the given condition, roots of the required equation are $3 \alpha$ and $3 \beta$.

$\therefore$ Sum of roots $=3 \alpha+3 \beta=3(\alpha+\beta)=-\frac{3 b}{a}$

Product of roots $=3 \alpha \cdot 3 \beta=9 \alpha \beta=\frac{9 c}{a}$

$\therefore$ Required equation

$ \begin{aligned} & =x^{2}-(\text{ sum of roots }) x+(\text{ product of roots })=0 \\ & \Rightarrow x^{2}-(\frac{-3 b}{a} x)+\frac{9 c}{a}=0 \\ & \Rightarrow a x^{2}+3 b x+9 c=0 . \end{aligned} $

~~ 9. Given, $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$. Then, $\alpha+\beta=-b / a, \alpha \beta=c / a$

Then, $\alpha \beta^{2}+\alpha^{2} \beta+\alpha \beta=\alpha \beta(\alpha+\beta)+\alpha \beta=(c / a)(-b / a)+c / a$

$ =-\frac{b c}{a^{2}}+\frac{c}{a}=\frac{-b c+a c}{a^{2}}=\frac{c(\boldsymbol{a}-\boldsymbol{b})}{\boldsymbol{a}^{2}} . $

~~ 10. Given, the roots of the given equation $12 x^{2}-m x+5=0$ are in the ratio $3: 2$. Let the roots of the given equation be $3 \alpha$ and $2 \alpha$. Then,

Sum of roots $=3 \alpha+2 \alpha=\frac{m}{12} \Rightarrow 5 \alpha=\frac{m}{12}$

and $(3 \alpha)(2 \alpha)=\frac{5}{12} \Rightarrow 6 \alpha^{2}=\frac{5}{12} \Rightarrow \alpha^{2}=\frac{5}{72}$

$\Rightarrow \alpha=\sqrt{\frac{5}{72}}$ $\therefore$ From $(i)$ and $(i i)$

$ \begin{aligned} & 5 \cdot \sqrt{\frac{5}{72}}=\frac{m}{12} \Rightarrow m=60 \sqrt{\frac{5}{72}}=60 \cdot \frac{\sqrt{5}}{6 \sqrt{2}}=10 \sqrt{\frac{5}{2}} \\ & =10 \cdot \frac{\sqrt{5}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{10}{2} \cdot \sqrt{10}=5 \sqrt{\mathbf{1 0}} . \end{aligned} $

~~ 11. Given equation is $a x^{2}+b x+c=0$.

$\therefore$ Roots are $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Given $[\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}]=-[\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}]$

$\Rightarrow-b+\sqrt{b^{2}-4 a c}=b+\sqrt{b^{2}-4 a c}$

$\Rightarrow 2 b=0 \Rightarrow b=0$. but $a \neq 0, c \neq 0$.

~~ 12. $2 x^{2}-7 x y+3 y^{2}=0$

$\Rightarrow 2(\frac{x}{y})^{2}-7(\frac{x}{y})+3=0$

$\therefore \frac{x}{y}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm 5}{4}=3, \frac{1}{2}$

$\therefore x: y=\mathbf{3}: \mathbf{1}$ and $\mathbf{1}: \mathbf{2}$.

~~ 13. Let $\alpha, \beta$ be the roots of the equation $a x^{2}+b x+c=0$. Then,

$ \begin{aligned} \alpha+\beta & =-\frac{b}{a} \\ \alpha \beta & =\frac{c}{a} \end{aligned} $

and $\quad \beta=\alpha^{1 / 3}$

$\therefore$ From (ii) and (iii), $\alpha .(\alpha)^{1 / 3}=\frac{c}{a} \Rightarrow \alpha^{4 / 3}=\frac{c}{a} \Rightarrow \alpha=(\frac{c}{a})^{3 / 4}$

$\therefore \quad \beta=((\frac{c}{a})^{3 / 4})^{1 / 3}=(\frac{c}{a})^{1 / 4}$

$\therefore$ Putting these values of $\alpha$ and $\beta$ in eqn. (i), we have

$(\frac{c}{a})^{3 / 4}+(c / a)^{1 / 4}=-\frac{b}{a}$

$\Rightarrow a \cdot a^{-3 / 4} c^{3 / 4}+a \cdot a^{-1 / 4} c^{1 / 4}=-b$

$\Rightarrow a^{1 / 4} c^{3 / 4}+a^{3 / 4} c^{1 / 4}+b=0$

$\Rightarrow(a c^{3})^{1 / 4}+(a^{3} c)^{1 / 4}+b=0$.

~~ 14. If the roots of the equation $a(b-c) x^{2}+b(c-a) x+c(a-b)$

$=0$ are equal, then Discriminant $(D)=0$, i.e.,

$\Rightarrow b^{2}(c-a)^{2}-4 a(b-c) c(a-b)=0$.

$\Rightarrow b^{2}(c^{2}+a^{2}-2 a c)-4 a c(a b-c a-b^{2}+b c)=0$

$\Rightarrow b^{2} c^{2}+b^{2} a^{2}-2 a b^{2} c-4 a^{2} b c+4 a^{2} c^{2}+4 a b^{2} c-4 a b c^{2}=0$

$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+4 a^{2} c^{2}+2 a b^{2} c-4 a^{2} b c-4 a b c^{2}=0$

$\Rightarrow(a b+b c-2 a c)^{2}=0 \Rightarrow a b+b c-2 a c=0$

$\Rightarrow a b+b c=2 a c \Rightarrow \frac{1}{c}+\frac{1}{a}=\frac{2}{b} \Rightarrow a, b, c$ are in H.P. ~~ 15. The equation $x^{2}+p x+\frac{p}{4}+\frac{1}{2}=0$ has real roots if the discriminant $D \geq 0$.

$\Rightarrow p^{2}-4(\frac{p}{4}+\frac{1}{2}) \geq 0 \Rightarrow p^{2}-p-2 \geq 0$

$\Rightarrow p^{2}-2 p+p-2 \geq 0 \Rightarrow p(p-2)+1(p-2) \geq 0$

$\Rightarrow(p-2)(p+1) \geq 0$

$\Rightarrow(p-2) \geq 0$ and $(p+1) \geq 0$

$\Rightarrow p \geq 2$ or $p \leq-1$

The condition $p \leq-1$ is not admissible as $0 \leq p \leq 5$.

Now $p \geq 2 \Rightarrow p$ can take up the value 2 or 3 or 4 or 5 from the given values. ${0,1,2,3,4,5}$

$\therefore$ Probability (Roots of given equation are real)

$ =\frac{\text{ Number of values } p \text{ can take }}{\text{ Given number of values }}=\frac{4}{6}=\frac{\mathbf{2}}{\mathbf{3}} . $

~~ 16. Let the roots of the quadratic equation $x^{2}+p x+q=0$ be $\alpha$ and $\beta$. According to the given condition, $A$ starts with a wrong value of $p$ and obtains the roots as 2 and 6 . But this time, the value of $q$ is correct.

$\therefore \quad q=$ Product of roots $=\alpha \beta=2 \times 6=12$.

According to the second condition, $B$ starts with a wrong value of $q$ and obtains the roots as 2 and -9 . But this time, the value of $p$ is correct.

$\therefore p=$ sum of roots $=\alpha+\beta=2+(-9)=-7$

$\therefore(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=(-7)^{2}-4.12=49-48=1$

$\Rightarrow \alpha-\beta=1$

$\therefore$ Solving equations $(i)$ and (ii), we get $\boldsymbol{\alpha}=-\mathbf{3}$ and $\boldsymbol{\beta}=-\mathbf{4}$.

~~ 17. $\alpha+\beta=6, \alpha \beta=6$

$\therefore(\alpha+\beta)^{2}=6^{2} \Rightarrow \alpha^{2}+\beta^{2}+2 \alpha \beta=36$

$\Rightarrow \alpha^{2}+\beta^{2}=36-2 \times 6=24$

Now, $\alpha^{3}+\beta^{3}+\alpha^{2}+\beta^{2}+\alpha+\beta$

$ \begin{aligned} & =(\alpha+\beta)(\alpha^{2}+\beta^{2}-\alpha \beta)+(\alpha^{2}+\beta^{2})+(\alpha+\beta) \\ & =6(24-6)+24+6=6 \times 18+30=\mathbf{1 3 8} \end{aligned} $

~~ 18. Since, $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+q=0$,

$\therefore \alpha+\beta=-p$ and $\alpha \beta=q$

Now, equation whose roots are $-\alpha^{-1}$ and $-\beta^{-1}$ is $x^{2}-$ (sum of the roots) $x+$ product of the roots $=0$

i.e., $\quad x^{2}-(-\alpha^{-1}-\beta^{-1}) x+(-\alpha^{-1})(-\beta^{-1})=0$

$ -\alpha^{-1}-\beta^{-1}=-(\frac{1}{\alpha}+\frac{1}{\beta})=-(\frac{\alpha+\beta}{\alpha \beta})=\frac{p}{q} $

and $(-\alpha^{-1})(-\beta^{-1})=\frac{1}{\alpha \beta}=\frac{1}{q}$

$\therefore$ Required equation $=x^{2}-\frac{p}{q} x+\frac{1}{q}=0$

$\Rightarrow q x^{2}-p x+1=0$.

~~ 19. $\log _4(x-1)=\log _2(x-3)$

$\Rightarrow \log _{2^{2}}(x-1)=\log _2(x-3)$ $\Rightarrow \frac{1}{2} \log _2(x-1)=\log _2(x-3)$

$[.$ Using $.\log _{m^{n}}(x)=\frac{1}{n} \log _{m} x]$

$\Rightarrow \log _2(x-1)=2 \log _2(x-3)$

$\Rightarrow \log _2(x-1)=\log _2(x-3)^{2}$

$\Rightarrow(x-1)=(x-3)^{2} \Rightarrow(x-1)=x^{2}-6 x+9$

$\Rightarrow x^{2}-7 x+10=0 \Rightarrow(x-2)(x-5)=0 \Rightarrow x=2$ or 5 .

$x=2$ is inadmissible as $\log _2(x-3)$ is not defined when $x=2$.

$\therefore \boldsymbol{x}=\mathbf{5}$.

~~ 20. Given, $e^{\sin x}-e^{-\sin x}-4=0$

Let $e^{\sin x}=y$. Then, the given equation becomes

$ \begin{aligned} & y-\frac{1}{y}=4 \Rightarrow y^{2}-4 y-1=0 \\ \Rightarrow & \quad y=\frac{4 \pm \sqrt{16+4}}{2}=2 \pm \sqrt{5} \\ \Rightarrow & e^{\sin x}=2 \pm \sqrt{5} \Rightarrow \sin x=\log _{e}(2 \pm \sqrt{5}) \\ \Rightarrow & \sin x=\log _{e}(2+\sqrt{5}) \end{aligned} $

$(\because(2-\sqrt{5})<0.$ and so $\log _{e}(2-\sqrt{5})$ is not defined $)$

Now $(2+\sqrt{5})>4 \Rightarrow \log _{e}(2+\sqrt{5})>1$

But the value of $\sin x$ lies between -1 and 1, both values inclusive, so $\sin x \neq \log _{e}(2+\sqrt{5})$

$\therefore$ There are no possible real roots of the given equation.

~~ 21. $5^{56}(\frac{1}{5})^{x}(\frac{1}{5})^{\sqrt{x}}>1$

$\Rightarrow 5^{56} \times 5^{-x} \times 5^{-\sqrt{x}}>1 \Rightarrow 5^{56-x-\sqrt{x}}>5^{0}$

$\Rightarrow 56-x-\sqrt{x}>0 \Rightarrow x+\sqrt{x}-56<0$

$\Rightarrow y^{2}+y-56<0$, where $y=\sqrt{x}$

$\Rightarrow(y+8)(y-7)<0 \Rightarrow-8<y<7 \Rightarrow-8<\sqrt{x}<7$

$\Rightarrow 0 \leq \sqrt{x}<7$ as $\sqrt{x}$ cannot be negative

$\Rightarrow 0 \leq x<49 \Rightarrow x \in[0,49)$

~~ 22. Given, $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$

$\Rightarrow \frac{(x+b)+(x+a)}{(x+a)(x+b)}=\frac{1}{c} \Rightarrow \frac{2 x+b+a}{x^{2}+(a+b) x+a b}=\frac{1}{c}$

$\Rightarrow 2 c x+(a+b) c=x^{2}+(a+b) x+a b$

$\Rightarrow x^{2}+(a+b-2 c) x+a b-a c-b c=0$

Let $\alpha, \beta$ be the roots of this equation. Then,

$\alpha+\beta=-(a+b-2 c)=0$ (Given)

$\Rightarrow a+b=2 c$ and $\alpha \beta=a b-a c-b c=a b-(a+b) c$

$ \begin{aligned} & =a b-(a+b) \frac{(a+b)}{2} \\ & =\frac{2 a b-(a^{2}+b^{2}+2 a b)}{2}=-(\frac{\boldsymbol{a}^{2}+\boldsymbol{b}^{\mathbf{2}}}{\mathbf{2}}) . \end{aligned} $

~~ 23. Let the roots of the equation $k x^{2}-5 x+6=0$ be $\alpha$ and $\beta$.

Then, $\alpha+\beta=5 / k$

$ \alpha \beta=6 / k $

Given $\alpha / \beta=\frac{2}{3} \Rightarrow \alpha=\frac{2}{3} \beta$

$\therefore$ From $(i)$ and (ii),

$\frac{2}{3} \beta+\beta=\frac{5}{k}$ and $\frac{2}{3} \beta^{2}=\frac{6}{k}$

$\Rightarrow \frac{5}{3} \beta=\frac{5}{k}$ and $\beta^{2}=\frac{9}{k} \Rightarrow \beta=\frac{3}{k}$ and $\beta^{2}=\frac{9}{k}$

$\Rightarrow \frac{9}{k^{2}}=\frac{9}{k} \Rightarrow 9 k^{2}-9 k=0 \Rightarrow k(k-1)=0 \Rightarrow k=0$ or 1

But $k=0$ does not satisfy the condition, so $\boldsymbol{k}=\mathbf{1}$.

~~ 24. $2|x|^{2}-5|x|+2=0$

$\Rightarrow(2|x|-1)(|x|-2)=0$

$\Rightarrow|x|=\frac{1}{2}, 2 \Rightarrow x= \pm \frac{1}{2}, \pm 2$

So, there are 4 solutions.

~~ 25. $\because p, q, r$ are in A.P.

$q=\frac{p+r}{2}$

$[\because p+r=2 q]$

For the real roots $q^{2}-4 p r \geq 0$

$[$ Disct. $\geq 0]$

$\Rightarrow(\frac{p+r}{2})^{2}-4 p r \geq 0$

$\Rightarrow p^{2}+r^{2}-14 p r \geq 0$

$\Rightarrow(\frac{p}{r})^{2}-14(\frac{p}{r})+1 \geq 0$

$\Rightarrow(\frac{p}{r}-7)^{2} \geq 48=|\frac{\boldsymbol{p}}{\boldsymbol{r}}-7| \geq \mathbf{4} \sqrt{\mathbf{3}}$.

~~ 26. Let $y=5+2 \sqrt{6}$. Then $\frac{1}{y}=5-2 \sqrt{6}$. Thus the given expression reduces to $y^{x^{2}-3}+(\frac{1}{y})^{x^{2}-3}=10$

Again let $y^{x^{2}-3}=t$. Then,

$ \begin{aligned} & t+\frac{1}{t}=10 \Rightarrow t^{2}-10 t+1=0 \\ & \Rightarrow \quad t=\frac{10 \pm \sqrt{100-4}}{2}=\frac{10 \pm \sqrt{96}}{2} \\ &=\frac{10 \pm 4 \sqrt{6}}{2}=5 \pm 2 \sqrt{6} \\ & \therefore(5+2 \sqrt{6})^{x^{2}-3}=5 \pm 2 \sqrt{6}=(5+2 \sqrt{6})^{ \pm 1} \\ & \Rightarrow x^{2}-3=1 \quad \text{ or } \quad x^{2}-3=-1 \\ & \Rightarrow x^{2}=4 \text{ or } \quad x^{2}=2 \\ & \Rightarrow x= \pm 2 \text{ or } \quad x= \pm \sqrt{2} \end{aligned} $

~~ 27. Given, $\alpha, \beta, \gamma$ are the roots of the $2 x^{3}-3 x^{2}+6 x+1=0$.

Since $S_1=-\frac{\text{ Coefficient of } x^{2}}{\text{ Coefficient of } x^{3}} \Rightarrow S_1=\alpha+\beta+\gamma=-(\frac{-3}{2})=\frac{3}{2}$ $S_2=\frac{\text{ Coefficient of } x}{\text{ Coefficient of } x^{3}} \Rightarrow S_2=\alpha \beta+\beta \gamma+\alpha \gamma=\frac{6}{2}=3$

$ S_3=-\frac{\text{ Coefficient of constant term }}{\text{ Coefficient of } x^{3}} \Rightarrow S_3=\alpha \beta \gamma=-\frac{1}{2} $

Now, $\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\beta \gamma+\gamma \alpha)$

$ =(\frac{3}{2})^{2}-2 \times 3=\frac{9}{4}-6=\frac{-\mathbf{1 5}}{4} . $

~~ 28. Given, $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}+4 x+1=0$.

$ \begin{aligned} & \therefore S_1.=\alpha+\beta+\gamma=0 \quad \text{ (Coefficient of } x^{2}=0) \\ & S_2=\alpha \beta+\beta \gamma+\alpha \gamma=4 \\ & S_3=\alpha \beta \gamma=-1 \\ & \therefore(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1} \\ &=\frac{1}{\alpha+\beta}+\frac{1}{\beta+\gamma}+\frac{1}{\gamma+\alpha} \\ &=\frac{1}{-\gamma}+\frac{1}{-\alpha}+\frac{1}{-\beta} \\ & \quad[\because \alpha+\beta+\gamma=0 \\ &\Rightarrow \alpha+\beta=-\gamma, \beta+\gamma=-\alpha, \gamma+\alpha=-\beta] \end{aligned} $

~~ 29. Let $(\alpha-d), \alpha,(\alpha+d)$ be the three roots of the given cubic equation $x^{3}-12 x^{2}+12 x-28=0$

$\therefore S_1=(\alpha-d)+\alpha+(\alpha+d)=12$

$\Rightarrow 3 \alpha=12 \Rightarrow \alpha=4$

and $S_3=(\alpha-d) \cdot \alpha \cdot(\alpha+d)=28$

$\Rightarrow(4-d) \cdot 4 .(4+d)=28$

$\Rightarrow 16-d^{2}=7 \Rightarrow d^{2}=9 \Rightarrow d= \pm \mathbf{3}$.

~~ 30. Let $\alpha, \beta$ be the roots of the equation $3 a x^{2}+3 b x+c=0$. We have to find the equation whose roots are $3 \alpha$ and $3 \beta$, which can be got by putting $y=3 x$ in the given equation, i.e., substituting $x$ for $\frac{y}{3}$ in the given equations.

$\therefore$ The required equation is : $3 a(\frac{y}{3})^{2}+3 b(\frac{y}{3})+c=0$

$\Rightarrow \quad \frac{a y^{2}}{3}+b y+c=0 \quad \Rightarrow \quad a y^{2}+3 b y+3 c=0$

$\Rightarrow a x^{2}+3 b x+3 c=0$.

~~ 31. As $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, so $\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$

The equation whose roots are $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ can be written as :

$ x^{2}-(\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}) x+(\frac{1-\alpha}{\alpha} \cdot \frac{1-\beta}{\beta})=0 $

$ \begin{aligned} & \text{ Now, } \frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{\beta-\alpha \beta+\alpha-\alpha \beta}{\alpha \beta}=\frac{\alpha+\beta-2 \alpha \beta}{\alpha \beta}=\frac{\alpha+\beta}{\alpha \beta}-2 \\ & =\frac{-b / a}{c / a}-2=-\frac{b}{c}-2=-\frac{b-2 c}{c} \\ & \text{ and }(\frac{1-\alpha}{\alpha} \cdot \frac{1-\beta}{\beta})=\frac{1-\alpha-\beta+\alpha \beta}{\alpha \beta}=\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta} \\ & =\frac{1}{\alpha \beta}-\frac{\alpha+\beta}{\alpha \beta}+1=\frac{a}{c}+\frac{b}{c}+1=\frac{a+b+c}{c} \end{aligned} $

$\therefore$ The required equation is

$ \begin{aligned} & x^{2}-[-\frac{(b+2 c)}{c}] x+[\frac{a+b+c}{c}]=0 \\ \Rightarrow & c x^{2}+(b+2 c) x+(a+b+c)=0 \end{aligned} $

Compairing eqn. (i) with the given equation $p x^{2}+q x+r=0$, we get $r=\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}$.

~~ 32. To find the equation whose roots are the negatives of the roots of the given equation, we replace $x$ by $(-x)$ in the given equation.

$\therefore$ Required equation is

$(-x)^{7}+3(-x)^{5}+(-x)^{3}-(-x)^{2}+7(-x)+2=0$

i.e, $-x^{7}-3 x^{5}-x^{3}-x^{2}-7 x+2=0$

i.e., $x^{7}+3 x^{5}+x^{3}+x^{2}+7 x-2=0$.

~~ 33. As, $\alpha, \gamma$ are the roots of the equation $A x^{2}-4 x+1=0$, so

$ \alpha+\gamma=\frac{4}{A} $

and $\quad \alpha \gamma=\frac{1}{A}$

Given, $\beta, \delta$ are the roots of the equation $B x^{2}-6 x+1=0$,

so $\beta+\delta=\frac{6}{B}$

and $\quad \beta \delta=\frac{1}{B}$

Given, $\alpha, \beta, \gamma, \delta$ are in H.P., so

$ \begin{aligned} & \beta=\frac{2 \alpha \gamma}{\alpha+\gamma}=\frac{2 / A}{4 / A}=\frac{1}{2} \\ & \gamma=\frac{2 \beta \delta}{\beta+\delta}=\frac{2 / B}{6 / B}=\frac{1}{3} . \end{aligned} $

Also $\beta$ is the root of the equation $B x^{2}-6 x+1=0$, so

$ \begin{aligned} & B \beta^{2}-6 \beta+1=0 \Rightarrow B \times \frac{1}{4}-6 \times \frac{1}{2}+1=0 \\ & \Rightarrow \quad \frac{B}{4}-2=0 \Rightarrow \frac{B}{4}=2 \Rightarrow B=8 \end{aligned} $

Given, $\gamma$ is the root of the equation $A x^{2}-4 x+1=0$, so

$ A \gamma^{2}-4 \gamma+1=0 $

$\Rightarrow A \times \frac{1}{9}-4 \times \frac{1}{3}+1=0 \Rightarrow \frac{A}{9}-\frac{1}{3}=0 \Rightarrow A=3$.

$A, B=\mathbf{3 , 8}$ respectively.

~~ 34. The given equation is $(x-a)(x-b)=c$

$\Rightarrow x^{2}-(a+b) x+(a b-c)=0$

As $\alpha, \beta$ are the roots of this equation, so

$ \alpha+\beta=a+b \text{ and } \alpha \beta=a b-c $

Let $\gamma, \delta$ be the roots of the equation $(x-\alpha)(x-\beta)+c=0$

i.e., $\gamma, \delta$ are the roots of the equation $x^{2}-(\alpha+\beta) x+(\alpha \beta+c)=0$

$ \begin{aligned} \therefore \quad \gamma+\delta & =\alpha+\beta=a+b \\ \gamma \delta & =\alpha \beta+c=a b-c+c=a b \end{aligned} $

$\therefore$ From $(i)$ and (ii) we can infer that the roots of the equation $(x-\alpha)(x-\beta)+c=0$ are $\boldsymbol{a}$ and $\boldsymbol{b}$.

~~ 35. Let the roots of the equation $x^{3}-a x^{2}+b x-c=0$ be

$(\alpha-1), \alpha,(\alpha+1)$

$\therefore S_2=(\alpha-1) \alpha+\alpha(\alpha+1)+(\alpha+1)(\alpha-1)=b$

$\Rightarrow \alpha^{2}-\alpha+\alpha^{2}+\alpha+\alpha^{2}-1=b$

$\Rightarrow 3 \alpha^{2}-1=b$

$\therefore$ Minimum value of $b=-1$, when $\alpha=\mathbf{0}$.

~~ 36. The given equations are written as :

$ \begin{aligned} & x^{2}+2 a x+a^{2}-1=0 \\ & x^{2}+2 b x+b^{2}-1=0 \end{aligned} $

If $\alpha$ is the common root of both the equations, $\alpha$ satisfies both the equations, so,

$ \begin{aligned} & \alpha^{2}+2 a \alpha+(a^{2}-1)=0 \\ & \alpha^{2}+2 b \alpha+(b^{2}-1)=0 \end{aligned} $

Solving equations (iii) and (iv) simultaneously

$ \frac{\alpha^{2}}{2 a(b^{2}-1)-2 b(a^{2}-1)}=\frac{\alpha}{(a^{2}-1)-(b^{2}-1)}=\frac{1}{2 b-2 a} $

$ \begin{aligned} & \Rightarrow \frac{\alpha^{2}}{2 a b^{2}-2 b a^{2}+2(b-a)}=\frac{\alpha}{(a^{2}-b^{2})}=\frac{1}{2(b-a)} \\ & \Rightarrow \frac{\alpha^{2}}{2 a b(b-a)+2(b-a)}=\frac{\alpha}{-(a+b)(b-a)}=\frac{1}{2(b-a)} \\ & \Rightarrow \frac{\alpha^{2}}{2(b-a)(a b+1)}=\frac{\alpha}{-(a+b)(b-a)}=\frac{1}{2(b-a)} \end{aligned} $

By the rule of cross-multiplication, the solution of two simultaneous equations :

$ \begin{gathered} a_1 x+b_1 y+c_1=0 \\ a_2 x+b_2 y+c_2=0 \text{ is } \\ \frac{x}{b_1 c_2-b_2 c_1}=\frac{y}{c_1 a_2-c_2 a_1}=\frac{1}{a_1 b_2-a_2 b_1} \end{gathered} $

$\therefore \alpha^{2}=a b+1$ and $\alpha=-\frac{1}{2}(a+b)$

$\Rightarrow(a b+1)=\frac{1}{4}(a+b)^{2}$

$\Rightarrow a^{2}+b^{2}+2 a b=4 a b+4$

$\Rightarrow a^{2}+b^{2}-2 a b=4 \quad \Rightarrow(a-b)^{2}=4 \Rightarrow|a-b|=\mathbf{2}$.

~~ 37. Given equation is $x^{3}-3 x+11=0$

If $\alpha, \beta, \gamma$ are the roots of the given equation, then

$ S_1=\alpha+\beta+\gamma=0 $

$\Rightarrow \beta+\gamma=-\alpha, \gamma+\alpha=-\beta$ and $\alpha+\beta=-\gamma$

$\therefore$ The equation whose roots are $(\alpha+\beta),(\beta+\gamma),(\gamma+\alpha)$ is the equation whose roots are $-\gamma,-\alpha,-\beta$.

$\therefore$ We can obtain the required equations by replacing $x$ by $(-x)$ in the given equation.

$\therefore$ Required equation is $(-x)^{3}-3(-x)+11=0$

i.e, $-x^{3}+3 x+11=0$

i.e., $x^{3}-3 x-11=\mathbf{0}$.

~~ 38. As $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, so

$\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$

Also, $(\alpha+x),(\beta+x)$ are the roots of the equation

$p x^{2}+q x+r=0$, then $\alpha+x+\beta+x=-\frac{q}{p}$

and $(\alpha+x)(\beta+x)=\frac{r}{p} \Rightarrow \alpha+\beta+2 x=-\frac{q}{p}$

$\Rightarrow \frac{-b}{a}+2 x=-\frac{q}{p} \Rightarrow \boldsymbol{K}=\frac{\mathbf{1}}{\mathbf{2}}(\frac{\boldsymbol{b}}{\boldsymbol{a}}-\frac{\boldsymbol{q}}{\boldsymbol{p}})$.

~~ 39. Let $x=\sqrt{6+\sqrt{6+\sqrt{6+\ldots . \infty}}}$

$\Rightarrow x=\sqrt{6+x}$

$\Rightarrow x^{2}=6+x \Rightarrow x^{2}-x-6=0$

$\Rightarrow(x-3)(x+2)=0$

$\Rightarrow x=3$ or -2

$\Rightarrow \boldsymbol{x}=\mathbf{3}$ as $x=-2$ does not satisfy the given equation.

~~ 40. The equation is $(x-a)(x-a-1)+(x-a-1)(x-a-2)$

$+(x-a)(x-a-2)=0$.

Let $(x-a)=y$, then the equation becomes

$y(y-1)+(y-1)(y-2)+y(y-2)=0$

$\Rightarrow y^{2}-y+y^{2}-3 y+2+y^{2}-2 y=0$

$\Rightarrow 3 y^{2}-6 y+2=0$

$\therefore$ Discriminant $=D=b^{2}-4 a c=36-4 \times 3 \times 2$

$ =36-24=12>0 $

$\therefore$ Roots are real and distinct.

SELF ASSESSMENT SHEET

If the sum as well as the product of roots of a quadratic equation is 9 , then the equation is:

(a) $x^{2}+9 x-18=0$

(b) $x^{2}-18 x+9=0$

(d) $x^{2}-9 x+9=0$.

~~ 2. If one root of the equation $\frac{x^{2}}{a}+\frac{x}{b}+\frac{1}{c}=0$ is reciprocal of the other, then which of the following is correct?

(a) $a=b$

(b) $b=c$

(d) $a=c$

(CDS 2012)

~~ 3. If one root of the equation $a x^{2}+x-3=0$ is -1 , then what is the other root?

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(d) 1

~~ 4. If the equation $(a^{2}+b^{2}) x^{2}-2(a c+b d) x+(c^{2}+d^{2})=0$ has equal roots, then which one of the following is correct?

(a) $a b=c d$

(b) $a d=b c$

(d) $a c=b d$

(CDS 2010)

~~ 5. If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then what is the value of $\alpha^{3}+\beta^{3}$ ?

(a) $\frac{b^{3}+3 a b c}{a^{3}}$

(b) $\frac{a^{3}-b^{3}}{3 a b c}$

(d) $\frac{b^{3}-3 a b c}{a^{3}}$

(CDS 2008)

~~ 6. If the sum of the roots of the equation $a x^{2}+b x+c=0$ is equal to the sum of their squares, then which one of the following is correct?

(a) $a^{2}+b^{2}=c^{2}$

(b) $a^{2}+b^{2}=a+b$

(d) $2 c+b=0$

~~ 7. One root of $x^{2}+k x-8=0$ is the square of the other, then the value of $k$ is :

(a) 2

(b) 8

(d) -2

(CAT 1995)

~~ 8. Let $p$ and $q$ be the roots of the quadratic equation $x^{2}-(\alpha-2) x-\alpha-1=0$. What is the minimum possible value of $p^{2}+q^{2}$ ?

(a) 0

(b) 3

(d) 5

(CAT 2003)

~~ 9. If $\alpha, \beta, \gamma$ are the roots of $x^{3}-2 x^{2}+3 x-4=0$, then the value of $\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}$ is :

(a) -7

(b) -5

(d) 0

(EAMCET 2007)

~~ 10. If $\alpha, \beta$ are the roots of the equations $x^{2}+4 x+3=0$, then the equation whose roots are $2 \alpha+\beta$ and $\alpha+2 \beta$ is

(a) $x^{2}-12 x-33=0$

(b) $x^{2}-12 x+35=0$

(d) $x^{2}+12 x+35=0$.

(J&K CET 2009)

~~ 11. The equation whose roots are twice the roots of the equation $x^{2}-3 x+3=0$ is

(a) $x^{2}-3 x+6=0$

(b) $x^{2}-4 x+8=0$

(d) $x^{2}-8 x+6=0$

~~ 12. If $x^{2}+m x+n=0$ and $x^{2}+p x+q=0$ have a common root, then the common root is

(a) $\frac{q-n}{m-p}$

(b) $\frac{q-n}{m+p}$

(d) None of these

ANSWERS

~~ 2. (d)

~~ 3. (c)

~~ 4. (b)

~~ 5. (c)

~~ 6. (c)

~~ 7. (d)

~~ 8. (d)

~~ 9. (a)

~~ 10. $(d)$

~~ 11. (c)

~~ 12. (a)

HINTS AND SOLUTIONS

Equation : $x^{2}-($ Sum of roots $) x+$ Product of roots $=0$ $\Rightarrow x^{2}-9 x+9=0$.

~~ 2. The equation can be written as $\frac{1}{a} x^{2}+\frac{1}{b} x+\frac{1}{c}=0$. i.e., $b c x^{2}+a c x+a b=0$.

Let $\alpha, 1 / \alpha$ be the roots of the given equation, then product of roots $=\alpha \times \frac{1}{\alpha}=\frac{a b}{b c} \Rightarrow \frac{a b}{b c}=1 \Rightarrow \boldsymbol{a}=\boldsymbol{c}$.

~~ 3. Let the other root of the equation $a x^{2}+x-3=0$ be $\alpha$.

As $(-1)$ is a root of the given equation, it satisfies the given equation, i.e., $a(-1)^{2}+(-1)-3=0 \Rightarrow a-1-3=0 \Rightarrow a=4$.

$\therefore$ The equation becomes $4 x^{2}+x-3=0$.

Now, product of roots $=\alpha \times(-1)=-\frac{3}{4}$

$\therefore \alpha=\frac{\mathbf{3}}{\mathbf{4}}$.

~~ 4. Equal roots $\Rightarrow$ Discriminant $=0$

$ \begin{aligned} & \Rightarrow 4(a c+b d)^{2}-4(a^{2}+b^{2})(c^{2}+d^{2})=0 \\ & \Rightarrow[a^{2} c^{2}+b^{2} d^{2}+2 a c b d]-[\not A^{2} c^{2}+b^{2} c^{2}+a^{2} d^{2}+b^{2} d^{2}]=0 \\ & \Rightarrow-[b^{2} c^{2}+a^{2} d^{2}-2 a c b d]=0 \\ & \Rightarrow(b c-a d)^{2}=0 \Rightarrow \boldsymbol{b} \boldsymbol{c}=\boldsymbol{a d} . \end{aligned} $

~~ 5. $\alpha+\beta=-b / a, \alpha \beta=c / a$.

$ \begin{aligned} & \alpha^{3}+\beta^{3}=(\alpha+\beta)(\alpha^{2}+\beta^{2}-\alpha \beta) \\ & =(\alpha+\beta){(\alpha+\beta)^{2}-2 \alpha \beta-\alpha \beta} \\ & =(\alpha+\beta)[(\alpha+\beta)^{2}-3 \alpha \beta] \\ & =-\frac{b}{a}[\frac{b^{2}}{a^{2}}-\frac{3 c}{a}] \end{aligned} $

~~ 6. Let $\alpha, \beta$ be the roots of the equation $a x^{2}+b x+c=0$. Then, $\alpha+\beta=-b / a, \alpha \beta=c / a$

Given, $\alpha+\beta=\alpha^{2}+\beta^{2}$

i.e., $\alpha+\beta=(\alpha^{2}+\beta)^{2}-2 \alpha \beta$

$\Rightarrow \frac{-b}{a}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a}$

$\Rightarrow-\frac{a b}{a^{2}}=\frac{b^{2}}{a^{2}}-\frac{2 a c}{a^{2}} \Rightarrow a b+b^{2}=\mathbf{2 a c}$.

~~ 7. Let $\alpha$ and $\alpha^{2}$ be the roots of the equation $x^{2}+k x-8=0$.

Then, product of roots $=\alpha \cdot \alpha^{2}=-8$

$ \alpha^{3}=-8 \Rightarrow \alpha=-2 $

$\therefore$ The root $(-2)$ satisfies the given equation, i.e.,

$(-2)^{2}+k .(-2)-8=0$

$4-2 k-8=0 \Rightarrow-2 k=4 \Rightarrow k=-\mathbf{2}$.

~~ 8. If $p$ and $q$ are the roots of the equation

$x^{2}-(\alpha-2) x-(\alpha+1)=0$.

$\therefore$ Sum of roots $=p+q=(\alpha-2)$

Product of roots $=p q=-\alpha-1$

$\therefore p^{2}+q^{2}=(p+q)^{2}-2 p q$

$=(\alpha-2)^{2}+2(\alpha+1)$

$=\alpha^{2}+4-4 \alpha+2 \alpha+2=(\alpha+1)^{2}+5$

$p^{2}+q^{2}$ will be minimum when $\alpha=0$.

$\therefore$ Minimum value of $p^{2}+q^{2}=\mathbf{5}$.

~~ 9. Given, $\alpha, \beta, \gamma$ are the roots of $x^{3}-2 x^{2}+3 x-4=0$. Then,

$ \begin{aligned} & S_1=\alpha+\beta+\gamma=2 \\ & S_2=\alpha \beta+\beta \gamma+\gamma \alpha=3 \\ & S_3=\alpha \beta \gamma=4 \end{aligned} $

Now, $(\alpha \beta+\beta \gamma+\gamma \alpha)^{2}$

$=\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}+2 \alpha \beta^{2} \gamma+2 \beta \gamma^{2} \alpha+2 \alpha^{2} \beta \gamma$

$=\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}+2 \alpha \beta \gamma(\beta+\gamma+\alpha)$

$\Rightarrow \alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}=(\alpha \beta+\beta \gamma+\gamma \alpha)^{2}-2 \alpha \beta \gamma(\alpha+\beta+\gamma)$ $=3^{2}-2 \times 4 \times 2=9-16=-7$.

~~ 10. Given, $\alpha, \beta$ are the roots of the quadratic equations $x^{2}+4 x+3=0$, $\Rightarrow$ Sum of roots $=\alpha+\beta=-4$

Product of roots $=\alpha \beta=3$

Given, $2 \alpha+\beta$ and $\alpha+2 \beta$ are the roots of the required equation, so

Required equation is

$ \begin{aligned} & =x^{2}-(2 \alpha+\beta+\alpha+2 \beta) x+(2 \alpha+\beta)(\alpha+2 \beta)=0 \\ & \text{ Sum of roots } \\ & \begin{aligned} \text{ Sum } & =2 \alpha+\beta+\alpha+2 \beta=3 \alpha+3 \beta=3(\alpha+\beta) \\ & =3 \times-4=-12 \end{aligned} \end{aligned} $

Product $=(2 \alpha+\beta)(\alpha+2 \beta)=2 \alpha^{2}+\beta \alpha+4 \alpha \beta+2 \beta^{2}$

$ \begin{aligned} & =2(\alpha^{2}+\beta^{2})+5 \alpha \beta \\ & =2(\alpha^{2}+\beta^{2}+2 \alpha \beta)+\alpha \beta \\ & =2(\alpha+\beta)^{2}+\alpha \beta=2 \times 16+3=35 \end{aligned} $

$\therefore$ Reqd. equation is $\boldsymbol{x}^{\mathbf{2}}-\mathbf{1 2 x}+\mathbf{3 5}=\mathbf{0}$.

~~ 11. Let $\alpha, \beta$ be the roots of the given equation $x^{2}-3 x+3=0$.

Given, $\alpha+\beta=+3, \alpha \beta=3$

By the given condition, the roots of the required equation are $2 \alpha$ and $2 \beta$.

$\therefore$ Sum of roots $=2 \alpha+2 \beta=2(\alpha+\beta)=2 \times 3=6$

Product of roots $=2 \alpha .2 \beta=4 \alpha \beta=4 \times 3=12$.

$\therefore$ Required equation is $\boldsymbol{x}^{2}-\mathbf{6 x}+\mathbf{1 2}=\mathbf{0}$

~~ 12. Let $\alpha$ be the common root of the equations $x^{2}+m x+n=0$ and $x^{2}+p x+q=0$. Then,

$\alpha^{2}+m \alpha+n=0$

$\alpha^{2}+p \alpha+q=0$

Solving (i) and (ii) simultaneously, we get

$\frac{\alpha^{2}}{m q-n p}=\frac{\alpha}{n-q}=\frac{1}{p-m} \Rightarrow \alpha=\frac{n-q}{p-m}=\frac{\boldsymbol{q}-\boldsymbol{n}}{\boldsymbol{m}-\boldsymbol{p}}$.

Inequalities

KEY FACTS

I. An inequality is a statement involving quantities (variables or constants) connected to each other with signs of inequality, i.e., $<, \leq,>, \geq$.

Thus, $x<y, x \leq y, x>y$ and $x \geq y$ represent inequalities.

II. Note on the use of brackets for open and closed intervals.

Sometimes the domains of variation, i.e., the intervals are denoted as follows:

(a) ’ $\boldsymbol{a}<\boldsymbol{x}<\boldsymbol{b}$ ’ is denoted by $(a, b)$ and is called the open interval of the variable $x$.

(b) ’ $\boldsymbol{a} \leq \boldsymbol{x} \leq \boldsymbol{b}$ ’ is denoted by $[a, b]$ and is called a closed interval of the variable $x$, as $x$ can take up values ’ $a$ ’ and ’ $b$ ’ also.

(c) Any ’ $x$ ’ is denoted by $(-\infty, \infty)$. Here, it should be noted that the symbols $-\infty, \infty$ are not numbers in any sense whatsoever.

(d) ’ $\boldsymbol{x} \geq \boldsymbol{a}$ ’ is denoted by $[a, \infty)$ and ’ $\boldsymbol{x} \leq \boldsymbol{b}$ ’ is denoted by $(-\infty, b]$.

(e) ’ $\boldsymbol{a} \leq \boldsymbol{x}<\boldsymbol{b}$ ’ is denoted by $[a, b)$, ’ $\boldsymbol{a}<\boldsymbol{x} \leq \boldsymbol{b}$ ’ is denoted by $(a, b]$. These are called semi-closed intervals.

III. Important Properties of Inequalities.

For all $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} \in \boldsymbol{R}$ (i) If $\boldsymbol{a}>\boldsymbol{b}$ and $\boldsymbol{b}>\boldsymbol{c}$, then $\boldsymbol{a}>\boldsymbol{c}$. Ex. $4>2$ and $2>-1 \Rightarrow 4>-1$	(ii) If $\boldsymbol{a} > \boldsymbol{b}$, then for all $\boldsymbol{c} \\ , \boldsymbol{a} + \boldsymbol{c} > \boldsymbol{b}+\boldsymbol{c}$ and $\boldsymbol{a}-\boldsymbol{c}>\boldsymbol{b}-\boldsymbol{c} \\ $. Ex. $7>4 \Rightarrow 7+2>4+2 \Rightarrow 9>6 \\ 7>4 \Rightarrow 7-2>4-2 \Rightarrow 5>2$
(iii) If $\boldsymbol{a}>\boldsymbol{b}$ and $\boldsymbol{c}>\mathbf{0}$, then $\boldsymbol{a} \boldsymbol{c}>\boldsymbol{b}$ and $\frac{\boldsymbol{a}}{\boldsymbol{c}}>\frac{\boldsymbol{b}}{\boldsymbol{c}}$. Ex. $4>2 \Rightarrow 4 \times 3>2 \times 3 \Rightarrow 12>6$ $4>2 \Rightarrow \frac{4}{2}>\frac{2}{2} \Rightarrow 2>1$
(v) If $\boldsymbol{a} \neq \mathbf{0}, \boldsymbol{b} \neq \mathbf{0}$ and $\boldsymbol{a}>\boldsymbol{b}>\mathbf{0}$, then $\frac{\mathbf{1}}{\boldsymbol{a}}<\frac{\mathbf{1}}{\boldsymbol{b}}$. Ex. $3>1>0 \Rightarrow \frac{1}{3}<1$

(vii) (a) If $m>0$, then for $\boldsymbol{a}>\boldsymbol{b}>\mathbf{0}$

$a^{m}>b^{m} \quad \bullet a^{-m}<b^{-m} \quad \bullet a^{1 / m}>b^{1 / m}$

Ex. Let $a=3, b=2$ and $m=3$, then

$3^{3}>2^{3}$ since $27>8$
$3^{-3}<2^{-3}$, i.e., $\frac{1}{3^{3}}<\frac{1}{2^{3}}$ since $\frac{1}{27}<\frac{1}{8}$
$3^{1 / 3}>2^{1 / 3}$ (b) Similarly, if $m>0$ and $a<b, a>0, b>0$, then
$a^{m}<b^{m} \quad \bullet a^{-m}>b^{-m} \quad \bullet a^{1 / m}<b^{1 / m}$

(viii) (a) If $a>1$ and $m>0$, then $\bullet a^{m}>1$ and $\bullet 0<a^{-m}<1$

Let $a=2, m=2 \Rightarrow 2^{2}=4>1$ and $0<2^{-2}<1$ as $0<\frac{1}{4}<1$

(b) If $0<a<1$ and $m>0$, then $0<a^{m}<1$ and $a^{-m}>1$

Let $a=\frac{1}{2}, m=2$, then $0<(\frac{1}{2})^{2}<1$ since $0<\frac{1}{4}<1$ and $(\frac{1}{2})^{-2}>1$ since $4>1$

(ix) (a) If $a>1$ and $m>n>0$, then $\boldsymbol{a}^{\boldsymbol{m}}>\boldsymbol{a}^{\boldsymbol{n}}$

Ex. $a=2, m=3, n=2$, then $2^{3}>2^{2}$

(b) If $0<a<1$ and $m>n>0$ then $a^{m}<a^{n}$

Ex. $a=\frac{1}{3}, m=3, n=2$, then $(\frac{1}{3})^{3}<(\frac{1}{3})^{2}$ since $\frac{1}{27}<\frac{1}{9}$

(x) If $0<\boldsymbol{a}<\mathbf{1}<\boldsymbol{b}$ and $\boldsymbol{r}$ is a positive rational number, then

$0<a^{r}<1<a^{-r}$ and $\bullet 0<b^{-r}<1<b^{r}$

Ex. If $0<\frac{1}{2}<1<3$ and $r=3$, then

$0<(\frac{1}{2})^{3}<1<(\frac{1}{2})^{-3}$ since $0<\frac{1}{8}<1<8$
$0<(3)^{-3}<1<(3)^{3}$ since $0<\frac{1}{27}<1<27$ (xi) (a) If $a>1$ and $x>y>0$, then $\log _{a} x>\log _{a} y$

Ex. $a=3, x=27, y=9$

$\Rightarrow \log _3 27>\log _3 9 \Rightarrow \log _3 3^{3}>\log _3 3^{2} \Rightarrow 3>2$

(b) If $0<a<1$ and $x>y>0$, then $\log _{a} x<\log _{a} y$

Ex. $a=1 / 3, \quad x=27, \quad y=9$

$\Rightarrow \log _{1 / 3} 27<\log _{1 / 3} 9 \Rightarrow \log _{1 / 3} 3^{3}<\log _{1 / 3} 3^{2}$

$\Rightarrow \log _{3^{-1}} 3^{3}<\log _{3^{-1}} 3^{2} \Rightarrow \frac{3}{-1}<\frac{2}{-1}$ since $-3<-2$ (xii) (a) If $a>1$, then $a^{f(x)}>\boldsymbol{a}^{g(x)} \Leftrightarrow f(x)>g(x)$

(b) If $0<a<1$, then $a^{f(x)}>a^{g(x)} \Leftrightarrow f(x)<g(x)$

(d) If $0<a<1$, then $\log _{a} f(x)>\log _{a} g(x) \Leftrightarrow 0<f(x)<g(x)$

(e) If $a>1$, then $\log _{a} x>p \Rightarrow x>a^{p}$

Ex. $\log _4 x>3 \Rightarrow x>4^{3} \Rightarrow x>64$

(f) If $0<a<1$, then $\log _{a} x>p \Rightarrow 0<x<a^{p}$

Ex. $\log _{1 / 2} x>3 \Rightarrow 0<x<(1 / 2)^{3}$

$ \Rightarrow 0<x<\frac{1}{8} $

(g) If $0<a<1$, then $0<\log _{a} x<p \Rightarrow a^{p}<x<1$

Ex. $0<\log _{1 / 3} x<2 \Rightarrow(\frac{1}{3})^{2}<x<1 \Rightarrow \frac{1}{9}<x<1$ (xiii) Recall Modulus Properties

(a) $a \leq|a|$

(b) $|a b|=|a||b|$

(d) $\bullet|a+b|<|a|+|b|$

$|a-b|>|| a|-| b \mid$

(e) $\bullet|a+b|=|a|+|b|$ if $a b>0$

$|a-b|=|a|-|b|$ if $a b<0$ (xiv) For any positive real number $a$,

(a) $|x|<a \Rightarrow-a<x<a \Rightarrow x \in(-a, a)$ Ex. $|x|<5 \Rightarrow-5 \leq x \leq 5$

(b) $|x| \leq a \Rightarrow-a \leq x \leq a \Rightarrow x \in[-a, a]$ Ex. $|x|<7 \Rightarrow x<-7$ or $x>7$

(d) $|x| \geq a \Rightarrow x \leq-a$ or $x \geq a \Rightarrow x \in(-\infty,-a] \cup[a, \infty)$

Note: i.e., square brackets mean that the point inside it is inclusive.

(xv) Let $r$ be a positive real number and a be fixed real number. Then,

(a) $|x-a|<r \Rightarrow a-r<x<a+r \Rightarrow x \in(a-r, a+r)$

Ex. $|x-2|<5 \Rightarrow 2-5<x<2+5$

(b) $|x-a| \leq r \Rightarrow a-r \leq x \leq a+r \Rightarrow x \in [a-r, a+r]$

Ex. $|x-3| \geq 8<x<7$

(d) $|x-a| \geq r \Rightarrow x \leq a-r$ or $x \geq a+r \Rightarrow x \in(-\infty, a-r] \cup[a+r, \infty)$

$\Rightarrow x<3-8$ or $x>3+8$ $\Rightarrow x<-5$ or $x>11$

(xvi) Let $\boldsymbol{a}, \boldsymbol{b}$ be positive real numbers. Then,

(a) $a<|x|<b \Rightarrow x \in(-b,-a) \cup(a, b)$

(b) $a \leq|x| \leq b \Rightarrow x \in[-b,-a] \cup[a, b]$

(d) $a \leq|x-c| \leq b \Rightarrow x \in[-b+c,-a+c] \cup[a+c, b+c]$

To find the solution set of a quadratic inequation of the form $x^{2}+a x+b<0$ or $x^{2}+a x+b>0$, the following steps are to be followed:

Step 1. Set inequality to zero

Step 3. Set each factor equal to zero to obtain critical points

Step 5. Use interval method for testing Step 2. Factorise the inequality

Step 4. Place critical points on the number line

Step 6. Write the solution set.

Ex. Find the solution set of $\boldsymbol{x}^{\mathbf{2}}-\mathbf{5 x}+\mathbf{6} \geq \mathbf{0}$.

Sol. $x^{2}-5 x+6 \geq 0 \Rightarrow(x-3)(x-2) \geq 0$.

The critical points are 3 and 2 .

Plotting 2 and 3 on the number line, we have it as shown below:

The number line is divided into three parts. Now we examine the sign of the expression $(x-3)(x-2)$ in each of these three intervals.

As $(x-3)(x-2) \geq 0$, the points 2 and 3 are included here.

When $x \leq 2,(x-3)(x-2) \geq 0 \Rightarrow x \in(-\infty, 2]$

When $x \geq 3,(x-3)(x-2)>0 \Rightarrow x \in[3, \infty)$

When $2<x<3,(x-3)(x-2)<0$

$\therefore \boldsymbol{x} \in(-\infty, 2] \cup[3, \infty)$

These solutions can be further remembered as given below:

(xvii) Quadratic Inequalities or Quadratic Inequations

If $a<b$, then

(a) $(x-a)(x-b)>0$ $\Rightarrow \quad x<a$ or $x>b$

(b) $(x-a)(x-b) \geq 0$ or $\quad x \in(-\infty, a) \cup(b, \infty)$ or $\boldsymbol{x} \in \boldsymbol{R}-[\boldsymbol{a}, \boldsymbol{b}]$ $\Rightarrow \quad x \leq a$ or $x \geq b$ or $x \in(-\infty, a] \cup[b, \infty)$ or $\boldsymbol{x} \in \boldsymbol{R}-(\boldsymbol{a}, \boldsymbol{b})$ Ex. $(x-3)(x-5)>0 \Rightarrow x<3$ or $x>5$ or $x \in(-\infty, 3) \cup(5, \infty)$ Ex. $(x-2)(x+5) \geq 0$ or $x \in R-[3,5]$ $\Rightarrow \quad(x-2)(x-(-5)) \geq 0$ Here $-5<2$ $\therefore x \leq-5$ or $x \geq 2$ or $x \in(-\infty,-5] \cup[2, \infty)$

(c) $(x-a)(x-b)<0$ $\Rightarrow a<x<b$ $\Rightarrow \quad x \in(a, b)$ Ex. $(x-4)(x+2)<0$ $\Rightarrow(x-4)(x-(-2))<0$ $\Rightarrow \quad-2<x<4 \quad(\because-2<4)$ $\Rightarrow \quad x \in(-2,4)$

(d) $(x-a)(x-b) \leq 0$ $\Rightarrow a \leq x \leq b$ $\Rightarrow x \in[a, b]$

Ex. $(x-2)(x-3) \leq 0$

$\Rightarrow 2 \leq x \leq 3$

$\Rightarrow \quad x \in[2,3]$.

IV. Important Inequalities

(a) (i) If $a>0$, then $a+\frac{1}{a} \geq 2$,

(ii) If $a<0$, then $a+\frac{1}{a} \leq-2$

(b) Arithmetic-Geometric Mean Inequality

For distinct positive real numbers,

Arithmetic mean $>$ Geometric mean, i.e.,

If $a_1, a_2, a_3, \ldots \ldots \ldots ., a_n$ are $n$ distinct positive real numbers, then

$\frac{a_1+a_2+\ldots \ldots \ldots+a_n}{n}>(a_1, a_2 \ldots \ldots . . . ., a_n)^{1 / n}$, i.e., A.M. $>$ G.M.

Also if $a_1=a_2=a_3 \ldots \ldots \ldots \ldots=\ldots \ldots \ldots \ldots=a_n$, then A.M. $=$ G.M.

Hence for positive real numbers, A.M. $\geq$ G.M.

(i) In particular,

$ \frac{a+b}{2} \geq \sqrt{a b}, \quad \frac{a+b+c}{3} \geq \sqrt[3]{a b c}, \quad \frac{a+b+c+d}{4} \geq(a b c d)^{1 / 4} $

(ii) Also A.M. $\geq$ G.M. $\geq$ H.M., i.e.,

$ \frac{a_1+a_2+\ldots \ldots . .+a_n}{n} \geq(a_1 \cdot a_2 \ldots \ldots . . a_n)^{1 / n} \geq \frac{n}{(\frac{1}{a_1}+\frac{1}{a_2}+\ldots \ldots \ldots \frac{1}{a_n})} $

(c) (i) If the sum of two positive quantities $a_1$ and $a_2$ is constant, then their product is greatest when $a_1=a_2$.

(ii) If the product of these quantities is a constant, then their sum is least when $a_1=a_2$.

In general, if $a_1, a_2, \ldots \ldots \ldots . ., a_n$ are $n$ positive quantities such that their sum $a_1+a_2+a_3+\ldots \ldots \ldots \ldots+a_n=C$ (constant), then the product $(a_1 \cdot a_2 \cdot a_3 \ldots \ldots \ldots . . a_n)$ is maximum

when $a_1=a_2=a_3=\ldots \ldots \ldots . .=a_n=\frac{C}{n}$ and the maximum product $=(\frac{C}{n})^{n}$.

Also, if $a_1, a_2, \ldots \ldots \ldots . . a_n$ are $n$ positive quantities such that their product $a_1 \cdot a_2 \ldots \ldots \ldots . a_n=P$ (constant), then their sum $(a_1+a_2+\ldots \ldots \ldots . .+a_n)$ is least when $a_1=a_2=a_3=$ $=a_n=(P)^{1 / n}$ and the least sum $=n(P^{1 / n})$.

(d) Weighted A.M. - G.M. inequality

If $a_1, a_2, a_3 \ldots \ldots \ldots . a_n$ are $n$ positive real numbers and $m_1, m_2, m_3 \ldots \ldots \ldots . m_n$ are $n$ positive rational numbers, then $\frac{m_1 a_1+m_2 a_2+\ldots \ldots \ldots+m_n a_n}{m_1+m_2+\ldots \ldots \ldots m_n}>(a_1^{m_1} a_2^{m_2} \ldots a_3^{m_n})^{\frac{1}{m_1+m_2+\ldots+m_n}}$

Weighted A.M. > Weighted G.M.

(e) If $a_1, a_2, a_3, \ldots \ldots . ., a_n$ are unequal positive real numbers and $m$ is a positive rational number different from 0 and 1 , then

(i) $(\frac{a_1^{m}+a_2^{m}+\ldots .+a_n^{m}}{n})<(\frac{a_1+a_2+\ldots .+a_n}{n})^{m}$ if $0<m<1$

The arithmetic mean of the $m$ th powers of $n$ positive quantities is less than the $m$ th power of their arithmetic mean if $\boldsymbol{m}$ lies between 0 and 1 . (ii) $(\frac{a_1^{m}+a_2^{m}+\ldots .+a_n^{m}}{n})>(\frac{a_1+a_2+\ldots .+a_n}{n})^{m}$ if $m<0$ or $m>1$

The arithmetic mean of the $m$ th powers of $n$ positive qualities is greater than the $m$ th power of their arithmetic mean if $\boldsymbol{m}<\mathbf{0}$ or $\boldsymbol{m}>\mathbf{1}$.

SOLVED EXAMPLES

~~ Ex. 1 . Solve the following linear inequations: (i) $4 x-12 \leq 0$ (ii) $-\mathbf{5 x}+\mathbf{1 0}<0$ (iii) $7 x-35 \geq 0$ (iv) $11 x+9>53$, where $x \in W$.

Sol. (i) $4 x-12 \leq 0 \Rightarrow 4 x \leq 12 \Rightarrow x \leq 3$

$\therefore$ Solution set of given inequation is $(-\infty, 3$ ]

(ii) $-5 x+10<0 \Rightarrow-5 x<-10$

$\Rightarrow \frac{-5 x}{-5}>\frac{-10}{-5} \quad[\because$ Dividing both sides of an inequality by a negative number reverses the inequality $]$

$\Rightarrow x>2 \quad \Rightarrow \quad$ Solution set is $(2, \infty)$

(iii) $7 x-35 \geq 0 \Rightarrow 7 x \geq 35 \Rightarrow x \geq 5$

$\Rightarrow$ Solution set is $[5, \infty)$

(iv) $11 x+9>53 \Rightarrow 11 x>44 \Rightarrow x>4$

$\therefore$ Solution set is $(4, \infty)$

Ex. 2 . Solve the inequality $4(\frac{1}{2}-p)+7 \geq 57$ over $R$ (set of real numbers).

~~ Sol. $4(\frac{1}{2}-p)+7 \geq 57 \Rightarrow 2-4 p+7 \geq 57$

$\Rightarrow-4 p \geq 57-9 \quad \Rightarrow-4 p \geq 48 \quad \Rightarrow 4 p \leq-48 \quad \Rightarrow p \leq-12$

$\therefore \boldsymbol{p} \in(-\infty,-\mathbf{1 2}]$

The solution set can be shown on the graph as:

~~ Ex. 3 . Solve the following inequation: $\frac{2 x-3}{4}+8 \geq 2+\frac{4 x}{3} ; x \in R$.

Sol. $\frac{2 x-3}{4}+8 \geq 2+\frac{4 x}{3} \Rightarrow 12(\frac{2 x-3}{4})+12 \times 8 \geq 12 \times 2+\frac{4 x}{3} \times 12$ (Multiplying each term by LCM $=12$ )

$\Rightarrow 3(2 x-3)+96 \geq 24+16 x \Rightarrow 6 x-9+96 \geq 24+16 x$

$\Rightarrow 6 x+87 \geq 24+16 x \Rightarrow 87-24 \geq 16 x-6 x \Rightarrow 63 \geq 10 x$

$\Rightarrow 10 x \leq 63 \Rightarrow x \leq 6.3$

$\therefore x \in(-\infty, 6.3]$

~~ Ex. 4 . Find the solution set of $-3<x-2 \leq 9-2 x ; x \in Z$ (set of integers).

Sol. $-3<x-2 \leq 9-2 x$

$\Rightarrow-3<x-2$ and $x-2 \leq 9-2 x \Rightarrow-3+2<x$ and $x+2 x \leq 9+2$

$\Rightarrow-1<x$ and $3 x \leq 11$ or $x \leq \frac{11}{3} \Rightarrow-1<x \leq \frac{11}{3}$

Since $x \in Z$, so the solution set $={\mathbf{0}, \mathbf{1}, \mathbf{2}, \mathbf{3}}$.

~~ Ex. 5 . Find the range of values of $x$, which satisfy the inequality: $\frac{-1}{5} \leq \frac{3 x}{10}+1<\frac{2}{5}: x \in R$

Sol. $-\frac{1}{5} \leq \frac{3 x}{10}+1<\frac{2}{5}$

$\Rightarrow 10 \times(-\frac{1}{5}) \leq 10 \times(\frac{3 x}{10}+1)<10 \times \frac{2}{5} \Rightarrow-2 \leq 3 x+10<4$

$\Rightarrow-2 \leq 3 x+10$ and $3 x+10<4 \Rightarrow-12 \leq 3 x$ and $3 x<-6$

$\Rightarrow-4 \leq x$ and $x<-2 \quad \Rightarrow-4 \leq x<-2$, i.e., $x \in[-4,-2)$

This solution set can be shown on a graph as:

~~ Ex. 6 . Solve the following pairs of inequations and also graph the solution set (i) $2 x-9<7$ and $3 x+9 \leq 25, x \in R$ (ii) $3 x-2>19$ or $3-2 x \geq-7, x \in R$

Sol. (i) Let $A={x: 2 x-9<7, x \in R}$

$ \begin{matrix} \therefore & 2 x-9<7 \quad \Rightarrow 2 x<16 \quad \Rightarrow x<8 \\ \Rightarrow & A={x: x<8, x \in R} \end{matrix} $

Let $B={x: 3 x+9 \leq 25, x \in R}$

$\therefore \quad 3 x+9 \leq 25 \Rightarrow 3 x \leq 16 \quad \Rightarrow x \leq \frac{16}{3}$

$\Rightarrow \quad B={x: x \leq \frac{16}{3}, x \in R}$

$\therefore \quad$ Required solution set $=A \cap B$

$ \begin{aligned} & ={x: x<8, x \in R} \cap{x: x \leq \frac{16}{3}, x \in R} \\ & ={x: x \leq \frac{16}{3}, x \in R}=x \in(-\infty, \frac{16}{3}] . \end{aligned} $

This solution set can be shown on a graph as:

(ii) Let $A={x: 3 x-2>19, x \in R}$

Then, $3 x-2>19 \Rightarrow 3 x>21 \Rightarrow x>7 \Rightarrow A={x: x>7, x \in R}$

Let $B={x: 3-2 x \geq-7, x \in R}$

Then, $3-2 x \geq-7 \Rightarrow 10 \geq 2 x \Rightarrow 2 x \leq 10 \Rightarrow x \leq 5 \Rightarrow B={x: x \leq 5, x \in R}$

Required solution set $=A$ or $B=A \cup B$

$ \begin{aligned} & ={x: x>7, x \in R} \cup{x: x \leq 5, x \in R} \\ & =x>7 \text{ or } x \leq 5=x \in(7, \infty) \text{ or } x \in(-\infty, 5] \end{aligned} $

This can be shown on a graph as:

~~ Ex. 7 . Solve the following inequations, $x \in R$. (i) $\frac{1}{x-3}<0$ (ii) $\frac{x+1}{x+2}>1$

Sol. (i) $\frac{1}{x-3}<0 \Rightarrow(x-3)<0 \quad[\because a / b<0$ and $a>0 \Rightarrow b<0]$

$\Rightarrow x<3 \Rightarrow \boldsymbol{x} \in(-\infty, 3)$

(ii) $\frac{x+1}{x+2}>1 \Rightarrow \frac{x+1}{x+2}-1>0 \Rightarrow \frac{x+1-x-2}{x+2}>0$

$\Rightarrow \frac{-1}{x+2}>0 \quad \Rightarrow x+2<0 \quad(\because a / b>0, a<0 \Rightarrow b<0)$

$\Rightarrow x<-2 \quad \therefore x \in(-\infty,-\mathbf{2})$

~~ Ex. 8 . Solve the following linear equations, $x \in R$ (i) $\frac{x-6}{x-11}>0$

(ii) $\frac{x-3}{x+5}>2$

(iii) $\frac{x-1}{x+3} \geq 2$

(iv) $\frac{5 x-6}{x+6} \leq 1$

Sol. (i) $\frac{x-6}{x-11}>0$

Equating $(x-6)$ and $(x-11)$ to zero, we obtain $x=6,11$ as the critical points.

Now we plot these points on the real number line as shown: $\underset{-\infty}{\stackrel{+}{-} \underbrace{-}_6} \underset{11}{+} \xrightarrow[\infty]{+}$

The real number line is divided into three regions. Now check the sign of the expression $\frac{x-6}{x-11}$ in all the three regions.

When $x<6$, both numerator and denominator are negative, so $\frac{x-6}{x-11}$ is + ve.

When $x>11$, both numerator and denominator are positive, so $\frac{x-6}{x-11}$ is +ve.

When $6<x<11$, the expression $\frac{x-6}{x-11}$ becomes - ve and hence $<0$.

So, the solution set of the given inequations is the union of regions containing positive signs.

$\therefore \quad \frac{x-6}{x-11}>0 \quad \Rightarrow x \in(-\infty, 6) \cup(11, \infty)$.

(ii) $\frac{\boldsymbol{x}-\mathbf{3}}{\boldsymbol{x}+\mathbf{5}}>\mathbf{2} \quad \Rightarrow \frac{x-3}{x+5}-2>0$

$\Rightarrow \frac{x-3-2 x-10}{x+5}>0 \Rightarrow \frac{-x-13}{x+5}>0$

$\Rightarrow \frac{x+13}{x+5}<0 \quad$ (Multiplying by -1 to make co-efficient of $x$ positive in the expression in the numerator)

Now, putting, $(x+13)$ and $(x+5)$ equal to zero, we get the critical points as $x=-13,-5$. Now plot these points on the real number line as shown and divide the line into three parts.

When $x$ lies between $-\infty$ and -13 , the expression $\frac{x+13}{x+5}$ becomes + ve.

Similarly, when $x$ lies between -5 and $\infty$, the expression becomes $+ve$.

The expression is negative or $(<0)$ when $x$ lies between -13 and -5 .

Hence $\frac{x+13}{x+5}<0 \Rightarrow x \in(-\mathbf{1 3},-\mathbf{5})$

(iii) $\frac{\boldsymbol{x}-\mathbf{1}}{\boldsymbol{x}+\mathbf{3}} \geq \mathbf{2} \quad \Rightarrow \frac{x-1}{x+3}-2 \geq 0$

$\Rightarrow \frac{x-1-2 x-6}{x+3} \geq 0 \Rightarrow \frac{-x-7}{x+3} \geq 0 \quad \Rightarrow \frac{x+7}{x+3} \leq 0$

The critical points are -7 and -3 , (on equating $x+7=0$ and $x+3=0$ )

When $x<-7, \frac{x+7}{x+3}$ becomes +ve

When $x>-3, \frac{x+7}{x+3}$ becomes $+ve$

The expression $\frac{x+7}{x+3}<0$ when $x$ lies between -7 and -3

Also $\frac{x+7}{x+3} \leq 0$ when -7 is also included.

$\therefore \frac{x+7}{x+3} \leq 0 \quad x \in[-7,-3)$

(iv) $\frac{\mathbf{5 x}-\mathbf{6}}{\boldsymbol{x}+\mathbf{6}} \leq \mathbf{1} \Rightarrow \frac{5 x-6}{x+6}-1 \leq 0$

$\Rightarrow \frac{5 x-6-x-6}{x+6} \leq 0 \Rightarrow \frac{4 x-12}{x+6} \leq 0 \Rightarrow \frac{4(x-3)}{x+6} \leq 0 \Rightarrow \frac{x-3}{x+6} \leq 0$

The critical points are $x=3,-6$. On the real number line they can be shown as:

When $x<-6$ or $x>3$, then the expresion $\frac{x-3}{x+6}$ becomes positive.

When $x$ lies between -6 and 3 (included), i.e., $-6<x \leq 3, \frac{x-3}{x+6} \leq 0$.

$\Rightarrow \frac{x-3}{x+6} \leq 0 \Rightarrow x \in(-6,3]$.

~~ Ex. 9 . Solve : (i) $|2 x-3| \leq \frac{1}{4}$ (ii) $|x-4| \geq 7$ (iii) $1 \leq|x-3| \leq 5$

Sol. (i) We know that $|x-a| \leq r \quad \Rightarrow \quad(a-r) \leq x \leq(a+r)$

(ii) Since $|x-a| \geq r \quad \Rightarrow x \leq a-r$ or $x \geq a+r$

$\therefore \quad|2 x-3| \leq \frac{1}{4} \Rightarrow(3-\frac{1}{4}) \leq 2 x \leq(3+\frac{1}{4}) \Rightarrow \frac{11}{4} \leq 2 x \leq \frac{13}{4}$

$\Rightarrow \frac{11}{8} \leq x \leq \frac{13}{8} \quad \Rightarrow \quad x \in[\frac{\mathbf{1 1}}{\mathbf{8}}, \frac{\mathbf{1 3}}{\mathbf{8}}]$.

$|x-4| \geq 7 \quad \Rightarrow x \leq 4-7$ or $x \geq 4+7 \quad \Rightarrow \quad x \leq-3$ or $x \geq 11$

$\Rightarrow x \in(-\infty,-3]$ or $x \in[11, \infty) \Rightarrow \boldsymbol{x} \in(-\infty,-3] \cup[11, \infty)$.

(iii) Since $a \leq|x-c| \leq b \Rightarrow x \in[-b+c,-a+c] \cup[a+c, b+c]$

$\therefore 1 \leq|x-3| \leq 5 \Rightarrow x \in[-5+3,-1+3] \cup[1+3,5+3] \Rightarrow x \in[-\mathbf{2}, \mathbf{2}] \cup[\mathbf{4}, \mathbf{8}]$.

~~ Ex. 10 . Solve $\frac{|x|-1}{|x|-2} \geq 0, x \in R, x \neq \pm 2$.

Sol. Let $|x|=y$. Then

$ \frac{|x|-1}{|x|-2} \geq 0 \Rightarrow \frac{y-1}{y-2} \geq 0 $

On equating $(y-1)$ and $(y-2)$ equal to zero, we have the critical points as $y=1,2$. Now using the real number line, we see that the expression $\frac{y-1}{y-2}$ is greater than equal to zero (positive) only when, $y \leq 1$ or $y>2$.

$ \begin{aligned} & \Rightarrow|x| \leq 1 \text{ or }|x|>2 \\ & \Rightarrow-1 \leq x \leq 1 \text{ or }(x<-2 \text{ or } x>2) \\ & \Rightarrow x \in[-\mathbf{1}, \mathbf{1}] \cup(-\infty,-\mathbf{2}) \cup(\mathbf{2}, \infty) \end{aligned} $

$ \begin{bmatrix} \because|x| \leq a & \Rightarrow-a \leq x \leq a \\ |x|>a & \Rightarrow x<-a \text{ or } x>a \end{bmatrix} $

~~ Ex. 11 . Solve the inequation $|\frac{2}{x-4}|>1, x \neq 4$.

Sol. $|\frac{2}{x-4}|>1, x \neq 4$

$\Rightarrow \frac{2}{|x-4|}>1 \Rightarrow 2>|x-4| \Rightarrow|x-4|<2$

$\Rightarrow 4-2<x<4+2 \Rightarrow 2<x<6 \quad[|x-a|<r \Rightarrow a-r<x<a+r]$

But $x \neq 4$

$\therefore x \in(2,4) \cup(4,6)$

Quadratic Inequalities

~~ Ex. 12 . Solve $x^{2}-5 x+4>0$.

Sol. $x^{2}-5 x+4>0 \Rightarrow(x-1)(x-4)>0$

Now equating $(x-1)$ and $(x-4)$ to zero, we get the critical points as 1 and 4 . Plot the points 1 and 4 on the real number line and then examine the sign of the expression in the three portions of the line divided by these points.

When $x<1$, i.e., $x \in(-\infty, 1)$, both the terms of the given expression are negative, hence $(x-1)(x-4)>0$.

Similarly when $x>4$, i.e., $x \in(4, \infty),(x-1)(x-4)>0$.

When $x \in(1,4)$, one term being +ve and other $-ve$, the expression $(x-1)(x-4)<0$.

$\therefore$ For $(x-1)(x-4)>0$, the required solution set is

$ x \in(-\infty, 1) \cup(4, \infty) $

~~ Ex. 13 . Find the range of values of $x$ for which $\frac{x^{2}+x+1}{x^{2}+2}<\frac{1}{3}, x$ being real.

Sol. $\frac{x^{2}+x+1}{x^{2}+2}-\frac{1}{3}<0$

$\Rightarrow \frac{3 x^{2}+3 x+3-x^{2}-2}{3(x^{2}+2)}<0 \Rightarrow \frac{2 x^{2}+3 x+1}{3(x^{2}+2)}<0$

Now we have to find the range of values in which $2 x^{2}+3 x+1<0$ as $3(x^{2}+2)$ is + ve for all real values of $x$.

Now $2 x^{2}+3 x+1<0 \quad \Rightarrow 2 x^{2}+2 x+x+1<0$

$\Rightarrow(2 x+1)(x+1)<0 \quad \Rightarrow(x+\frac{1}{2})(x+1)<0$.

Critical points are -1 and $-\frac{1}{2}$. Plotting the critical points on the real number line and observing the sign of the expression $(2 x+1)(x+1)$ in the intervals formed, we see that the expression $(x+\frac{1}{2})(x+1)$ is negative or less than zero in the interval $(-1,-\frac{1}{2})$, i.e., $-1<x<-\frac{1}{2}$.

$x \in(-1,-\frac{1}{2})$

~~ Ex. 14 . Find the range of values of $x$ which satisfy $x^{2}+6 x-27>0,-x^{2}+3 x+4>0$ simultaneously.

Sol. $x^{2}+6 x-27>0 \Rightarrow(x+9)(x-3)>0 \quad \Rightarrow \quad x=-9,3$

By the method of intervals we see that $(x+9)(x-3)$ is positive when $x<-9$

and $x>3$

$x^{2}+6 x-27>0 \Rightarrow x \in(-\infty,-9) \cup(3, \infty)$

Now $-x^{2}+3 x+4>0 \Rightarrow x^{2}-3 x-4<0 \Rightarrow(x-4)(x+1)<0$

$\therefore \quad(x-4)(x+1)=0 \quad \Rightarrow \quad x=-1,4$

$\therefore \quad$ The expression $(x-4)(x+1)$ is negative when $x$ lies between -1 and 4 .

$\therefore \quad x \in(-1,4)$

$\therefore \quad$ (i) and $(i i) \quad \Rightarrow \quad x \in(3,4)$.

~~ Ex. 15 . Prove that :

(i) $x+\frac{1}{x} \geq 2$, if $x>0$ (ii) $x+\frac{1}{x} \leq-2$, if $x<0$ (iii) $|x+\frac{1}{x}| \geq 2$, if $x \neq 0$.

Sol. (i) We have $(\sqrt{x}-\frac{1}{\sqrt{x}})^{2} \geq 0$ (Being a perfect square)

$\Rightarrow x+\frac{1}{x}-2 \sqrt{x} \cdot \frac{1}{\sqrt{x}} \geq 0 \Rightarrow x+\frac{1}{x}-2 \geq 0 \Rightarrow x+\frac{\mathbf{1}}{\boldsymbol{x}} \geq \mathbf{2}$

(ii) Since $x<0$, we cannot take $\sqrt{x}$ and proceed

Let $x=-a$, since $x$ is negative and ’ $a$ ’ must be positive.

$\therefore(\sqrt{a}-\frac{1}{\sqrt{a}})^{2} \geq 0 \quad \Rightarrow a+\frac{1}{a}-2 \geq 0 \Rightarrow a+\frac{1}{a} \geq 2$

Now replace $a$ by $-x$, so $-x-\frac{1}{x} \geq 2 \Rightarrow \boldsymbol{x}+\frac{\mathbf{1}}{\boldsymbol{x}} \leq \mathbf{2}$

(iii) To prove $|x+\frac{1}{x}| \geq 2$, we can prove $(x+\frac{1}{x})^{2} \geq 4$

We have $(x-\frac{1}{x})^{2} \geq 0 \quad \Rightarrow \quad x^{2}+\frac{1}{x^{2}}-2 \geq 0 \quad \Rightarrow \quad x^{2}+\frac{1}{x^{2}} \geq 2$

$ \Rightarrow \quad x^{2}+\frac{1}{x^{2}}+2 \geq 4 \Rightarrow(x+\frac{1}{x})^{2} \geq 4 $

~~ Ex. 16 . Prove that for any three positive reals numbers $a, b, c, a^{2}+b^{2}+c^{2} \geq a b+b c+c a$.

Sol. $a^{2}+b^{2}+c^{2}>a b+b c+c a$

To prove $a^{2}+b^{2}+c^{2}-a b-b c-c a \geq 0$

Let $\quad S=a^{2}+b^{2}+c^{2}-a b-b c-c a$

Then $\quad S=\frac{1}{2}(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a)$

$ =\frac{1}{2}(a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 b c+c^{2}+a^{2}-2 c a)=\frac{1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}] \geq 0 $

As the RHS is the sum of squares which are positive $\Rightarrow S \geq 0$. Also the equality, i.e. $=0$ holds when $a=b=c$.

~~ Ex. 17 . If $x, y, z$ are all positive real numbers. Prove that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9$

(IIT 1984)

Sol. We apply the AM, GM inequality here

$ \begin{aligned} & \text{ AM of } x, y, z=\frac{x+y+z}{3}, \quad \text{ GM of } x, y, z=\sqrt[3]{x y z} \\ & \text{ AM of } \frac{1}{x}, \frac{1}{y}, \frac{1}{z}=\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}, \quad \text{ GM of } \frac{1}{x}, \frac{1}{y}, \frac{1}{z}=\sqrt[3]{\frac{1}{x y z}} \\ & \therefore \quad \frac{x+y+z}{3} \geq \sqrt[3]{x y z} \quad \text{ (Applying A.M. } \geq \text{ G.M.) } \\ & \text{ and } \quad \frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3} \geq \sqrt[3]{\frac{1}{x y z}} \end{aligned} $

Multiplying (i) and (ii), we get

$ \frac{(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}{9} \geq \sqrt[3]{x y z \cdot \frac{1}{x y z}} \Rightarrow(x+y+z)(\frac{\mathbf{1}}{\boldsymbol{x}}+\frac{\mathbf{1}}{\boldsymbol{y}}+\frac{\mathbf{1}}{\boldsymbol{z}}) \geq \mathbf{9} $

~~ Ex. 18 . If $x, y$ are positive real numbers such that $x+y=1$, prove that $(1+\frac{1}{x})(1+\frac{1}{y}) \geq 9$.

Sol. For positive real numbers, $x, y$ applying $AM \geq GM$, we have

$ \begin{gathered} \frac{x+y}{2} \geq \sqrt{x y} \\ \Rightarrow \frac{1}{2} \geq \sqrt{x y} \quad(\because x+y=1) \\ \Rightarrow 1 \geq 2 \sqrt{x y} \Rightarrow 1 \geq 4 x y \Rightarrow 2 \geq 8 x y \Rightarrow 1+1 \geq 8 x y \\ \Rightarrow 1+x+y \geq 8 x y \quad \Rightarrow 1+x+y+x y \geq 9 x y \Rightarrow(1+x)(1+y) \geq 9 x y \\ \Rightarrow \frac{(1+x)(1+y)}{x y} \geq 9 \Rightarrow(\frac{x+1}{x})(\frac{y+1}{y}) \geq 9 \Rightarrow(\mathbf{1 + \frac { 1 } { x }})(\mathbf{1}+\frac{\mathbf{1}}{\boldsymbol{y}}) \geq \mathbf{9} . \end{gathered} $

~~ Ex. 19 . If $a, b, c, x, y, z$ are all positive, then prove that $(a b+x y) \cdot(a x+b y) \geq 4 a b x y$.

Sol. Applying AM - GM inequality, we have

$ \frac{a b+x y}{2} \geq \sqrt{a b \cdot x y} \Rightarrow(a b+x y) \geq 2 \sqrt{a b \cdot x y} $

and $\frac{(a x+b y)}{2} \geq \sqrt{a x \cdot b y} \Rightarrow(a x+b y) \geq 2 \sqrt{a b \cdot x y}$

Multiplying ( $i$ ) and (ii), we get

$(a b+x y)(a x+b y) \geq 4 \sqrt{a b x y} \sqrt{a b x y} \quad \Rightarrow \quad(a b+x y)(a x+b y) \geq 4 a b x y$.

~~ Ex. 20 . If $a, b, c$ are all positive, $a+b+c=1$ and $(1-a)(1-b)(1-c) \geq K(a b c)$, then find the value of $K$.

Sol. Given, $a+b+c=1$, so that $a+b=1-c, b+c=1-a, c+a=1-b$

Now, $\quad AM \geq GM \Rightarrow(a+b) \geq 2 \sqrt{a b} \Rightarrow(1-c) \geq 2 \sqrt{a b}$

Similarly, $(b+c) \geq 2 \sqrt{b c} \Rightarrow(1-a) \geq 2 \sqrt{b c}$

$ (c+a) \geq 2 \sqrt{c a} \Rightarrow(1-b) \geq 2 \sqrt{c a} $

Multiplying (i), (ii) and (iii), we get

$ \begin{aligned} & \Rightarrow \quad(1-a)(1-b)(1-c) \geq 8 \sqrt{a b} \sqrt{b c} \sqrt{c a} \\ & \Rightarrow \quad(1-a)(1-b)(1-c) \geq 8 a b c \Rightarrow K=\mathbf{8} . \end{aligned} $

~~ Ex. 21 . Prove that 2.4.6 ………. $(2 n)<(n+1)^{n}$.

Sol. Applying AM - GM, inequality, we have

$ \begin{aligned} & \frac{2+4+6+\ldots .+2 n}{n}>(2.4 .6 \ldots . .2 n)^{1 / n} \\ \Rightarrow & \frac{1}{n}[\frac{n}{2}(2+2 n)]>(2.4 .6 \ldots \ldots .2 n)^{\frac{1}{n}} \quad[\text{ Sum of an AP }=\frac{n}{2}[\text{ First term }+ \text{ Last term }]] \\ \Rightarrow & (n+1)>(2.4 .6 \ldots . .2 n)^{1 / n} \Rightarrow(2.4 .6 \ldots . .2 n)^{\frac{1}{n}}<(n+1) \Rightarrow(2.4 .6 . \ldots . .2 n)<(n+1)^{n} . \end{aligned} $

~~ Ex. 22 . For positive real numbers $x, y$, $z$, prove that

$ 2(x^{3}+y^{3}+z^{3}) \geq x y(x+y)+y z(y+z)+z x(z+x) $

Sol. For positive real numbers $x, y, z$, we have $AM \geq GM$

$ \begin{matrix} \Rightarrow \quad \frac{x^{2}+y^{2}}{2} \geq(x^{2} \cdot y^{2})^{1 / 2} \Rightarrow x^{2}+y^{2} \geq 2 x y & \\ \Rightarrow x^{2}+y^{2}-x y \geq x y & {[\because x>y \Rightarrow x+a>y+a \forall a \in R]} \\ \Rightarrow(x+y)(x^{2}+y^{2}-x y) \geq x y(x+y) & {[\because x>y \Rightarrow a x>a y \forall a>0]} \\ \Rightarrow(x^{3}+y^{3}) \geq x y(x+y) & \end{matrix} $

Similarly, we can show that,

$ y^{3}+z^{3} \geq y z(y+z) $

and $z^{3}+x^{3} \geq z x(z+x)$

Adding (i), (ii) and (iii), we have

$\Rightarrow \quad x^{3}+y^{3}+y^{3}+z^{3}+z^{3}+x^{3} \geq x y(x+y)+y z(y+z)+z x(z+x)$

$\Rightarrow \quad 2(x^{3}+y^{3}+z^{3}) \geq x y(x+y)+y z(y+z)+z x(z+x)$

~~ Ex. 23 . Find the minimum value of $\log _{n} m+\log _{m} n$, where $m>1$ and $n>1$.

Sol. When $m>1$ and $n>1$, then, we have

$\log _{n} m>0$ and $\log _{m} n>0$

Now, $AM>GM$ for positive real numbers

$ \begin{aligned} & \therefore \quad \frac{1}{2}(\log _{n} m+\log _{m} n) \geq(\log _{n} m \cdot \log _{m} n)^{1 / 2} \Rightarrow \frac{1}{2}(\log _{n} m+\log _{m} n) \geq(\frac{\log _{e} m}{\log _{e} n} \cdot \frac{\log _{e} n}{\log _{e} m})^{1 / 2} \\ & \Rightarrow \quad(\log _{n} m+\log _{m} n) \geq 2 \times 1=2 \\ & \therefore \quad \text{ Minimum value of } \log _{n} m+\log _{m} n \text{ is } 2 \end{aligned} $

~~ Ex. 24 . If $a, b, c$ are positive real numbers, then show that $(a+1)^{7}(b+1)^{7}(c+1)^{7}>7^{7} a^{4} b^{4} c^{4}$.

(IIT 2004)

Sol. LHS $=(a+1)^{7}(b+1)^{7}(c+1)^{7}$

$ \begin{aligned} & =[(a+1)(b+1)(c+1)]^{7} \\ & =[1+a+b+c+a b+b c+c a+a b c]^{7}>(a+b+c+a b+b c+c a+a b c)^{7} \end{aligned} $

Now using the AM, GM inequality, i.e., $AM \geq GM$, we have

$ \begin{aligned} & \frac{a+b+c+a b+b c+c a+a b c}{7} \geq(a . b . c . a b . b c . c a . a b c)^{1 / 7} \\ & \Rightarrow \quad \frac{1}{7^{7}}(a+b+c+a b+b c+c a+a b c)^{7} \geq(a^{4} b^{4} c^{4}) \quad \text{ [Raising both sides to power 7] } \\ & \Rightarrow \quad(a+b+c+a b+b c+c a+a b c)^{7} \geq 7^{7}(a^{4} b^{4} c^{4}) \end{aligned} $

From $(i)$ and $(i i)$

$(a+1)^{7}(b+1)^{7}(c+1)^{7}>7^{7} a^{4} b^{4} c^{4}$.

~~ Ex. 25 . If $a, b, c$ are the sides of $a \triangle A B C$, then show that $\frac{1}{2}<\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leq 1$.

Sol. We have $a>0, b>0, c>0$, so

$ \begin{aligned} & (a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0 \\ & \Rightarrow \quad a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 b c+c^{2}+a^{2}-2 c a \geq 0 \\ & \Rightarrow \quad 2 a^{2}+2 b^{2}+2 c^{2} \geq 2 a b+2 b c+2 c a \quad \Rightarrow \quad a^{2}+b^{2}+c^{2} \geq a b+b c+c a \\ & \Rightarrow \quad 1 \geq \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \Rightarrow \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leq 1 \end{aligned} $

Now for the other part let us using the triangle inequality,

i.e., the sum of two sides of a triangle is greater than the third side, we have

$\therefore a+b>c, \quad b+c>a, \quad c+a>b . \quad$ Also, $a>0, b>0, c>0$

$\Rightarrow a+b-c>0, \quad b+c-a>0, \quad c+a-b>0$

$\Rightarrow a(b+c-a)+b(c+a-b)+c(a+b-c)>0$

$\Rightarrow a b+a c-a^{2}+b c+b a-b^{2}+c a+c b-c^{2}>0$

$\Rightarrow 2(a b+b c+a c)>a^{2}+b^{2}+c^{2} \Rightarrow \frac{a b+b c+a c}{a^{2}+b^{2}+c^{2}} \geq \frac{1}{2}$

$\therefore \quad$ From $(i)$ and $(i i)$

$ \frac{1}{2}<\frac{a b+b c+a c}{a^{2}+b^{2}+c^{2}} \leq 1 . $

~~ Ex. 26 . If $A, B, C$ are the angles of an acute angled triangle, then show that $\tan A \cdot \tan B \cdot \tan C \geq 3 \sqrt{3}$.

Sol. $A, B, C$ being the angles of an acute angled triangle, $\tan A>0, \tan B>0, \tan C>0$.

Also,

$ A+B+C=\pi $

$ \begin{matrix} \Rightarrow & A+B=\pi-C \Rightarrow \tan (A+B)=\tan (\pi-C) \\ \Rightarrow & \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C \Rightarrow \tan A+\tan B+\tan C=\tan A \cdot \tan B \cdot \tan C \end{matrix} $

Now applying $AM \geq GM$, we have

$ \begin{aligned} & \frac{\tan A+\tan B+\tan C}{3} \geq(\tan A \cdot \tan B \cdot \tan C)^{1 / 3} \\ \Rightarrow \quad & (\tan A+\tan B+\tan C)^{3} \geq 27(\tan A \cdot \tan B \cdot \tan C) \\ \Rightarrow \quad & (\tan A \cdot \tan B \cdot \tan C)^{3} \geq 27(\tan A \cdot \tan B \cdot \tan C) \quad(\text{ From }(i)) \\ \Rightarrow \quad & (\tan A \cdot \tan B \cdot \tan C)^{2} \geq 27 \Rightarrow \tan A \cdot \tan B \cdot \tan C \geq \mathbf{3} \sqrt{\mathbf{3}} . \end{aligned} $

~~ Ex. 27 . Show that If $m>1$, then the sum of the $m$ th powers of $n$ even numbers is greater than $n(n+1)^{m}$.

Sol. Since $m>1$, using the property:

AM of $m$ th power of $n$ quantities $>m$ th power of the AM of $n$ quantities, we have

$ \begin{aligned} & \quad \frac{2^{m}+4^{m}+6^{m}+\ldots+(2 n)^{m}}{n}>(\frac{2+4+6+\ldots .+2 n}{n})^{m} \\ & \Rightarrow \quad \frac{2^{m}+4^{m}+6^{m}+\ldots .+(2 n)^{m}}{n}>\frac{1}{n^{m}} \cdot 2^{m}(1+2+3+\ldots .+n)^{m} \\ &=\frac{1}{n^{m}} 2^{m} \cdot(\frac{n}{2}(1+n))^{m}=\frac{1}{n^{m}} 2^{m}(\frac{n^{m}}{2^{m}}(n+1)^{m})=(n+1)^{m} \\ & \Rightarrow \quad \mathbf{2}^{m}+\mathbf{4}^{m}+\mathbf{6}^{m}+\ldots(\mathbf{2 n})^{m}>\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})^{m} . \end{aligned} $

PRACTICE SHEET

The solution set for the inequality $2 x-10<3 x-15$ over the set of real numbers is

(a) $(0,5)$

(b) $(5, \infty)$

(d) $(-5,0)$

~~ 2. The range of values of $x$ which satisfy the inequality $-2 \frac{2}{3} \leq x+\frac{1}{3}<3 \frac{1}{3}: x \in R$ is

(a) $-3 \leq x \leq 3$

(b) $-3 \leq x<3$

(d) $3<x<\infty$

~~ 3. $A={x: 11 x-5>7 x+3, x \in R}$ and

$B={x: 18 x-9>15+12 x, x \in R}$.

The range of the set $A \cap B$ is

(a) $[-\infty, 4)$

(b) $(0,4)$

(d) $(-4,4)$

~~ 4. If the inequality $\frac{6 x-5}{4 x+1}<0$ exists over a set $R$ of real numbers, then $x \in$

(a) $(\frac{1}{4}, \frac{5}{6})$

(b) $(-\infty, \frac{1}{4}) \cup(\frac{5}{6}, \infty)$

(d) $(\frac{5}{6}, \infty)$

~~ 5. The solution set of $\frac{x+3}{x-2} \leq 2$ is

(a) $(-\infty, 2) \cup(7, \infty)$

(b) $(-\infty, 2] \cup(7, \infty)$

(d) $(-\infty, 2] \cup[7, \infty)$

(Kerala PET 2011)

~~ 6. If $\frac{2 x-3}{3 x-7}>0$ over a set of real numbers $(R)$, then the solution set of $x$ is

(a) $(\frac{3}{2}, \frac{7}{3})$

(b) $(\frac{-3}{2}, \frac{7}{3})$

(d) $(-\infty, \frac{3}{2}) \cup(\frac{7}{3}, \infty)$

~~ 7. The solution set of the system of inequations $\frac{x}{2 x+1} \geq \frac{1}{4}$ and $\frac{6 x}{4 x-1}<\frac{1}{2}$ is

(a) $(-\frac{1}{2}, \frac{1}{8})$

(b) $(\frac{-1}{8}, \frac{1}{4})$

(d) Null set

~~ 8. If $r$ is real number such that $|r|<1$ and if $a=5(1-r)$, then

(a) $-5<a<5$

(b) $0<a<10$

(d) $-10<a<10$

~~ 9. For $\frac{|x-1|}{x+2}<1, x$ lies in the interval

(a) $(-\infty,-4)$

(b) $(-\infty,-2) \cup(-\frac{1}{2}, \infty)$

(d) $(-\infty, 1) \cup[2,3]$

~~ 10. If $|x-4|<9$, then

(a) $|x|<13$

(b) $|x-13|<0$

(d) $-13<x<13$

~~ 11. If $|x+3| \geq 7$, then $x \in$

(a) $(-10,4)$

(b) $(-\infty,-10] \cup[4, \infty)$

(d) $[-10, \infty]$

~~ 12. If $\frac{-1}{|x|-2} \geq 1$, where $x \in R, x \neq \pm 2$ then the solution set of $x$ is

(a) $(-\infty, 2) \cup(1, \infty)$

(b) $(-2,-1]$

(d) $(-2,-1] \cup[1,2)$

~~ 13. The set of values for which $x^{3}+1 \geq x^{2}+x$ is

(a) $x \geq 0$

(b) $x \leq 0$

(d) $-1 \leq x \leq 1$

~~ 14. Solve $(|x-1|-3)(|x+2|-5)<0$. Then

(a) $-7<x<-2$

(b) $3<x<4$

(d) Both (a) and (b)

~~ 15. If $-x^{2}+3 x+4>0$, then which of the following is correct?

(a) $x \in(-1,4)$

(b) $x \in[-1,4]$

(d) $x \in(-\infty, 1] \cup[4, \infty)$

(NDA/NA 2003)

~~ 16. Which of the following values of $x$ do not satisfy the inequality $x^{2}-3 x+2>0$ at all.

(a) $1 \leq x \leq 2$

(b) $-1 \geq x \geq-2$

(d) $0 \geq x \geq-2$

(CAT)

~~ 17. What values of $x$ satisfy $x^{2 / 3}+x^{1 / 3}-2 \leq 0$ ?

(a) $-8 \leq x \leq 8$

(b) $-8 \leq x \leq 1$

(d) $1 \leq x \leq 8$

(CAT 2006)

~~ 18. Find the complete set of values that satisfy the inequalities ||$x|-3|<2$ and ||$x|-2|<3$.

(a) $(-5,5)$

(b) $(-5,-1) \cup(1,5)$

(d) $(-1,1)$

(CAT 2012)

~~ 19. Find the range of real values of $x$ for which $\frac{x-1}{4 x+5}<\frac{x-3}{4 x-3}$.

(a) $(-\infty, \frac{-5}{4}) \cup(\frac{3}{4}, \infty)$

(b) $(\frac{-3}{4}, \frac{5}{4})$

(d) $(\frac{-5}{4}, \frac{3}{4})$

~~ 20. If $x$ is real and $x^{2}+3 x+2>0, x^{2}-3 x-4 \leq 0$, then which of the following is correct?

(a) $-1 \leq x<4$

(b) $2 \leq x \leq 4$

(d) $-1 \leq x<1$ or $2<x \leq 4$

(NDA/NA 2008)

~~ 21. The set of all real numbers $x$, for which $x^{2}-|x+2|+x>0$, is

(a) $(-\infty,-2) \cup(2, \infty)$

(b) $(\sqrt{2}, \infty)$

(d) $(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)$

(IIT 2002)

~~ 22. If $\log _{e}(x^{2}-16) \leq \log _{e}(4 x-11)$, then

(a) $-1 \leq x \leq 5$

(b) $x<-1$ or $x>5$

(d) $x<-4$ or $x>4$

(WBJEE 2012)

~~ 23. If $x, y, z$ are positive real numbers, then $(x^{2}+y^{2}+z^{2}) \geq$

(a) $x y z$

(b) $x^{3}+y^{3}+z^{3}$

(d) $x+y+z$

~~ 24. If $a_1, a_2, \ldots . a_n$ are positive numbers such that $a_1 \cdot a_2 \cdot a_3 \ldots . a_n$ $=1$, then their sum is

(a) a positive integer

(b) divisible by $n$

(d) none of these

(IIT 1991)

~~ 25. If $a^{2}+b^{2}+c^{2}=1, x^{2}+y^{2}+z^{2}=1$, where $a, b, c, x, y, z$ are positive reals then $a x+b y+c z$ is $(a) \geq 1$

(b) $\geq 2$

(d) None of these

~~ 26. If $x, y, z$ are three positive numbers, then the minimum value of $\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}$ is

(a) 1

(b) 2

(d) 6

(AMU 2010)

~~ 27. If $a_n \geq 1$ for all $n \in N(n \geq 3)$, then the minimum value of $\log _{a_2} a_1+\log _{a_3} a_2+\log _{a_4} a_3+\ldots . .+\log _{a_1} a_n$ is

(a) 0

(b) 1

(d) None of these

~~ 28. If $a, b, c, x, y, z$ are all positve real numbers, then $(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}) \geq$

(a) 8

(b) $64 a b c$

(d) 9

~~ 29. For positive real numbers $a, b, c$, the least value of $a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}$ is

(a) 0

(b) 1

(d) 6

~~ 30. If $a, b, c, d$ are four distinct positive real numbers and if $3 s=a+b+c+d$, then

(a) $a b c d>81(s-a)(s-b)(s-c)(s-d)$

(b) $a b c d<9(s-a)(s-b)(s-c)(s-d)$

(d) $a b c d<27(s-a)(s-b)(s-c)(s-d)$

~~ 31. If $(\frac{n+1}{2})^{n}(\frac{2 n+1}{3})^{n}>(n !)^{k}$, then $k=$

(a) 1

(b) 3

(d) 4

If $a, b, c$ be the lengths of the sides of a triangle and $(a+b+c)^{3} \geq k(a+b-c)(b+c-a)(c+a-b)$, then $k$ equals

(a) 1

(b) 3

(d) 27

~~ 33. Let $a, b, c$ be the lengths of the sides of a right angled triangle, the hypotenuse having the length $c$, then $a+b$ is $(a)>2 c$

(b) $<\sqrt{2} c$

~~ 34. If $a, b, c$ are the sides of a triangle, then

$ \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \text{ is } $

(a) $\leq 3$

(b) $\geq 2$

~~ 35. For three distinct positive real numbers $a, b, c$

$(1+a^{3})(1+b^{3})(1+c^{3})$ is greater than

(a) $a b c$

(b) $(1+a b c)$

(d) $(1+a b c)^{2}$

~~ 36. If $a, b, c, d$ are positive reals such that $a+b+c+d=2$, then $M=(a+b)(c+d)$ satisfies the relation

(a) $0 \leq M \leq 1$

(b) $1 \leq M \leq 2$

(d) $3 \leq M \leq 4$

(IIT 2000)

~~ 37. If $a_1, a_2, a_3 \ldots \ldots a_n$ are positive real numbers whose product is a fixed number ’ $c$ ‘, then the minimum value of $a_1+a_2 \ldots$. $+a _{n-1}+2 a_n$ is

(a) $n(2 c)^{1 / n}$

(b) $(n+1) c^{1 / n}$

(d) $(n+1)(2 c)^{1 / n}$

(IIT 2002)

~~ 38. If $a, b, c$ are the lengths of the sides of a non-equilateral triangle, then $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$ is

(a) $>\frac{s}{a+b+c}$

(b) $>\frac{2 s}{a b c}$

(d) None of these, where $s=\frac{a+b+c}{2}$.

~~ 39. Let $a, b, c$ be positive numbers, then

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ is

(a) $\geq \frac{3}{2}$

(b) $\geq 4$

(d) None of these

~~ 40. Let $a, b, c$ be positive numbers lying in the interval $(0,1]$, then $\frac{a}{1+b+c a}+\frac{b}{1+c+a b}+\frac{c}{1+a+b c}$ is $(a)<0$

(b) $>0$

~~ 41. If $\log _{10}(x^{3}+y^{3})-\log _{10}(x^{2}+y^{2}-x y) \leq 2$, then the maximum value of $x y$ for all $x \geq 0, y \geq 0$ is

(a) 1200

(b) 2500

(d) 3500

(DCE 2009)

~~ 42. The minimum value of $3^{\sin ^{2} x}+3^{\cos ^{2} x}, x \in R$ is

(a) $\frac{1}{2}$

(b) 3

(d) $2 \sqrt{3}$

~~ 43. When $a$ and $b$ are positive reals, prove that $(a^{4}+b^{4})(a^{5}+b^{5})$ is $(a)>2(a^{9}+b^{9})$

(b) $<2(a^{9}+b^{9})$ $(c)=2(a^{9}+b^{9})$

(d) None of these

~~ 44. The minimum value of the sum of real numbers $a^{-5}, a^{-4}$, $3 a^{-3}, 1, a^{8}$ and $a^{10}$ with $a>0$ is

(a) $\frac{1}{2}$

(b) 6

(d) $\frac{10}{3}$

(IIT 2011)

~~ 45. If $a, b, c$ are the sides of a non-equilateral triangle, then the expression $(b+c-a)(c+a-b)(a+b-c)-a b c$ is

(a) negative

(b) non-negative

(d) non-positive

(IIT 1980)

~~ 46. If $a, b, c$ are distinct positive integers, then $a x^{b-c}+b x^{c-a}$ $+c x^{a-b}$ is $(a)>3$

(b) $>\frac{1}{a+b+c}$

(d) $>a+b+c$

~~ 47. If $a, b, c$ are positive real numbers such that $a+b+c=p$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is greater than

(a) 16

(b) $9 p$

(d) $\frac{16}{p}$

~~ 48. If $a, b, c$ are the lengths of the sides of a non-equilateral triangle, then, $\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}$ is

(a) $>\frac{9}{a+b+c}$

(b) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

(d) None of these

~~ 49. If $a_1, a_2, \ldots ., a_n$ are distinct positive real numbers such that $a_1+a_2+\ldots . .+a_n=1$, then $\frac{1}{a_1}+\frac{1}{a_2}+\ldots .+\frac{1}{a_n}$ is $(a)>n$

(b) $>n^{2}$

(d) $>n^{3}$

~~ 50. If $A, B, C$ are the angles of an acute angled triangle, then $\cot ^{A} / 2 \cdot \cot ^{B} / 2 \cdot \cot C / 2 \geq$

(a) $4 \sqrt{3}$

(b) 8

(d) 3

ANSWERS

1. $(b)$	2. $(b)$	3. $(c)$	4. $(c)$	5. $(c)$	6. $(d)$	7. $(d)$	8. $(b)$
9. $(b)$	10. $(c)$	11. $(b)$	12. $(d)$	13. $(c)$	14. $(d)$	15. $(a)$	16. $(a)$
17. $(b)$	18. $(b)$	19. $(d)$	20. $(d)$	21. $(d)$	22. $(a)$	23. $(c)$	24. $(c)$
25. $(c)$	26. $(d)$	27. $(d)$	28. $(d)$	29. $(c)$	30. $(a)$	31. (c)	32. (d)
33. $(b)$	34. $(c)$	35. $(c)$	36. $(a)$	37. $(a)$	38. $(b)$	39. (a)	40. $(c)$
41. $(b)$	42. $(d)$	43. $(b)$	44. (c)	45. $(a)$	46. $(d)$	47. (c)	48. (c)
49. $(a)$	50. $(c)$

HINTS AND SOLUTIONS

$2 x-10<3 x-15 \Rightarrow-10+15<3 x-2 x$

$\Rightarrow 5<x \Rightarrow x>5 \Rightarrow \boldsymbol{x} \in(\mathbf{5}, \infty)$.

~~ 2. $-2 \frac{2}{3} \leq x+\frac{1}{3}<3 \frac{1}{3} \Rightarrow \frac{-8}{3} \leq x+\frac{1}{3}<\frac{10}{3}$

$\Rightarrow \frac{-8}{3}-\frac{1}{3} \leq x<\frac{10}{3}-\frac{1}{3}$

$\Rightarrow \frac{-9}{3} \leq x<\frac{9}{3} \Rightarrow-\mathbf{3} \leq \boldsymbol{x}<\mathbf{3}$.

~~ 3. $A={x: 11 x-5>7 x+3, x \in R}$

$\Rightarrow 11 x-5>7 x+3 \Rightarrow 4 x>8 \Rightarrow x>2$

$\Rightarrow A={x: x>2, x \in R}$

$B={x: 18 x-9 \geq 15+12 x, x \in R}$

$\Rightarrow 18 x-9 \geq 15+12 x \Rightarrow 6 x \geq 24 \Rightarrow x \geq 4$

$\Rightarrow B={x: x \geq 4, x \in R}$

$\therefore A \cap B={x: x>2, x \in R} \cap{x: x \geq 4, x \in R}$

$\Rightarrow x \geq 4 \Rightarrow x \in[4, \infty)$

~~ 4. $\frac{6 x-5}{4 x+1}<0$ on putting $(6 x-5)$ and $(4 x+1)$ equal to zero, we get the critical points as $x=\frac{5}{6}$ and $x=-\frac{1}{4}$. On the real number line they can be shown as:

When $x<-\frac{1}{4}$, both numerator and denominator of the expression $\frac{6 x-5}{4 x+1}$ are negative making the expression positive.
When $x$ lies between $-\frac{1}{4}$ and $\frac{5}{6}$, the expression $\frac{6 x-5}{4 x+1}$ becomes negative, hence less than zero.
When $x>\frac{5}{6}$, both $(6 x-5)$ and $(4 x+1)$ are positive, making the expression $\frac{6 x-5}{4 x+1}$ positive and hence greater than zero. $\therefore \quad$ For $\frac{6 x-5}{4 x+1}<0, x \in(-\frac{1}{4}, \frac{\mathbf{5}}{\mathbf{6}})$.

~~ 5. $\frac{x+3}{x-2} \leq 2 \Rightarrow \frac{x+3}{x-2}-2 \leq 0 \Rightarrow \frac{x+3-2 x+4}{x-2} \leq 0$

$\Rightarrow \frac{7-x}{x-2} \leq 0 \Rightarrow \frac{x-7}{x-2} \geq 0$ Here $x \neq 2$

The critical points on putting $(x-7)$ and $(x-2)$ are obtained as equal to zero are 7 and 2 .

The real number line is divided into 3 parts as shown below by these two critical points.

When $x<2$, the expression $\frac{x-7}{x-2}>0$.

When $x$ lies between 2 and 7, the expression $\frac{x-7}{x-2}<0$.

When $x \geq 7$, the expression $\frac{x-7}{x-2} \geq 0$.

$\therefore \quad$ The inequality $\frac{x+3}{x-2} \leq 2$ holds when $x<2$ or $x \geq 7$, i.e., $x \in(-\infty, 2) \cup[7, \infty)$.

~~ 6. Similar to Q. 5 . Also see Solved Example 8 (ii).

~~ 7. $\frac{x}{2 x+1} \geq \frac{1}{4}$ and $\frac{6 x}{4 x-1}<\frac{1}{2}$

$\Rightarrow \quad \frac{x}{2 x+1}-\frac{1}{4} \geq 0$ and $\frac{6 x}{4 x-1}-\frac{1}{2}<0$

$\Rightarrow \quad \frac{4 x-2 x-1}{4(2 x+1)} \geq 0$ and $\frac{12 x-4 x+1}{2(4 x-1)}<0$

$\Rightarrow \quad \frac{2 x-1}{2 x+1} \geq 0$ and $\frac{8 x+1}{4 x-1}<0$

Now for $\frac{2 x-1}{2 x+1} \geq 0, \quad x \neq \frac{1}{2}$

The critical points are $\frac{-1}{2}$ and $\frac{1}{2}$ which are plotted on the real number line as shown below:

Examining the expression for the three parts of the number line, i.e., $x<-\frac{1}{2}, \frac{-1}{2}<x \leq \frac{1}{2}, x \geq \frac{1}{2}$, we see that $\frac{2 x-1}{2 x+1} \geq 0$ when $x<-\frac{1}{2}$ or $x \geq \frac{1}{2}$

$\therefore \quad x \in(-\infty,-\frac{1}{2}) \cup[\frac{1}{2}, \infty)$

Now for $\frac{8 x+1}{4 x-1}<0, \quad x \neq \frac{1}{4}$

The critical points are $-\frac{1}{8}, \frac{1}{4}$ which are plotted on the real number line as shown below:

The expression $\frac{8 x+1}{4 x-1}<0$, when $x \in(\frac{-1}{8}, \frac{1}{4})$

$\therefore \quad$ It is clear from (i) and (ii) that the intersection of the solution sets given by (i) and (ii), i.e.,

$\lbrace(-\infty,-\frac{1}{2}) \cup[\frac{1}{2}, \infty)] \cap(-\frac{1}{8}, \frac{1}{4})=\phi \text{. } $

~~ 8. $|r|<1$

$\Rightarrow \quad-1<r<1$

$\Rightarrow 1>-r>-1 \Rightarrow-1<-r<1$

$(\because a>b \Rightarrow a x<b x$, where $x<0)$

$\Rightarrow \quad 0<1-r<2$

(Adding $1, c<a<b \Rightarrow c+x<a+x<b+x \forall x \in R$ )

$\Rightarrow 0 \times 5<5(1-r)<5 \times 2$

$(c<a<b \Rightarrow c x<a x<b x \forall x \in R, x>0)$

$\Rightarrow \quad 0<5(1-r)<10$

$\Rightarrow \mathbf{0}<\boldsymbol{a}<\mathbf{1 0}$.

~~ 9. Case I: When $x-1<0$, i.e., $x<1$, then $|x-1|=-(x-1)=1-x$

$\therefore \quad$ Given expression $=\frac{1-x}{x+2}<1$

$\Rightarrow \frac{1-x}{x+2}-1<0 \Rightarrow \frac{1-x-x-2}{x+2}<0$

$\Rightarrow \frac{-2 x-1}{x+2}<0 \Rightarrow \frac{2 x+1}{x+2}>0$

Now critical points are $x=-2,-\frac{1}{2}$.

Examining the expression $\frac{2 x+1}{x+2}$ in the intervals marked by the critical points on the number line, we see that $\frac{2 x+1}{x+2}>0$,

when $x \in(-\infty,-2)$ or $x \in(-\frac{1}{2}, \infty)$. Here $x<1$, therefore

the solution set is $x \in(-\infty,-2) \cup(-\frac{1}{2}, 1)$

Case II : When $x-1 \geq 0$, i.e., when $x \geq 1$

$|x-1|=(x-1)$

$\therefore \quad$ Given expression $=\frac{x-1}{x+2}<1$

$\Rightarrow \frac{x-1}{x+2}-1<0 \Rightarrow \frac{x-1-x-2}{x+2}<0 \Rightarrow \frac{-3}{x+2}<0$

This is possible only when $x+2>0 \Rightarrow x>-2$

Here $x \geq 1 \quad \therefore x \in[1, \infty)$

$\therefore \quad$ From (i) and (ii) $x \in(-\infty,-2) \cup(-\frac{1}{2}, 1) \cup[1, \infty)$

$\Rightarrow x \in(-\infty,-2) \cup(-\frac{1}{2}, \infty)$

~~ 10. $|x-4|<9 \Rightarrow-9<x-4<9$

$\Rightarrow-9+4<x<9+4 \Rightarrow \mathbf{- 5}<\boldsymbol{x}<\mathbf{1 3}$.

~~ 11. $|x+3| \geq 7$

$\Rightarrow \quad x+3 \leq-7$ or $x+3 \geq 7$

$\Rightarrow x \leq-10$ or $x \geq 4 \Rightarrow x \in(-\infty,-10]$ or $x \in[4, \infty)$

$\Rightarrow x \in(-\infty,-10] \cup[4, \infty)$.

~~ 12. $\frac{-1}{|x|-2} \geq 1 \Rightarrow \frac{-1}{|x|-2}-1 \geq 0$

$\Rightarrow \frac{-1-(|x|-2)}{|x|-2} \geq 0 \quad \Rightarrow \quad \frac{1-|x|}{|x|-2} \geq 0$

$\Rightarrow \frac{|x|-1}{|x|-2} \leq 0 \Rightarrow \frac{y-1}{y-2} \leq 0$, where $y=|x|$

Examining $\frac{y-1}{y-2}$ on the real number line shown below:

We see that $\frac{y-1}{y-2} \leq 0$ when $1 \leq y<2$

$\Rightarrow 1 \leq|x|<2 \Rightarrow|x| \geq 1$ and $|x|<2$

$\Rightarrow x \leq-1$ or $x \geq 1$ and $-2<x<2$

$\Rightarrow x \in(-2,-1] \cup[1,2)$.

~~ 13. $x^{3}+1 \geq x^{2}+x \Rightarrow x^{3}+1-x^{2}-x \geq 0$

$\Rightarrow x^{3}-x^{2}+1-x \geq 0 \quad \Rightarrow \quad x^{2}(x-1)-(x-1) \geq 0$

$\Rightarrow(x^{2}-1)(x-1) \geq 0 \Rightarrow(x+1)(x-1)^{2} \geq 0$

As $(x-1)^{2}$ is + ve so the given expression is $\geq 0$, when $x+1 \geq 0 \Rightarrow \boldsymbol{x} \geq-\mathbf{1}$.

~~ 14. $(|x-1|-3)(|x+2|-5)<0$

The product of the two factors is $<0$, i.e., negative when one of the factors is positive and one negative.

Case I : $(|x-1|-3)>0$ and $(|x+2|-5)<0$

$\Rightarrow|x-1|>3$ and $|x+2|<5$

$\Rightarrow-(x-1)>3,(x-1)>3$ and $-(x+2)<5, x+2<5$

$\Rightarrow \quad-x>2, x>4$ and $-x<7, x<3$

$\Rightarrow x<-2, x>4$ and $x>-7, x<3$

Combining we get $-7<x<-2$

Case II : $(|x-1|-3)<0$ and $(|x+2|-5)>0$

$\Rightarrow|x-1|<3$ and $|x+2|>5$

$\Rightarrow \quad-(x-1)<3,(x-1)<3$ and $(x+2)>5,-(x+2)>5$

$\Rightarrow-x<2, x<4$ and $x>3,-x>7$

$\Rightarrow x>-2, x<4$ and $x>3, x<-7$

$\Rightarrow 3<x<4$

$\therefore \quad-7<x<-2$ and $3<x<4$ is the solution.

~~ 15. $-x^{2}+3 x+4>0$

$\Rightarrow x^{2}-3 x-4<0 \Rightarrow(x-4)(x+1)<0$

$\Rightarrow \quad(x-4)<0,(x+1)>0$ or $(x-4)>0,(x+1)<0$

$\Rightarrow x<4, x>-1$ or $x>4, x<-1$

$\Rightarrow-1<x<4$.

~~ 16. $x^{2}-3 x+2>0 \Rightarrow(x-2)(x-1)>0$

$\Rightarrow(x-2)>0,(x-1)<0$ or $(x-2)<0,(x-1)>0$

$\Rightarrow x>2, x<1$ or $x<2, x>1 \Rightarrow x<1$ or $x>2$

$\therefore$ No value of $x$ which lies between these extremes, i.e., 1 and 2 satisfies the inequality, i.e., $1 \leq x \leq 2$ is the solution set not satisfying the inequality $x^{2}-3 x+2>0$ at all.

~~ 17. $x^{2 / 3}+x^{1 / 3}-2 \leq 0$

$\Rightarrow y^{2}+y-2 \leq 0$, where $y=x^{1 / 3}$

$\Rightarrow(y-1)(y+2) \leq 0$

$\Rightarrow \quad(y-1) \leq 0,(y+2) \geq 0$ or $(y-1) \geq 0,(y+2) \leq 0$

$\Rightarrow y \leq 1, y \geq-2$ or $y \geq 1, y \leq-2$

$\Rightarrow-2 \leq y \leq 1 \Rightarrow-2 \leq x^{1 / 3} \leq 1$

$\Rightarrow-8 \leq x \leq 1$.

~~ 18. Let $|x|=p, \quad$ where $p \geq 0$

So $|p-3|<2$ and $|p-2|<3$

$\Rightarrow 1<p<5$ and $-1<p<5$

$ (\therefore|x-a|<r \Rightarrow a-r<x<a+r) $

Therefore, the conditions (i), (ii) and (iii) are satisfied by $1<p<5$, i.e. $1<|x|<5$, i.e., $|x|>1$ and $|x|<5$

i.e., $x<-1$ or $x>1$ and $-5<x<5$

$\Rightarrow x \in(-5,-1) \cup(1,5)$.

~~ 19. $\frac{x-1}{4 x+5}<\frac{x-3}{4 x-3}$ or $\frac{x-1}{4 x+5}-\frac{x-3}{4 x-3}<0$

$\Rightarrow \quad \frac{(4 x-3)(x-1)-(x-3)(4 x+5)}{(4 x+5)(4 x-3)}<0$

$\Rightarrow \quad \frac{(4 x^{2}-3 x-4 x+3)-(4 x^{2}-12 x+5 x-15)}{(4 x+5)(4 x-3)}<0$

$\Rightarrow \frac{4 x^{2}-7 x+3-4 x^{2}+7 x+15}{(4 x+5)(4 x-3)}<0$

$\Rightarrow \frac{18}{(4 x+5)(4 x-3)}<0$

$\Rightarrow(4 x+5)(4 x-3)<0$ as $18>0$

$\Rightarrow 16(x+\frac{5}{4})(x-\frac{3}{4})<0$

Putting each factor equal to zero, the critical points are $-\frac{5}{4}, \frac{3}{4}$. Plotting on the real number line, we have the following figure:

Thus the expression $16(x+\frac{5}{4})(x-\frac{3}{4})$ is positive when $x<-\frac{5}{4}$ or $x>\frac{3}{4}$

$\therefore \quad$ The required range $=(-\frac{5}{4}, \frac{3}{4})$.

~~ 20. $x^{2}-3 x+2>0$

$\Rightarrow \quad(x-1)(x-2)>0$

$\Rightarrow \quad x \in(-\infty, 1) \cup(2, \infty)$

$\Rightarrow \quad-1 \leq x \leq 4$

$\therefore \quad$ Combining the solution sets in (i) and (ii), we have $-1<x<1$ or $2<x \leq 4$

~~ 21. Case I: When $x+2 \geq 0$, then $x \geq-2$

$\Rightarrow|x+2|=x+2$

$\therefore \quad x^{2}-|x+2|+x>0=x^{2}-(x+2)+x>0$

$\Rightarrow x^{2}-2>0 \Rightarrow x^{2}>2$

$\Rightarrow \quad x<-\sqrt{2}$ or $x>\sqrt{2}$

$\Rightarrow x \in(-2,-\sqrt{2}) \cup(\sqrt{2}, \infty) \quad[\because x \geq-2]$

Case II: When $x+2<0$, then $x<-2$

$\Rightarrow|x+2|=-(x+2)$

$\therefore x^{2}-|x+2|+x>0 \Rightarrow x^{2}+(x+2)+x>0$

$\Rightarrow x^{2}+2 x+2>0 \Rightarrow(x^{2}+2 x+1)+1>0$

$\Rightarrow(x+1)^{2}+1>0$, which is true for all value of $x$.

$\therefore x<-2 \Rightarrow x \in(-\infty,-2)$

From (i) and (ii)

$ \begin{aligned} & x \in(-2,-\sqrt{2}) \cup(\sqrt{2}, \infty) \cup(-\infty,-2) \\ \Rightarrow \quad & x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty) \end{aligned} $

~~ 22. $\log _{e}(x^{2}-16)<\log _{e}(4 x-11)$

$ \begin{aligned} & (x^{2}-16) \leq 4 x-11 \\ & (\because a>1, \log _{a} f(x)>\log _{a}(g x) \Rightarrow f(x)>g(x)>0) \\ & \Rightarrow \quad x^{2}-4 x-5 \leq 0 \Rightarrow(x-5)(x+1)<0 \\ & \Rightarrow \quad-1 \leq x \leq 5 \\ & (\because \text{ If } a<b, \text{ then }(x-a)(x-b) \leq 0 \Rightarrow a \leq x \leq b) \end{aligned} $

~~ 23. Since $x>0, y>0, z>0$, therefore

$ \begin{aligned} & (x-y)^{2} \geq 0,(y-z)^{2} \geq 0,(z-x)^{2} \geq 0 \\ \Rightarrow & x^{2}+y^{2}-2 x y \geq 0 \\ & y^{2}+z^{2}-2 y z \geq 0 \\ & z^{2}+x^{2}-2 z x \geq 0 \\ \Rightarrow & x^{2}+y^{2} \geq 2 x y, y^{2}+z^{2} \geq 2 y z, z^{2}+x^{2} \geq 2 z x \\ \Rightarrow & x^{2}+y^{2}+y^{2}+z^{2}+z^{2}+x^{2}>2 x y+2 y z+2 z x \\ \Rightarrow & 2(x^{2}+y^{2}+z^{2}) \geq 2(x y+y z+z x) \\ \Rightarrow & x^{2}+y^{2}+z^{2} \geq x y+y z+z x . \end{aligned} $

~~ 24. For positive numbers we know that $AM \geq GM$

$\therefore \quad(\frac{a_1 + a_2 + a_3+\ldots + a_n}{n}) \geq(a_1 \cdot a_2 \cdot a_3 \ldots a_n)^{\frac{1}{n}} \\ $

$ \Rightarrow \quad(\frac{a_1+a_2+a_3+\ldots + a_n}{n}) \geq 1^{\frac{1}{n}} \\ $

$ \Rightarrow \quad(\frac{a_1+a_2+a_3+\ldots + a_n}{n}) \geq 1 \\ $

$ \Rightarrow \boldsymbol a_1 + \boldsymbol a_2 + \boldsymbol a_3 + …. + \boldsymbol a_n \geq n $

~~ 25. Applying $AM \geq GM$, we have

$ \begin{aligned} & \frac{a+x}{2} \geq \sqrt{a x} \Rightarrow a+x \geq 2 \sqrt{a x} \\ & \Rightarrow \quad(a+x)^{2} \geq 4 a x \Rightarrow a^{2}+x^{2}+2 a x \geq 4 a x \\ & \Rightarrow \quad a^{2}+x^{2} \geq 2 a x \end{aligned} $

Similarly, $b^{2}+y^{2} \geq 2 b y$

$ c^{2}+z^{2} \geq 2 c z $

$\therefore \quad a^{2}+x^{2}+b^{2}+y^{2}+c^{2}+z^{2} \geq 2 a x+2 b y+2 c z$

$\Rightarrow(a^{2}+b^{2}+c^{2})+(x^{2}+y^{2}+z^{2}) \geq 2(a x+b y+c z)$

$\Rightarrow \quad 2 \geq 2(a x+b y+c z) \quad \Rightarrow \quad 1 \geq a x+b y+c z$

$\Rightarrow a x+b y+c z \leq 1$.

~~ 26. $x>0, y>0, z>0$

$\Rightarrow \quad \frac{x}{y}, \frac{y}{x}, \frac{y}{z}, \frac{z}{y}, \frac{x}{z}, \frac{z}{x}$ are all positive numbers.

$\therefore \quad$ Applying AM - GM inequality, we have

$\Rightarrow \frac{1}{2}(\frac{x}{y}+\frac{y}{x}) \geq(\frac{x}{y} \cdot \frac{y}{x})^{\frac{1}{2}} \Rightarrow \frac{x}{y}+\frac{y}{x} \geq 2$

Similarly, $\frac{y}{z}+\frac{z}{y} \geq 2, \frac{x}{z}+\frac{z}{x} \geq 2$

$\therefore \quad \frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x} \geq 2+2+2=6$

$\Rightarrow \frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z} \geq 6$

$\Rightarrow \quad \frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z} \geq 6$

$\therefore$ The minimum value of $\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}$ is 6 .

~~ 27. As $a_n \geq 1 \forall n \in N$, therefore

$\log _{a_2} a_1 \geq 0, \log _{a_3} a_2 \geq 0, \ldots \ldots ., \log _{a_1} a_n \geq 0$

For positive numbers, $AM \geq GM$

$ \begin{gathered} \Rightarrow \quad \frac{1}{n}[\log _{a_2} a_1+\log _{a_3} a_2+\ldots \ldots+\log _{a_1} a_n] \\ \quad \geq(\log _{a_2} a_1 \cdot \log _{a_3} a_2 \ldots . . \log _{a_n} a _{n-1} \log _{a_1} a_n)^{\frac{1}{n}} \\ \Rightarrow \quad \frac{1}{n}[\log _{a_2} a_1+\log _{a_3} a_2+\ldots \ldots \log _{a_1} a_n] \\ \quad \geq[\frac{\log _{e} a_1}{\log _{e} a_2} \cdot \frac{\log _{e} a_2}{\log _{e} a_3} \ldots \ldots . \frac{\log _{e} a _{n-1}}{\log _{e} a_n} \frac{\log _{e} a_n}{\log _{e} a_1}]^{\frac{1}{n}}=1^{\frac{1}{n}}=1 \end{gathered} $

$\therefore \quad[\log _{a_2} a_1+\log _{a_3} a_2+\ldots . . \log _{a_1} a_n] \geq n$

The minimum value of the given sum is $n$, but as $n>3, n$ cannot take up the values $\mathbf{0}, \mathbf{1}$ or 2 .

~~ 28. $a, b, c, x, y, z$ being all positive real numbers

$\Rightarrow \quad \frac{x}{a}, \frac{y}{b}, \frac{z}{c}, \frac{a}{x}, \frac{b}{y}, \frac{c}{z}$ are all positive real numbers.

Now applying the AM - GM inequality, we have

$ \begin{aligned} & \frac{1}{3}[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}] \geq(\frac{x}{a} \cdot \frac{y}{b} \cdot \frac{z}{c})^{\frac{1}{3}} \text{ and } \frac{1}{3}[\frac{a}{x}+\frac{b}{y}+\frac{c}{z}] \geq(\frac{a}{x} \cdot \frac{b}{y} \cdot \frac{c}{z})^{\frac{1}{3}} \\ & \therefore \quad \frac{1}{3} \times \frac{1}{3}[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}][\frac{a}{x}+\frac{b}{y}+\frac{c}{z}] \\ & \geq(\frac{x}{a} \cdot \frac{y}{b} \cdot \frac{z}{c})^{\frac{1}{3}}(\frac{a}{x} \cdot \frac{b}{y} \cdot \frac{c}{z})^{\frac{1}{3}} \\ & \Rightarrow[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}][\frac{a}{x}+\frac{b}{y}+\frac{c}{z}] \geq 9(\frac{x}{a} \cdot \frac{y}{b} \cdot \frac{z}{c} \cdot \frac{a}{x} \cdot \frac{b}{y} \cdot \frac{c}{z})^{\frac{1}{3}} \\ & \Rightarrow \quad[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}][\frac{a}{x}+\frac{b}{y}+\frac{c}{z}] \geq 9.1^{\frac{1}{3}} \end{aligned} $

~~ 29. $a>0, b>0, c>0 \Rightarrow \log a, \log b, \log c$ are all defined.

Also $a^{\log b-\log c}, b^{\log c-\log a}, c^{\log a-\log b}$ are all positive quantities.

$\therefore$ Applying $AM \geq GM$, we have

$ \begin{aligned} & \frac{1}{3}[a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}] \\ & \quad \geq[a^{\log b-\log c} \cdot b^{\log c-\log a} \cdot c^{\log a-\log b}]^{\frac{1}{3}} \end{aligned} $

Let $x=a^{\log b-\log c} \cdot b^{\log c-\log a} \cdot c^{\log a-\log b}$

$\Rightarrow \log x=(\log b-\log c) \log a+(\log c-\log a) \log b$ $+(\log a-\log b) \log c$

Now, $\log _{e} x=0 \Rightarrow x=e^{0}=1$.

$\therefore \quad$ (i) $\Rightarrow \quad \frac{1}{3}[a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}] \geq 1$

$\Rightarrow a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b} \geq 3$

$\Rightarrow$ The least value of $a^{\log b-\log c}+b^{\log c-\log a}+c^{\log a-\log b}$ is 3.

~~ 30. $3 s=a+b+c+d \Rightarrow 3 s-b-c-d=a$

$\Rightarrow \quad a=(s-b)+(s-c)+(s-d)$

For distinct positive real numbers $AM>GM$

$ \begin{aligned} & \Rightarrow \frac{1}{3}[(s-b)+(s-c)+(s-d)]>{(s-b)(s-c)(s-d)}^{\frac{1}{3}} \\ & \Rightarrow(s-b)+(s-c)+(s-d)>3{(s-b)(s-c)(s-d)}^{\frac{1}{3}} \\ & \Rightarrow a>3{(s-b)(s-c)(s-d)}^{\frac{1}{3}} \end{aligned} $

Similarly, $b>3{(s-a)(s-c)(s-d)}^{\frac{1}{3}}$

$ \begin{aligned} & c>3{(s-a)(s-b)(s-d)}^{\frac{1}{3}} \\ & d>3{(s-a)(s-b)(s-c)}^{\frac{1}{3}} \\ \therefore \quad & (i) \times(i i) \times(i i i) \times(i v) \end{aligned} $

$\Rightarrow a b c d>81{(s-a)^{3}(s-b)^{3}(s-c)^{3}(s-d)^{3}}^{\frac{1}{3}}$

$\Rightarrow \quad a b c d>81(s-a)(s-b)(s-c)(s-d)$.

~~ 31. $\frac{1^{2}+2^{2}+3^{2}+\ldots . n^{2}}{n} \geq(1^{2} .2^{2} \cdot 3^{2} \ldots . n^{2})^{\frac{1}{n}}$

$\Rightarrow \quad \frac{n(n+1)(2 n+1)}{6 n} \geq(1.2 .3 \ldots . . n)^{\frac{2}{n}}$

$\Rightarrow(\frac{n+1}{2})(\frac{2 n+1}{3}) \geq(n !)^{2 / n}$

$(\frac{n+1}{2})^{n}(\frac{2 n+1}{3})^{n} \geq(n !)^{2}$

$\Rightarrow \quad K=\mathbf{2}$.

~~ 32. By triangle inequality, we have sum of lengths of two sides of $a \Delta>$ length of third side

$\Rightarrow a+b>c, b+c>a, c+a>b$

$\Rightarrow a+b-c>0, b+c-a>0, c+a-b>0$

All being positive quantities, we apply $AM>GM$

$\Rightarrow \frac{(a+b-c)+(b+c-a)+(c+a-b)}{3}$

$ \begin{aligned} & \geq[(a+b-c)(b+c-a)(c+a-b)]^{\frac{1}{3}} \\ & \Rightarrow \quad(\frac{a+b+c}{3})^{3} \geq[(a+b-c)(b+c-a)(c+a-b)] \\ & \quad(a+b+c)^{3} \geq 27(a+b-c)(b+c-a)(c+a-b) \\ & \Rightarrow \quad K=\mathbf{2 7 .} \end{aligned} $

~~ 33. Since $a, b, c$ are the lengths of the sides of a right triangle having hypotenuse of length $c$, therefore $a^{2}+b^{2}=c^{2}$ (Pythagoras, Theoram)

Now, $\quad a>0, b>0 \Rightarrow(a-b)^{2} \geq 0$

$ \begin{aligned} & \Rightarrow a^{2}+b^{2} \geq 2 a b \\ & \Rightarrow 2(a^{2}+b^{2}) \geq a^{2}+b^{2}+2 a b \\ & \Rightarrow 2(a^{2}+b^{2}) \geq(a+b)^{2} \\ & \Rightarrow 2 c^{2} \geq(a+b)^{2} \Rightarrow(a+b)^{2} \leq 2 c^{2} \\ & \Rightarrow(a+b) \leq \sqrt{2} c . \end{aligned} $

~~ 34. Since $a, b, c$ are the sides of a triangle,

$a+b-c>0, b+c-a>0, c+a-b>0$

Let $x=b+c-a, y=c+a-b, z=a+b-c$

Then $x+y=2 c, y+z=2 a, z+x=2 b$.

$ \begin{matrix} \therefore \quad & \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \\ & =\frac{y+z}{2 x}+\frac{z+x}{2 y}+\frac{x+y}{2 z}=\frac{1}{2}[\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}] \\ & =\frac{1}{2}[\frac{y}{x}+\frac{x}{y}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}] \\ & {[(\frac{y}{x}+\frac{x}{y})(\frac{y}{z}+\frac{z}{y})(\frac{x}{z}+\frac{y}{z})] \geq 6} \\ & \quad[\because \forall a>0, a+\frac{1}{a} \geq 2] \\ & \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq \frac{6}{2} \geq \mathbf{3} . \end{matrix} $

~~ 35. $(1+a^{3})(1+b^{3})(1+c^{3})=1+a^{3}+b^{3}+c^{3}+a^{3} b^{3}+b^{3} c^{3}+$ $c^{3} a^{3}+a^{3} b^{3} c^{3}$

Now for distinct positive reals, $a, b, c, AM>GM$

$ \begin{aligned} & \frac{a^{3}+b^{3}+c^{3}}{3}>(a^{3} b^{3} c^{3})^{1 / 3} \\ \Rightarrow & a^{3}+b^{3}+c^{3}>3 a b c \end{aligned} $

$ (\because a, b, c>0 \Rightarrow a^{3}, b^{3}, c^{3}>0) $

Also $\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}}{3}>(a^{3} b^{3} \cdot b^{3} c^{3} \cdot c^{3} a^{3})^{1 / 3}$

$\Rightarrow a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}>3 a^{2} b^{2} c^{2}$

$\therefore \quad$ Putting the values from (ii) and (iii) on the RHS of (i), we have

$ \begin{aligned} & (1+a^{3})(1+b^{3})(1+c^{3})>1+3 a b c+3 a^{2} b^{2} c^{2}+a^{3} b^{3} c^{3} \\ \therefore \quad & (1+a^{\mathbf{3}})(\mathbf{1}+\boldsymbol{b}^{\mathbf{3}})(\mathbf{1}+\boldsymbol{c}^{\mathbf{3}})>(\mathbf{1}+\boldsymbol{a b c})^{\mathbf{3} .} \quad=(1+a b c)^{3} \end{aligned} $

~~ 36. Let $x=a+b, y=c+d$. Then, $x+y=1$ and $M=x y$.

If the sum of two quantities is a constant, then their product is maximum when both the quantities are equal, i.e., $x=y$ $\Rightarrow x=1, y=1 \Rightarrow x y=1$.

$\therefore \quad$ Maximum value of $M$ is 1

Also, $a, b, c, d$ being positive real numbers

$ \begin{aligned} & (a+b)(c+d) \geq 0 \\ \therefore \quad & \mathbf{0} \leq \boldsymbol{M} \leq \mathbf{1} \end{aligned} $

Alternatively: For positive quantities, $AM \geq GM$

$ \begin{aligned} & \frac{1}{2}{(a+b)+(c+d)} \geq{(a+b)(c+d)}^{1 / 2} \\ \Rightarrow & M^{\frac{1}{2}} \leq \frac{1}{2} \times 2 \Rightarrow M^{\frac{1}{2}} \leq 1 \Rightarrow M \leq 1 \end{aligned} $

Also, $M=(a+b)(c+d) \geq 0$

$\therefore \quad \mathbf{0} \leq \boldsymbol{M} \leq \mathbf{1}$.

~~ 37. For positive real numbers, $AM \geq GM$

$ \begin{aligned} & \Rightarrow \quad \frac{(a_1+a_2+a_3+\ldots \ldots+2 a_n)}{n} \geq(a_1 \cdot a_2 \cdot a_3 \ldots \ldots .2 a_n)^{\frac{1}{n}} \\ & \Rightarrow \quad(a_1+a_2+a_3+\ldots \ldots+2 a_n) \geq n(c \cdot 2)^{\frac{1}{n}}=n(2 c)^{\frac{1}{n}} \end{aligned} $

Hence the least value is $\boldsymbol{n}(\mathbf{2 c})^{\frac{1}{\boldsymbol{n}}}$.

~~ 38. $a>0, b>0, c>0 \Rightarrow a b>0, b c>0, c a>0$

$ \begin{aligned} & \therefore \quad \frac{(a b)^{2}+(b c)^{2}}{2} \geq((a b)^{2}(b c)^{2})^{\frac{1}{2}} \\ & \Rightarrow \quad a^{2} b^{2}+b^{2} c^{2} \geq 2(a^{2} b^{4} c^{2})^{1 / 2} \\ & \Rightarrow \quad a^{2} b^{2}+b^{2} c^{2} \geq 2(a b c) \cdot b \end{aligned} $

Similarly, $b^{2} c^{2}+c^{2} a^{2} \geq 2(a b c) . c$

(i) $+($ ii $)+($ iii)

$ c^{2} a^{2}+a^{2} b^{2} \geq 2(a b c) a $

$\Rightarrow 2(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}) \geq 2 a b c(a+b+c)$

$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq a b c(a+b+c)$

Dividing both sides by $a^{2} b^{2} c^{2}$, we have

$\Rightarrow \frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} \geq \frac{a+b+c}{a b c}$

$\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geq \frac{2 s}{a b c}$

$\Rightarrow \quad \frac{2 s}{a b c} \leq \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$.

~~ 39. We know that $a+\frac{1}{a} \geq 2 \therefore a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c} \geq 2+2+2 \geq 6$

$\therefore \quad \frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x} \geq 6$

$\therefore \quad \frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{b+c}{c+a}+\frac{c}{b+c}+\frac{a+b}{c+a}+\frac{c+a}{a+b} \geq 6$

$\Rightarrow \quad \frac{b+c+2 a}{b+c}+\frac{a+b+2 c}{a+b}+\frac{c+a+2 b}{c+a} \geq 6$

$\Rightarrow 1+\frac{2 a}{b+c}+1+\frac{2 c}{a+b}+1+\frac{2 b}{c+a} \geq 6$

$\Rightarrow \frac{2 a}{b+c}+\frac{2 c}{a+b}+\frac{2 b}{c+a} \geq 3$

$\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2}$

Equality holds for $\boldsymbol{a}=\boldsymbol{b}=\boldsymbol{c}$.

~~ 40. $(1+a b)=(1-a)(1-b)+a+b$

$\Rightarrow \quad 1+c+a b=(1-a)(1-b)+a+b+c \geq a+b+c$

$\therefore \quad 1+a+b c=(1-b)(1-c)+a+b+c \geq a+b+c$

$1+b+a c=(1-a)(1-c)+a+b+c \geq a+b+c$

$\therefore \quad \frac{a}{1+b+c a}+\frac{b}{1+c+a b}+\frac{c}{1+a+b c}$

$\leq \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}$

$(\because a \geq b \Rightarrow \frac{1}{a} \leq \frac{1}{b})$

$\Rightarrow \frac{a}{1+b+c a}+\frac{b}{1+c+a b}+\frac{c}{1+a+b c} \leq \frac{a+b+c}{a+b+c}=1$

$\Rightarrow \frac{a}{1+b+c a}+\frac{b}{1+c+a b}+\frac{c}{1+a+b c} \leq 1$.

~~ 41. $\log _{10}(x^{3}+y^{3})-\log _{10}(x^{2}+y^{2}-x y) \leq 2$

$\Rightarrow \log _{10} \frac{(x^{3}+y^{3})}{(x^{2}+y^{2}-x y)} \leq 2$

$\Rightarrow \quad \log _{10}[\frac{(x+y)(x^{2}+y^{2}-x y)}{(x^{2}+y^{2}-x y)}] \leq 2$

$\Rightarrow \log _{10}(x+y) \leq 2$

$\Rightarrow(x+y)<10^{2} \Rightarrow x+y \leq 100$

Now $x>0, y>0 \Rightarrow AM \geq GM$

$\therefore \quad \frac{x+y}{2} \geq \sqrt{x y}$ $\Rightarrow \quad \frac{100}{2} \geq \frac{x+y}{2} \geq \sqrt{x y}$

$\Rightarrow \sqrt{x y} \leq 50 \Rightarrow x y \leq 2500$.

Maximum value of $x y=2500$

~~ 42. For all $x \in R, 3^{\sin ^{2} x}>0,3^{\cos ^{2} x}>0$

$\therefore \quad AM \geq GM$

$\Rightarrow \quad \frac{1}{2}[3^{\sin ^{2} x}+3^{\cos ^{2} x}] \geq(3^{\sin ^{2} x} \cdot 3^{\cos ^{2} x})^{\frac{1}{2}}$

$\Rightarrow[3^{\sin ^{2} x}+3^{\cos ^{2} x}] \geq 2(3^{\sin ^{2} x+\cos ^{2} x})^{\frac{1}{2}}$

$\Rightarrow \quad 3^{\sin ^{2} x}+3^{\cos ^{2} x} \geq 2 \sqrt{3} \quad(\therefore \sin ^{2} x+\cos ^{2} x=1)$

~~ 43. Given, $x>0, y>0$ and $x \neq y$

$\Rightarrow x^{4}>0, y^{4}>0, x^{5}>0, y^{5}>0, x^{9}>0, y^{9}>0$.

For distinct positive reals, we know that

$ \begin{aligned} & \quad AM>GM \\ \therefore \quad & \frac{1}{2}(x^{4}+y^{4})>(x^{4} \cdot y^{4})^{\frac{1}{2}} \\ \Rightarrow \quad & (x^{4}+y^{4})>2(x^{2} y^{2}) \\ \text{ Also } & \frac{1}{2}(x^{5}+y^{5})>(x^{5} y^{5})^{\frac{1}{2}} \\ \Rightarrow \quad & (x^{5}+y^{5})>2(x^{5 / 2} y^{5 / 2}) \end{aligned} $

Now multiplying (i) and (ii), we have

$ \begin{aligned} & (x^{4}+y^{4})(x^{5}+y^{5})>4(x^{2} y^{2})(x^{5 / 2} y^{5 / 2}) \\ \Rightarrow \quad & (x^{4}+y^{4})(x^{5}+y^{5})>4 x^{9 / 2} y^{9 / 2} \end{aligned} $

Also, $\frac{1}{2}(x^{9}+y^{9})>(x^{9} \cdot y^{9})^{\frac{1}{2}}$

$ \Rightarrow \quad(x^{9}+y^{9})>2 x^{9 / 2} y^{9 / 2} $

$ \Rightarrow \quad 2(x^{9}+y^{9})>4 x^{9 / 2} y^{9 / 2} $

$\therefore \quad$ From (iii) and (iv), we have

$ (x^{4}+y^{4})(x^{5}+y^{5})<2(x^{9}+y^{9}) . $

~~ 44. These real numbers can be taken as 8 quantities,

$a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, 1, a^{8}, a^{10}$.

All are positive as $a>0$

$\therefore \quad AM \geq GM$ (as all quantities are not distinct)

$\therefore \quad \frac{1}{8}[a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^{8}+a^{10}]$

$ \geq(a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^{8} \cdot a^{10})^{1 / 8} $

$\Rightarrow \frac{1}{8}[a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^{8}+a^{10}] \geq 1^{1 / 8}$

$\Rightarrow \quad(a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^{8}+a^{10}) \geq 8$

$\therefore \quad$ Minimum value of the sum is 8 .

~~ 45. Let $x=b+c-a, y=c+a-b, z=a+b-c$

Since $a, b, c$ are the sides of a non-equilateral triangle, by the triangle inequality $a+b>c, b+c>a, c+a>b$

$\Rightarrow a+b-c>0, b+c-a>0, c+a-b>0$

$\Rightarrow z>0, x>0, y>0$

Also, $x, y, z$ are distinct.

$\therefore \quad AM>GM$

$ \begin{aligned} & c=\frac{x+y}{2}>\sqrt{x y} \\ & b=\frac{x+z}{2}>\sqrt{x z} \\ & a=\frac{y+z}{2}>\sqrt{y z} \end{aligned} $

$\therefore a b c>\sqrt{x y} \cdot \sqrt{x z} \cdot \sqrt{y z}$

$\Rightarrow a b c>x y z \Rightarrow a b c>(b+c-a)(c+a-b)(a+b-c)$

$\Rightarrow(b+c-a)(c+a-b)(a+b-c)$

$-a b c<0$, i.e., negative.

~~ 46. Using weighted $AM-GM$ inequality, i.e.,

$\frac{m_1 a_1+m_2 a_2+\ldots .+m_n a_n}{m_1+m_2+\ldots . .+m_n} \geq(a_1^{m_1} \cdot a_2^{m_2} \ldots . a_n^{m_n})^{\frac{1}{m_1+m_2+\ldots+m_n}}$,

we have

$\Rightarrow \frac{a x^{b-c}+b x^{c-a}+c x^{a-b}}{a+b+c}>((x^{b-c})^{a} \cdot(x^{c-a})^{b} \cdot(x^{a-b})^{c})^{\frac{1}{a+b+c}}$

$\Rightarrow \frac{a x^{b-c}+b x^{c-a}+c x^{a-b}}{a+b+c}>{(x^{a b-a c} \cdot x^{c b-a b} \cdot x^{a c-b c})}^{\frac{1}{a+b+c}}$

$\Rightarrow \frac{a x^{b-c}+b x^{c-a}+c x^{a-b}}{a+b+c}>(x^{a b-a c+c b-a b+a c-b c})^{\frac{1}{a+b+c}}$

$ =(x^{0})^{\frac{1}{a+b+c}}=x^{0}=1 $

$\therefore a x^{b-c}+b x^{c-a}+c x^{a-b}>a+b+c$.

~~ 47. The AM of $m$ th powers of $n$ positive numbers is greater than the $m$ th power of their AM if $m<0$ or $m>1$. So for $m=-1$,

$\frac{a^{-1}+b^{-1}+c^{-1}}{3}>(\frac{a+b+c}{3})^{-1}$

$\Rightarrow \frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})>(\frac{p}{3})^{-1}$

(Given: $a+b+c \equiv p$ )

$\Rightarrow(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})>\frac{9}{p}$.

~~ 48. Since $a, b, c$ are the sides of a non-equilateral triangle, $a>0, b>0, c>0$. Also by triangle inequality.

$a+b>c \Rightarrow a+b-c>0$

$b+c>a \Rightarrow b+c-a>0$

$c+a>b \Rightarrow c+a-b>0$

$\therefore \frac{(a+b-c)^{-1}+(b+c-a)^{-1}+(c+a-b)^{-1}}{3}$

$ > \lbrace {\frac{(a+b-c)+(b+c-a)+(c+a-b)}{3}}\rbrace^{-1} $

$[\because$ AM of the $m$ th powers of $n$ positive quantities is greater than the $m$ th power of their AM if $m<0$ or $m>1$ ]

$ \begin{aligned} & \therefore \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}>3 \times[\frac{a+b+c}{3}]^{-1} \\ & \Rightarrow \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}>3 \times \frac{3}{a+b+c} \\ & \Rightarrow \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}>\frac{9}{a+b+c} \end{aligned} $

Also, $\frac{1}{a+b-c}+\frac{1}{b+c-a}=\frac{b+c-a+a+b-c}{{b+(a-c)}{b-(a-c)}}$

$ \begin{aligned} &=\frac{2 b}{b^{2}-(a-c)^{2}}>\frac{2 b}{b^{2}}=\frac{2}{b} \\ & \because \quad b^{2}-(a-c)^{2}<b^{2} \Rightarrow \frac{1}{b^{2}-(a-c)^{2}}>\frac{1}{b^{2}} \\ & \therefore \quad \frac{1}{a+b-c}+\frac{1}{b+c-a}>\frac{2}{b} \end{aligned} $

Similarly, $\frac{1}{b+c-a}+\frac{1}{c+a-b}>\frac{2}{c}$

$ \frac{1}{a+b-c}+\frac{1}{c+a-b}>\frac{2}{a} $

Adding (i), (ii) and (iii)

$ \begin{aligned} & \Rightarrow \frac{2}{a+b-c}+\frac{2}{b+c-a}+\frac{2}{c+a-b}>\frac{2}{a}+\frac{2}{b}+\frac{2}{c} \\ & \Rightarrow \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}>\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \end{aligned} $

Hence ( $a$ ) and (b) are both correct option.

~~ 49. AM of $m$ th powers $>m$ th of AM, when $m<0$ or $m>1$

$ \begin{aligned} & \Rightarrow \frac{(a_1)^{-1}+(a_2)^{-1}+(a_3)^{-1}+\ldots .+(a_n)^{-1}}{n} \\ & >(\frac{a_1+a_2+a_3+\ldots .+a_n}{n})^{-1} \\ & \Rightarrow \quad \frac{1}{n}(\frac{1}{a_1}+\frac{1}{a_2}+\ldots .+\frac{1}{a_n})>\frac{n}{a_1+a_2+\ldots .+a_n} \\ & \Rightarrow(\frac{1}{a_1}+\frac{1}{a_2}+\ldots .+\frac{1}{a_n})>\frac{n^{2}}{a_1+a_2+\ldots .+a_n} \\ & \Rightarrow \quad \frac{1}{a_1}+\frac{1}{a_2}+\ldots .+\frac{1}{a_n}>n^{2}[\because a_1+a_2+\ldots .+a_n=1] \end{aligned} $

~~ 50. $A, B, C$ being the angles of an acute angled triangle, $A+B+C=\pi$

$\because A, B, C$ are acute angles, $\tan A>0, \tan B>0, \tan C>0$

Also, $A+B+C=\pi \Rightarrow \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}$ $\therefore \cot (\frac{A}{2}+\frac{B}{2})=\cot (\frac{\pi}{2}-\frac{C}{2})=\tan \frac{C}{2}$

$\Rightarrow \frac{\cot \frac{A}{2} \cot \frac{B}{2}-1}{\cot \frac{A}{2}+\cot \frac{B}{2}}=\frac{1}{\cot \frac{C}{2}}$

$\Rightarrow \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}$

$[\because \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}]$

Also as $\cot \frac{A}{2}>0, \cot \frac{B}{2}>0, \cot \frac{C}{2}>0$

Applying AM - GM inequality, we have

$ \begin{aligned} & \frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{3} \geq(\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2})^{\frac{1}{3}} \\ & \Rightarrow(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2})^{3} \geq 27(\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2}) \\ & \Rightarrow(\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2})^{3} \geq 27(\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}) \\ & \Rightarrow(\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2})^{2} \geq 27 \\ & \Rightarrow \cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2} \geq 3 \sqrt{3} \text{. } \end{aligned} $

SELF ASSESSMENT SHEET

The number of positive integral values of $m$ satisfying the inequalities $8 m+35>75$ and $5 m+18<53$ is

(a) 4

(b) 1

(d) 2

~~ 2. Given $a>0, b>0, a>b$ and $c \neq 0$, the inequality which is not always correct is

(a) $a-c>b-c$

(b) $\frac{a}{c^{2}}>\frac{b}{c^{2}}$

(d) $a c>b c$

~~ 3. If $x$ is an integer that satisfies $9<4 x-1 \leq 19$, then $x$ is an element of which of the following sets?

(a) ${3,4}$

(b) ${2,3,4}$

(d) ${2,3,4,5}$

(NDA/NA 2008)

~~ 4. The set of all $x$ satisfying the inequality $\frac{4 x-1}{3 x+1} \geq 1$ is

(a) $[2, \infty)$

(b) $(-\infty,-\frac{1}{3}) \cup[2, \infty)$

(d) $(-\infty,-\frac{2}{3}] \cup[4, \infty)$

(Kerala PET 2006)

~~ 5. Find the set of values of $x$ satisfying $|\frac{5-x}{3}|<2$

(a) $1<x<11$

(b) $-1<x<11$

(d) None of these

~~ 6. The solution of $4 x^{2}+4 x+1>0$ is

(a) All real numbers except $-\frac{1}{2}$ (b) All real numbers

(d) None of these

~~ 7. The set of values of $x$ for which the inequalities $x^{2}-3 x-10$ $<0,10 x-x^{2}-16>0$ hold simultaneously is

(a) $(-2,5)$

(b) $(2,8)$

(d) $(2,5)$

(EAMCET 2007)

~~ 8. For three distinct positive numbers $p, q$ and $r$, if $p+q+r$ $=a$, then

(a) $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>\frac{18}{a}$

(b) $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>\frac{9}{a}$

(d) $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<\frac{6}{a}$

~~ 9. $1.3 .5 \ldots . .(2 n-1)$ is

(a) $<n$

(b) $<n^{n}$ $(c)>n$

(d) None of these

~~ 10. Let $a_1, a_2, \ldots . a_n$ be positive real numbers such that $a_1 a_2 a_3$ $\ldots \ldots a_n=1$. Then $(1+a_1)(1+a_2) \ldots . .(1+a_n)$ is

(a) $\geq 2$

(b) $\geq 2^{n-1}$

(d) $\leq 2^{n-1}$

~~ 11. If three positive real numbers, $a, b, c$ are such that $a+b$ $+c=1$, then the minimum value of $\frac{(1-a)(1-b)(1-c)}{a b c}$ is

(a) 2

(b) 3

(d) 8

~~ 12. The minimum value of the expression $(\frac{3 b+4 c}{a})+(\frac{4 c+a}{3 b})+(\frac{a+3 b}{4 c})(a, b, c$ are $+ve)$ is

(a) 1

(b) 4

(d) 8

(AMU 2005)

ANSWERS

~~ 2. $($ d)

~~ 3. (c)

~~ 4. (b)

~~ 5. (b)

~~ 6. (a)

~~ 7. (d)

~~ 8. (b)

~~ 9. (b)

~~ 10. (c)

~~ 11. (d)

~~ 12. (c)

HINTS AND SOLUTIONS

$8 m+35>75$ and $5 m+18<53$

$\Rightarrow 8 m>40 \Rightarrow 5 m<35$

$\Rightarrow m>5 \quad \Rightarrow m<7$

$\therefore$ There is only one integral value of $m$, i.e., $m=6$ satisfying the given inequalities.

~~ 2. $a>b \Rightarrow a c>b c$ when $c>0$, but

$a>b \Rightarrow a c<b c$ when $c<0$.

$\therefore a c>b c$ is not always correct.

~~ 3. Given, $9<4 x-1 \leq 19$

$\Rightarrow 9<4 x-1$ and $4 x-1 \leq 19$

$\Rightarrow 9+1<4 x$ and $4 x \leq 19+1$

$\Rightarrow x>\frac{5}{2}$ and $x \leq 5$

$\therefore \frac{5}{2}<x \leq 5$

$\because x \in Z \therefore x={3,4,5}$

~~ 4. $\frac{4 x-1}{3 x+1} \geq 1 \Rightarrow \frac{4 x-1}{3 x+1}-1 \geq 0$

$\Rightarrow \frac{4 x-1-3 x-1}{3 x+1} \geq 0 \Rightarrow \frac{x-2}{3 x+1} \geq 0 \Rightarrow x \neq-\frac{1}{3}$

Critical points are $x=-\frac{1}{3}, 2$. Place the points on the number line.

When $x<-\frac{1}{3}, \frac{x-2}{3 x+1}>0$
When $x \geq 2, \quad \frac{x-2}{3 x+1} \geq 0$
When $-\frac{1}{3}<x \leq 2, \quad \frac{x-2}{3 x+1}$ is - ve.

$\therefore x \in(-\infty,-\frac{1}{3}) \cup[2, \infty)$.

~~ 5. $|\frac{5-x}{3}|<2 \Rightarrow|5-x|<6$

$\Rightarrow-6<5-x<6 \Rightarrow-11<-x<1 \Rightarrow 11>x>-11$

$(\because a<b<c \Rightarrow m a>m b>m c$, when $m$ is - ve. Here $m=-\mathbf{1})$

$\Rightarrow-1<x<11$.

~~ 6. $4 x^{2}+4 x+1>0 \Rightarrow(2 x+1)^{2}>0$

$\Rightarrow(2 x+1)<0$ or $(2 x+1)>0$

$\Rightarrow x>-\frac{1}{2}$ or $x<-\frac{1}{2}$

$\therefore$ The inequality holds for all real numbers except $x=-\frac{1}{2}$.

~~ 7. $x^{2}-3 x-10<0$

$\Rightarrow(x+2)(x-5)<0 \Rightarrow(x-(-2))(x-5)<0$

$\Rightarrow x \in(-2,5)$

(If $a<b$, then $(x-a)(x-b)<0 \Rightarrow a<x<b)$

$10 x-x^{2}-16>0 \Rightarrow x^{2}-10 x+16<0 \Rightarrow(x-2)(x-8)<0$

$\Rightarrow x \in(2,8)$

$($ If $a<0$, then $(x-a)(x-b)<0 \Rightarrow a<x<b)$

$\therefore x \in(-2,5) \cap(2,8)$

$\Rightarrow x \in(2,5)$.

~~ 8. $p>0 \Rightarrow \frac{1}{p}>0 ; q>0 \Rightarrow \frac{1}{q}>0 ; r>0 \Rightarrow \frac{1}{r}>0$

For distinct positive numbers, $AM>GM$

$\therefore \frac{p+q+r}{3}>(p q r)^{\frac{1}{3}} \Rightarrow(p+q+r)>3(p q r)^{\frac{1}{3}}$

Also, $\frac{1}{3}(\frac{1}{p}+\frac{1}{q}+\frac{1}{r})>(\frac{1}{p} \cdot \frac{1}{q} \cdot \frac{1}{r})^{\frac{1}{3}}$

$\Rightarrow(\frac{1}{p}+\frac{1}{q}+\frac{1}{r})>3(\frac{1}{p} \cdot \frac{1}{q} \cdot \frac{1}{r})^{\frac{1}{3}}$

From $(i) \times(i i)$

$\Rightarrow(p+q+r)(\frac{1}{p}+\frac{1}{q}+\frac{1}{r})>9(p q r)^{\frac{1}{3}} \times(\frac{1}{p q r})^{\frac{1}{3}}$

$\Rightarrow a(\frac{1}{p}+\frac{1}{q}+\frac{1}{r})>9(1)^{\frac{1}{3}} \Rightarrow(\frac{1}{p}+\frac{1}{q}+\frac{1}{r})>\frac{9}{a}$.

~~ 9. Here $AM>GM$, as all are distinct positive numbers.

$ \begin{aligned} & (\frac{1+3+5+\ldots .+(2 n-1)}{n}>(1.3 .5 \ldots \ldots(2 n-1))^{\frac{1}{n}}. \\ & \Rightarrow \frac{\frac{n}{2}[1+2 n-1]}{n}>(1.3 .5 \ldots .(2 n-1))^{\frac{1}{n}} \\ & \Rightarrow(1.3 .5 \ldots .(2 n-1)^{\frac{1}{n}}<n. \\ & \Rightarrow 1.3 .5 \ldots .(2 n-1)<n^{n} . \end{aligned} $

~~ 10. $(\frac{1+a_1}{2}) \geq \sqrt{a_1} \Rightarrow(1+a_1) \geq 2 \sqrt{a_1}$

$ \begin{aligned} & (\frac{1+a_2}{2}) \geq \sqrt{a_2} \Rightarrow(1+a_2) \geq 2 \sqrt{a_2} \\ & (\frac{1+a_n}{2}) \geq \sqrt{a_n} \Rightarrow(1+a_n) \geq 2 \sqrt{a_n} \\ & \begin{aligned} & \therefore(1+a_1)(1+a_2) \ldots \ldots(1+a_n) \\ & \geq 2 \sqrt{a_1} \times 2 \sqrt{a_2} \times \ldots . \times 2 \sqrt{a_n} \\ & \Rightarrow(1+a_1)(1+a_2) \ldots(1+a_n) \geq 2^{n} \sqrt{(a_1 \cdot a_2 \ldots a_n)} \\ &=2^{n} \cdot \sqrt{1}=2^{n} . \end{aligned} \end{aligned} $

~~ 11. $a+b+c=1 \Rightarrow b+c=1-a$

$a+c=1-b$

$a+b=1-c$

Now, $a>0, b>0, c>0 \Rightarrow AM \geq GM$

$ \begin{matrix} \Rightarrow & \frac{b+c}{2} \geq \sqrt{b c} \Rightarrow(1-a) \geq 2 \sqrt{b c} \\ & \frac{a+c}{2} \geq \sqrt{a c} \Rightarrow(1-b) \geq 2 \sqrt{a c} \\ & \frac{a+b}{2} \geq \sqrt{a b} \Rightarrow(1-c) \geq 2 \sqrt{a b} \\ \therefore \quad & (1-a)(1-b)(1-c) \geq 8 \sqrt{b c} \cdot \sqrt{a c} \cdot \sqrt{a b} \\ \Rightarrow & (1-a)(1-b)(1-c) \geq 8 a b c \\ \Rightarrow & \frac{(1-a)(1-b)(1-c)}{a b c} \geq 8 \end{matrix} $

$\therefore \quad$ Minimum value is $\mathbf{8}$.

~~ 12. $\frac{3 b+4 c}{a}+\frac{4 c+a}{3 b}+\frac{a+3 b}{4 c}$

$=\frac{3 b}{a}+\frac{4 c}{a}+\frac{4 c}{3 b}+\frac{a}{3 b}+\frac{a}{4 c}+\frac{3 b}{4 c}$

$=(\frac{3 b}{a}+\frac{a}{3 b})+(\frac{4 c}{a}+\frac{a}{4 c})+(\frac{4 c}{3 b}+\frac{3 b}{4 c})$

$\because a>0, b>0, c>0$

$\therefore \frac{a}{3 b}, \frac{3 b}{a}, \frac{4 c}{a}, \frac{a}{4 c}, \frac{4 c}{3 b}, \frac{3 b}{4 c}$ are all greater than zero.

Now $\forall a>0,(a+\frac{1}{a}) \geq 2$

$\therefore(\frac{3 b}{a}+\frac{a}{3 b})+(\frac{4 c}{a}+\frac{a}{4 c})+(\frac{4 c}{3 b}+\frac{3 b}{4 c}) \geq 2+2+2=6$

$\therefore$ Minimum value of the expression is 6 .

Relations

KEY FACTS

1. Cartesian product: Let $A$ and $B$ be two non-empty sets. Then the set of all possible ordered pairs $(x, y)$ such that the first component $x$ of the ordered pairs is an element of set $A$, and the second component $y$ is an element of set $B$, is called the cartesian product of the sets $A$ and $B$. It is denoted by $\boldsymbol{A} \times \boldsymbol{B}$ read as “A cross $\boldsymbol{B}$ “.

$ A \times B= \lbrace (a, b), a \in A \text{ and } b \in B \rbrace $

Also, $n(A \times B)=n(A) \times n(B)=p q$ if set $A$ has $p$ elements and set $B$ has $q$ elements.

Notes: 1. The cartesian product $A \times B$ is not the same as $B \times A$.

In $A \times B$, the set $A$ is named first so its elements will appear as the first components of the ordered pairs.

In $B \times A$, the set $B$ is named first, so its elements will appear as the first components of the ordered pairs.

2. If either $A$ or $B$ is a null set, then we define $A \times B$ to be a null set.

If $A= \lbrace a, b \rbrace$ and $B=\phi$ then $A \times B=\phi$

3. If either $A$ or $B$ is an infinite set and the other is a non-empty set, then $A \times B$ is also an infinite set.

4. If $A$ and $B$ are two non-empty sets having $n$-elements in common, then $A \times B$ and $B \times A$ have $n^{2}$ elements in common.

5. If $A=B$, then $A \times B=A \times A$ and is denoted by $A^{2}$.

Examples:

Ex. 1 . If $A= \lbrace a, b \rbrace$ and $B= \lbrace 1,2,3 \rbrace$, then

$A \times B= \lbrace (a, 1),(b, 1),(a, 2),(b, 2),(a, 3),(b, 3) \rbrace$

$B \times A= \lbrace (1, a),(2, a),(3, a),(1, b),(2, b),(3, b) \rbrace$.

Ex. 2 . If $A= \lbrace 2,4,6 \rbrace$, then

$A \times A= \lbrace (2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6) \rbrace$.

2. Relation:

If $A$ and $B$ are any two non-empty sets, then any subset of $A \times B$ is defined as a relation from $A$ to $B$.

For example,

Suppose $A= \lbrace 1,2,3 \rbrace$ and $B= \lbrace 1,2,3,4 \rbrace$. Then $ \lbrace (2,3),(2,4),(1,3) \rbrace$ is a relation in $A \times B$. Many more relations (subsets) can be selected at random from our product set $A \times B$.

3. Domain, codomain and range of a relation:

Let $R$ be a relation from set $A$ to set $B$. Then, the set of first element of the ordered pairs in $\boldsymbol{R}$ is called the domain and the set of second elements of the ordered pairs in $\boldsymbol{R}$ is called the range. The second set $B$ is called the codomain of $R$.

Thus for a relation $R= \lbrace (a, b) ; a, b \in R \rbrace$,

Domain $= \lbrace a:(a, b) \in R \rbrace$ and Range $= \lbrace b:(a, b) \in R \rbrace$

For example,

If $A= \lbrace 16,25,36,49 \rbrace$ and $B= \lbrace 1,4,5,6 \rbrace$ and $\boldsymbol{R}$ be the relation “is square of” from $A$ to $B$, then $R= \lbrace (a, b): a=b^{2}, a \in A, b \in B \rbrace$

$\therefore R= \lbrace (16,4),(25,5),(36,6) \rbrace$. Then,

Domain of $R= \lbrace 16,25,36 \rbrace$, Range of $R= \lbrace 4,5,6 \rbrace$ and Codomain of $R= \lbrace 1,4,5,6 \rbrace$.

4. Number of relations that are possible from a set $A$ of $m$ elements to another set $B$ of $n$ elements.

Number of elements in set $A=m$

Number of elements in set $B=n$

$\therefore$ Number of elements in set $(A \times B)=m n$

$\therefore$ Number of subsets of $(A \times B)=\mathbf{2}^{m n}$

Since every subset of $A \times B$ is a relation from $A$ to $B$, therefore,

Remember

If $n(A)=m$

$n(B)=n$

Then, $\boldsymbol{n}(\boldsymbol{A} \times \boldsymbol{B})=\boldsymbol{m} \boldsymbol{n}$

Number of relations possible

from $A$ to $B=2^{m n}$

Number of relations possible from $A$ to $B=\mathbf{2}^{\boldsymbol{m} \boldsymbol{n}}$

Note: 1. If $R_1$ and $R_2$ are two relations from $A$ to $B$, then $R_1 \cup R_2, R_1 \cap R_2$ and $R_1-R_2$ are also relations from $A$ to $B$.

2. Since $\phi \leq A$, i.e., null set is a subset of every set, $\phi$ is a relation from $A$ to $B$. Also domain $(\phi)=\phi$ and $Range(\phi)=\phi$

5. Inverse of a relation:

For any binary relation $R$, a second relation can be constructed by merely interchanging first and second components in every ordered pair.

The relation thus obtained is called the inverse of the first one and designated as $\boldsymbol{R}^{\mathbf{- 1}}$. Thus,

$ R^{-1}= \lbrace (y, x):(x, y) \in R \rbrace $

For example,

Then inverse of the husband-wife relation is wife-husband relation. ~~
Let $R= \lbrace (2,1),(3,2),(4,3),(4,5) \rbrace$. Then,

$ R^{-1} \lbrace (1,2),(2,3),(3,4),(5,4) \rbrace \text{. } $

So $(R^{-1})^{-1}=R$.

6. Types of relations:

Let $A$ be a non-empty set. Then, $a$ relation $R$ on $A$ is said to be

Reflexive if $(a, a) \in R$ for each $a \in A$, i.e., if a $R$ a for each $a \in A$.

For example, the relation “is as strong as” is reflexive since every member of a particular set will be as strong as himself, but the relation “is the mother of” is not reflexive as a person cannot be his/her own mother.

Symmetric if $(a, b) \in R \Rightarrow(b, a) \in R$ for all $a, b \in A$, i.e., if $a R b b$ a for all $a, b \in A$.

For example, the relation “weighs the same as” is symmetric as if $x$ weighs same as $y$. Then $y$ weighs same as $x$, but the relation “is less than” is not symmetric as: if $x$ is less than $y$, then $y$ is not less than $x$.

-Transitive if $(a, b) \in R,(b, c) \in R \Rightarrow(a, c) \in R$ for all $a, b, c \in A$, i.e., if $a R b$ and $b R$ then $a R c$.

For example, the relation “equals to” is a transitive relation for if $x=y$ and $y=z$, then $x=z$, but the relation “is perpendicular to” on $a$ set of coplanar lines is not transitive for if line $a$ is perpendicular to line $b$ and line $b$ is perpendicular to line $c$, then line $a$ is not perpendicular to line $c$.

Equivalence: $A$ relation $R$ on a set $A$ is said to be an equivalence relation if it is reflexive, symmetric and transitive.

Equivalence Relation

For example,

(i) “Equality” is an equivalence relation because

$x=x$
$x=y \Rightarrow y=x$
$x=y, y=z \Rightarrow x=z$. (ii) “Is parallel to” on a set $A$ of coplanar lines is an equivalence relation since: for all the lines $a, b, c \in A$.
$a|a \quad \bullet a| b \Rightarrow b|a \quad \bullet \quad a| b, b|c \Rightarrow a| c$.

SOLVED EXAMPLES

~~ Ex. 1 . Let $R$ be a relation from $A= \lbrace 1,2,3,4,5,6 \rbrace$ to $B= \lbrace 1,3,5 \rbrace$ which is defined as " $x$ is less than $y$ “. Write $\boldsymbol{R}$ as a set of ordered pairs. Also state the domain, range and codomain of $\boldsymbol{R}$.

Sol. $R= \lbrace a, b: a<b, a \in A, b \in B \rbrace$, where $A= \lbrace 1,2,3,4,5,6 \rbrace$ and $B= \lbrace 1,3,5 \rbrace$.

$\therefore \quad R= \lbrace (1,3),(1,5),(2,3),(2,5),(3,5),(4,5) \rbrace$

Domain of $R= \lbrace 1,2,3,4 \rbrace$

Range of $R= \lbrace 3,5 \rbrace$

Codomain of $R= \lbrace 1,3,5 \rbrace$.

~~ Ex. 2 . Let $A= \lbrace a, b, c, d \rbrace$ and $B= \lbrace x, y, z \rbrace$. Which of the following are relations from $A$ to $B$ ? (i) $\quad \lbrace (a, y),(a, z),(c, x),(d, y) \rbrace$ (ii) $ \lbrace (a, x),(b, y),(c, x),(a, d) \rbrace$ (iii) $ \lbrace (a, x),(y, d),(x, c) \rbrace$ (iv) $ \lbrace (y, a),(z, a),(z, c),(y, d) \rbrace$ (v) $ \lbrace (a, x),(x, a),(b, y),(y, b) \rbrace$ (vi) $ \lbrace (a, x),(b, y),(c, z), z \rbrace$

(vii) $ \lbrace a, b, x, y, z \rbrace$

Sol. (i) Yes.

(ii) No, because in the ordered pair ( $a, d), a \in A$ and $d \notin B$.

(iii) No, because in $(y, d), y \in B$.

(iv) No. because here the first entries in all the ordered pairs are in the set $B$.

(v) No.

(vi) No, because the element $z$ is not an ordered pair.

(vii) No, because the elements of the set are not ordered pairs.

~~ Ex. 3 . Determine the domain and range of the following relations:

(i) $ \lbrace (-3,1),(-1,1),(1,0),(3,0) \rbrace$

(ii) $ \lbrace {(x, y): \text{x is a multiple of 3 and y is a multiple of 5}}\rbrace$

(iii) $ \lbrace (x, x^{2}): x.$ is a prime number less than 15 $\rbrace$

Sol. (i) Domain $= \lbrace -3,-1,1,3 \rbrace$, Range $= \lbrace 0,1\rbrace$

(ii) Domain $= \lbrace \text{x: x is a multiple of 3}\rbrace$=$ \lbrace {3 n: n \in Z}\rbrace$

Range = $ \lbrace \text {y: y is a multiple of 5}\rbrace$=$ \lbrace {5 n: n \in Z\rbrace}$

(iii) Relation $= \lbrace (x, x^{2}): x.$ is a prime number less than 15 $ \rbrace $

$ = \lbrace {(2,4),(3,9),(5,25),(7,49),(11,121),(13,169)\rbrace} $

Domain $= \lbrace 2,3,5,7,11,13\rbrace$, Range $= \lbrace 4,9,25,49,121,169 \rbrace$

~~ Ex. 4 . Let $\boldsymbol{N}$ be the set of natural numbers. Describe the following relation in words giving its domain and the range. $\quad \lbrace (1,1),(16,2),(81,3),(216,4) \rbrace$

Sol. The given relation stated in words is

$R= \lbrace (x, y): x$ is the fourth power of $y ; x \in N, y \in \lbrace 1,2,3,4 \rbrace \rbrace$.

~~ Ex. 5 . $I$ is the set of integers. Describe the following relations in words, giving its domain and range.

$ \lbrace (0,0),(1,-1),(2,-2),(3,-3) \ldots \rbrace $

Sol. $R= \lbrace (0,0),(1,-1),(2,-2),(3,-3) \ldots \rbrace= \lbrace (x, y): y=-x, x \in W \rbrace$

Domain $= \lbrace 0,1,2,3, \ldots \rbrace=$.$W , Range = \lbrace \ldots,-3,-2,-1,0 \rbrace$

~~ Ex. 6 . Choose the correct option.

The relation $R= \lbrace (1,1),(2,2),(3,3),(1,2),(2,3),(1,3) \rbrace$ on a set $A= \lbrace 1,2,3 \rbrace$ is

(a) reflexive, transitive but not symmetric

(b) reflexive, symmetric but not transitive

(d) reflexive but neither symmetric nor transitive.

(NDA/NA 2010)

Sol. $R= \lbrace (1,1),(2,2),(3,3),(1,2),(2,3),(1,3) \rbrace$ on set $A= \lbrace 1,2,3 \rbrace$.

(i) Since $(1,1),(2,2),(3,3) \in R \Rightarrow R$ is reflexive.

(ii) $(1,2) \in R$ but $(2,1) \notin R \Rightarrow R$ is not symmetric.

(iii) $(1,2) \in R,(2,3) \in R$ and $(1,3) \in R \Rightarrow R$ is transitive.

$\therefore \quad$ Option $(a)$ is the right answer.

~~ Ex. 7 . Let $X$ be the set of all graduates in India. Elements $x$ and $y$ in $X$ are said to be related, if they are graduates of the same university. Is this given relation an equivalence relation?

(NDA/NA 2010)

Sol. $R= \lbrace (x, y) \rbrace: x$ and $y$ are graduates of same university, $x, y \in \lbrace $ All graduates of India $ \rbrace$.

$\boldsymbol{R}$ is reflexive as $(x, x) \in R$, since $\boldsymbol{x}$ and $\boldsymbol{x}$ are graduates from the same university.

$\boldsymbol{R}$ is symmetric as $(x, y) \in R \Rightarrow x$ and $y$ are graduates from the same university.

$\Rightarrow y$ and $x$ are graduates from the same university

$\Rightarrow(y, x) \in R$.

$\boldsymbol{R}$ is transitive as $(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow x$ and $y$ are graduates from the same university and $y$ and $z$ are graduates from the same university

$\Rightarrow x$ and $z$ are graduates from the same university

$\Rightarrow(x, z) \in R$.

$R$ being reflexive, symmetric and transitive is an equivalence relation.

~~ Ex. 8 . Show that the relation $R$ in the set $A$ of all the books in a library of a school given by $R= \lbrace (x, y): x$ and $y$ have the same number of pages $ \rbrace$ is an equivalence relation.

Sol. Given: $A= \lbrace \text{ All books in a library of a school }\rbrace$

$\boldsymbol{R}=(\boldsymbol{x}, \boldsymbol{y}): x$ and $y$ have the same number of pages

Reflexivity: $(x, x) \in R \Rightarrow R$ is reflexive on $A$

Symmetric: Since books $x$ and $y$ have the same number of pages, so $(x, y) \in R$.

Since books $y$ and $x$ have the same number of pages, so $(y, x) \in R$.

$\Rightarrow R$ is symmetric on $A$.

Transitivity: Books $x, y, z$ have the same number of pages $\Rightarrow(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow(x, z) \in R \Rightarrow R$ is transitive on $A$.

Hence, $R$ is an equivalence relation.

~~ Ex. 9 . Show that the relation ’ $\cong$ ’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation.

Sol. Reflexive : $A \cong A$

Symmetric : $\quad$ if $A \cong B$ then $B \cong A$

Transitive : $\quad$ if $A \cong B$ and $B \cong C$,

then $\quad A \cong C$ True

True

Therefore, the relation ’ $\cong$ ’ is an equivalence relation.

~~ Ex. 10 . Show that the relation ’ $\subset$ ’ with respect to sets is not an equivalence relation.

Sol. Reflexive: If $A$ is a set, then $A \subset A$.

False

Symmetric: If $A$ and $B$ are sets and $A \subset B$, then $B \subset A$.

False

Transitive: If $A, B$ and $C$ are sets and if $A \subset B$ and $B \subset C$ then $A \subset C$

True

Hence ’ $C$ ’ is not an equivalence relation since it possesses only the transitive property.

~~ Ex. 11 . Show that the relation " $\geq$ " on the set of real numbers is not an equivalence relation.

Sol. Reflexive : $a \geq a \quad$ True since $a=a$

Symmetric : If $a \geq b$, then $b \geq a$. This statement is true for the case $a=b$, but false in the other instances.

Transitive : If $a \geq b$, and $b \geq c$, then $a \geq c$.

True

The relation " $\geq$ " is not an equivalence relation since it lacks the property of symmetry.

~~ Ex. 12 . Write ‘yes’ if each of the following relations is an equivalence relation. If it is not, then write whether it is reflexive or symmetric or transitive. (i) is parallel to (ii) is perp. to (iii) is greater than (iv) is a factor of (v) is a multiple of

Sol. (i) ‘is parallel to’ is reflexive, because any line is parallel to itself.

Symmetric, because if line $l$ is parallel to line $m$, then line $m$ is parallel to line $l$.

Transitive, because if $l | m$ and $m | n$ then $l | n$.

Therefore,’ is parallel to’ is an equivalence relation.

(ii) No, because this relation is only symmetric.

(iii) No, because this relation is only transitive. It is neither reflexive nor symmetric.

(iv) No, because this relation is only reflexive and transitive. It is not symmetric. If $x$ is a factor of $y$, and $x \neq y$, $y$ cannot be a factor of $x$.

(v) No, because the relation is reflexive and transitive but not symmetric.

~~ Ex. 13 . If $R$ is a relation in $N \times N$ defined by $(a, b) R(c, d)$ if and only if $a d=b c$, show that $R$ is an equivalence relation.

Sol. (i) $\boldsymbol{R}$ is reflexive. For all $(a, b) \in N \times N$ we have $(a, b) R(a, b)$ because

$a b=b a$

$\Rightarrow \quad R$ is reflexive.

(ii) $\boldsymbol{R}$ is symmetric. Suppose $(a, b) R(c, d)$ Then $(a, b) R(c, d) \Rightarrow a d=b c \Rightarrow c b=d a$

$\Rightarrow \quad(c, d) R(a, b) \Rightarrow R$ is symmetric.

(iii) $R$ is transitive. Suppose $(a, b) R(c, d)$ and $(c, d) R(e, f)$. Then

$a d=b c$ and $c f=d e$

$\Rightarrow \quad(a d)(c f)=(b c)(d e)$

$\Rightarrow \quad a f=b e \Rightarrow(a, b) R(e, f) \Rightarrow R$ is transitive.

Since $R$ is reflexive, symmetric and transitive, therefore, $R$ is an equivalence relation on $N \times N$.

~~ Ex. 14 . Let $A= \lbrace $ real numbers $ \rbrace$

Let $R= \lbrace (a, b): a, b \in A$ and $a-b<5 \rbrace$

Is $R$ an equivalence relation? Justify your answer.

Sol. (i) Reflexive: For all $a \in A, a-a=0<5 \Rightarrow a R a$.

$\therefore \quad R$ is reflexive. (ii) Symmetric: For example let $a=2, b=8$, then $a-b=2-8=-6<5 \Rightarrow a=a R b$ But, $6-a=8-2=6 \nless 5 \Rightarrow b \not R a$.

Hence $R$ is not symmetric.

(iii) Transitive: If $a=4, b=0, c=-4$, then $a-b=4-0=4<5 \Rightarrow a R b$

and $b-c=0-(-4)=4<5 \Rightarrow b R c$

But $a-c=4-(-4)=8<5 \Rightarrow a \not K c$.

$\therefore \quad R$ is not transitive.

$\Rightarrow R$ is not an equivalence relation.

~~ Ex. 15 . If $R= \lbrace a, b: a-b$ is even and $a, b \in Z \rbrace$, then $R$ is an equivalence relation.

Sol. Reflexive: For all $a \in Z, a-a=0$, an even integer $\Rightarrow a R a \Rightarrow R$ is reflexive

Symmetric. For all $a, b \in Z$, if $a-b$ is an even integer, then $b-a=-(a-b)$ is an even integer.

i.e., $a R b \Rightarrow b R a \Rightarrow R$ is symmetric.

Transitive. If $a-b$ is an even integer and $b-c$ is an even integer,

then $(a-b)+(b-c)$ is an even integer $\Rightarrow a-c$ is an even integer.

i.e., $a R b$ and $b R c \Rightarrow a R c$.

Hence $R$ is transitive.

$\Rightarrow R$ is an equivalence relation.

~~ Ex. 16 . Let $Z$ be the set of all integers and $R$ be the relation on $Z$ defined as $R= \lbrace (a, b): a, b \in Z)$ and $(a-b)$ is divisible by 5 .

Prove that $R$ is an equivalence relation.

Sol. Reflexive: Since 5 divides $a-a$ for all $a \in Z$, therefore, $R$ is reflexive.

Symmetric: $(a, b) \in R \Rightarrow 5$ divides $a-b$

$\Rightarrow 5$ divides $b-a \Rightarrow b-a \in R$

$\therefore \quad R$ is symmetric.

Transitive: $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow a-b$ and $b-c$ are both divisible by 5

$\Rightarrow a-b+b-c$ is divisible by 5

$\Rightarrow(a-c)$ is divisible by 5

$\Rightarrow(a, c) \in R$

$\therefore \quad R$ is transitive.

Since $R$ is reflexive, symmetric and transitive, therefore, $R$ is an equivalence relation.

~~ Ex. 17 . Show that the relation $S$ in the set $R$ of real numbers, defined as $S= \lbrace (a, b): a, b \in R.$ and $.a \leq b^{3} \rbrace$ is neither reflexive, nor symmetric, nor transitive.

Sol. (i) Since $a \leq a^{3}$ is not true for all $a \in R$

For example, if $a=\frac{1}{4}$, then $a<a^{3}$, i.e., $a \leq a^{3}$ is not true.

So, $R$ is not reflexive.

(ii) $(a, b) \in R$ need not imply that $(b, a) \in R$

For example, $(1,2) \in R$ but $(2,1) \in R$ because $1 \leq 2^{3}$ but $2 \not 1^{3}$.

$\therefore \quad R$ is not symmetric. (iii) $(a, b) \in R$ and $(b, c) \in R$ need not imply that $(a, c) \in R$.

For example, $(80,5) \in R$ and $(5,2) \in R$ but $(80,2) \in R$

Since, $80<5^{3}, 5<2^{3}$ but $80 \nless 2^{3}$.

$\therefore \quad R$ is not transitive.

~~ Ex. 18 . Show that the relation $S$ in the set $A= \lbrace x \in z: 0 \leq x \leq 12 \rbrace$ given by $S= \lbrace (a, b): a, b \in z,|a-b|$ is divisible by 4$ \rbrace$ is an equivalence relation. Find the set of all elements related to 1 .

Sol. Reflexive: For all $a \in A,|a-b|=0$ is divisible by $4 \Rightarrow(a, a) \in S$.

$\therefore \quad S$ is reflexive.

Symmetric: Let $a, b \in A$. Then $(a, b) \in S \Rightarrow|a-b|$ is divisible by 4

$\Rightarrow|b-a|$ is divisible by $4 \Rightarrow(b, a) \in S$

$\therefore S$ is symmetric.

Transitive: Let $a, b, c \in A,(a, b) \in S$ and $(b, c) \in S$

$\Rightarrow|a-b|$ is divisible by 4 and $|b-c|$ is divisible by 4 .

$\Rightarrow(a-b)$ and $(b-c)$ are divisible by 4 .

$\Rightarrow(a-b)+(b-c)=(a-c)$ is divisible by 4 .

$\Rightarrow|a-c|$ is divisible by $4 \Rightarrow(a, c) \in S$.

$\therefore \quad S$ is transitive.

Since $S$ is reflexive, symmetric and transitive, therefore, $S$ is an equivalence relation.

~~ Ex. 19 . Check whether the relation $R$ defined in the set $ \lbrace 1,2,3,4,5,6 \rbrace$ as $R= \lbrace (a, b): b=a+1 \rbrace$ is reflexive, symmetric or transitive.

Sol. Reflexive:

$ \begin{aligned} R & = \lbrace (a, b): b=a+1 \rbrace \\ & = \lbrace (a, a+1): a, a+1 \in \lbrace 1,2,3,4,5,6 \rbrace \rbrace \\ & = \lbrace (1,2),(2,3),(3,4),(4,5),(5,6) \rbrace \end{aligned} $

$\Rightarrow R$ is not reflexive since $(a, a) \notin R$ for all $a$.

Symmetric: $R$ is not symmetric as $(a, b) \in R$ but $(b, a) \notin R$

Transitive: $R$ is not transitive as $(a, b) \in R$ and $(b, c) \in R$ but $(a, c) \notin R$

e.g., $(1,2) \in R(2,3) \in R$ but $(1,3) \notin R$

PRACTICE SHEET

Let $A= \lbrace 2,3,4,6 \rbrace$ and let $R$ be a relation on $A$ defined as $R= \lbrace (a, b) ; a \in A, b \in A$, a divides $B \rbrace$ Then which of following is not $R$ ?

(a) $R= \lbrace (2,2) \rbrace,(2,4),(3,6),(3,6)$

(b) $R= \lbrace (2,2),(3,3),(2,4),(4,6) \rbrace$

(d) $R= \lbrace (2,2),(3,3),(4,4),(2,4),(2,6),(3,6),(6,6) \rbrace$

~~ 2. Which of the following are relations from $B$ to $A$ where $A= \lbrace a, b, c, d \rbrace$ and $B= \lbrace x, y, z \rbrace$ ?

(i) $ \lbrace (z, b),(y, c),(x, a) \rbrace$

(ii) $ \lbrace (x, b),(y, a) \rbrace$

(iii) $ \lbrace (b, y),(z, a),(x, c) \rbrace$

(iv) $ \lbrace (x, a),(x, b),(y, c),(z, d) \rbrace$

(a) (i) and (iii)

(b) Only (iv)

(d) All of the above

~~ 3. Given $A= \lbrace -2,-1,0,1,2 \rbrace$, which of the following relations on $A$ have both domain and range equal to $A$ ?

(i) $R$ : “is equal to”

(ii) $R$ : “is the multiplicative inverse of”

(iii) $R$ : “is the additive inverse of”

(iv) $R$ : “is less than”

(a) Only $(i)$

(b) Only (iv)

(d) $(i)$ and $($ iii)

~~ 4. The relation $R$ in the set $ \lbrace 1,2,3 \rbrace$ given by $R= \lbrace (1,2)$, $(2,1) \rbrace$ is:

(a) Reflexive only

(b) Reflexive and symmetric only

(d) Transitive only

~~ 5. If $n(A)=5$ and $n(B)=7$, then the number of relations on $A \times B$ is

(a) $2^{25}$

(b) $2^{35}$

(d) 35

(Kerala PET 2012)

~~ 6. Let $R= \lbrace (3,3),(6,6),(9,9),(12,12),(6,12),(3,12)$, $(3,6) \rbrace$ be a relation on set $A= \lbrace 3,6,9,12 \rbrace$. The relation is

(a) Reflexive only

(b) Reflexive and symmetric only

(d) An equivalence relation.

(AIEEE 2005)

~~ 7. Which of the following is an equivalence relation defined on set $A= \lbrace 1,2,3 \rbrace$

(a) $ \lbrace (1,1),(1,2),(1,3),(2,2),(2,3),(3,1),(3,2),(3,3) \rbrace$

(b) $ \lbrace (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2) \rbrace$

(d) $ \lbrace (1,2),(2,1),(1,3),(3,1) \rbrace$

~~ 8. Given the relation $R= \lbrace (1,2),(2,3) \rbrace$ on the set $A= \lbrace 1,2,3 \rbrace$, the minimum number of ordered pairs which can be added to $R$ to make it an equivalence relation is

(a) 2

(b) 3

(d) 7

(AMU 2008)

~~ 9. Let $Z$ be the set of integers. Then the relation $R= \lbrace (a, b)$; $a, b \in Z$ and $(a+b)$ is even $ \rbrace$ defined on $Z$ is

(a) Only symmetric

(b) Symmetric and transitive only

(d) None of the above

~~ 10. If $R$ be a relation defined as $a R b \Leftrightarrow|a|<b$, then $R$ is

(a) Reflexive only

(b) Symmetric only

(d) Reflexive and transitive but not symmetric

(Odisha JEE 2012)

~~ 11. Let the relation $R$ defined on the set of natural numbers $N$ be : $R= \lbrace (a, b): b$ is divisible by $\forall a, b \in N \rbrace$. Then $R$ is

(a) Reflexive and symmetric only

(b) Symmetric and transitive only

(d) An equivalence relation

~~ 12. Let $R$ be a relation defined on the set $A$ of all triangles such that $R= \lbrace (T_1, T_2): T_1.$ is similar to $.T_2 \rbrace$. Then $R$ is

(a) Reflexive only

(b) Transitive only

(d) An equivalence relation. ~~ 13. Let $R$ be a relation defined as a $R b$ if $|a-b|>0$, then the relation is

(a) Reflexive only

(b) Symmetric only

(d) Symmetric and transitive

(VITEEE 2008)

~~ 14. On the set $R$ of all real numbers, $a$ relation $R$ is defined by $R= \lbrace (a, b): 1+a b>0 \rbrace$. Then $R$ is

(a) Reflexive and symmetric only (b) Reflexive and Transitive only

(d) An equivalence relation

~~ 15. Which of the following relations is only symmetric?

(a) “less than equal to” on a set of Real numbers

(b) “is a multiple of” on the set of positive integers

(d) “is the father of” on a set of family members.

~~ 16. If $R$ is a relation defined on the set of natural numbers $N$ such that $(a, b) R(c, d)$ if and only if $a+d=b+c$, then $R$ is

(a) Symmetric and transitive but not reflexive

(b) Reflexive and transitive but not symmetric

(d) An equivalence relation

~~ 17. Let $I$ be the set of integers and $R$ be a relation on $I$ defined by $R= \lbrace (x, y):(x-y)$ is divisible by $11, x, y \in I \rbrace$. Then $R$ is

(a) An equivalence relation

(b) Symmetric only

(d) Transitive only ~~ 18. Let $W$ denote the words in the English dictionary. Let the relation $R$ be defined by $R= \lbrace (x, y) \in W \times W$ : the words $x$ and $y$ have at least one letter in common $ \rbrace$. Then $R$ is

(a) Reflexive and transitive, not symmetric

(b) Reflexive and symmetric and not transitive

(d) Reflexive, symmetric and transitive

(AIEEE 2006)

~~ 19. On a set $N$ of all natural numbers is defined the relation $R$ by $a R b$ iff the $G C D$ of $a$ and $b$ is 2 , then $R$ is

(a) Reflexive and Transitive

(b) Symmetric and Transitive

(d) Not reflexive, not symmetric, not transitive

(Kerala PET 2007)

~~ 20. Let $N$ be the set of integers. $A$ relation $R$ on $N$ is defined as $R= \lbrace (x, y): x y>0, x, y \in N \rbrace$. Then, which of the following is correct?

(a) $R$ is symmetric but not reflexive

(b) $R$ is reflexive but not symmetric

(d) $R$ is an equivalence relation

(NDA/NA2007)

~~ 21. Let $R$ be a relation on the set of integers given by $a=2^{k} . b$ for some integer $k$. Then $R$ is

(a) reflexive but not symmetric

(b) reflexive and transitive but not symmetric

(d) symmetric and transitive but not reflexive

(Kerala CEE 2006)

~~ 22. Consider the following relations $R= \lbrace (x, y) \mid x, y$ are real numbers and $x=w y$ for some rational number $w \rbrace$; $S= \lbrace .(\frac{m}{n}, \frac{p}{q}) \rvert, m, n, p.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p n \rbrace$. Then,

(a) $R$ is an equivalence relation but $S$ is not an equivalence relation

(b) Neither $R$ nor $S$ is an equivalence relation

(d) $R$ and $S$ are both equivalence relations

(AIEEE 2010)

~~ 23. Let $R$ be a relation on the set $N$, defined by $ \lbrace (x, y): 2 x-y$ $=10 \rbrace$ then $R$ is

(a) Reflexive

(b) Symmetric

(d) None of the above

(AMU 2012)

~~ 24. If $A$ and $B$ are two equivalence relations defined on a set $C$, then

(a) $A \cap B$ is an equivalence relation

(b) $A \cap B$ is not an equivalence relation

(d) $A \cup B$ is not an equivalence relation

(UPSEE 2011)

~~ 25. For any two real number $a$ and $b$, we defined $a R b$ if and only if $\sin ^{2} a+\cos ^{2} b=1$. The relation $R$ is

(a) reflexive but not symmetric

(b) symmetric but not transitive

(d) an equivalence relation

ANSWERS

| 1. $(b)$ | 2. $(c)$ | 3. $(d)$ | 4. $(c)$ | 5. $(b)$ | | 6. (c) | 7. (c) | 8.(d) | 9. (c) | 10.(c) | | 11. $(c)$ | 12. $(d)$ | 13. $(d)$ | 14. $(a)$ | 15. $(c)$ | | 21. $(c)$ | 22. $(c)$ | 23. $(a)$ | 24. $(a)$ | 25. $(d)$ |

HINTS AND SOLUTIONS

The relation $= \lbrace (2,2),(3,3),(2,4),(4,6) \rbrace$ is not $R$ as the one of the ordered pairs $(4,6)$ does not satisfy the condition " $a$ divides $b$ " as 4 does not divide 6 .

~~ 2. The set of ordered pairs $ \lbrace (b, y),(z, a),(x, c) \rbrace$ does not state a relation from $B$ to $A$ as the ordered pair $(b, y)$ has the first element ’ $b$ ’ from set $A$, whereas it should be from set $B$.

~~ 3. Let us examine the domain and range of the each relation individually:

(i) $R$ : “is equal to” means $R= \lbrace (a, b): a=b, a \in A, b \in A \rbrace$

$\therefore \quad R= \lbrace (-2,-2),(-1,-1),(0,0),(1,1),(2,2) \rbrace$

$\therefore$ Domain of $R= \lbrace -2,-1,0,1,2 \rbrace$ and Range of $R= \lbrace -2,-1,0,1,2 \rbrace$

Hence both are equal and equal to $A$.

(ii) $R$ : “is the multiplicative inverse of " means

$R= \lbrace (a, b): a b=1, a \in A, b \in A \rbrace$

$\therefore \quad R= \lbrace (-1,-1),(1,1) \rbrace$

Here domain $= \lbrace -1,-1 \rbrace$, range $= \lbrace -1,1 \rbrace$.

(iii) $R$ : “is the additive inverse of “means

$R= \lbrace (a, b): a=-b, a \in A, b \in A \rbrace$

$\therefore \quad R= \lbrace (-2,2),(-1,1),(0,0),(1,-1),(2,-2) \rbrace$

Hence domain $= \lbrace -2,-1,0,1,2 \rbrace$ and range $= \lbrace 2,1,0,-1,-2 \rbrace$

Both are equal and equal to $A$.

(iv) $R$ : “is less than " means $R= \lbrace (a, b): a<b, a \in A, b \in A \rbrace$

$\therefore \quad R= \lbrace (-2,-1),(-2,0),(-2,1),(-2,2),(-1,0),(-1,1)$,

$(-1,2),(0,1),(0,2),(1,2) \rbrace$

Here domain $= \lbrace -2,-1,0,1 \rbrace$ and range $= \lbrace -1,0,1,2 \rbrace$.

$\therefore$ Relations given in (i) and (iii) satisfy the given condition.

~~ 4. As $(1,1),(2,2) \notin R$ so $R$ is not reflexive. $A$ relation $a R b$ is said to symmetric if $(a, b) \in R \Rightarrow(b, a) \in R$.

Here $(1,2) \in R$ and $(2,1) \in R$ so it is symmetric.

Also, as $(1,2) \in R$, but $(2,3),(1,3) \notin R$, so $R$ is not transitive.

~~ 5. $n(A)=5$ and $n(B)=7$

$\therefore n(A \times B)=5 \times 7=35$

Total number of relations from $A$ to $B=$ Total number of subsets of $(A \times B)=\mathbf{2}^{\mathbf{3 5}}$.

~~ 6. Here $A= \lbrace 3,6,9,12 \rbrace$ and

$R= \lbrace (3,3),(6,6),(9,9),(12,12),(6,12),(3,12),(3,6) \rbrace$

$\because(3,3),(6,6),(9,9),(12,12)$ all belong to $R$

$\Rightarrow \lbrace (a, a): \in R \forall a \in A \rbrace$

$\Rightarrow R$ is reflexive

$(3,6) \in R$ but $(6,3) \notin R$. Also $(6,12) \in R$ but $(12,6) \notin R$.

So $(a, b) \in R \nRightarrow(b, a) \in R$

Hence $R$ is not symmetric

Now $(3,6) \in R,(6,12) \in R$ and $(3,12) \in R \Rightarrow(3,12) \in R$ $(x, y) \in R,(y, z) \in R \Rightarrow(x, z) \in R$

Hence $R$ is transitive.

~~ 7. Option (c) satisfies all the conditions of an equivalence relation.

$(1,1),(2,2),(3,3) \in R$

$\Rightarrow(a, a) \in R \forall a \in A \Rightarrow R$ is reflexive

$(3,1) \in R$ and $(1,3) \in R$

$\Rightarrow(a, b) \in R \Rightarrow(b, a) \in R \forall a, b \in A \Rightarrow$ symmetric

$(1,3) \in R,(3,1) \in R$ and $(1,1) \in R \Rightarrow(a, b) \in R,(b, c)$

$\in R \Rightarrow(a, c) \in R \forall a, b, c \in A \Rightarrow R$ is transitive.

~~ 8. $A= \lbrace 1,2,3 \rbrace$.

$R= \lbrace (1,2),(2,3) \rbrace$

To make $R$ an equivalence relation, it should be:

(i) Reflexive: So three more ordered pairs $(1,1),(2,2)$, $(3,3)$ should be added to $R$ to make it reflexive.

(ii) Symmetric : As $R$ contains $(1,2)$ and $(2,3)$ so two more ordered pairs $(2,1)$ and $(3,2)$ should be added to make it symmetric.

(iii) Transitive: $(1,2) \in R,(2,3) \in R$. So to make $R$ transitive $(1,3)$ should be added to $R$. Also to maintain the symmetric property $(3,1)$ should then be added to $R$.

So, $R= \lbrace (1,1),(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)$, $(1,3),(3,1) \rbrace$ is an equivalence relation. So minimum 7 ordered pairs are to be added.

~~ 9. $R$ is reflexive as $(a, a) \in R$. $a+a=2 a$ is even.

$R$ is symmetric as $(a, b) \in R \Rightarrow(b, a) \in R$ as $a+b=b+a$ $=$ even (Commutative law)

$R$ is transitive as $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow(a+b)$ is even and $(b+c)$ is even

$\Rightarrow(a+b)+(b+c)$ is even

$\Rightarrow(a+2 b+c)$ is even $\Rightarrow(a+c)$ is even as ( $2 b$ is even)

$\Rightarrow(a, c) \in R$

$\therefore \quad R$ is an equivalence relation on the set of integers.

~~ 10. - There exists a real number -2 , such that $|-2|$ is not less than -2 as $2 \pm-2$.

Thus for all negative, real numbers $x,|x| \not x$.

Hence $(x, x) \notin R \forall$ real numbers.

Hence $R$ is not reflexive.

$R$ is not symmetric since there exist real numbers -2 and 3 such that $|-2| \leq 3$ but $|3| \leq-2$,

i.e., $(-2,3) \in R \nRightarrow(3,-2) \in R$.

$R$ is transitive since $\forall$ real numbers $a, b, c$

$(a, b) \in R,(b, c) \in R \Rightarrow|a| \leq b$ and $|b| \leq c$

$\Rightarrow|a| \leq b \leq|b| \leq c$

$(\because$ For all $x,|x| \leq x)$

$\Rightarrow|a| \leq c$

$\Rightarrow(a, c) \in R$

~~ 11. $R$ is reflexive as every natural number is divisible by itself. So $(a, a) \in R$.

$R$ is not symmetric as $(a, b) \in R$ does not imply $(b, a) \in R$, i.e., if $b$ is divisible by $a$, then $a$ is not divisible by $b$.

For example $(4,8) \in R$ as 8 is divisible by 4 , but $(8,4) \notin R$ as 4 is not divisible by 8 .

$R$ is transitive as $(a, b) \in R,(b, c) \in R \Rightarrow(a, c) \in R$

$(a, b) \in R \Rightarrow b$ is divisible by $a \Rightarrow b=m a$ where $m \in N$

$(b, c) \in R \Rightarrow c$ is divisible by $b \Rightarrow c=n b \quad n \in N$

$c=n \times m a \Rightarrow c=n m a$

$c$ is divisible by $a \Rightarrow(a, c) \in R$.

~~ 12. Every triangle is similar to itself, so $(T_1, T_1) \in R \Rightarrow R$ is reflexive.

$(T_1, T_2) \in R \Rightarrow T_1 \sim T_2 \Rightarrow T_2 \sim T_1, \Rightarrow(T_2, T_1) \in R \Rightarrow R$ is symmetric

$.\begin{matrix} (T_1, T_2) \in R \Rightarrow T_1 \sim T_2 \\ (T_2, T_3) \in R \Rightarrow T_2 \sim T_3\end{matrix} \rbrace \Rightarrow T_1 \sim T_3 \Rightarrow(T_1, T_3) \in R \Rightarrow R$ is transitive.

$\therefore \quad R$ is an equivalence relation.

~~ 13. $|a-a|=|0|=0$ so $(a, a) \notin R \Rightarrow R$ is not reflexive

$(a, b) \in R \Rightarrow|a-b|>0 \Rightarrow|b-a|>0 \Rightarrow(b, a) \in R$

$ (\because|a-b|=|b-a|) $

$\Rightarrow R$ is symmetric

$ (a, b) \in R \Rightarrow|a-b|>0 \text{ and }(b, c) \in R \Rightarrow|b-c|>0 $

$\forall$ real numbers $a, b, c$.

$\therefore \quad|a-b|>0$ and $|b-c|>0 \Rightarrow|a-c|>0$

$\Rightarrow(a, c) \in R \Rightarrow R$ is transitive.

~~ 14. $(a, a) \in R \Rightarrow 1+a . a=1+a^{2}>0 \forall$ real numbers $a$

$\Rightarrow R$ is reflexive

$(a, b) \in R \Rightarrow 1+a b>0 \Rightarrow 1+b a>0 \Rightarrow(b, a) \in R$

$\Rightarrow R$ is symmetric

We observe that $(1, \frac{1}{2}) \in R$ and $(\frac{1}{2},-1) \in R$

but $(1,-1) \notin R$ as $1+1 \times(-1)=0 \ngtr 0$

$\Rightarrow R$ is not transitive.

~~ 15. (a) Let $a, b, c \in A$ where $A$ is a set of real numbers.

Then $R= \lbrace (a, b): a \leq b, a, b \in A \rbrace$ is :

Reflexive: $a \leq a \Rightarrow(a, a) \in R$

(Yes)

Symmetric: $a \leq b \Rightarrow(a, b) \in R$, but

$a \leq b \Rightarrow a<b$ or $a=b$

$\Rightarrow b=a$ but $b \nless a$ so $(b, a) \notin R$

(No)

Transitive: $.\begin{matrix} (a, b) \in R \Rightarrow a \leq b \\ (b, c) \in R \Rightarrow b \leq c\end{matrix} \rbrace \Rightarrow a \leq c \Rightarrow(a, c) R$ (Yes)

(b) Let $A=$ set of positive integers. Then,

$R= \lbrace (a, b): a$ is $a$ multiple of $b, a, b \in R \rbrace$

Reflexive: Every positive integer is a multiple itself, so,

$ (a, a) \in R $

(Yes)

Symmetric: $(a, b) \in R \Rightarrow a$ is $a$ multiple of $b \nRightarrow b$ is a multiple of $a \Rightarrow(b, a) \notin R$

(No)

Transitive : $(a, b) \in R \Rightarrow a$ is a multiple of $b$

$\Rightarrow a=m b \forall m \in N$.

$(b, c) \in R \Rightarrow b$ is a multiple of $c \Rightarrow b=n c \forall n \in N$

$\therefore \quad a=m \times n c \Rightarrow a=m n c \forall m, n \in N$

$\Rightarrow a$ is a multiple of $c \Rightarrow(a, c) \in R$.

If $l_1 \perp l_2$ then $l_2 \perp l_1$, therefore given relation is symmetric
$l_1 \perp l_2$ and $l_2 \perp l_3 \nRightarrow l_1 \perp l_3$, so given relation is not transitive.

(d) - A person cannot be his own father, so relation is not reflexive.

If $a$ is father of $b$, then $b$ cannot be father of $a$, so relation is not symmetric.
If $a$ is father of $b, b$ is father of $c$, then $a$ cannot be father of $c$, so relation is not transitive.

$\therefore \quad$ From the given options (c) is only symmetric.

~~ 16. We can check the given properties as follows:

Reflexive: Let $(a, b) \in N \times N$. Then $(a, b) \in N$

$\Rightarrow a+b=b+a$ (Communtative law of Addition)

$\Rightarrow(a, b) R(b, a)$

$\Rightarrow \quad(a, b) R(a, b)$

$\Rightarrow R$ is reflexive.

Symmetric: Let $(a, b),(c, d) \in N \times N$ such that $(a, b) R(c, d)$. Then

$(a, b) R(c, d) \Rightarrow a+d=b+c \Rightarrow b+c=a+d$

$\Rightarrow c+b=a+d$ (By commutativity of addition on $N$ )

$\Rightarrow \quad(c, d) R(a, b)$

$\therefore \quad R$ is symmetric.

Transitive : Let $(a, b),(c, d),(e, f) \in N \times N$ such that $(a, b) R(c, d)$ and $(c, d) R(e, f)$. Then,

$(a, b) R(c, d) \Rightarrow a+d=b+c$

$(c, d) R(e, f) \Rightarrow c+f=d+e \rbrace$

$\Rightarrow(a+d)+(c+f)=(b+c)+(d+e)$

$\Rightarrow a+c=b+e \Rightarrow(a, b) R(e, f)$

$\therefore \quad(a, b) R(c, d)$ and $(c, d) R(e, f) \Rightarrow(a, b) R(e, f)$ on $N \times N$ so $R$ is transitive.

Hence $R$ is an equivalence relation on $N \times N$.

~~ 17. - For all $a \in I, a-a=0$, which is divisible by 11 .

Thus, $(a, a) \in R$ for all $a \in N \Rightarrow R$ is reflexive

Let $(a, b) \in R(a-b)$ is divisible by 11

$\Rightarrow-(a-b)$ is divisible by 11

$\Rightarrow(b-a)$ is divisible by 11

$\Rightarrow(b, a) \in R$

$\Rightarrow R$ is symmetric.

Let $(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow(x-y)$ is divisible by 11 and $(y-z)$ is divisible by 11

$\Rightarrow(x-y)+(y-z)$ is divisible by 11

$\Rightarrow(x-z)$ is divisible by 11

$\Rightarrow(x, z) \in R$

$\Rightarrow R$ is transitive

$\therefore \quad R$ is an equivalence relation.

~~ 18. $\cdot$ Let $x \in W$.

$(x, x) \in R$, since the words ’ $x$ ’ and ’ $x$ ’ have all letters in common

$\Rightarrow$ ’ $x$ ’ and ’ $x$ ’ have at least one letter in common

$\Rightarrow R$ is reflexive.

Let $x, y, z \in W$.

Then $(x, y) \in R \quad \Rightarrow$ ’ $x$ ’ and ’ $y$ ’ have at least one letter in common

$\Rightarrow$ ’ $y$ ’ and ’ $x$ ’ have at least one letters common

$\Rightarrow(y, x) \in R$

$\Rightarrow R$ is symmetric.

Let $x, y, z \in W$

Then $(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow$ ’ $x$ ’ and ’ $y$ ’ have at least one letter common and

’ $y$ ’ and ’ $z$ ’ have at least one letter common

which does not necessarily mean that ’ $x$ ’ and ’ $z$ ’ have at lest one letter common.

$\therefore \quad R$ is not transitive

For example, let $x=$ ‘AND’, $y=$ ‘NOT’, $z=$ ‘PET’

$x$ and $y$ have ’ $N$ ’ common

$y$ and $z$ have ’ $T$ ’ common

but $x$ and $z$ have no common letter.

~~ 19. - Let $a \in N$. Then

$(a, a) \notin R$ as the GCD of ’ $a$ ’ and ’ $a$ ’ is ’ $a$ ’ not 2 .

$R$ is not reflexive

Let $a, b \in N$. Then,

$(a, b) \notin R \Rightarrow$ GCD of ’ $a$ ’ and ’ $b$ ’ is 2

$\Rightarrow$ GCD of ’ $b$ ’ and ’ $a$ ’ is 2

$\Rightarrow(b, a) \in R$

$\therefore \quad \boldsymbol{R}$ is symmetric

$\quad$ Let $a, b, c \in N$. Then,

$(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow$ GCD of $a$ and $b$ is 2 and GCD of $b$ and $c$ is 2

$\Rightarrow GCD$ of $a$ and $c$ is 2

$R$ is not transitive

For example, let $a=4, b=10, c=12$

$GCD$ of $(4,10)=2$

$GCD$ of $(10,12)=2$

But $GCD$ of $(4,12)=4$.

~~ 20. $R= \lbrace (x, y): x y>0 x, y \in N \rbrace$

$\quad x, x \in N \Rightarrow x^{2}>0 \Rightarrow R$ is reflexive
$\quad x, y \in N$ and $(x, y) \in R \Rightarrow x y>0$

$\Rightarrow y x>0(y, x) \in R$

$\Rightarrow R$ is symmetric

$\quad x, y, z \in N$ and $(x, y) \in R,(y, z) \in R$

$\Rightarrow x y>0$ and $y z>0$

$\Rightarrow x z>0 \Rightarrow(x, z) \in R$

$\Rightarrow R$ is transitive.

$\therefore \quad R$ is an equivalence relation.

~~ 21. Given, $a R b=a=2^{k} . b$ for some integer.

Reflexive: $a R a \Rightarrow a=2^{0}$. $a$ for $k=0$ (an integer). True Symmetric: $a R b a=2^{k} b \Rightarrow b=2^{-k}$. $a$

$\Rightarrow b R a$ as $k,-k$ are both integers. True

Transitive: $a R b \Rightarrow a=2^{k_1} b$

$ b R c \Rightarrow b=2^{k_2} c $

$\therefore \quad a=2^{k_1} \cdot 2^{k_2 c}=2^{k_1}+k_2 c, k_1+k_2$ is an integer.

$\therefore \quad a R b, b R c \Rightarrow a R c$ True

$\therefore \quad R$ is an equivalence relation.

~~ 22. $R= \lbrace (x, y) \mid x, y \in R, x=w y, w$ is a rational number $ \rbrace$

Reflexive: $x R x \Rightarrow x=w x \Rightarrow w=1$, (a rational number)

Hence $R$ is reflexive.

Symmetric: $x R y \nRightarrow y R x$ as

$0 R 1 \Rightarrow 0=(0) .1$ where 0 is a rational number but $1 R 0 \Rightarrow 1=(w) 0$ which is not true for any rational number.

$\therefore R$ is not an equivalence relation.

$S= \lbrace .(\frac{m}{n}, \frac{p}{q}) \rvert, m, n, p, q \in I, n, q \neq 0.$ and $.q m=p n \rbrace$.

Reflexive $\frac{m}{n} S \frac{m}{n} \Rightarrow m n=n m$ (True)

Symmetric $\frac{m}{n} S \frac{p}{q} \Rightarrow m q=p n \Rightarrow p n=m q \Rightarrow \frac{p}{q} S \frac{m}{n}$

Transitive $\frac{m}{n} S \frac{p}{q}$ and $\frac{p}{q} S \frac{r}{s}$

(True)

$ \Rightarrow m q=p n \text{ and } p s=q r $

$ \begin{aligned} & \Rightarrow m q \cdot p s=p n \cdot q r \Rightarrow m s=n r \\ & \Rightarrow \frac{m}{n}=\frac{r}{s} \Rightarrow \frac{m}{n} S \frac{r}{s} \text{ (True) } \end{aligned} $

$\therefore \quad S$ is an equivalence relation.

~~ 23. Given, $ \lbrace (x, y): 2 x-y=10 \rbrace$

Reflexive, $x R x=2 x-x=10 \Rightarrow x=10 \Rightarrow y=10$

$\therefore$ Point $(10,10) \in N \Rightarrow R$ is reflexive.

SELF ASSESSMENT SHEET

Let $A= \lbrace 1,2,3,4 \rbrace, B= \lbrace 5,6,7,8 \rbrace$. Then $R= \lbrace (1,5)$, $(1,7),(2,6) \rbrace$ is a relation from set $A$ to $B$ defined as :

(a) $R= \lbrace (a, b): a, b$ are odd $ \rbrace$

(b) $R= \lbrace (a, b): a, b$ are even $ \rbrace$

(d) $R= \lbrace (a, b): b / a$ is odd $ \rbrace$

~~ 2. The range of the relation $R$ defined by $R= \lbrace (x+1, x+5)$ $: x \in \lbrace 0,1,2,3,4,5 \rbrace \rbrace$ is

(a) $ \lbrace 1,2,3,4,5,6 \rbrace$

(b) $ \lbrace 5,6,7,8,9,10 \rbrace$

(d) $ \lbrace 0,1,2,3,4,5 \rbrace$

~~ 3. The relation “is parallel to” on a set $S$ of all straight lines in a plane is :

(a) Symmetric only

(b) Reflexive and Transitive only

(d) An equivalence relation

~~ 4. Let $R$ be the relation in the set $ \lbrace 1,2,3,4 \rbrace$ given by

$R= \lbrace (1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2) \rbrace$. Then $R$ is

(a) Reflexive and symmetric but not transitive

(b) Reflexive and transitive but not symmetric

(d) An equivalence relations

~~ 5. Let $A= \lbrace 2,3,5,6 \rbrace$. Then, which of the following relations is transitive only?

(a) $R= \lbrace (2,3),(6,6) \rbrace$ ~~ 25. Given, $a R b \Rightarrow \sin ^{2} a+\cos ^{2} b=1$

Reflexive: $a R a \Rightarrow \sin ^{2} a+\cos ^{2} a=1 \forall a \in R$ (True)

Symmetric: $a R b \sin ^{2} a+\cos ^{2} b=1$

$ \begin{aligned} & \Rightarrow 1-\cos ^{2} a+1-\sin ^{2} b=1 \\ & \Rightarrow \sin ^{2} b+\cos ^{2} a=1 \\ & \Rightarrow b R a \forall a, b \in R \text{ (True) } \end{aligned} $

Transitive: $a R a$ and $b R c$

$ \begin{aligned} \Rightarrow & \sin ^{2} a+\cos ^{2} b=1 \text{ and } \sin ^{2} b+\cos ^{2} c=1 \\ & \text{ Adding these two equations we get } \\ & \sin ^{2} a+\cos ^{2} b+\sin ^{2} b+\cos ^{2} c=2 \\ \Rightarrow & \sin ^{2} a+\cos ^{2} c=1 \Rightarrow a R c \text{ (True) } \end{aligned} $

$ \therefore \quad \text{ Adding these two equations we get } $

$\therefore \quad R$ is an equivalence relation. (b) $R= \lbrace (2,2),(3,3),(5,5),(6,6),(2,3),(3,5),(2,6) \rbrace$.

(d) $R= \lbrace (2,3),(3,5),(2,5) \rbrace$.

~~ 6. Let $A= \lbrace (2,5,11) \rbrace, B= \lbrace 3,6,10 \rbrace$ and $R$ be a relation from $A$ to $B$ defined by $R= \lbrace (a, b): a$ and $b$ are co-prime $ \rbrace$. Then $R$ is

(a) $ \lbrace (2,3),(2,6),(5,10),(5,6) \rbrace$

(b) $ \lbrace (2,6),(2,10),(5,10) \rbrace$

(d) $ \lbrace (2,10),(5,3),(5,6),(11,10) \rbrace$.

~~ 7. If $R$ is a relation on a finite set $A$ having $n$ elements, then the number of relations on $A$ is

(a) $n^{2}$

(b) $2^{n}$

(d) $2^{n^{2}}$

(AMU 2006)

~~ 8. The relation ‘is less than’ on a set of natural numbers is

(a) Only reflexive

(b) Only symmetric

(d) An equivalence relation.

~~ 9. Let $R$ be $a$ relation on the set of all real numbers $R$ $R= \lbrace (a, b) \in R \times R: a^{2}+b^{2}=1 \rbrace$, then $R$ is

(a) Equivalence

(b) Only transitive

(d) None of these

~~ 10. Let $A= \lbrace 1,2,3 \rbrace$ and $R= \lbrace (1,2),(1,1),(2,3) \rbrace$ be relation on $A$.

What minimum number of ordered pairs must be added with the elements of $R$ so that it may become transitive?

(a) 1

(b) 2

(d) 3

ANSWERS

$(d)$

~~ 2. (b)

~~ 3. $(d)$

~~ 4. $(b)$

~~ 5. $(d)$

~~ 6. (c)

~~ 7. $(d)$

~~ 8. (c)

~~ 9. (c)

~~ 10. (a)

HINTS AND SOLUTIONS

- Since $(2,6) \in R$, the relation " $a$ and $b$ are odd” does not exist.

Since $(1,5)$ and $(1,7) \in R$, the relation " $a$ and $b$ are even” does not exist.
None of the ordered pairs in $R$ are prime numbers.
$5 / 1=5,7 / 1=7,6 / 2=3$, quotients being all odd numbers, the relation $b / a$ is odd exists.

~~ 2. The range of the given relation is defined by the second element, i.e., $(x+5)$ in the ordered pair $(x+1, x+5)$ defining the relation. $\because x \in \lbrace 0,1,2,3,4,5 \rbrace$, therefore Range $= \lbrace 0+5,1+5$,

$2+5,3+5,4+5,5+5 \rbrace= \lbrace 5,6,7,8,9,10 \rbrace$.

~~ 3. Let $R= \lbrace (x, y):$ line $x$ is parallel to line $y, x y \in$ set of coplanar straight lines $ \rbrace$.

Every line is parallel to itself. So, if $x \in S$, then $(x, x) \in R$

$\Rightarrow R$ is reflexive

If $(x, y) \in R \Rightarrow x|y \Rightarrow y| x \Rightarrow(y, x) \in R$

$\Rightarrow R$ is symmetric

$(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow x | y$ and $y | z$

$\Rightarrow x | z \Rightarrow(x, z) \in R$

$\Rightarrow R$ is transitive

$\therefore \quad R$ being reflexive, symmetric and transitive, it is an equivalence relation.

~~ 4. Let $A= \lbrace 1,2,3,4 \rbrace$

$\because(1,1),(2,2),(3,3)$ and $(4,4) \in R \Rightarrow R$ is reflexive
$\because(1,2) \in R$ but $(2,1) \notin R ;(1,3) \in R$ and $(3,1) \notin R$; $(3,2) \in R$ and $(2,3) \notin R \Rightarrow R$ is not symmetric
$(1,3) \in R$ and $(3,2) \in R$ and $(1,2) \in R$

$\Rightarrow R$ is transitive.

~~ 7. Set $A$ has $n$ elements $\Rightarrow n(A)=n$

$\Rightarrow A \times A$ has $n \times n=n^{2}$ elements

$\therefore$ Number of relations on $A=$ Number of subsets of $A \times A=\mathbf{2}^{n^{2}}$

~~ 8. Let $N$ be the set of natural numbers. Then

$R= \lbrace (a, b): a<b, a, b \in N \rbrace$ $A$ natural number is not less than itself

$\Rightarrow(a, a) \notin R$ where $a \in N$

$\Rightarrow R$ is not reflexive

$\forall a, b \in N,(a, b) \in R \Rightarrow a<b \nRightarrow b<a \Rightarrow(b, a) \notin R$

$\Rightarrow R$ is not symmetric.

$\forall a, b, c \in N,(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow a<b$ and $b<c$

$\Rightarrow a<c(a, c) \in R$

$\Rightarrow R$ is transitive.

~~ 9. $\bullet \forall a \in R, a^{2}+a^{2} \neq 1 \Rightarrow(a, a) \notin R \Rightarrow R$ is not reflexive.

For example, $0^{2}+0^{2}=0,1^{2}+1^{2}=1,(\frac{1}{2})^{2}+(\frac{1}{2})^{2}=\frac{1}{2}$ and so on.

$\forall a, b \in R,(a, b) \in R \Rightarrow a^{2}+b^{2}=1 \Rightarrow b^{2}+a^{2}=1$

$\Rightarrow(b, a) \in R$

$\Rightarrow R$ is symmetric

$\forall a, b, c \in R,(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow a^{2}+b^{2}=1$ and $b^{2}+c^{2}=1$ which does not necessarily mean

$a^{2}+c^{2}=1 \Rightarrow(a, c) \notin R$

For example, let $a=0, b=1, c=0$

$0^{2}+1^{2}=1$ and $1^{2}+0^{2}=1$

But $\mathbf{0}^{\mathbf{2}}+\mathbf{0}^{\mathbf{2}} \neq \mathbf{1}$.

~~ 10. For the relation $R$ to become transitive:

$(1,2) \in R$ and $(2,3) \in R$ should imply $(1,3) \in R$

$\therefore$ Minimum one ordered pair $(1,3)$ should be added to $R$.

Plane Geometry -Triangles

KEY FACTS

I. Definitions

A triangle is a three sided closed figure formed by three non-collinear points.

The three points $P, Q$ and $R$ in the given figure are called the vertices, line segments joining the three vertices, i.e., $P Q, Q R$ and $P R$ are called the sides and $\angle P, \angle Q$ and $\angle R$ are the

interior angles of the triangle.

If the sides of a triangle are produced as shown in the given diagram, then the angles $\angle P R C, \angle Q R D, \angle P F Q, \angle R Q E, \angle Q P A$ and $\angle R P B$ are the exterior angles of $\triangle A B C$.

II. Types of Triangles:

a. By sides:

Scalene Triangle $a \neq b \neq c$
(All the sides are unequal)

b. By angles:

Acute Angled Triangle	Right Angled Triangle
All angles are acute, i.e., $\angle A<\angle 90^{\circ}, \angle B<90^{\circ}$, $\angle C<90^{\circ}$	One of the angles is a right angle. The other two are complementary to each other	One of the angles is an obtuse angle.

III. Some Important Properties of Triangles:

a. The sum of the three interior angles of a triangle is always $180^{\circ}$, i.e., $\angle B A C+\angle A B C+\angle B C A=180^{\circ}$.

b. (i) If the sides of a triangle are produced in order then, the sum of the three (ordered) exterior angles of a triangle is $360^{\circ}$, i.e., in both the figures, $\angle 1+\angle 2+$ $\angle 3=360^{\circ}$

(ii) Themeasure of an exteriorangle is equalto the sum of the measures of the interior opposite angles, i.e., in figure (ii) $\angle \mathbf{3}=\angle \mathbf{4}+\angle \mathbf{5}$.

(iii) The measure of an exterior angle is greater than the measure of each of the interior opposite angles, i.e., in figure (ii) $\angle 3>\angle 4$ and $\angle 3>\angle 5$.

(i)

(iv) The sum of the measure of exterior angle at a vertex and its adjacent interior angle is $180^{\circ}$.

IV. Triangle Inequalities:

(i) Sum of any two sides of a triangle is always greater than the third side.

(ii) The difference of any two sides is always less than the third side.

(iii) If two sides of a triangle are not equal, then the angle opposite to the greater side is greater and vice versa.

(iv) Let $a, b, c$ be the three sides of a triangle $\triangle A B C$ where $A B=c$ is the longest side (say). Then,

if $c^{2}<a^{2}+b^{2}$, then the triangle is acute angled.
if $c^{2}=a^{2}+b^{2}$, then the triangle is right angled.

if $c^{2}>a^{2}+b^{2}$, then the triangle is obtuse angled.

V. Sine Rule: In a $\triangle A B C$, if $a, b, c$ be the three sides opposite to the angles $A, B$ and $C$ respectively, then

$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} $

VI. Cosine Rule: In a $\triangle A B C$, if $a, b, c$ be the sides opposite to the angles $A, B$ and $C$ respectively, then

$ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}, \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} $

VII. Some Important Definitions:

(i) Altitude of a triangle is the perpendicular drawn from a vertex to the opposite side (produced if necessary). Every triangle has three altitudes.

$A D$ is the altitude on side $B C$.

Orthocentre is the point of intersection of the three altitudes of a triangle. Hence, $O$ is the orthocentre of $\triangle A B C$, where $\angle B O C=180^{\circ}-\angle A, \angle A O B=180^{\circ}-\angle C, \angle C O A=180^{\circ}-\angle B$. (ii) Median is the straight line segment joining the mid-point of any side to the opposite vertex. Every triangle has three medians and a median bisects the area of a $\Delta$.

VIII. Important Theorems on Triangles

Basic Proportionality Theorem (BPT): Any line parallel to one side of a triangles divides the other two sides proportionally.

Thus, if $S T$ is drawn parallel to side $Q R$ of $\triangle P Q R$, then

$ \frac{P S}{S Q}=\frac{P T}{T R} \quad \text{ or } \frac{P S}{P Q}=\frac{P T}{P R} $

Converse of Mid-point Theorem: The straight line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Thus, a line drawn through $D$, the mid-point of side $A B$ of $\triangle A B C$, parallel to $B C$ bisects $A C$, i.e., $E$ is the mid-points of $A C$, i.e., $A E=E C$.

~~ 3. Apollonius Theorem: In a triangle, the sum of the squares of any two sides of a triangle is equal to twice the sum of the square of the median to the third side and square of half the third side,

i.e., $A B^{2}+A C^{2}=2(A D^{2}+(\frac{B C}{2})^{2})=2(A D^{2}+B D^{2})=2(A D^{2}+C D^{2})$.

~~ 4. Interior Angle Bisector Theorem: In a triangle, the angle bisector of an angle divides the opposite side to the angle in the ratio of the remaining two sides.

Thus, if $A D$ is the internal bisector of $\angle A$ of $\triangle A B C$, then $\frac{B D}{C D}=\frac{A B}{A C}$ and $B D \times A C=D C \times A B=A D^{2}$

~~ 5. External Angle Bisector Theorem: In a triangle, the angle bisector of any exterior angle of a triangle divides the side opposite to the external angle in the ratio of the remaining two sides, i.e., If $C E$ is the bisector of external angle $A C D$, then $\frac{B E}{A E}=\frac{B C}{A C}$.

$\therefore \quad Area(\triangle A B D)=Area(\triangle A C D)=\frac{1}{2} Area(\triangle A B C)$.

Centroid is point of intersection or point of concurrence of the three medians of a triangle. Also it divides each median in the ratio 2:1 (vertex : base) $G$ is the centroid of $\triangle A B C$, i.e., the point of concurrence of medians $A D, B E$ and $C F$. Also,

$ \frac{A G}{G D}=\frac{B G}{G E}=\frac{C G}{G F}=\frac{2}{1} $

~~ 6. Angle Bisector: A line segment from a vertex of triangle which bisects the angle at the vertex is called the angle bisector. $P D, Q E$ and $R F$ are angle bisectors of $\angle P, \angle Q$ and $\angle R$ respectively of $\triangle P Q R$ such that $\angle Q P D=\angle R P D, \angle P Q E=\angle R Q E$ and $\angle P R F=\angle Q R F$.

Incentre is the point of intersection of the angle bisectors of a triangle and it is equidistant from all the three sides of the triangle, i.e., perpendicular distance

between the side and incentre is equal for all the three sides.

~~ 7. Perpendicular Bisector: A line segment bisecting a side of a triangle at right angles is called a perpendicular bisector. The point of concurrence of the perpendicular bisectors is called the circumcentre. Here $O$ is the circumcentre of $\triangle A B C$. The circumcentre is equidistant from the vertices of a triangle, i.e., $O A=O B=O C$.

Note: The orthocentre, centroid, incentre and circumcentre coincide in case of an equilateral triangle.

~~ 8. Pythagoras’ Theorem: The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

Here $\triangle A B C$ is right angled at $C$. So $A C$ is the hypotenuse. Hence, according to Pythagoras, Theorem $A C^{2}=A B^{2}+B C^{2}$.

Converse of Pythagoreas’ Theorem: If the square on one side of a triangle is equal

to the sum of the squares on the other two sides, the triangle is right angled.

If $a, b, c$, be the sides opposite to $\angle A, \angle B$ and $\angle C$ respectively of a $\triangle A B C$ and $c^{2}=a^{2}+b^{2}$, then $\triangle A B C$ is right angled at $C$.

Such numbers or triplets $a, b, c$ which satisfy the condition $a^{2}+b^{2}=c^{2}$ are called Pythagorean triplets. Some examples of Pythagorean triplets are: $(3,4,5) ;(5,12,13),(7,24,25)$, etc.

~~ 9. $45^{\circ}-45^{\circ}-90^{\circ}$ triangle: If the angles of a triangle are $45^{\circ}, 45^{\circ}$ and $90^{\circ}$, then the hypotenuse (the longest side) is $\sqrt{2}$ times any smaller side, i.e.,

In $\triangle A B C, \angle B=90^{\circ}, \angle A=\angle C=45^{\circ}$, then

$A B=B C$ and $A C=\sqrt{2} A B$ or $\sqrt{2} B C$.

The converse of the above theorem: If the ratio of the sides of a triangle is $1: 1: \sqrt{2}$, then it is a $45^{\circ}-45^{\circ}$ $-90^{\circ}$ triangle.

~~ 10. $30^{\circ}-60^{\circ}-90^{\circ}$ triangle: If the angles of a triangle are $30^{\circ}, 60^{\circ}$ and $90^{\circ}$, then the side opposite to the $30^{\circ}$ angle is half of the hypotenuse and side opposite to $60^{\circ}$ is $\frac{\sqrt{3}}{2}$ times the hypotenuse, i.e., in $\triangle A B C$, where $\angle B=90^{\circ}$ and $\angle C=30^{\circ}, \angle A=60^{\circ}$, then

$A B=\frac{1}{2} A C$ and $BC=\frac{\sqrt{3}}{2} A C$, then

i.e., the ratio of the sides is $1: \sqrt{3}: 2$. The converse of the above also holds true.

IX. Congruency of Triangles: Two triangles are congruent to each other

(i) if each of the three sides and three angles of one triangle are equal to the respective sides and angles of the other.

$\triangle A B C \cong \triangle P Q R \text{ if } \\ $

$A B =P Q \angle A=\angle P \\ $

$A C =P R \text{ and } \angle B=\angle Q \\ $

$B C =Q R \angle C=\angle R $

Tests of Congruency:

(i) SAS Axiom (Side-angle-side): If the two sides and the included angle of one triangle are respectively equal to the two sides and the included angle of the other, the triangles are congruent.

$ \begin{aligned} & A B=P Q \\ & A C=P R \\ & B C=Q R \end{aligned} $

Note: The two equal sides must be opposite to angles which are known to be equal.

(ii) ASA or AAS Axiom (Two angles, corresponding side): If two angles and one side of a triangle are respectively equal to two angles and the corresponding side of the other triangles, the triangles are congruent. The side may be in included side.

$ \begin{aligned} & \angle X=\angle D \\ & \angle Z=\angle E \quad \therefore \triangle X Y Z \cong \triangle D E F(AAS) \\ & X Y=D F \end{aligned} $

(iii) SSS Axiom (Three sides): If three sides of one triangle are respectively equal to the corresponding three sides of the other triangle, the triangles are congruent :

$ .\begin{matrix} A B=P Q \\ A C=P R \\ B C=Q R \end{matrix} \rbrace \quad \Rightarrow \quad \triangle A B C \cong \triangle P Q R $

(iv) RHS Axiom (Right angle-Hypotenuse-side): If the hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and corresponding side of the other right angled triangle, the two triangles are congruent.

$ \therefore \quad A C=P R, B C=Q R \text{ and } \angle B=\angle Q=90^{\circ} \Rightarrow \triangle A B C \cong P Q R $

X. Similarity of Triangles:

Two triangles are said to be similar, if their corresponding angles are equal, and their corresponding sides are proportional.

Thus, $\triangle A_1 B_1 C_1$ is similar to $\triangle A_2 B_2 C_2$ or $\triangle A_1 B_1 C_1 \sim \Delta A_2 B_2 C_2$ if

(i) $\angle A_1=\angle A_2 ; \angle B_1=\angle B_2 ; \angle C_1=\angle C_2$

(ii) $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$

Tests for Similarity:

(i) A-A axiom of similarily: If two angles of one triangle are equal to the corresponding two angles of the other triangle, then the triangles are said to be similar.

Note : If two pairs of corresponding angles in two triangles are equal, then the third pair will obviously be equal.

$\angle A B C=\angle D E F, \angle A C B=\angle D F E \Rightarrow \triangle A B C \sim \triangle D E F$. (AA similarity)

(ii) S-A-S axiom of similarity: If two triangles have a pair of corresponding angles equal and the sides including them proportional, then the triangles are similar.

In $\triangle s A B C$ and $D E F, \angle A=\angle D, \frac{A B}{D E}=\frac{A C}{D F}=k$, then $\triangle A B C \sim \triangle D E F$.

(iii) S-S-S axiom of similarity: If two triangles have their pairs of corresponding sides proportional, then the triangles are similar.

If $\frac{B C}{E F}=\frac{A B}{D E}=\frac{A C}{D F}$, then, $\triangle A B C \sim \triangle D E F$

XI. Properties of Similar Triangles:

For a pair of similar triangles,

Ratio of sides $=$ Ratio of Altitudes

$ \begin{aligned} & =\text{ Ratio of Medians } \\ & =\text{ Ratio of angle bisectors } \\ & =\text{ Ratio of in-radii } \\ & =\text{ Ratio of circums-radii } . \end{aligned} $

~~ 2. Ratio of areas = Ratio of squares of corresponding sides.

$ \frac{\text{ Area }(\triangle A B C)}{\text{ Area }(\triangle D E F)}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}=\frac{B C^{2}}{E F^{2}} $

~~ 3. In a right angled triangle, the triangles on each side of the altitude drawn from the vertex to the right angle to the hypotenuse are similar to the original triangle and to each other too.

i.e., $\triangle B C A \sim \triangle B D C \sim \triangle C D A$.

XII. Some Useful Results for a Triangle:

In a $\triangle A B C$, if the bisectors of $\angle A B C$ and $\angle A C B$ meet at $O$, then $\angle B \boldsymbol{O C}=\mathbf{9 0}^{\circ}+\frac{\mathbf{1}}{\mathbf{2}} \angle \boldsymbol{A}$.

~~ 2. In a $\triangle A B C$, if the sides $A B$ and $A C$ are produced to $D$ and $E$ respectively and the bisectors of $\angle D B C$ and $\angle E C B$ intersect at $O$, then $\angle B \boldsymbol{O C}=\mathbf{9 0}^{\circ}-\frac{\angle \boldsymbol{A}}{\mathbf{2}}$.

~~ 3. In a $\triangle A B C$, if $A D$ is the angle bisector of $\angle B A C$ and $A E \perp B C$, then $\angle \boldsymbol{D} \boldsymbol{A E}$ $=\frac{1}{2} \angle A B C-\angle A C B$.

~~ 4. In an acute angled triangle, $A B C, A D$ is the perpendicular dropped on $B C$, then $A C^{2}=A B^{2}+B C^{2}-2 B D . D C$.

~~ 5. In an obtuse angled triangle, if $A D$ is the perpendicular dropped on $B C$ produced, then $A C^{2}=A B^{2}+B C^{2}+2 B D . D C$.

~~ 6. In a right angled $\Delta$, the median to the hypotenuse is half the hypotenuse.

$B N$ is the median from $B$ on $A C$ such that $A N=N C$.

Then, $B N=\frac{1}{\mathbf{2}} \boldsymbol{A C}=\boldsymbol{A N}=\boldsymbol{N C}$.

~~ 7. If in a right angled $\triangle P Q R, \angle Q=90^{\circ}, P R$ is the hypotenuse, and a perpendicular $Q S$ is dropped on the hypotenuse from the right angle vertex $Q$, then (i) $\boldsymbol{Q S}=\frac{\boldsymbol{P Q} \times \boldsymbol{Q R}}{\boldsymbol{P R}}$ (ii) $\frac{1}{Q S^{2}}=\frac{1}{P Q^{2}}+\frac{1}{Q R^{2}}$

XIII. Area Formulae for a Triangle:

General Formula: Area of a triangle $=\frac{1}{2} \times$ base $\times$ height

~~ 2. Area of a scalene triangle $=\sqrt{\boldsymbol{s}(\boldsymbol{s}-\boldsymbol{a})(\boldsymbol{s}-\boldsymbol{b})(\boldsymbol{s}-\boldsymbol{c})}$, where $a, b, c$ are the sides of the triangle and $s$ is the semi-perimeter of the triangle, i.e., $s=\frac{a+b+c}{2}$.

Also, Area of $a=r \times s=\frac{a b c}{4 \boldsymbol{R}}$, where $r \to$ in radius, $R \to$ circumradius.

~~ 3. Area of a right angled triangle $=\frac{1}{2} \times$ base $\times$ height.

For a right angled triangle,

(i) Inradius $=\frac{\boldsymbol{A} \boldsymbol{B}+\boldsymbol{B} \boldsymbol{C}-\boldsymbol{A} \boldsymbol{C}}{\mathbf{2}}$, where $\triangle A B C$ is $rt \angle d$ at $\angle B$

(ii) Inradius $=\frac{\text{ Area }}{\text{ Semi-perimeter }}$

(iii) Circum radius $=\frac{\text{ Hypotenuse }}{2}$

~~ 4. Area of an isosceles triangle $=\frac{b}{4} \sqrt{4 a^{2}-b^{2}}$, where $a$ is the length of one of the equal sides, $b$ is the length of third side.

~~ 5. For an equilateral triangle,

Area $=\frac{\sqrt{3}}{4}(\text{ side })^{2}, \quad$ Inradius $=\frac{\text{ Side }}{2 \sqrt{3}}, \quad$ Circumradius $=\frac{\text{ Side }}{\sqrt{3}}$

$\therefore$ Circumradius $=\mathbf{2} \times$ Inradius.

Note: (i) For the given perimeter of a triangle, the area of an equilateral triangle is maximum.

(ii) For the given area of a triangle, the perimeter of an equilateral triangle is minimum.

~~ 6. Two triangles on equal (or same bases and lying between same parallel lines have equal area.)

SOLVED EXAMPLES

~~ Ex. 1 . Let $O$ be any point inside a triangle $A B C$. Let $L, M$ and $N$ be the points on $A B, B C$ and $C A$ respectively, where perpendicular from $O$ meet these lines. Show that :

$A L^{2}+B M^{2}+C N^{2}=A N^{2}+C M^{2}+B L^{2}$

Sol. Join $O$ to $A, B$ and $C$

In right $\triangle S O A L, O B M$ and $O C N$.

$O L^{2}+A L^{2}=O A^{2}$

$O M^{2}+B M^{2}=O B^{2}$

$O N^{2}+C N^{2}=O C^{2}$

$(\begin{matrix} \text{ Pythagoras’ } \\ \text{ Theoram }\end{matrix} )$

$\therefore$ Adding (i), (ii) and (iii), we get $O L^{2}+A L^{2}+O M^{2}+B M^{2}+O N^{2}+C N^{2}=O A^{2}+O B^{2}+O C^{2}$

$A L^{2}+B M^{2}+C N^{2}=(O A^{2}+O B^{2}+O C^{2})-(O L^{2}+O M^{2}+O N^{2})$

Similarly in right $\triangle s O A N, O B L$ and $O M C$,

$O N^{2}+A N^{2}=O A^{2}$

$O L^{2}+B L^{2}=O B^{2}$

$O M^{2}+C M^{2}=O C^{2}$

$\therefore \quad$ Adding $(i v),(v)$ and $(v i)$

$O N^{2}+A N^{2}+O L^{2}+B L^{2}+O M^{2}+C M^{2}=O A^{2}+O B^{2}+O C^{2}$

$\Rightarrow A N^{2}+B L^{2}+C M^{2}=(O A^{2}+O B^{2}+O C^{2})-(O N^{2}+O L^{2}+O M^{2})$

From (1) and (2)

$A L^{2}+B M^{2}+C N^{2}=A N^{2}+B L^{2}+C M^{2}$

~~ Ex. 2 . A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. Prove that the sum of these perpendiculars is independent of the location of the given point.

Sol. Let $P$ be any point in the equilateral $\triangle A B C$ with each side $=S$ units.

Let $P E, P F$ and $P G$ be the lengths of the perpendiculars from $P$ on $A B, A C$ and $B C$ respectively. Then.

Area of $\triangle A B C=Area(\triangle P A B)+Area(\triangle P B C)+Area(\triangle P A C)$

$ \begin{aligned} & =\frac{1}{2} \times A B \times P E+\frac{1}{2} \times A C \times P F+\frac{1}{2} \times B C \times P G \\ & =\frac{1}{2} \times S \times(P E+P F+P G) \end{aligned} $

where $A B=B C=C A=S$

Also, let $h$ be the height be $\triangle A B C$, then

Area of $\triangle A B C=\frac{1}{2} \times S \times h$

$\therefore \quad \frac{1}{2} \times S \times h=\frac{1}{2} \times S \times(P E+P F+P G)$

$\Rightarrow \quad h=(P E+P F+P G)$

$\Rightarrow$ Sum of the perpendiculars $=$ height of equilateral $\Delta=$ a constant

$\Rightarrow$ Sum of perpendiculars is independent of location of $P$.

~~ Ex. 3 . In a $\triangle A B C$, the lengths of sides $B C, C A$ and $A B$ are $a, b$ and $c$ respectively. Median $A D$ drawn from $A$ is perpendicular to $B C$. Express $b$ in terms of $a$ and $c$.

Sol. By Appolonius theorem, in $\triangle A B C$,

$ \begin{aligned} & A B^{2}+A C^{2}=2(A D^{2}+B D^{2}) \\ & c^{2}+b^{2}=2(c^{2}-\frac{a^{2}}{4}+\frac{a^{2}}{4})(\because \text{ In rt. } \triangle A B D, A D=\sqrt{A B^{2}-B D^{2}}) \\ & \Rightarrow b^{2}+c^{2}=a^{2}-2 c^{2} \\ & \Rightarrow b^{2}=a^{2}-3 c^{2} \\ & \Rightarrow b=\sqrt{a^{2}-3 c^{2}} \end{aligned} $

~~ Ex. 4 . If the medians $A D, B E$ and $C F$ of $\triangle A B C$ meet at $G$, prove that $G$ is the centroid of $\triangle D E F$ also.

Sol. Since $D$ and $E$ are the mid-points of sides $B C$ and $A C$ of $\triangle A B C$, therefore, $D E | B A$ and $D E=\frac{1}{2} B A$

(By mid-point theorem)

$\Rightarrow D E | F A$ and $D E=F A$

$(\because F A=\frac{1}{2} B A)$

Also, $D F | A C$ and $D F=\frac{1}{2} A C$

(By mid-point theorem)

$\Rightarrow D F | A E$ and $D F=A E$.

$\therefore \quad D E A F$ is a parallelogram, whose diagonals $A D$ and $F E$ intersect at $P$.

Since the diagonals of a parallelogram bisect each other, therefore, $A P=P D$ and $F P=P E$

$\Rightarrow P$ is the mid-point of $F E$

$\Rightarrow D P$ is the median of $\triangle D E F$.

Similarly it can be shown that, $F D E C$ is a parallelogram and hence $R$ is the mid-point of $E D$ and hence $F R$ is the median of $\triangle D E F$.

$\therefore \quad$ Medians $D P$ and $F R$ intersect at point $G$, where $G$ is the centroid of $\triangle D E F$.

~~ Ex. 5 . $A B C$ is an acute angled triangle. $C D$ is the altitude through $C$. If $A B=8$ units, $C D=6$ units, find the distance between the mid-points of $A D$ and $B C$.

Sol. Let $E$ be the mid-point of $A D$ and $F$ the mid-point of $B C$.

Draw $F R \perp A B$.

$\because C D \perp A B \Rightarrow F R | C D$.

Also, $F$ being the mid-point of $B C$ and $F R | C D \Rightarrow R$ is the mid-point of $B D$

(By converse of mid-point theorem).

Now by basic proportionality theorem, $F R=\frac{1}{2} C D=F R=3$ units.

Also, $E R=E D+D R=\frac{1}{2}(A D+D B)=\frac{1}{2} \times A B=4$ units.

$\therefore \quad$ In $\triangle F E R, E F=\sqrt{F R^{2}+E R^{2}}=\sqrt{9+16}=\sqrt{25}=\mathbf{5}$ units.

~~ Ex. 6 . In the given figure, $\angle M O N=\angle M P O=\angle N Q O=90^{\circ}$ and $O Q$ is the bisector of $\angle M O N$ and $Q N=10$, $O R=40 / 7$. Find $O P$.

(CDS 2012)

Sol. In $\triangle O M P, \angle M O P=45^{\circ}$

(OQbisects $\angle M O N$ )

$\therefore \angle O M P=45^{\circ}$.

$\Rightarrow O P=P M=x$ (say)

(sides opp. equal $\angle s$ are equal)

Also in $O Q N, \angle Q O N=45^{\circ}$ ( $O Q$ bisects $\angle M O N$ )

$\Rightarrow \angle Q N O=45^{\circ} \Rightarrow O Q=O N=10$ (sides opp. equal $\angle s$ are equal)

$\therefore \quad Q R=O Q-O R=10-\frac{40}{7}=\frac{30}{7}$.

$\triangle P M R \sim \triangle Q N S$, since $\angle M P R=\angle N Q R=90^{\circ}$ and $\angle M R P=\angle Q R P($ vert. opp. $\angle s)$

$\Rightarrow \frac{P M}{P R}=\frac{Q N}{Q R} \Rightarrow \frac{x}{\frac{40}{7}-x}=\frac{10}{\frac{30}{7}}=\frac{7}{3}$

$\Rightarrow \frac{7 x}{40-7 x}=\frac{7}{3} \Rightarrow 21 x=280-49 x \Rightarrow x=4$.

~~ Ex. 7 . In the figure $C D, A E$ and $B F$ are one-third of their respective sides. It is given that $A N_2: N_2 N_1: N_1 D$ $=3: 3: 1$ and similarly for the lines $B E$ and $C F$. Show that the area of $\triangle N_1 N_2 N_3$ is $\frac{1}{7} Area(\triangle A B C)$.

Sol. $Ar(\triangle N_1 N_2 N_3)=Area(\triangle A B C)-[(Area(\triangle C B F)+Area(\triangle A B E)+Area(\triangle A D C)]$

$ +(Area(\Delta N_1 C D)+Area(N_2 A E)+Area(\Delta N_3 F B)) $

$\because A E=\frac{1}{3} A C$ and height of $\triangle A B E$ and $\triangle A B C$ is same, keeping $A C$ as the base, Area of $\triangle A B E=\frac{1}{3} Area(\triangle A B C) \quad(.$ Area $.=\frac{1}{2} \times b \times h)$

Similarly, Area $(\triangle C B F)=Area(\triangle A D C)=\frac{1}{3} Area(\triangle A B C)$

In $\triangle C B F, F N_3: N_3 N_1: N_1 C=3: 3: 1 \Rightarrow$ Area of $\triangle N_1 C D=\frac{1}{7}$ Area $(\triangle B F C)$

$\Rightarrow$ Area $(\triangle N_1 C D)=\frac{1}{7} \times \frac{1}{3} Area(\triangle A B C)=\frac{1}{21} Area(\triangle A B C)$

Similarly, Area $(\Delta N_2 A E)=Area(\triangle N_3 F B)=\frac{1}{21} Area(\triangle A B C)$

$\therefore \quad$ Area $(\triangle N_1 N_2 N_3)=Area(\triangle A B C)-3 \cdot \frac{1}{3}$ Area $(\triangle A B C)+3 \cdot \frac{1}{21}$ Area $(\triangle A B C)=\frac{1}{7}$ Area $(\triangle \boldsymbol{A B C})$

~~ Ex. 8 . In the diagram $A B$ and $A C$ are the equal sides of an isosceles triangle $A B C$, in which is inscribed equilateral triangle $D E F$. Designate angle $B F D$ by $a$, angle $A D E$ by $b$ and angle $F E C$ by $c$. Then show that $a=\frac{b+c}{2}$.

Sol. For $\triangle B D F, \quad$ ext $\angle A D F=\angle B+a$

$\Rightarrow b+60^{\circ}=\angle B+a$

Similarly $a+60^{\circ}=c+\angle C$

$\therefore \quad Eq(i)-Eq(i i) \Rightarrow b-a=a-c+\angle B-\angle C$

$\because \quad A B=A C \Rightarrow \angle B=\angle C$ (Isosceles $\triangle$ property)

$\therefore \quad b-a=a-c \Rightarrow b+c=2 a \Rightarrow a=\frac{\boldsymbol{b}+\boldsymbol{c}}{\mathbf{2}}$.

~~ Ex. 9 . In the given figure, line $D E$ is parallel to line $A B . C D=3$ while $D A=6$. Which of the following must be true? I. $\triangle C D E \sim \triangle C A B$ II. $\frac{\text{ Area of } \triangle C D E}{\text{ Area of } \triangle C A B}=(\frac{C D}{C A})^{2}$ III. If $A B=4$, then $D E=2$

Sol. I. Since $D E | A B, \angle C D E=\angle C A B$

$\angle C E D=\angle C B A$

Corresponding angles

$\therefore \quad \triangle C D E \sim \triangle C A B$ (AA similarly)

II. Since ratio of the areas of similar triangles is the square of the ratio of the corresponding sides.

$ \therefore \quad \frac{Area(\triangle C D E)}{Area(\triangle C A B)}=(\frac{C D}{C A})^{2} $

III. $C A=C D+D A=3+6=9$.

$\therefore \triangle C D E \sim \triangle C A B \Rightarrow \frac{C D}{C A}=\frac{D E}{A B}$

$\Rightarrow \frac{3}{9}=\frac{D E}{4} \Rightarrow D E=\frac{12}{9}=\frac{4}{3}$

$\therefore \quad$ If $A B=4$, then $D E=\frac{4}{3}$

$\therefore \quad$ I and II are true and III is false.

~~ Ex. 10 . If the angles of a triangle are in the ratio $1: 2: 3$, then find the ratio of the corresponding opposite sides.

Sol. If the angles are in the ratio $1: 2: 3$, then the angles of the triangle are $30^{\circ}, 60^{\circ}, 90^{\circ}$. Therefore, the triangle is a right angled triangle.

The side ratios opposite to the angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ is $B C: A B: A C$, i.e., $A C \sin 30^{\circ} ; A C \sin 60^{\circ}: A C$,

i.e., $-\frac{1}{2} \times A C: \frac{\sqrt{3}}{2} \times A C: A C$, i.e., $\mathbf{1}: \sqrt{\mathbf{3}}: \mathbf{2}$.

~~ Ex. 11 . The angles of a triangle are in the ratio $8: 3: 1$. What is the ratio of the longest side of the triangle to the next longest side?

Sol. The angles are $\frac{8}{12} \times 180^{\circ}, \frac{3}{12} \times 180^{\circ}, \frac{1}{12} \times 180^{\circ}$ i.e., $120^{\circ}, 45^{\circ}$ and $15^{\circ}$.

Also we know that the longest side is opposite the greatest angle and so on.

$\therefore \quad$ Let the longest side opposite the greatest angle $120^{\circ}$ be $x$ and let the next longest side opposite angle $45^{\circ}$ be $y$. Then, by the sine rule

$ \begin{aligned} \frac{\sin 120^{\circ}}{x} & =\frac{\sin 45^{\circ}}{y} \\ \Rightarrow \quad \frac{x}{y} & =\frac{\sin 120^{\circ}}{\sin 45^{\circ}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{6}}{2} . \end{aligned} $

~~ Ex. 12 . In $\triangle A B C, a=2 x, b=3 x+2, c=\sqrt{12}$ and $\angle c=60^{\circ}$. Find $x$.

Sol. Here we use the law of cosines. So $c^{2}=a^{2}+b^{2}-2 a b \cos C$

$\Rightarrow \quad 12=(2 x)^{2}+(3 x+2)^{2}-2(2 x)(3 x+2) \cos 60^{\circ}$

$\Rightarrow \quad 12=4 x^{2}+9 x^{2}+12 x+4-(12 x^{2}+8 x) \times \frac{1}{2}$

$\Rightarrow 7 x^{2}+8 x-8=0$

$x=\frac{-8 \pm \sqrt{64+224}}{14}=\frac{-8 \pm \sqrt{288}}{14}(.$ Recall that roots of a quadratic eq. $a x^{2}+b x+c$ are $.x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a})$.

Since the side of a triangle must be positive, therefore,

$ x=\frac{-8+\sqrt{288}}{14} \simeq \mathbf{0 . 6 4} $

~~ Ex. 13 . The bisectors of the angles of a triangle $A B C$ meet $B C, C A$ and $A B$ at $X, Y$ and $Z$ respectively. Prove that $\frac{B X}{X C} \cdot \frac{C Y}{Y A} \cdot \frac{A Z}{Z B}=1$.

Sol. We make use of the bisector theorem here, i.e., the bisector (internal or external) of an angle of triangle divides the opposite side in the ratio of the sides containing the angle.

$A X$ bisects $\angle A \Rightarrow \frac{B X}{X C}=\frac{A B}{A C}$

$B Y$ bisects $\angle B \Rightarrow \frac{C Y}{Y A}=\frac{B C}{B A}$

$C Z$ bisects $\angle C \Rightarrow \frac{A Z}{Z B}=\frac{C A}{C B}$

$\therefore \quad$ Multiplying $(i),(i i)$ and (iii), we get

$ \frac{B X}{X C} \cdot \frac{C Y}{Y A} \cdot \frac{A Z}{Z B}=\frac{A B}{A C} \cdot \frac{B C}{B A} \cdot \frac{C A}{C B}=\mathbf{1} $

~~ Ex. 14 . As given in the diagram alongside, $P$ is any point on $A B, P S \perp B D$ and $P R \perp A C . A F \perp B D$ and $P Q \perp A F$. Then show that $P R+P S$ is equal to $A F$.

Sol. $\triangle P T R \sim \triangle A T Q$ (AA similarity)

$ \begin{aligned} & (\because \angle A T Q=\angle P T R(\text{ vert. opp. } \angle s) \text{ and } \angle P R T=\angle A Q T=90^{\circ}) \\ & \therefore \quad \frac{P R}{A Q}=\frac{P T}{A T} \end{aligned} $

Also, diagonals of a rectangle are equal and bisect each other

$\Rightarrow E A=E B \Rightarrow \angle E A B=\angle E B A \Rightarrow \angle T A P=\angle P B S$

$\therefore$ From $(i)$ and (ii), $\angle T P A=\angle T A P \Rightarrow P T=A T$ (Isosceles triangle property)

$\Rightarrow P R=A Q$

Also, $P Q | B D$ and $P S \perp B D$ and $Q F \perp B D \Rightarrow P S=Q F$

$\therefore \quad P R+P S=A Q+Q F=A F$.

PRACTICE SHEET

In the given figure, what is the sum of the angles formed around $A, B$, $C$ except at angles of $\triangle A B C$.

(a) $360^{\circ}$

(b) $720^{\circ}$

(d) $1000^{\circ}$

(CDS 2010)

~~ 2. In the given figure, $\angle B=\angle C=55^{\circ}$ and $\angle D=25^{\circ}$. Then,

(a) $B C<C A<C D$

(b) $B C>C A>C D$

(d) $B C>C A, C A<C D$

~~ 3. In-centre of a triangle lies in the interior of:

(a) An isosceles triangle only

(b) Any triangle

(d) A right triangle only

~~ 4. $G$ is the centroid of $\triangle A B C$ with height $h$ units. If a line $D E$ parallel to $B C$ cuts $\triangle A B C$ at a height $h / 4$ from $B C$, find the distance $G G^{\prime}$ in terms of $A G$ if $G^{\prime}$ is the centroid of $\triangle A D E$.

(a) $\frac{1}{2} A G$

(b) $\frac{3}{4} A G$

(d) $\frac{2}{3} A G$

~~ 5. In an equilateral triangle if $a, b$ and $c$ denote the lengths of the perpendicular from $A, B$ and $C$ respectively on the opposite sides, then

(a) $a>b>c$

(b) $a>b<c$

(d) $a=b=c$

~~ 6. In the given figure, $L M$ is parallel to $Q R$. If $L M$ divides the $\triangle P Q R$ such that the area of trapezium $L M R Q$ is two times the area of $\triangle P L M$, then what is $\frac{P L}{L Q}$ equal to?

(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{\sqrt{3}}$

(d) $\frac{1}{3}$

(CDS 2011)

~~ 7. If the medians of two equilateral triangles are in the ratio $3: 2$, then what is the ratio of their sides?

(a) $1: 1$

(b) $2: 3$

(d) $\sqrt{3}: \sqrt{2}$

(CDS 2009)

~~ 8. If $A$ is the area of the right angled triangle and $b$ is one of the sides containing the right angle, then what is the length of the altitude on the hypotenuse?

(a) $\frac{2 A b}{\sqrt{b^{4}+4 A^{2}}}$

(b) $\frac{2 A^{2} b}{\sqrt{b^{4}+4 A^{2}}}$

(d) $\frac{2 A^{2} b^{2}}{\sqrt{b^{4}+A^{2}}}$

(CDS 2008)

~~ 9. In the given figure, $A B C$ is an equilateral triangle of side length $30 cm . X Y$ is parallel to $B C, X P$ is parallel to $A C$ and $Y Q$ is parallel to $A B$. If $(X Y+X P+Y Q)$ is $40 cm$, then what is $P Q$ equal to?

(a) $5 cm$

(b) $12 cm$

(d) None of these

~~ 10. If $A D$ is the median of $\triangle A B C$, then

(a) $A B^{2}+A C^{2}=2 A D^{2}+2 B D^{2}$

(b) $A B^{2}+A C^{2}=2 A D^{2}+B D^{2}$

(d) $A B^{2}+A C^{2}=A D^{2}+2 B D^{2}$

~~ 11. Let $A B C$ be a triangle of area $16 cm^{2} . X Y$ is drawn parallel to $B C$ dividing $A B$ in the ratio $3: 5$. If $B Y$ is joined, then the area of triangle $B X Y$ is

(a) $3.5 cm^{2}$

(b) $3.7 cm^{2}$

(d) $4.0 cm^{2}$

~~ 12. From a point $O$ in the interior of a $\triangle A B C$ if perpendiculars $O D, O E$ and $O F$ are drawn to the sides $B C, C A$ and $A B$ respectively, then which of the following statements is true?

(a) $A F^{2}+B D^{2}+C E^{2}=A E^{2}+C D^{2}+B F^{2}$

(b) $A B^{2}+B C^{2}=A C^{2}$

(d) $A F^{2}+B D^{2}+C E^{2}=O D^{2}+O E^{2}+O F^{2}$

~~ 13. In an equilateral triangle $A B C$, the side $B C$ is trisected at $D$. Then $A D^{2}$ is equal to

(a) $\frac{9}{7} A B^{2}$

(b) $\frac{7}{9} A B^{2}$

(d) $\frac{4}{5} A B^{2}$

~~ 14. In a $\triangle A B C, A B=10 cm, B C=12 cm$ and $A C=14 cm$. Find the length of median $A D$. If $G$ is the centroid, find the length of $G A$.

(a) $\frac{5}{3} \sqrt{7}, \frac{5}{9} \sqrt{7}$

(b) $5 \sqrt{7}, 4 \sqrt{7}$

(d) $4 \sqrt{7}, \frac{8}{3} \sqrt{7}$

~~ 15. If $a, b, c$ are the sides of a triangle and $a^{2}+b^{2}+c^{2}=b c$ $+c a+a b$, then the triangle is:

(a) equilateral

(b) isosceles

(d) obtuse angled

(CAT 2000)

~~ 16. $H$ is the orthocentre of $\triangle A B C$ whose altitudes are $A D, B E$ and $C F$. Then the orthocentre of $\triangle H B C$ is

(a) $F$

(b) $E$

(d) $D$

~~ 17. In the given figure, $\angle B C A=120^{\circ}$ and $A B=c, B C=a$, $A C=b$. Then

(a) $c^{2}=a^{2}+b^{2}+b a$

(b) $c^{2}=a^{2}+b^{2}-b a$

(d) $c^{2}=a^{2}+b^{2}+2 a b$

~~ 18. In the $\triangle A B C, A B=2 cm, B C=3 cm$ and $A C=4 cm$. $D$ is the middle-point of $A C$. If a square is constructed on the side $B D$, what is the area of the square?

(a) $4.5 cm^{2}$

(b) $2.5 cm^{2}$

(d) None of these

(CDS 2009)

~~ 19. In the given figure, (not drawn to scale), $P$ is a point on $A B$ such that $A P: P B=4: 3 . P Q$ is parallel to $A C$ and $Q D$ is parallel to $C P$. In $\triangle A R C, \angle A R C=90^{\circ}$ and in $\triangle P Q S$, $\angle P S Q=90^{\circ}$. The length of $Q S=6 cm$. What is the ratio of $A P: P D$ ?

(a) $10: 3$

(b) $2: 1$

(d) $8: 3$

(CAT 2003)

~~ 20. In $\triangle L M N, L O$ is the median. Also $L O$ is the bisector of $\angle M L N$. If $L O=3 cm$, and $L M=5 cm$, then find the area of $\triangle L M N$.

(a) $12 cm^{2}$

(b) $10 cm^{2}$

(d) $6 cm^{2}$

(CAT 2009)

~~ 21. A point within an equilateral triangle whose perimeter is 30 $m$ is $2 m$ from one side and $3 m$ from another side. Find its distance from third side.

(a) $\sqrt{5}-3$

(b) $5 \sqrt{3}-5$

(d) $5 \sqrt{3}-3$

~~ 22. A city has a park shaped as a right angled triangle. The length of the longest side of this park is $80 m$. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.

(a) $4000 m$

(b) $4800 m$

(d) $6400 m$

(XAT 2012)

~~ 23. In a triangle $A B C, A D$ is the angle bisector of $\angle B A C$ and $\angle B A D=60^{\circ}$. What is the length of $A D$ ?

(a) $\frac{b+c}{b c}$

(b) $\frac{b c}{b+c}$

(d) $\frac{(b+c)^{2}}{b c}$

~~ 24. In a $\triangle A B C$, the internal bisector of angle $A$ meets $B C$ at $D$. If $A B=4, A C=3$ and $\angle A=60^{\circ}$, then the length of $A D$ is

(a) $2 \sqrt{3}$

(b) $\frac{12 \sqrt{3}}{7}$

(d) $\frac{6 \sqrt{3}}{7}$

(CAT 2002)

~~ 25. Suppose the medians $P P^{\prime}$ and $Q Q^{\prime}$ of $\triangle P Q R$ intersect at right angles. If $Q R=3$ and $P R=4$, then the length of side $P Q$ is

(a) $\sqrt{3}$

(b) $\sqrt{2}$

(d) $\sqrt{6}$

~~ 26. In the given triangle $A B C$, the length of sides $A B$ and $A C$ is same (i.e., $b=c$ ) and $60^{\circ}<A<90^{\circ}$. Then

(a) $b<a<b \sqrt{3}$

(b) $c<a<c \sqrt{2}$

(d) $\frac{c}{3}<a<3 c$

~~ 27. In the given figure, $P$ and $Q$ are the mid-points of $A C$ and $A B$. Also, $P G$ $=G R$ and $H Q=H R$. What is the ratio of the Area of $\triangle P Q R$ : Area of $\triangle A B C$

(a) $1: 2$

(b) $2: 3$

(d) $3: 5$

~~ 28. The lengths of the sides $a, b, c$ of a $\triangle A B C$ are connected by the relation $a^{2}+b^{2}=5 c^{2}$. The angle between medians drawn to the sides ’ $a$ ’ and ’ $b$ ’ is

(a) $60^{\circ}$

(b) $45^{\circ}$

(d) None of these

~~ 29. $A B C$ is a triangle with $\angle B A C=60^{\circ}$. A point $P$ lies on onethird of the way from $B$ to $C$ and $A P$ bisects $\angle B A C . \angle A P C$ equals

(a) $90^{\circ}$

(b) $45^{\circ}$

(d) $120^{\circ}$

(XAT 2007)

~~ 30. A rectangle inscribed in a triangle has its base coinciding with the base $\boldsymbol{b}$ of the triangle. If the altitude of the triangle is $\boldsymbol{h}$, and the altitude $\boldsymbol{x}$ of the rectangle is half the base of the rectangle, then

(a) $x=\frac{1}{2} h$

(b) $x=\frac{b h}{h+b}$

(d) $x=\sqrt{\frac{h b}{2}}$

(FMS 2011)

~~ 31. In the given figure $A B=B C$

$=C D=D E=E F=F G$

$=G A$. Then $\angle D A E$ is approximately:

(a) $15^{\circ}$

(b) $20^{\circ}$

(d) $25^{\circ}$

(CAT 2000)

~~ 32. If $A B C$ is a triangle in which $\angle B=2 \angle C$. $D$ is a point on side $B C$ such that $A D$ bisects $\angle B A C$ and $A D=C D$, then $\angle B A C=$

(a) $62^{\circ}$

(b) $72^{\circ}$

(d) $84^{\circ}$

~~ 33. In the figure shown here, $Q S=S R, Q U=S U, P W=W S$ and $S T | R V$. What is the value of $\frac{\text{ Area of } \triangle P S X}{\text{ Area of } \triangle P Q R}$ ?

(a) $\frac{1}{5}$

(b) $\frac{1}{3}$

(d) $\frac{1}{7}$

~~ 34. In a $\triangle A B C, A D, B E$ and $C F$ are the medians drawn from the vertices $A, B$ and $C$ respectively. Then study the following statements and choose the correct option. I. $3(A B+B C+A C)>2(A D+B E+C F)$

II. $3(A B+B C+A C)<2(A D+B E+C F)$

III. $3(A B+B C+A C)<4(A D+B E+C F)$

IV. $3(A B+B C+A C)>4(A D+B E+C F)$

(a) I and IV are true

(b) I and III are true

(d) II and III are true

~~ 35. In a $\triangle A B C, A B=A C . P$ and $Q$ are points on $A C$ and $A B$ respectively such that $C B=B P=P Q=Q A$. Then $\angle A Q P=$

(a) $\frac{2 \pi}{7}$

(b) $3 \pi$

(d) $\frac{4 \pi}{7}$

~~ 36. In a triangle, the ratio of the distance between a vertex and the orthocentre and the distance of the circumcentre from the side opposite the vertex is

(a) $3: 1$

(b) $4: 1$

(d) $\sqrt{2}: 1$

~~ 37. In a $\triangle A B C$, angle $A$ is twice angle $B$. Then,

(a) $a^{2}=b(a+c)$

(b) $a^{2}=\sqrt{b c}$

(d) $a^{2}=b+c$

~~ 38. Let $A B C$ be a triangle. Let $D, E, F$ be points respectively on segments $B C, C A, A B$ such that $A D, B E$ and $C F$ concur at point $K$. Suppose $B D / D C=B F / F A$ and $\angle A D B=\angle A F C$, then

(a) $\angle A B E=\angle C A D$

(b) $\angle A B E=\angle A F C$

(d) $\angle A B E=\angle B C F$

(RMO 2011)

~~ 39. Let $A B C$ be a triangle in which $A B=A C$ and let $I$ be its in-centre. Suppose $B C=A B+A I$. $\angle B A C$ equals.

(a) $45^{\circ}$

(b) $90^{\circ}$

(d) $75^{\circ}$

(RMO 2009)

~~ 40. In a $\triangle A B C$, let $D$ be the mid-point of $B C$. If $\angle A D B=45^{\circ}$ and $\angle A C D=30^{\circ}$. then $\angle B A D$ equals.

(a) $45^{\circ}$

(b) $60^{\circ}$

(d) $15^{\circ}$

(RMO 2005)

ANSWERS

1. (c)	2. $(d)$	3. $(b)$	4. (c)	5. $(d)$	6. $(b)$	7. $(c)$	8. (a)	9. (d)	10. (a)
11. $(c)$	12. $(a)$	13. $(b)$	14. $(d)$	15. (a)	16. (c)	17. (a)	18. (b)	19. (c)	20. (a)
21. (b)	22. (c)	23. $(b)$	24. (b)	25. (c)	26. (b)	27. (a)	28. (c)	29. (d)	30. (c)
31. $(d)$	32. (b)	33. (a)	34. $(b)$	35. (c)	36. $(c)$	37. (c)	38. (a)	39. $(b)$	40. $(c)$

HINTS AND SOLUTIONS

~~ 1. $\begin{aligned} \angle A & =360^{\circ}-\text{ Ext. } \angle A \\ \angle B & =360^{\circ}-\text{ Ext. } \angle B \\ \angle C & =360^{\circ}-\text{ Ext. } \angle C \end{aligned} $

We know,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow 360^{\circ}-$ Ext. $\angle A+360^{\circ}-$

Ext. $\angle B+360^{\circ}-$ Ext. $\angle C=180^{\circ}$

$\Rightarrow$ Ext. $\angle A+$ Ext. $\angle B+$ Ext. $\angle C=1080^{\circ}-180^{\circ}=\mathbf{9 0 0}^{\circ}$.

~~ 2. $\angle B=\angle C=55^{\circ} \Rightarrow A C=A B$

$\Rightarrow \angle B A C=180^{\circ}-(\angle B+\angle C)=180^{\circ}-110^{\circ}=70^{\circ}$

Now in $\triangle A C D, \angle A C D=180^{\circ}-55^{\circ}=125^{\circ}$

$\therefore \angle C A D=180^{\circ}-(125^{\circ}+25^{\circ})=30^{\circ}$

$\Rightarrow \angle C A D>\angle C D A \Rightarrow C D>A C$

$(\because$ In a given triangle, the greater angle has greater side opposite to it)

Also $\angle B A C>\angle A B C \Rightarrow B C>A C$

$\therefore \boldsymbol{B C}>\boldsymbol{C A}$ and $\boldsymbol{C A}<\boldsymbol{C D}$.

~~ 4. $A F$ is the median to $B C$ from $A$ in $\triangle A B C$. $G$ is the centroid of $\triangle A B C$

$\Rightarrow A G=\frac{2}{3} A F \Rightarrow A F=\frac{3}{2} A G$.

Now $A H$ is the median to $D E$ from $A$ in $\triangle A D E . G^{\prime}$ is the centroid of $\triangle A D E$

$\Rightarrow A G^{\prime}=\frac{2}{3} A H$

Since $D E | B C, A H=\frac{3}{4} A F$

(By basic proportionality theorem)

$\therefore A G^{\prime}=\frac{2}{3} \times \frac{3}{4} A F=\frac{1}{2} A F=\frac{1}{2} \times \frac{3}{2} A G=\frac{3}{4} A G$

$\therefore G G^{\prime}=A G-A G^{\prime}=A G-\frac{3}{4} A G=\frac{\mathbf{1}}{\mathbf{4}} \boldsymbol{A} \boldsymbol{G}$.

~~ 6. $\because L M | Q R$

$.\begin{matrix} \angle P L M=\angle P Q R \\ \angle P M L=\angle P R Q\end{matrix} \rbrace$ Corresponding $\angle s$

$\Rightarrow \triangle P Q R \sim \triangle P L M$

$\therefore \frac{\text{ Area of } \triangle P L M}{\text{ Area of } \triangle P Q R}=\frac{P L^{2}}{P Q^{2}}$

Area of $\triangle P Q R=$ Area of $\triangle P L M+$ Area of trap. $L M R Q$

Given, Area of trapezium $L M R Q=2$ Area of $\triangle P L M$

$\therefore$ Area of $\triangle P Q R=$ Area of $\triangle P L M+2$ Area of $\triangle P L M$

$=3$ Area of $\triangle P L M$

$\therefore$ From $(i)$, we have $\frac{\text{ Area of } \triangle P L M}{3 \times \text{ Area of } \triangle P L M}=\frac{P L^{2}}{P Q^{2}}$

$\Rightarrow \frac{P L^{2}}{P Q^{2}}=\frac{1}{3} \Rightarrow \frac{P L}{P Q}=\frac{1}{\sqrt{\mathbf{3}}}$.

~~ 7. Median of an equilateral triangle $=\frac{\sqrt{3}}{2} \times$ side

Let the sides of the two equilateral triangles be $a_1$ and $a_2$ respected. Then,

$\frac{\frac{\sqrt{3}}{2} a_1}{\frac{\sqrt{3}}{2} a_2}=\frac{3}{2} \Rightarrow \frac{a_1}{a_2}=\frac{\mathbf{3}}{\mathbf{2}}$.

~~ 8. Let $A B C$ be the given right angled triangle, right angled at $B$. Let $B C=b$.

Then, Area of $\triangle A B C=\frac{1}{2} \times B C \times A B$

$\Rightarrow A=\frac{1}{2} \times b \times A B \Rightarrow A B=\frac{2 A}{b}$

In $\triangle A B C, A C^{2}=A B^{2}+B C^{2}$

$ A C^{2}=\frac{4 A^{2}}{b^{2}}+b^{2} $

$\Rightarrow A C=\sqrt{\frac{4 A^{2}}{b^{2}}+b^{2}}$

Again in $\triangle A B C$,

$ A=\frac{1}{2} \times A C \times B D $

$\Rightarrow A=\frac{1}{2} \times \sqrt{\frac{4 A^{2}+b^{4}}{b^{2}}} \times B D$

$\Rightarrow B D=\frac{\mathbf{2} \boldsymbol{A} \boldsymbol{b}}{\sqrt{\mathbf{4} \boldsymbol{A}^{2}+\boldsymbol{b}^{\mathbf{4}}}}$.

~~ 9. $A B | Y Q \Rightarrow \angle X B P=\angle Y Q C$

(corresponding angles) and

$X P | A C \Rightarrow \angle X P B=\angle Y C Q$

$\triangle A B C$ being an equilateral triangle,

$\angle B=\angle C=60^{\circ}$

$\Rightarrow \angle X P B=\angle Y Q C=60^{\circ}$

$\Rightarrow \angle X P B=\angle Q Y C=60^{\circ}$

$\Rightarrow \triangle X B P$ and $\triangle Y Q C$ are equilateral triangles.

Now, $X Y | B C \Rightarrow \frac{A X}{A B}=\frac{X Y}{B C} \Rightarrow A X=X Y \quad(\because A B=B C)$

Also, $X Y+X P+Y Q=40 \Rightarrow A X+X B+Y Q=40$

$(\because A X=X Y, X P=X B)$

$\Rightarrow A B+Y Q=40 \Rightarrow Y Q=40-A B=40-30=10$

$\therefore X P=Y Q=10 cm$.

$\Rightarrow B P=Q C=10 cm(\triangle X B P$ and $\triangle Y Q C$ are equilateral)

$\Rightarrow P Q=B C-(B P+Q C)=30-10-10=\mathbf{1 0} \mathbf{~ c m}$.

~~ 10. Let $A D$ be the median of $\triangle A B C$

$\therefore B D=D C$

Let $A E$ be the perpendicular from $A$ on $B C$.

Then,

In rt. $\triangle A B E$,

$ \begin{aligned} A B^{2} & =B E^{2}+A E^{2} \\ & =(B D-E D)^{2}+A E^{2} \\ & =B D^{2}+E D^{2}+A E^{2}-2 B D \cdot E D \\ & =B D^{2}+A D^{2}-2 B D \cdot E D \end{aligned} $

rt. $\triangle A C E$,

$ \begin{aligned} A C^{2} & =A E^{2}+E C^{2} \\ & =A E^{2}+(E D+D C)^{2} \\ & =A E^{2}+E D^{2}+D C^{2}+2 E D \cdot D C \\ & =A D^{2}+D C^{2}+2 E D \cdot D C \\ & =A D^{2}+B D^{2}+2 E D \cdot B D \end{aligned} $

$ (\because B D=D C) $

Adding (i) and (ii), we get

$A B^{2}+A C^{2}=2 A D^{2}+2 B D^{2}$.

~~ 11. Let the areas of $\triangle A X Y$ and $\triangle B X Y$ be $A_1$ and $A_2$ respectively. The heights of the $\triangle A X Y$ and $\triangle B X Y$ are equal, so $\frac{A_1}{A_2}=\frac{\frac{1}{2} \times A X \times \text{ height }}{\frac{1}{2} \times B X \times \text{ height }}$

$\Rightarrow \frac{A_1}{A_2}=\frac{A X}{B X}=\frac{3}{5}$

Also, $\triangle A X Y \sim \triangle A B C$

$\Rightarrow \frac{\text{ Area of } \triangle A X Y}{\text{ Area of } \triangle A B C}=(\frac{A X}{A B})^{2}$

$\Rightarrow \frac{A_1}{16}=(\frac{3 x}{3 x+5 x})^{2} \Rightarrow \frac{A_1}{16}=\frac{9}{64}$

$\Rightarrow A_1=\frac{9}{64} \times 16=\frac{9}{4}$

$\therefore$ From (i) and (ii) $A_2=\frac{5}{3} \times A_1=\frac{5}{3} \times \frac{9}{4}=\frac{15}{4}=\mathbf{3 . 7 5} \mathbf{c m}^{2}$.

~~ 12. Join $O A, O B$ and $O C$.

By Pythagoras, theorem,

In $\triangle A O F, A F^{2}=A O^{2}-O F^{2}$

In $\triangle B O D, B D^{2}=B O^{2}-O D^{2}$

In $\triangle C O E, C E^{2}=C O^{2}-O E^{2}$

Adding (i), (ii) and (iii), we get

$A F^{2}+B D^{2}+C E^{2}=A O^{2}-O F^{2}+B O^{2}-O D^{2}+C O^{2}-O E^{2}$

$ \begin{aligned} & =\mathscr{O O}^{2}-O E^{2}+B O^{2}-O F^{2}+C O^{2}-O D^{2} \\ & =\boldsymbol{A \boldsymbol { E } ^ { 2 } + \boldsymbol { B F } ^ { 2 }}+\boldsymbol{C D}^{\mathbf{2}} . \end{aligned} $

~~ 13. $A B C$ being an equilateral triangle, $A B=B C=A C$.

Also $B C$ being trisected at $D$

$\Rightarrow B D=\frac{1}{3} B C$ Let $A F$ be drawn perpendicular to $B C \Rightarrow B F=F C$.

Also $E$ is given as the other point of trisection, so $B D=D E=E C$ Also, $B D=2 D F$

Now $A B^{2}=B F^{2}+A F^{2}$ and

$ A D^{2}=D F^{2}+A F^{2} $

Now $A B^{2}=B F^{2}+A F^{2}$

$ \begin{aligned} & \Rightarrow A B^{2}=(\frac{1}{2} B C)^{2}+A F^{2} \\ & \Rightarrow A B^{2}=\frac{1}{4} B C^{2}+A F^{2}(\because B C=A B) \\ & \Rightarrow A B^{2}=\frac{1}{4} A B^{2}+A F^{2} \Rightarrow A F^{2}=\frac{3}{4} A B^{2} \end{aligned} $

Also, $A D^{2}=D F^{2}+A F^{2}$

$ \begin{aligned} = & (\frac{1}{2} \times \frac{1}{3} B C)^{2}+A F^{2} \quad(\text{ From }(i) \text{ and }(i i)) \\ = & (\frac{1}{6} B C)^{2}+A F^{2} \\ = & \frac{1}{36} B C^{2}+A F^{2} \\ = & \frac{1}{36} A B^{2}+\frac{3}{4} A B^{2} \\ & (\because B C=A B \text{ and putting the value from (iii)) } \\ = & \frac{28}{36} A B^{2} \Rightarrow A D^{2}=\frac{7}{9} A B^{2} . \end{aligned} $

~~ 14. Applying the Apollosnius theorem,

$A B^{2}+A C^{2}=2(B D^{2}+A D^{2})$

$\Rightarrow 100+196=2(36+A D^{2})$

$\Rightarrow 2 A D^{2}=296-72=224$

$\Rightarrow A D^{2}=112 \Rightarrow A D=4 \sqrt{7}$

As $G$ is the centroid, so $\frac{A G}{G D}=\frac{2}{1}$

$\Rightarrow A G=\frac{2}{3} A D=\frac{2}{3} \times 4 \sqrt{7}=\frac{8}{3} \sqrt{7}$.

~~ 15. $a^{2}+b^{2}+c^{2}=a b+b c+c a$

$\Rightarrow a^{2}+b^{2}+c^{2}-a b-b c-c a=0$

$\Rightarrow 2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a=0$

$\Rightarrow(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$

Sum of perfect squares $=0 \Rightarrow$ Each term of the sum is zero

$\Rightarrow(a-b)=0=(b-c)=(c-a)$

$\Rightarrow a=b=c$

$\Rightarrow$ The triangle is equilateral.

~~ 16. $H$ is the orthocentre of $\triangle A B C$

$\Rightarrow A D \perp B C, B E \perp C A, C F \perp A B$

In $\triangle B H C$,

$A D \perp B C \Rightarrow H D \perp B C$

$C F \perp A B \Rightarrow B F \perp H C$ (Produced)

$\Rightarrow$ The altitudes $H D$ and $B F$ of $\triangle H B C$ intersect in $A$.

$\Rightarrow A$ is the orthocentre of $\triangle H B C$.

~~ 17. By the cosine rule, we have,

$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$\Rightarrow \cos 120^{\circ}=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$\Rightarrow-\frac{1}{2}=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$\Rightarrow c^{2}=a^{2}+b^{2}+a b$.

~~ 18. In $\triangle A B C$,

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$

$=\frac{A C^{2}+A B^{2}-B C^{2}}{2 \times A C \times A B}$

$=\frac{4^{2}+2^{2}-3^{2}}{2 \times 4 \times 2}$

(a)

$=\frac{16+4-9}{16}=\frac{11}{16}$

In $\triangle B A D$,

$\cos A=\frac{A D^{2}+A B^{2}-B D^{2}}{2 \times A D \times A B}$

$\Rightarrow \frac{11}{16}=\frac{4+4-B D^{2}}{2 \times 2 \times 2}=\frac{8-B D^{2}}{8}$

$\Rightarrow 11=16-2 B D^{2} \Rightarrow 2 B D^{2}=5 \Rightarrow B D^{2}=\mathbf{2 . 5}$

$\therefore$ Area of square on $B D=(B D)^{2}=2.5 cm^{2}$.

~~ 19. $P Q | A C$

$\Rightarrow \frac{C Q}{Q B}=\frac{A P}{P B}=\frac{4}{3}$

Also, $Q D | C P$

$\Rightarrow \frac{P D}{D B}=\frac{C Q}{Q B}=\frac{4}{3}$

(From $(i))$

$ \frac{P D}{D B}=\frac{4}{3} \Rightarrow \frac{P D}{P B}=\frac{P D}{P D+D B}=\frac{4}{4+3}=\frac{4}{7} $

$\therefore$ From $(i)$ and (ii)

$ \begin{aligned} & \frac{A P}{P B}=\frac{4}{3} \text{ and } \frac{P B}{P D}=\frac{7}{4} \\ \therefore & \frac{A P}{P B} \times \frac{P B}{P D}=\frac{4}{3} \times \frac{7}{4} \Rightarrow \frac{A P}{P D}=\frac{7}{3} \Rightarrow A P: P D=7: 3 . \end{aligned} $

~~ 20. $L O$ being the median on $M N$,

$M O=O N$

Also, $L O$ being the internal bisector of $\angle M L N$, $\frac{L M}{L N}=\frac{M O}{O N}=1$

(By bisector theorem)

$\Rightarrow L M=L N$

$\Rightarrow \triangle L M N$ is isosceles triangle.

Now $\triangle L O M \cong \triangle L O N$ (By SSS)

$\Rightarrow \angle L O M=\angle L O N=90^{\circ}$ (cpct)

$\therefore$ In $\triangle L O M, M O=\sqrt{L M^{2}-L O^{2}}$

$ =\sqrt{25-9}=\sqrt{16}=4 cm . $

Area of $\triangle L M N=\frac{1}{2} \times M N \times L O$

$ =\frac{1}{2} \times 8 \times 3 cm^{2}=\mathbf{1 2} cm^{2} . $

~~ 21. Given, $A B C$ is anequilateral triangle such that $A B=B C=C A=10 m$ If $O$ is any point in the $\triangle A B C$, then

Area of $\triangle A B C$

$=$ Area $(\triangle O A B)+Area(\triangle O A C)$

Area $(\triangle O B C)$

$=\frac{1}{2} \times A B \times O R+\frac{1}{2} \times A C \times O P+\frac{1}{2} \times A C \times O Q$

$=\frac{1}{2} \times A B \times(O R+O P+O Q) \quad(\because A B=B C=C A)$

$=\frac{1}{2} \times 10 \times(2+3+O Q)$

$\because$ Area of an equilateral $\Delta=\frac{\sqrt{3}}{4}$ (side $)^{2}$

$\therefore \frac{\sqrt{3}}{4} \times(10)^{2}=\frac{1}{2} \times 10 \times(5+O Q)$

$\Rightarrow 5 \sqrt{3}=5+O Q \Rightarrow O Q=\mathbf{5} \sqrt{\mathbf{3}}-\mathbf{5}$.

~~ 22. $C E, C D$ and $C F$ are the required paths such that the longest side of the park $A B$ is divided into four equal segments.

$ A E=E D=D F=F B=20 m . $

Let $A C=b, B C=a$

Then, by Apollonius theorem in $\triangle A C D$

$ A C^{2}+C D^{2}=2(C E^{2}+20^{2}) $

$\Rightarrow \frac{1}{2}(b^{2}+C D^{2})=C E^{2}+20^{2}$

Similarly in $\triangle C D B,(C B^{2}+C D^{2})=2(C F^{2}+20^{2})$

$\Rightarrow C F^{2}+20^{2}=\frac{1}{2}(a^{2}+C D^{2})$

Adding (i) and (ii),

$ \begin{aligned} & C E^{2}+C F^{2}+2 \times 20^{2}=\frac{1}{2}(a^{2}+b^{2}+2 \times C D^{2}) \\ & \Rightarrow C E^{2}+C F^{2}=\frac{1}{2}(80^{2}+2 \times 40^{2})-2 \times 20^{2} \\ & (\because A C^{2}+C B^{2}=A B^{2} \Rightarrow a^{2}+b^{2}=80^{2}) \end{aligned} $

$(\because C D=40, \because$ line joining the vertex at the right $\angle$ to the mid-point of the hypotenuse is half the hypotenuse)

$ \begin{aligned} \Rightarrow C E^{2}+C F^{2} & =\frac{1}{2}(6400+3200)-800 \\ & =\mathbf{4 0 0 0 .} \end{aligned} $

Now $C E^{2}+C F^{2}+C D^{2}=4000+40^{2}=\mathbf{5 6 0 0}$.

~~ 23. Let $A D=p$

Area of $\triangle A B C=\frac{1}{2} b c \sin \angle B A C=\frac{1}{2} b c \sin 120^{\circ}$

$ =\frac{1}{2} b c \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} b c $

Area of $\triangle B A D=\frac{1}{2} c p \sin 60^{\circ}$

$ =\frac{\sqrt{3}}{4} c p $

Area of $\triangle C A D=\frac{1}{2} b p \sin 60^{\circ}=\frac{\sqrt{3}}{4} b p$

Now Area $(\triangle A B C)=Area(\triangle B A D)+Area(\triangle C A D)$

$\Rightarrow \frac{\sqrt{3}}{4} b c=\frac{\sqrt{3}}{4} c p+\frac{\sqrt{3}}{4} b p \Rightarrow b c=p(b+c) \Rightarrow p=\frac{\boldsymbol{b} \boldsymbol{c}}{\boldsymbol{b}+\boldsymbol{c}}$.

~~ 24. Let $B C=x$ and $A D=y$, then as per bisector theorem,

$ \begin{aligned} & \frac{B D}{D C}=\frac{A B}{A C}=\frac{4}{3} \\ & \therefore B D=\frac{4 x}{7} \text{ and } D C=\frac{3 x}{7} \end{aligned} $

Now in $\triangle A B D$ using cosine rule, $\cos 30^{\circ}=\frac{A B^{2}+A D^{2}-B D^{2}}{2 \times A B \times A D}$

$ =\frac{16+y^{2}-\frac{16 x^{2}}{49}}{2 \times 4 \times y} \Rightarrow \frac{\sqrt{3}}{2}=\frac{16+y^{2}-\frac{16 x^{2}}{49}}{2 \times 4 \times y} $

$\Rightarrow 4 \sqrt{3 y}=16+y^{2}-\frac{16 x^{2}}{49}$

In $\triangle A C D$ using the cosine rule,

$\cos 30^{\circ}=\frac{A C^{2}+A D^{2}-C D^{2}}{2 \times A C \times A D}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{9+y^{2}-\frac{9 x^{2}}{49}}{2 \times 3 \times y}$

$\Rightarrow 3 \sqrt{3 y}=9+y^{2}-\frac{9 x^{2}}{49}$

Also in $\triangle A B C, \cos 60^{\circ}=\frac{A B^{2}+A C^{2}-B C^{2}}{2 \times A B \times A C}$

$\Rightarrow \frac{1}{2}=\frac{16+9-B C^{2}}{2 \times 4 \times 3} \Rightarrow 12=25-B C^{2}$

$\Rightarrow B C^{2}=13 \Rightarrow x^{2}=13$

$\therefore$ Subtracting eqn (ii) from (i), we get $\sqrt{3} y=7-\frac{7 x^{2}}{49}=7-\frac{7 \times 13}{49}=7-\frac{13}{7}=\frac{49-13}{7}=\frac{36}{7}$

$\Rightarrow y=\frac{36}{7 \times \sqrt{3}}=\frac{\mathbf{1 2} \sqrt{\mathbf{3}}}{\mathbf{7}}$.

~~ 25. Join $P^{\prime} Q^{\prime}$.

$P^{\prime}, Q^{\prime}$ being the mid-points of $Q R$ and $P R$ respectively, we have $P^{\prime} Q^{\prime} | P Q$ and $P^{\prime} Q^{\prime}=\frac{1}{2} P Q$

(By the mid-point theoram)

Let $O P^{\prime}=a, O Q^{\prime}=b, O P=c, O Q=d$ and $P Q=x$

Then, $P^{\prime} Q^{\prime}=\frac{1}{2} x$.

$\therefore$ In rt. $\triangle O P^{\prime} Q^{\prime}$,

$ a^{2}+b^{2}=\frac{x^{2}}{4} $

In rt. $\triangle O P^{\prime} Q$,

$ a^{2}+d^{2}=\frac{9}{4} $

In rt. $\triangle O Q P$,

$ c^{2}+d^{2}=x^{2} $

In rt. $\triangle O Q^{\prime} P$,

$ b^{2}+c^{2}=4 $

$\therefore$ Eq $(i)-Eq(i i)+Eq(i i i)-Eq(i v)$

$\Rightarrow a^{2}+b^{2}-(a^{2}+d^{2})+(c^{2}+d^{2})-(b^{2}+c^{2})=\frac{x^{2}}{4}-\frac{9}{4}+x^{2}-4$

$\Rightarrow 0=\frac{5 x^{2}}{4}-\frac{25}{4} \Rightarrow \frac{5 x^{2}}{4}=\frac{25}{4} \Rightarrow x^{2}=5 \Rightarrow \boldsymbol{x}=\sqrt{\mathbf{5}}$.

~~ 26. When $\angle A=60^{\circ}$ and $b=c$, then $\triangle A B C$ is equilateral

$\Rightarrow a=b=c$

when $A=90^{\circ}$, then $\triangle A B C$ is an isosceles right angled triangle and $a=\sqrt{2} b$ or $a=\sqrt{2} c$.

$\therefore 60^{\circ}<A<90^{\circ}=c<a<\sqrt{2} c$.

~~ 27. $P$ and $Q$ being the mid-points of $A C$ and $A B$ respectively, $P Q | B C$ and $P Q=\frac{1}{2} B C$

Let $A F \perp B C$ be drawn such that it intersects $P Q$ and $B C$ in $E$ and $F$ respectively.

$P Q | B C \Rightarrow \frac{A E}{E F}=\frac{A P}{P C}=1 \Rightarrow A E=E F$

Also let $R I \perp P Q$ be drawn such that it intersect $B C$ and $P Q$ in $J$ and $I$ respectively. $G$ and $H$ being the mid-points of sides $P R$ and $R Q$ of $\triangle P Q R, G H | P Q$ and $G H=\frac{1}{2} P Q$ (By midpoint Theorem)

Also, $G H | P Q \Rightarrow \frac{R J}{J I}=\frac{R G}{G P}=1 \Rightarrow R J=J I$

(By Basic Proportionality Theorem)

But $E F=J I$

$\therefore A E=E F=R J=J I$

$\therefore A F=A E+E F=R J+J I=R I=h$ (say)

Then, $\frac{\text{ Area }(\triangle P Q R)}{\text{ Area }(\triangle A B C)}=\frac{\frac{1}{2} \times P Q \times h}{\frac{1}{2} \times B C \times h}=\frac{P Q}{B C}=\frac{\mathbf{1}}{\mathbf{2}}$.

~~ 28. $A D$ being the median to $B C$,

$A B^{2}+A C^{2}=2(B D^{2}+A D^{2})$ (Apollonius Th.)

$\Rightarrow c^{2}+b^{2}=2(\frac{a^{2}}{4}+A D^{2})$

$\Rightarrow 2 c^{2}+2 b^{2}=4(\frac{a^{2}}{4}+A D^{2})$

$\Rightarrow 4 A D^{2}=2 c^{2}+2 b^{2}-a^{2}$

$\Rightarrow A D=\frac{1}{2} \sqrt{2 c^{2}+2 b^{2}-a^{2}}$

Now $G$ divides $A D$ in ratio $2: 1 \quad \therefore A G=\frac{2}{3} A D$

$\Rightarrow \quad A G^{2}=\frac{4}{9} A D^{2}=\frac{4}{9} \times \frac{1}{4}(2 c^{2}+2 b^{2}-a^{2})$

$=\frac{1}{9}(2 c^{2}+2 b^{2}-a^{2})$

Similarly, $\quad G B^{2}=\frac{1}{9}(2 c^{2}+2 a^{2}-b^{2})$

$ \begin{aligned} A G^{2}+G B^{2} & =\frac{1}{9}[2 c^{2}+2 b^{2}-a^{2}+2 c^{2}+2 a^{2}-b^{2}] \\ & =\frac{1}{9}[a^{2}+b^{2}+4 c^{2}] \\ & =\frac{1}{9}[5 c^{2}+4 c^{2}]=\frac{9 c^{2}}{9}=c^{2}=B C^{2} \end{aligned} $

$\Rightarrow A G \perp G B \Rightarrow$ Angle between the medians drawn to sides $\boldsymbol{a}$ and $\boldsymbol{b}$ is $\mathbf{9 0}^{\circ}$.

~~ 29. Let $B P=x$. Then $P C=2 x$

$\because A P$ bisects $\angle B A C$,

By the angle bisector theorem,

$\frac{A B}{A C}=\frac{B P}{P C}=\frac{1}{2}$

Using the sine formula, $\frac{A C}{\sin B}=\frac{B A}{\sin C}$

$\Rightarrow \frac{\sin C}{\sin B}=\frac{B A}{A C}=\frac{1}{2}$

$\Rightarrow \frac{\sin C}{\sin B}=\frac{\frac{1}{2}}{1} \Rightarrow \sin C=\frac{1}{2}$ and $\sin B=1$

$\Rightarrow \angle C=30^{\circ}$ and $\angle B=90^{\circ}$

$\therefore$ In $\triangle A P C, A P C=180^{\circ}-(30^{\circ}+30^{\circ})=\mathbf{1 2 0}^{\circ}$.

~~ 30. Let $A D$ be the height of the given triangle $A B C$, where $A D=h$ and $B C=b$ Also let height $H G$ of rectangle $E F G H$ equal $x$ so that $H E=G F=2 x$

Now $\triangle B G H \sim \triangle B D A$

$\Rightarrow \frac{B G}{B D}=\frac{H G}{A D}$

$\Rightarrow \frac{B G}{B D}=\frac{x}{h} \Rightarrow B G=\frac{x}{h} B D$

Also, $\triangle C F E \sim \triangle C D A$

$\Rightarrow \frac{C F}{C D}=\frac{E F}{A D}=\frac{x}{h}$

$\Rightarrow C F=\frac{x}{h} C D$

$B G+C F=B C-G F=b-2 x$

$\therefore$ From (i), (ii) and (iii)

$b-2 x=\frac{x}{h}(B D+C D) \Rightarrow b-2 x=\frac{x}{h} B C=\frac{x}{h} b$

$\Rightarrow b h-2 x h=x b \Rightarrow b h=x b+2 x h=x(b+2 h)$

$\Rightarrow x=\frac{\boldsymbol{b h}}{\boldsymbol{b}+\mathbf{2 h}}$.

~~ 31. Let $\angle E A D=\alpha$. Then,

In $\triangle A B C, A B=B C \Rightarrow \angle B C A=\alpha \quad$ sides opp. equal In $\triangle A G F, A G=G F \Rightarrow \angle A F G=\alpha]$ angles are equal

$\therefore$ For $\triangle A B C$, ext $C B D=2 \alpha$

In $\triangle C B D, C B=C D \Rightarrow \angle C D B=2 \alpha$

For $\triangle A F G$, ext $\angle F G C=2 \alpha$

$\because$ In $\triangle G F E, G F=E F \Rightarrow \angle F E G=\angle F G E=2 \alpha$

For $\triangle E A F$, ext. $\angle E F D=3 \alpha$

$\because E F=E D \therefore \angle E D F=\angle E F D=3 \alpha \Rightarrow \angle E D P=\alpha$

For $\triangle C A D$, ext. $\angle D C E=3 \alpha$

In $\triangle E C D, D C=E D \Rightarrow \angle D E C=\angle D C E=3 \alpha$

$\Rightarrow \angle F E D=\angle D E C-\angle F E C=3 \alpha-2 \alpha=\alpha$.

$\therefore$ In $\triangle E F D, \alpha+3 \alpha+3 \alpha=180^{\circ}$

$\Rightarrow 7 \alpha=180^{\circ} \Rightarrow \alpha=\frac{180^{\circ}}{7}=26^{\circ}$ or approximately $25^{\circ}$.

~~ 32. In $\triangle A B C$, let $B P$ bisect $\angle A B C$

Let $\angle C=x \Rightarrow \angle B=2 x$

$\therefore \angle P B C=\angle A B P=x$

In $\triangle P B C$,

$ \angle P B C=\angle P C B=x $

$\Rightarrow P C=P B$ (sides opposite equal angles are equal) Now in $\triangle A P B$ and $\triangle B P C$ $A B=C D$ (Given)

$P B=P C$ (Proved above)

$\angle A B P=\angle D C P=x$

$\therefore \triangle A P B \cong \triangle D P C$

$\Rightarrow \angle B A P=\angle P D C=2 y$ and $A P=D P$

$\therefore$ In $\triangle A P D, A P=D P \Rightarrow \angle P D A=\angle P A D=y$

$\therefore \angle D P A=180^{\circ}-2 y$

Also from $\triangle D P C, \angle D P C=180^{\circ}-(x+2 y)$

$\therefore$ From $(i)$ and (ii), $\angle D P A+\angle D P C=180^{\circ}$

$ \begin{aligned} & \Rightarrow 180^{\circ}-2 y+180^{\circ}-(x+2 y)=180^{\circ} \\ & \Rightarrow x+4 y=180^{\circ} \end{aligned} $

Also in $\triangle A B C, \angle A+\angle B+\angle C=180^{\circ}$

$ \begin{aligned} & 2 y+2 x+x=180^{\circ} \\ & 3 x+2 y=180^{\circ} \end{aligned} $

$\therefore(2)-3 \times(1) \Rightarrow 3 x+2 y-(3 x+12 y)=180^{\circ}-3 \times 180^{\circ}$

$\Rightarrow-10 y=-360^{\circ} \Rightarrow y=36^{\circ}$

$\therefore \angle B A C=2 y=2 \times 36^{\circ}=\mathbf{7 2}^{\circ}$.

~~ 33. Area $(\triangle P S X)=Area(P U S)-Area(S U X)$

In $\triangle P X S, W Y | S X$

$\Rightarrow \frac{P Y}{Y X}=\frac{P W}{W S}=1$

(Given, $P W=W S$ )

$\Rightarrow P Y=Y X$

In $\triangle R U Y$,

$S X | R Y=\frac{U X}{X Y}=\frac{U S}{S R}=\frac{1}{2}$

$(\because Q S=S R$ and $Q V=U S)$

$ =U X=\frac{1}{2}(X Y) $

$\therefore$ In $\triangle P U S, U X=\frac{1}{2} X Y=\frac{1}{2}(\frac{1}{2} P X)=\frac{1}{4} P X$

$P U=U X+P X=\frac{1}{4} P X+P X=\frac{5}{4} P X$

$\therefore \quad \frac{\text{ Area of } \triangle S U X}{\text{ Area of } \triangle P U S}=\frac{1}{5}$

Now Area of $\triangle P U S=\frac{1}{4}($ Area of $\triangle P Q R)$

Area of $\triangle S U X=\frac{1}{4} \times \frac{1}{5} \times$ Area of $\triangle P Q R$

$ =\frac{1}{20} \text{ Area of } \triangle P Q R $

$\therefore \frac{\text{ Area of } \triangle P S X}{\text{ Area of } \triangle P Q R}=\frac{\frac{1}{4} \text{ Area of } \triangle P Q R-\frac{1}{20} \text{ Area of } \triangle P Q R}{\text{ Area of } \triangle P Q R}$

$ =\frac{5-1}{20}=\frac{1}{5} \text{. } $

~~ 34. $G$ being the centroid of $\triangle A B C$,

$\frac{A G}{G D}=\frac{B G}{G E}=\frac{C G}{G F}=\frac{2}{1}$

In $\triangle A B D$,

$A B+B D>A D$

$\Rightarrow A B+\frac{B C}{2}>A D$

In $\triangle B E C$,

$ B C+C E>B E \Rightarrow B C+\frac{A C}{2}>B E $

In $\triangle A F C$,

$ A C+A F>C F \Rightarrow A C+\frac{A B}{2}>C F $

Adding (i), (ii) and (iii), we get

$ \begin{aligned} & A B+B C+A C+\frac{B C}{2}+\frac{A C}{2}+\frac{A B}{2}>A D+B E+C F \\ \Rightarrow & \frac{2 A B+2 B C+2 A C+B C+A C+A B}{2}>A D+B E+C F \\ \Rightarrow & 3(A B+B C+A C)>2(A D+B E+C F) \Rightarrow \boldsymbol{I} \text{ is true } \end{aligned} $

Also, in $\triangle B G C$,

$B G+G C>B C$

$(\because \frac{B G}{G E}=\frac{2}{1}.$ and $\frac{C G}{G F}=\frac{2}{1} \Rightarrow B G=\frac{2}{3} B E$ and $.C G=\frac{2}{3} C F)$

$\Rightarrow \frac{2}{3} B E+\frac{2}{3} C F>B C \Rightarrow 2 B E+2 C F>3 B C$

Similarly in $\triangle B G A$,

$B G+G A>A B$

$\Rightarrow \frac{2}{3} B E+\frac{2}{3} A D>A B \Rightarrow 2 B E+2 A D>3 A B$

In $\triangle C G A$,

$ C G+G A>A C $

$\Rightarrow \frac{2}{3} C F+\frac{2}{3} A D>A C$

$\Rightarrow 2 C F+2 A D>3 A C$

Adding (iii), (iv) and (v), we get

$ \begin{aligned} & 2 B E+2 C F+2 B E+2 A D+2 C F+2 A D \\ & >3 B C+3 A B+3 A C \\ & \Rightarrow 4(A D+B E+C F)>3(A B+B C+A C) \\ & \Rightarrow 3(A B+B C+A C)<4(A D+B E+C F) \Rightarrow \text{ III is true. } \end{aligned} $

~~ 35. Let $\angle A Q P=\alpha$

In $\triangle A Q P, A Q=Q P$

$ \begin{aligned} \Rightarrow \angle Q A P & =\angle Q P A=\frac{1}{2}(\pi-\alpha) \\ & =\frac{\pi}{2}-\frac{\alpha}{2} \end{aligned} $

$\angle P Q B=\pi-\alpha$

In $\triangle P Q B$,

$ P Q=P B \Rightarrow \angle P B Q=\angle P Q B=\pi-\alpha $

$ \therefore \angle Q P B=\pi-2(\pi-\alpha)=2 \alpha-\pi $

In $\triangle A B C$,

$ \begin{aligned} & \angle B A C+\angle A B C+\angle A C B=\pi \\ & \Rightarrow \angle Q P A+\angle A C B+\angle A C B=\pi \\ &(\because \angle Q A P=\angle B A C, A B=A C \Rightarrow \angle A C B=\angle A B C) \end{aligned} $

$\Rightarrow \frac{\pi}{2}-\frac{\alpha}{2}+2 \angle A C B=\pi$

$\Rightarrow 2 \angle A C B=\pi-\frac{\pi}{2}+\frac{\alpha}{2}=\frac{\pi}{2}+\frac{\alpha}{2} \Rightarrow \angle A C B=\frac{\pi}{4}+\frac{\alpha}{4}$

$\therefore$ In $\triangle B P C, B P=B C$

$\Rightarrow \angle B P C=\angle B C P=\angle A C B=\frac{\pi}{4}+\frac{\alpha}{4}$

Now $A P C$ being a straight line,

$\angle A P Q+\angle B P Q+\angle B P C=\pi$

$\Rightarrow \frac{\pi}{2}-\frac{\alpha}{2}+2 \alpha-\pi+\frac{\pi}{4}+\frac{\alpha}{4}=\pi$

$\Rightarrow \frac{7 \alpha}{4}=\frac{5 \pi}{4} \Rightarrow \alpha=\frac{5 \pi}{7}$.

~~ 36. Let $A B C$ be the given triangle whose circumcentre is $O$. Produce $C O$ to meet the circle at $G \Rightarrow C O G$ is the diameter of the circle.

$\Rightarrow \angle G A C=\angle G B C=90^{\circ}$ (Angles in a semicircle)

Let $A F$ and $B E$ be the perpendiculars from vertex $A$

and $B$ respectively on sides $B C$ and $A C$.

$\therefore H$ is the orthocentre of $\triangle A B C$. We need to find the ratio $A H: O D$, where $O D$ is the perpendicular distance of the circumcentre $O$ from side $B C$.

$O D \perp B C \Rightarrow B D=D C$ (Perpendicular from the centre of the circle bisects the chord)

Also, $G B \perp B C$ and $O D \perp B C \Rightarrow O D | G B$.

$\therefore$ In $\triangle B G C$, by the midpoint theorem, $O$ and $D$ being the mid-points of sides $G C$ and $B C, O D | G B$ and $O D=\frac{1}{2} G B$

Now $G A$ and $B E$ are both perpendiculars to $A C$

$\Rightarrow G A|B E \Rightarrow G A| B H$

Also, $G B|A F \Rightarrow G B| A H$

$\Rightarrow G A H B$ is a parallelogram

$\Rightarrow G B=A H=2 O D$

$\therefore \frac{A H}{O D}=\frac{2 O D}{O D}=\frac{\mathbf{2}}{\mathbf{1}}$.

~~ 37. In $\triangle s A B C$ and $D A C$

$\angle A B C=\angle D A C$

( $\because A D$ bisects $\angle A$ and $\angle A=2 \angle B$ )

Also, $\angle A C B=\angle D C A$

$\Rightarrow \triangle A B C \sim D A C$ (AA similarity) $\frac{A C}{D C}=\frac{B C}{A C} \Rightarrow A C^{2}=B C \cdot D C$

$\Rightarrow D C=\frac{A C^{2}}{B C}=\frac{b^{2}}{a}$

(In a $\triangle A B C, A B=c, B C=a, A C=b$ )

Also, $A D$ being the angle bisector of $\angle A$,

$ \begin{aligned} & \frac{B D}{C D}=\frac{A B}{A C}=\frac{c}{b} \\ \therefore & \frac{B D}{C D}+1=\frac{c}{b}+1 \\ \Rightarrow & \frac{B D+C D}{C D}=\frac{c+b}{b} \\ \Rightarrow & \frac{B C}{C D}=\frac{C+b}{b} \Rightarrow \frac{a}{C D}=\frac{c+b}{b} \Rightarrow C D=\frac{a b}{b+c} \end{aligned} $

$\therefore$ From (i) and (ii), $\frac{b^{2}}{a}=\frac{a b}{b+c} \Rightarrow \boldsymbol{a}^{2}=\boldsymbol{b}(\boldsymbol{b}+\boldsymbol{c})$.

~~ 38. Since $\frac{B D}{D C}=\frac{B F}{F A}$

By basic proportionality theorem, we have $F D | A C$.

Also, $\angle B D K=\angle A D B=\angle A F C$

$ =180^{\circ}-\angle B F K $

$\therefore \angle B D K+\angle B F K=180^{\circ}$

$\Rightarrow B D K F$ is a cyclic quadrilateral

$\Rightarrow \angle F B K=\angle F D K$

(Angles in the same segment)

$\therefore \angle \boldsymbol{A B E}=\angle F B K=\angle F D K=\angle F D A=\angle \boldsymbol{D} \boldsymbol{A} \boldsymbol{C}$

$ (\because F D | A C, \text{ alt. } \angle s \text{ are equal) } $

~~ 39. $A I$ and $B I$ are the bisectors of $\angle C A B$ and $\angle C B A$ respectively,

$\therefore$ In $\triangle A B C$,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \angle A+\angle B=180^{\circ}-\angle C$

$\frac{\angle A}{2}+\frac{\angle B}{2}=90^{\circ}-\frac{\angle C}{2}$

Also, in $\triangle A I B$,

$\angle A I B=180^{\circ}-(\frac{\angle A}{2}+\frac{\angle B}{2})$

$=180^{\circ}-(90^{\circ}-\frac{\angle C}{2})=90^{\circ}+\frac{\angle C}{2}$

Extend $C A$ to $D$ such that $A D=A I$.

Then, $B C=A B+A I \Rightarrow B C=C A+A D=C D$

(By hypothesis, $A B=A C, A D=A I$ )

$\Rightarrow \angle C D B=\angle C B D=90^{\circ}-\frac{\angle C}{2}$

(Since in $\triangle C D B, C D=C B$ and $\angle C+\angle D+\angle B=180^{\circ}$ )

Thus, $\angle A I B+\angle A D B=90^{\circ}+\frac{\angle C}{2}+90^{\circ}-\frac{\angle C}{2}=180^{\circ}$

$\Rightarrow A I B D$ is a cyclic quadrilateral

Also $\angle A D I=\angle A B I=\frac{\angle B}{2}$ (angles in the same segment)

$\therefore$ In $\triangle D A I, \angle D A I,=180^{\circ}-2(\angle A D I)=180^{\circ}-\angle B$

$ (\because A D=A I \Rightarrow \angle A D I=\angle A I D) $

Thus, $\angle C A I=B \Rightarrow A=2 B$

(CAD is a straight angle)

Since, $A C=A B \Rightarrow \angle B=\angle C$

$\therefore$ In $\triangle A B C, \angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow 4 \angle B=180^{\circ} \Rightarrow \angle B=45^{\circ} \Rightarrow \angle A=\mathbf{9 0}^{\circ}$.

~~ 40. Draw $B P \perp A C$ and join $P$ to $D$.

In $\triangle B P C$,

$\angle P B C=180^{\circ}-(\angle B P C+\angle B C P)$

$=180^{\circ}-(90^{\circ}+30^{\circ})=180^{\circ}-120^{\circ}=60^{\circ}$

In $\triangle B P C$,

$\sin 30^{\circ}=\frac{B P}{B C} \Rightarrow \frac{B P}{B C}=\frac{1}{2} \Rightarrow B P=\frac{1}{2} B C=B D$ Now in $\triangle B P D, B P=B D \Rightarrow \angle B D P=\angle P B D=60^{\circ}$ $\Rightarrow \angle B P D=60^{\circ} \Rightarrow \triangle B P D$ is equilateral

$\therefore P B=P D$ and $\angle A D P=60^{\circ}-45^{\circ}=15^{\circ}$

In $\triangle A D C$, ext. $\angle A D B=\angle A C D+\angle D A C$

$45^{\circ}=30^{\circ}+\angle D A C \Rightarrow \angle D A C=15^{\circ}$

$\therefore$ In $\triangle A P D, \angle A D P=\angle P A D \Rightarrow P D=P A$

We have $P D=P A=P B$

$\Rightarrow P$ is the circumcentre of $\triangle A D B$ as circumcentre is equidistant from the vertices of a $\triangle$.

$\Rightarrow B A D=\frac{1}{2} \angle B P D$

( $\because$ Angle subtended by an are at the centre of a circle is half the angle subtended by the same are at any other point on the remaining part of the circumference.)

$\Rightarrow B A D=\frac{1}{2} \times 60^{\circ}=\mathbf{3 0}^{\circ}$.

7 Quadrilaterals

KEY FACTS

1. Polygon: A closed plane figure bounded by line segments is called a polygon. Poly means ‘many’, so a polygon is a closed many sided figure. $\square \triangle \square \square$ are all polygons.

Polygons are classified as follows with reference to the number of sides they have:

5 sides pentagon	6 sides hexagon	7 sides heptagon	8 sides octagon	9 sides nonagon	10 sides decagon	12 sides dodecagon

2. Some important terms used with polygons

Vertex: It is the point where two adjacent sides of a polygon meet.

For example, $A, B, C, D, E$.

Consecutive sides: Are those sides which have a vertex in common, viz., $A B, B C ; B C, C D ; C D, D E ; D E E A$.
Diagonal: The line joining any two non-adjacent or non-consecutive vertices of a polygon, viz., $A C, B D, C E, B E$.
Perimeter: The sum of the lengths of all sides of a polygon, i.e., $A B+B C+C D+D E+E A$.
Convex polygon: In a convex polygon each interior angle is less than $180^{\circ}$.
Concave polygon: In this type of a polygon at least one of the interior angles is a reflex angle, i.e., greater than $180^{\circ}$. It is also called a re-entrant polygon.
Regular polygon: A polygon is regular if:

(i) it is convex

(ii) all of its sides are equal

(iii) all of its interior angles are equal

(iv) all of its exterior angles are equal.

The figure shows a regular hexagon $A B C D E F$, with all equal interior angles marked ’ $\boldsymbol{a}$ '

and all exterior angles marked ’ $\boldsymbol{b}$ ‘. The sides are equal, i.e., $A B=B C=C D=D E=E F=F A$.

Some common examples of a regular polygon are: an equilateral triangle, a square.

Sum of the angles of a polygon:

a. Sum of the interior angles of a convex polygon of $n$ sides $=(2 n-4) r t . \angle s=(2 n-4) \times 90^{\circ}$

b. If the sides of a convex polygon are produced in order, the sum of the exterior angles so formed is 4 rt. $\angle$ s, i.e., $360^{\circ}$.

In a regular polygon: $\boldsymbol{a}$. Each exterior angle $=\frac{360^{\circ}}{\text{ Number of sides }}$ b. Each interior angle $=180^{\circ}-$ Exterior angle

$ =\frac{(2 n-4) \times 90^{\circ}}{n}, \text{ where } n \text{ is the number of sides. } $

Quadrilateral: A quadrilateral is a plane figure bounded by four straight line segments.

Ex. $A B C D$ is a quadrilateral. The four line segments bounding it, i.e., $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{D A}$ are called its sides. $\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}, \boldsymbol{D}$ are the vertices and $\boldsymbol{A C}$ and $\boldsymbol{B D}$, the line segments joining the opposite vertices are the diagonals of the given quadrilateral.

Properties:

Sum of the four interior angles $=360^{\circ}$.
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram.

Ex. PQRS is a parallelogram.

The sum of the opposite sides of a quadrilateral circumscribed about a circle is always equal.

$ A B+D C=A D+B C $

Area of quadrilateral $=\frac{1}{2} \times$ one of the diagonals $\times$ sum of the perpendiculars drawn on the diagonal from opposite vertices

Ex. Area of quad. $A B C D=\frac{1}{2} \times A C \times(B F+D E)$

Special Types of Qudrilaterals and their Properties

PARALLELOGRAM:

A quadrilateral in which opposite sides are equal and parallel is called a parallelogram.

Properties:

Both pairs of opposite sides are equal, i.e., $A B=D C, A D=B C$.
Opposite angles are equal, i.e., $\angle A=\angle C, \angle B=\angle D$.
Sum of any two adjacent angles is $180^{\circ}$, i.e.,

$ \angle A+\angle B=\angle B+\angle C=\angle C+\angle D=\angle D+\angle A=180^{\circ} \text{. } $

Diagonals bisect each other, i.e., $A P=P C, B P=P D$.
Each diagonal divides a parallelogram into two congruent triangles, i.e., $\triangle A B C \cong \triangle A D C, \triangle D A B \cong \triangle B C D$.
In case of a parallelogram, the diagonals need not be of equal length, need not bisect at right angles and need not bisect the angles at the vertices.
Lines joining the mid-points of the adjacent sides of a parallelogram is a parallelogram.
The parallelogram that is inscribed in a circle is a rectangle. ABCD is a rectangle.

The parallelogram that inscribed in acircle is a rectangle. ABCD is a rectangle.

The parallelogram that is circumscribed about a circle is a rhombus. $A B C D$ is a rhombus.
1. Area of a parallelogram $=$ base $\times$ height

Area of a parallelogram $=$ product of any two adjacent sides $\times$ sine of included angle Property 9

$ =a b \sin \theta $

Perimeter of a parallelogram $=2$ (Sum of any two adjacent sides)
$A C^{2}+B D^{2}=2(A B^{2}+B C^{2})$
Parallelograms that lie on the same base and between the same parallel lines are equal in area.

RECTANGLE:

A rectangle is a parallelogram in which all the four angles at the vertices are right angles, i.e., $=90^{\circ}$.

Properties:

Opposite sides are equal and parallel.
All angles are each equal to $90^{\circ}$.
Diagonals are equal and bisect each other but are not necessarily at right angles.
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

rhombus.

Thus, $P Q R S$ is a rhombus.

The quadrilateral formed by the intersection of the angle bisectors of a parallelogram is a rectangle.
Area of a rectangle $=$ length $\times$ breadth $=\boldsymbol{l} \times \boldsymbol{b}$
Diagonals of a rectangle $=\sqrt{\text{ length }^{2}+\text{ breadth }^{2}}=\sqrt{l^{2}+b^{2}}$
Perimeter of a rectangle $=2($ length + breadth $)=(l+b)$

RHOMBUS:

A parallelogram with all sides equal (adjacent sides equal) is called a rhombus.

Properties:

Opposite sides are parallel and all sides are equal.
Opposite angles are equal.
Diagonals bisect each other at right angles but they are not necessarily equal.
Diagonals bisect the vertex angles.

Sum of any two adjacent angles is $180^{\circ}$.
Figure formed by joining the mid-points of the adjacent sides of a rhombus is a rectangle.
- Area of rhombus $=$ base $\times$ height

Area of rhombus $=\frac{1}{2} \times$ product of diagonals
Area of rhombus $=$ Product of adjacent sides $\times$ sine of included angle.

SQUARE:

A parallelogram whose all sides are equal and whose all angles are each equal to right angle is a square. Thus each square is a parallelogram, a rectangle and a rhombus.

Properties:

All sides are equal and parallel.
All angles are each equal to $90^{\circ}$.
Diagonals are equal and bisect each other at right angles.
Diagonal of an inscribed square is equal to the diameter of the inscribing circle.
Side of the circumscribed square is equal to the diameter of the inscribed circle.
The figure formed by joining the mid-points of the adjacent sides of a square is a square.

Property 5

Area of a square $=(\text{ side })^{2}=\frac{(\text{ diagonal })^{2}}{2}$
Diagonal $=$ side $\sqrt{2}$
Perimeter $=4 \times$ side

TRAPEZIUM:

A quadrilateral which has only one pair of opposite sides parallel and other pair of opposite sides not parallel is called a trapezium or a trapezoid.

The parallel sides $\boldsymbol{A B}$ and $\boldsymbol{D C}$ of the trapezium $A B C D$ shown in the given figure are called the bases of the trapezium.
The pair of angles that contain the same base, i.e., $\angle \boldsymbol{A}$ and $\angle \boldsymbol{B} ; \angle \boldsymbol{C}$ and $\angle D$ are called base angles.
The line joining the mid-points of the non-parallel sides is called the midline or median, i.e., here is $E F$.
If the non parallel sides of a trapezium are equal, it is an isosceles trapezium.

Properties:

The median is half the sum of the parallel sides, i.e., $E F=\frac{1}{2}(A B+D C)$.
In case of an isosceles trapezium, the diagonals are also equal to each other, i.e., $A C=B D$
Diagonals intersect each other proportionally in the ratio of lengths of parallel sides.

$\frac{A O}{O C}=\frac{B O}{O D}=\frac{A B}{C D}$

$(AC=BD)$

By joining the mid-points of the adjacent sides of a trapezium four similar triangles are obtained.
If a trapezium is inscribed in a circle (cyclic trapezium), then it is an isosceles trapezium.
Area of a trapezium $=\frac{1}{2} \times$ height $\times$ (sum of parallel sides)
Also, $A C^{2}+B D^{2}=B C^{2}+A D^{2}+2 A B \cdot C D$

KITE:

A quadrilateral in which two pairs of adjacent sides are equal is called a kite.

Properties:

Adjacent sides are equal, i.e., $A D=D C ; A B=B C$.
Shorter diagonal is bisected by the longer diagonal, i.e., $O A=O C$.
The diagonals are perpendicular to each other, i.e., $A C \perp B D$.
The angles at the vertices of the shorter diagonal are equal, i.e., $\angle A=\angle C$.
The longer diagonal, i.e., $B D$ here, divides the kite into two congruent triangles, i.e., $\triangle A B D \cong \triangle D B C$.
Area of a kite $=\frac{1}{2} \times$ product of diagonals.

CYCLIC QUADRILATERAL:

A quadrilateral whose all four vertices lie on a circle is called cyclic quadrilateral.

Properties:

The sum of pair of opposite angles of a cyclic quadrilateral is $180^{\circ}$, i.e., $\angle D A B+\angle B C D=180^{\circ} ; \angle A D C+\angle A B C=180^{\circ}$
If a side of a cyclic quadrilateral is produced, then the exterior angle so formed is equal to the interior opposite angle, i.e., $\angle C B E=\angle A D C$.
- Area of a cyclic quadrilateral $=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ is

Cyclic Quadrilateral the semi-perimeter, i.e., $s=\frac{a+b+c+d}{2}$, and $a, b, c, d$ denote the lengths of the sides of the quadrilateral.

Area of a cyclic quadrilateral in which a circle can be inscribed $=\sqrt{\boldsymbol{a} \times \boldsymbol{b} \times \boldsymbol{c} \times \boldsymbol{d}}$, where $a, b, c$ and $d$ are the lengths of the sides of the quadrilateral.
1. Apollonius’ Theorem:

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side, together with twice the square on the median which bisects the third side.

$\triangle A B C$ is the given triangle where the median $A E$ is drawn from $A$ to $E$.

According to the given theorem,

$ A B^{2}+A C^{2}=2 B E^{2}+2 A E^{2} $

SOLVED EXAMPLES

~~ Ex. 1 . $A B C D$ is a parallelogram. $A B$ is produced to $E$ such that $B E=A B$. Prove that $E D$ bisects $B C$.

Sol. $A B C D$ is a parallelogram.

$\Rightarrow A B|C D \Rightarrow A E| C D, C B$ is the transversal $\Rightarrow \angle 3=\angle 4 \quad$ (alt. $\angle s$ are equal)

Now in $\triangle s O C D$ and $O B E$,

$ \begin{aligned} C D & =A B=B E \\ \angle 2 & =\angle 1 \\ \angle 3 & =\angle 4 \end{aligned} $

(Vertically opposite $\angle s$ )

(Proved above)

$\therefore \triangle O C D \cong \triangle O B E$

$\Rightarrow B O=O C \Rightarrow O$ is the mid-point of $B C$

$\Rightarrow E D$ bisects $B C$.

~~ Ex. 2 . $A B C D$ is a parallelogram. $P$ is a point on $A D$ such that $A P=\frac{1}{3} A D$ and $Q$ is a point on $B C$ such that $C Q$ $=\frac{1}{3} B C$. Prove that $A Q C P$ is a parallelogram.

Sol. $A B C D$ is a parallelogram

$\Rightarrow A D=B C$ and $A D | B C$

$\Rightarrow \frac{1}{3} A D=\frac{1}{3} B C \quad$ and $\quad A D | B C$

$\Rightarrow A P=C Q$ and $A P | C Q$

$\Rightarrow A P C Q$ is a parallelogram.

~~ Ex. 3 . Prove that the angle bisectors of a parallelogram form a rectangle.

Sol. Let $A B C D$ be the given parallelogram, whose angle bisectors form the quadrilateral $L M N O$ which needs to be proved a rectangle.

$A B | D C$ and $A D$ is the transversal

$\Rightarrow \angle A+\angle D=180^{\circ}$

(Co-interior angles)

$\Rightarrow \frac{1}{2} \angle A+\frac{1}{2} \angle D=90^{\circ}$

( $A O$ and $D O$ are angle bisectors of $\angle A$ and $\angle D$ respectively)

$\Rightarrow \angle D A O+\angle A D O=90^{\circ}$

$\therefore$ In $\triangle A D O$,

$\angle D A O+\angle A D O+\angle A O D=180^{\circ}$

$\Rightarrow \angle A O D=180^{\circ}-90^{\circ}=90^{\circ}$

$\Rightarrow \angle N O L=\angle A O D=90^{\circ} \quad$ (vert. opp. $\angle s$ )

Similarly, we can show,

$\angle O N M=\angle N M L=\angle M L O=90^{\circ}$

Hence, $\angle M N O$ is a rectangle.

~~ Ex. 4 . Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.

Sol. Given, a triangle $A B C$ with $E$ and $F$ respectively as mid-points of $A B$ and $A C$. The line $A D$ from vertex $A$ meets $E F$ in $G$ and we are to prove that $A G=G D$.

Draw a line $M A N$ through $A$ parallel to $B C$.

Since, $E$ and $F$ are mid-points of $A B$ and $A C$ respectively, $E F | B C$. (Mid pt. Theorem)

Given, $M A N | B C$

$\Rightarrow M A N|E F| B C$

Thus, by the intercept theorem, the intercepts $A E$ and $E B$ made by the transversal $A B$

on the parallel lines $M A N, E F$ and $B C$ are equal, so the intercepts made by the transversal $A D$ on these three parallel lines will also be equal, i.e., $A G=G D$.

~~ Ex. 5 . In $\triangle A B C$ and $\triangle D E F, A B=D E, A B | D E, B C=E F$ and $B C | E F$. Vertices $A, B$ and $C$ are joined to vertices $D, E$ and $F$ respectively. Show that $\triangle A B C \cong \triangle D E F$.

Sol. Given: $A B=D E, A B | D E$

As one pair of opposite sides is equal and parallel

$\Rightarrow A B E D$ is a parallelogram.

Given: $B C=E F$ and $B C | E F$

$\Rightarrow B C F E$ is a parallelogram

Now, $A B E D$ is a parallelogram $\Rightarrow A D | B E$ and $A D=B E$ $B C F E$ is a parallelogram $\Rightarrow C F | B E$ and $C F=B E$

$\therefore$ From $(i)$ and (ii),

$ A D=C F \quad \text{ and } \quad A D | C F $

$\Rightarrow A D F C$ is a parallelogram.

$\Rightarrow A C=D F$.

Now in $\triangle S A B C$ and $D E F$

$A B=D E$ (Given), $B C=E F$ (Given) and $A C=D F$ (Proved above)

$\therefore \triangle A B C \cong \triangle D E F$

~~ Ex. 6 . If $A B C D$ is a rectangle and $P, Q, R$ and $S$ are the mid-points of the sides $A B, B C, C D$ and $D A$ respectively, then quadrilateral $P Q R S$ is a rhombus.

Sol. $A B C D$ is the given rectangle. Join $A C$. In $\triangle A B C, P$ and $Q$ are the mid-points of sides $A B$ and $B C$ respectively.

$\therefore$ By mid-point theorem,

$P Q | A C$ and $P Q=\frac{1}{2} A C$

In $\triangle A D C, R$ and $S$ are the mid-points of $C D$ and $A D$ respectively

$\therefore S R | A C$ and $S R=\frac{1}{2} A C$

From $(i)$ and (ii), we get

$P Q | S R$ and $P Q=S R \Rightarrow P Q R S$ is a parallelogram

$A B C D$ is a rectangle, $A D=B C \Rightarrow \frac{1}{2} A D=\frac{1}{2} B C \Rightarrow A S=B Q$

$\therefore$ In $\triangle s A P S$ and $B P Q$

$ \begin{matrix} A S & =B Q & & (\text{ Proved above }) \\ A P & =P B & & (P \text{ is mid-point of } A B) \\ \angle S A P & =\angle Q B P=90^{\circ} & & \\ \therefore \quad & & (S A S) \\ \Rightarrow \quad & & & \end{matrix} $

$\therefore \quad P Q R S$ is a parallelogram with adjacent sides equal

$\Rightarrow P Q R S$ is a rhombus.

~~ Ex. 7 . $W X Y Z$ is a square of side length $30 . V$ is a point on $X Y$ and $P$ is a point inside the square with $P V$ perpendicular to $X Y . P W=P Z=P V-5$. Find $P V$.

Sol. $P Z=P W$

$\Rightarrow P$ lies half-way between $Z Y$ and $W X$. Let $P Z=P W=x \Rightarrow P V=x+5$

$\because P V \perp X Y$ (Given)

$\therefore \quad V$ is also the mid-point of $X Y$

$ \begin{matrix} \Rightarrow & & V X & =V Y=15 \\ & & & \\ & & W X-Q X=W X-P V=30-(5+x)=25-x . \\ P Q & =15 \\ \therefore & P W & =\sqrt{W Q^{2}+P Q^{2}} \\ \Rightarrow & & x & =\sqrt{(25-x)^{2}+15^{2}} \end{matrix} $

$ \begin{matrix} \Rightarrow & x^{2}=625-50 x+x^{2}+225 \\ \Rightarrow & 50 x=850 \Rightarrow x=17 \\ \therefore & P V=x+5=\mathbf{2 2} . \end{matrix} $

~~ Ex. 8 . $A B C D$ is a trapezium in which side $A B$ is parallel to side $D C$ and $E$ is the mid-point of side $A D$. If $F$ is a point on side $B C$ such that segment $E F$ is parallel to side $D C$. Prove that $E F=\frac{1}{2}(A B+D C)$.

Sol. Let $A B C D$ be the given trapezium in which $A B | D C$. In $\triangle A D C, E$ is the mid-point of $A D$ and $E G | D C$ $\Rightarrow G$ is the mid-point of $A C$

$\Rightarrow E G=\frac{1}{2} D C$

( $\therefore$ Line segment joining the mid-points of to sides of a triangle is parallel to the third side and half of it).

Also, $A B|D C, E F| D C \Rightarrow E F|A B \Rightarrow G F| A B$

$\therefore$ In $\triangle A B C, G F | A B$ and $G$ is the mid-point of $A C$

$\Rightarrow F$ is the mid-point of $B C$

$\Rightarrow G F=\frac{1}{2} A B$

(By mid-point theoram)

$\therefore$ From $(i)$ and $(i i), G E+G F=\frac{1}{2} D C+\frac{1}{2} A B$

$\Rightarrow E F=\frac{1}{2}(A B+D C)$.

~~ Ex. 9 . A point $O$ in the interior of a rectangle $A B C D$ is joined with each of the vertices $A, B, C$ and $D$. Then, show that $O A^{2}+O C^{2}=O B^{2}+O D^{2}$.

Sol. As the diagonals of a rectangle are equal and bisect each other, $A M=D M$ Applying Applonius theorem,

In $\triangle A O C, O A^{2}+O C^{2}=2(O M^{2}+A M^{2})$

In $\triangle B O D, O B^{2}+O D^{2}=2(O M^{2}+D M^{2})$

$\because A M=D M$

$\therefore O A^{2}+O C^{2}=O B^{2}+O D^{2}$.

~~ Ex. 10 . The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio $1: 2$. Compute the ratio of the smaller base of the trapeziod to the larger base.

(Median of a trapezoid means the line joining the mid-points of the $\mathbf{2}$ non-parallel sides of the trapezoid)

Sol. $A B C D$ is the given trapezoid, whose median or midline is $E F$.

Let $A H$ be drawn perpendicular to $D C$

$E G | D H$

$(\because E F | D C)$

$\therefore$ By Basic Proportionality Theorem,

$ \frac{A E}{E D}=\frac{A G}{G H} \Rightarrow \frac{A G}{G H}=1 \quad(\because E \text{ is the mid-point of } A D) $

$\Rightarrow A G=G H$

Let $A G=G H=h$ and $A B=x, E F=y$ and $C D=z$.

Then, $\quad$ Area $(A B F E)=h / 2(x+y)$

Area $(E F C D)=h / 2(y+z)$

$ \text{ Area }(A B C D)=h(x+z) $

Now given, $\frac{Area(A B F E)}{Area(E F C D)}=\frac{1}{2} \Rightarrow \frac{h / 2(x+y)}{h / 2(y+z)}=\frac{1}{2}$

$\Rightarrow \quad 2 x+2 y=y+z \quad \Rightarrow 2 x+y=z$

Also, $\quad \frac{\text{ Area }(A B F E)}{\text{ Area }(A B C D)}=\frac{1}{3} \quad \Rightarrow \frac{h / 2(x+y)}{h(x+z)}=\frac{1}{3}$

$\Rightarrow \quad 3 x+3 y=2 x+2 z \Rightarrow x+3 y=2 z$

$\Rightarrow$ From $(i)$ and $(i i)$

$ \begin{aligned} x+3 y & =2(2 x+y) \Rightarrow x+3 y=4 x+2 y \\ \Rightarrow \quad 3 x & =y \end{aligned} $

$\Rightarrow$ Putting in $(i)$, we get

$ 2 x+3 x=z \quad \Rightarrow \quad 5 x=z \quad \Rightarrow \quad x: z=1: 5 $

~~ Ex. 11 . Let $A B C D$ be a cyclic quadrilateral. Show that the incentres of the triangles $A B C, B C D, C D A$ and $D A B$ form a rectangle.

Sol. We know that incentre is the point of concurrence of the internal bisectors of the angles of a triangle.

Let $P, Q, R$ and $S$ be the incentres of the triangles $A B C, B C D, D A C$ and $D A B$ respectively.

$\Rightarrow P$ is point of intersection of $\angle B A C$ and $\angle A B C$

In $\triangle A P B, \angle A P B=180^{\circ}-[\angle P A B+\angle P B A]$

$\angle A P B=180^{\circ}-[\frac{\angle B A C}{2}+\frac{\angle A B C}{2}]$

But in $\triangle A B C, \angle A B C+\angle B A C+\angle B C A=180^{\circ}$

$\Rightarrow \frac{1}{2}(\angle A B C+\angle B A C)=90^{\circ}-\frac{\angle B C A}{2}$

From (i) and (ii), $\angle A P B=180^{\circ}-[90^{\circ}-\frac{\angle B C A}{2}]=90^{\circ}+\frac{\angle B C A}{2}$

Similarly, in $\triangle A S B, \angle A S B=180^{\circ}-[\angle S A B+\angle S B A]$

$\Rightarrow \angle A S B=180^{\circ}-[\frac{\angle D A B}{2}+\frac{\angle A B D}{2}]$

Also in $\triangle A D B$,

$ \begin{aligned} & \angle A D B+\angle D A B+\angle A B D=180^{\circ} \\ \Rightarrow & \frac{\angle D A B}{2}+\frac{\angle A B D}{2}=90^{\circ}-\frac{\angle A D B}{2} \end{aligned} $

$\therefore$ From $(i v)$ and $(v)$

$ \angle A S B=180^{\circ}-(90^{\circ}-\frac{\angle A D B}{2})=90^{\circ}+\frac{\angle A D B}{2} $

But $\angle B C A=\angle A D B$

(Angles in the same segment are equal)

$\therefore$ From (iii) and (vi)

$ \angle A P B=\angle A S B $

$\Rightarrow A S P B$ is a cyclic quadrilateral.

$ \begin{matrix} \Rightarrow & \angle S P B+\angle S A B & =180^{\circ} & \\ \Rightarrow & \angle S P B & =180^{\circ}-\angle S A B & \\ \angle S P B & =180^{\circ}-\angle A / 2 & (\because S A \text{ bisects } \angle A) \end{matrix} $

Similarly, $\angle B P Q=180^{\circ}-C / 2$

$ \begin{aligned} \Rightarrow \quad \angle S P Q & =360^{\circ}-(\angle S P B+\angle B P Q) \\ & =360^{\circ}-(180^{\circ}-\frac{\angle A}{2}+180^{\circ}-\frac{\angle C}{2}) \\ & =\frac{\angle A+\angle C}{2}=\frac{180^{\circ}}{2}=90^{\circ} \quad(\because A B C D \text{ is cylic quadrilateral }) \end{aligned} $

Thus, in quadrilateral $S P Q R, \angle P=90^{\circ}$

Similarly it can be shown that $\angle S=\angle Q=\angle R=90^{\circ}$

$\Rightarrow$ Quadrilateral is a rectangle.

~~ Ex. 12 . Let $A B C D$ be a quadrilateral. Let $X$ and $Y$ be the mid-points of $A C$ and $B D$ respectively and the lines through $X$ and $Y$ respectively parallel to $B D$ and $A C$ meet in $O$. Let $P, Q, R, S$ be the mid-points of $A B$, $B C, C D$ and $D A$ respectively. Prove that:

(a) Quadrilateral $A P O S$ and $A P X S$ have the same area.

(b) The areas of quadrilaterals $A P O S, B Q O P, C R O Q$ and $D S O R$ are all equal.

Sol. $A B C D$ is the given quadrilateral.

$P, Q, R$ and $S$ are the mid-points of sides $A B, B C, C D$ and $D A$ respectively.

$O X | B D$ and $O Y | A C$

(a) In $\triangle A B D$, the line joining the mid-points $P$ and $S$ of sides $A B$ and $A D$ respectively is parallel to the third side.

$\therefore P S | B D$.

Now $P S | B D$ and $O X|B D \Rightarrow O X| P S$.

$\because$ Area of triangles on the same base and between the same parallel lines are equal,

So, Area $(\triangle P X S)=Area(\triangle P O S)$

$\Rightarrow$ Area $(\triangle P A S)+Area(\triangle P X S)=Area(\triangle P A S)+Area(\triangle P O S)$

$\Rightarrow$ Area $(A P X S)=$ Area $(A P O S)$

(b) Area $(A P X S)=Area(\triangle A P X)+Area(\triangle A S X)$

$ =\frac{1}{2} Area(\triangle A B X)+\frac{1}{2} Area(\triangle A X D) $

( $\because X P$ is the median of $\triangle A X B$ and $X S$ is the median of $\triangle A D X$ and a median divides a triangle into two triangles of equal area)

$\because X$ is the mid-point of $A C, X B$ is the median of $\triangle A B C$ and $X D$ is the median of $\triangle A D C$.

So, Area $(A P X S)=\frac{1}{2} \times \frac{1}{2} Area(\triangle A B C)+\frac{1}{2} \times \frac{1}{2} Area(\triangle A D C)$

$ \begin{aligned} & =\frac{1}{4}[Area(\triangle A B C)+Area(\triangle A D C)] \\ & .=\frac{1}{4} \text{ Area (quad. } A B C D) \end{aligned} $

Now in part $(a)$, we have proved Area $(A P X S)=$ Area $(A P O S)$

$\Rightarrow Area(A P O S)=\frac{1}{4}$ Area (quad. $.A B C D)$

Similarly, it can be shown by symmetry that

$Area(B P O Q)=Area(C R O Q)=Area(D S O R)=\frac{1}{4} Area(A B C D)$

$\Rightarrow Area(A P O S)=Area(B P O Q)=Area(C R O Q)=Area(D S O R)$.

~~ Ex. 13 . The diagonals $A C$ and $B D$ of a cyclic quadrilateral $A B C D$ intersect at $P$. Let $O$ be the circumcentre of $\triangle A P B$ and $H$ be the orthocentre of $\triangle C P D$. Show that the points $H, P, O$ are collinear.

Sol. Let $O$ be the circumcentre (point of intersection of the perpendicular bisectors of the sides of a $\triangle$ ) of $\triangle A P B$ Join $A$ to $O$ and draw $O F \perp A P$

$O$ being the circumcentre of $\triangle A P B$, it is equidistant from the three vertices $A, B$ and $P$

$\therefore A O=O P \Rightarrow \triangle A O P$ is isosceles $\Rightarrow O F$ bisects $\angle A O P$

$\Rightarrow \angle F O P=\frac{1}{2} \angle A O P$

Also, $O$ being equidistant form the three vertices, we can consider a circle passing through the vertices $A, P$ and $B$ with $O$ as its centre.

Then, $\angle A B P=\frac{1}{2} \angle A O P$…

Also, $\quad \angle A B D=\angle A C D$

$\Rightarrow \quad \angle A B P=\angle P C D$

$\therefore$ From $(i),(i i)$ and (iii)

$ \angle F O P=\angle P C D $

Now in $\triangle F O P$ and $\triangle E P C$

$ \begin{matrix} \angle F O P & =\angle P C E & \text{ (Proved above) } \\ & \angle O F P & =\angle P E C=90^{\circ} \\ \Rightarrow \quad & & \\ \angle F O P & \sim \angle E P C & (A A \text{ similarity) } \end{matrix} $

$\Rightarrow \quad \angle F P O=\angle E P C$, which can only be equal if they are vertically opposite angles.

$\Rightarrow E O$ is a straight line.

$\Rightarrow H, P$ and $O$ are collinear.

~~ Ex. 14 . Let $A B C D$ be a convex quadrilateral. $P, Q, R$ and $S$ are the mid-points of $A B, B C, C D$ and $D A$ respectively such that the triangles $A Q R$ and $C S P$ are equilateral. Prove that $A B C D$ is a rhombus and determine its angles.

(RMO 2005)

Sol. In $\triangle D C B$, by mid-point theoram,

$ Q R=\frac{1}{2} B D $

Also, in $\triangle A B D$, by mid-point theoram,

$ \begin{aligned} P S & =\frac{1}{2} B D \\ \Rightarrow Q R & =P S . \end{aligned} $

Also, given $\triangle A Q R$ and $\triangle C S P$ are equilateral $\Delta s$ and $Q R=P S$

$\Rightarrow \triangle A Q R$ is congruent to $\triangle C S P$

$\Rightarrow A R=R Q=A Q=C S=C P=P S$

Also, $\triangle C E F \sim \triangle C S P \Rightarrow \angle C E F=60^{\circ} \quad(\triangle C S P$ is equilateral)

$\triangle A Q R$ being equilateral $\Rightarrow \angle A Q R=60^{\circ}$

$\therefore \quad \angle C E F=\angle A Q R \Rightarrow$ alternate angles are equal $\Rightarrow C S | Q A$

So, $C S=Q A$ (from (i)) and $C S | Q A \Rightarrow C S A Q$ is a parallelogram

$\Rightarrow S A | C Q$ and $S A=C Q \Rightarrow A D | B C$ and $A D=B C \Rightarrow A B C D$ is a parallelogram.

Let diagonals $A C$ and $B D$ bisect each other at $K$. Then,

$D K=\frac{B D}{2}=Q R=C S=A R$.

$\Rightarrow$ In $\triangle A D C$, medians, $A R, D K$ and $C S$ are all equal

$\Rightarrow \triangle A D C$ is equilateral $\Rightarrow A D=A B$.

$\therefore A B C D$ is a parallelogram with equal adjacent sides

$\Rightarrow A B C D$ is a rhombus.

~~ Ex. 15 . If a triangle and a convex quadrilateral are drawn on the same base and no part of the quadrilateral is outside the triangle, show that the perimeter of the triangle is greater than the perimeter of the quadrilateral.

(SAT 2000)

Sol. Let $A B C$ be the given triangle and $B C D E$, the given quadrilateral.

Produce $E D$ to meet $A B$ in $F$ and $A C$ in $G$.

Then, in $\triangle B E F, B E<F E+B F \quad$ (Triangle inequality)

In $\triangle D G C, \quad D C<D G+G C \quad$ (Triangle inequality)

Now perimeter of quadrilateral $B E D C$

$=B E+E D+D C+B C$

$<B F+F E+E D+D G+G C+B C$

$=B F+F G+G C+B C<B F+A F+A G+G C+B C$

$(\because$ In $\triangle A F G, F G<A F+A G)$

$=A B+A C+B C$

$=$ Perimeter of $\triangle A B C$.

$\therefore$ Perimeter of quad. $B E D C<$ Perimeter of $\triangle A B C$.

~~ Ex. 16 . $A B C D$ is a cyclic quadrilateral such that $A C \perp B D$. $A C$ meet $B D$ at $E$. Prove that $E A^{2}+E B^{2}+E C^{2}+E D^{2}$ $=4 R^{2}$, where $R$ is the radius of the circle.

Sol. Let $O$ be the centre of the circle, circumscribing the cyclic quadrilateral $A B C D$.

Let $O G$ and $O H$ be the perpendiculars from $O$ on $B D$ and $A C$ respectively.

Then $O G$ bisects $B D$, i.e., $B G=G D$

and $O H$ bisects $A C$, i.e., $A H=H C$

( $\because$ The perpendicular from the centre of the chord, bisects the chord)

Now,

$ \begin{aligned} B E^{2}+E D^{2} & =(B G+G E)^{2}+(G D-G E)^{2} \\ & =B G^{2}+G E^{2}+2 B G \cdot G E+G D^{2}+G E^{2}-2 G D \cdot G E \\ & =2 B G^{2}+2 G E^{2} \end{aligned} $

Also, $\quad A E^{2}+E C^{2}=(A H-E H)^{2}+(C H+E H)^{2}$

$ \begin{aligned} & =A H^{2}+E H^{2}-2 A H \cdot E H+C H^{2}+E H^{2}-2 C H \cdot E H \\ & =2 A H^{2}+2 E H^{2} \end{aligned} $

$\therefore E A^{2}+E B^{2}+E C^{2}+E D^{2}=2 A H^{2}+2 E H^{2}+2 B G^{2}+2 G E^{2}$

$ \begin{aligned} & =2(A H^{2}+G O^{2}+B G^{2}+O H^{2}) \quad(\therefore E H=G O, G E=O H) \\ & =2(A O^{2}+B O^{2})=2(R^{2}+R^{2})=\mathbf{4} \boldsymbol{R}^{2} \end{aligned} $

PRACTICE SHEET

LEVEL-1

Which one of the following statements is correct?

If the diagonals of a quadrilateral bisect each other at right angles, then the quadrilateral is:

(a) A rhombus but not a square

(b) A square but not a rhombus

(d) A rectangle but not a square

(CDS 2005)

~~ 2. If the diagonals of a quadrilateral are equal and bisect each other at right angles, then the quadrilateral is a

(a) rectangle

(b) square

(d) trapezium (CDS 2012-II)

~~ 3. If the bisectors of the angles $A$ and $B$ of a quadrilateral $A B C D$ meet at $O$, then $\angle A O B$ is equal to :

(a) $\angle C+\angle D$

(b) $\frac{1}{2}(\angle C+\angle D)$

(d) $\frac{1}{3} \angle C+\frac{1}{2} \angle D$

(CDS 2004)

~~ 4. If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following?

(a) Rectangle

(b) Square

(d) Parallelogram

(CDS 2006)

~~ 5. $A B C D$ is a cyclic quadrilateral such that $\angle A+\angle B$ $=2(\angle C+\angle D)$. If $\angle C>30^{\circ}$, then which one of the following is correct?

(a) $\angle D \geq 90^{\circ}$ (b)

(b) $\angle D<90^{\circ}$

(CDS 2006)

~~ 6. In a parallelogram $A B C D, M$ is the mid-point of $B D$ and $B M$ bisects $\angle B$. Then what is $\angle A M B$ equal to?

(a) $45^{\circ}$

(b) $60^{\circ}$

(d) $120^{\circ}$

~~ 7. If $A B C D$ is a rhombus, then

(CDS 2006)

(a) $A C^{2}+B D^{2}=6 A B^{2}$

(b) $A C^{2}+B D^{2}=5 A B^{2}$

(d) $A C^{2}+B D^{2}=3 A B^{2}$

~~ 8. In a trapezium $A B C D, A B$ is parallel to $C D$ and the diagonals intersect each other at $O$. In this case, the ratio $O A / O C$ is equal to:

(a) $\frac{O B}{O D}$

(b) $\frac{B C}{C D}$

(d) $\frac{A C}{B D}$

~~ 9. $A B C D$ is a trapezium where $A B$ and $C D$ are non-parallel sides. If the vertices $A, B, C$ and $D$ are concyclic, then

(a) $A B$ is also parallel to $C D$

(b) $A B=\frac{1}{2} C D$

(d) $A B=C D$

~~ 10. $X, Y$ are the mid-points of opposite sides $A B$ and $D C$ of a parallelogram $A B C D . A Y$ and $D X$ are joined intersecting in $P$. $C X$ and $B Y$ are joined intersecting in $Q$. Then $P X Q Y$ is

(a) rectangle

(b) rhombus

(d) parallelogram

~~ 11. If the sum of the diagonals of a rhombus is $12 cm$ and its perimeter is $8 \sqrt{5} cm$, then the lengths of the diagonals are:

(a) $6 cm$ and $6 cm$

(b) $7 cm$ and $5 cm$

(d) $9 cm$ and $3 cm$

~~ 12. In the given figure $A E=B C$ and $A E | B C$ and the three sides $A B, C D$ and $E D$ are equal in length. If $\angle A=102^{\circ}$, then find the measure of $\angle B C D$.

(a) $138^{\circ}$

(b) $162^{\circ}$

(d) $150^{\circ}$

~~ 13. If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, and the ratio of the perimeter of the rectangle and that of the parallelogram is $k$, then

(a) $k>1$

(b) $k<1$

(d) Cannot be determined

~~ 14. $A B C D$ is a parallelogram. The diagonals $A C$ and $B D$ intersect at the point $O$. If $E, F, G$ and $H$ are the mid-points of $A O$, $D O, C O$ and $B O$ respectively, then the ratio of $(E F+F G$ $+G H+H E)$ to $(A D+D C+C B+B A)$ is

(a) $1: 1$

(b) $1: 2$

(d) $1: 4$

~~ 15. A cyclic parallelogram having unequal adjacent sides is necessarily a:

(a) Square

(b) Rectangle

(d) Trapezium

(MAT)

LEVEL-2

~~ 16. In a quadrilateral $A B C D, \angle B=90^{\circ}$ and $A D^{2}=A B^{2}+B C^{2}$ $+C D^{2}$. Then $\angle A C D$ is equal to:

(a) $90^{\circ}$

(b) $60^{\circ}$

(d) None of these

(MAT 2003)

~~ 17. In a cyclic quadrilateral $P Q R S, \angle P$ is double its opposite angle and difference between the other two angles is onethird of $\angle P$. The minimum difference between any two angles of this quadrilateral is

(a) $30^{\circ}$

(b) $10^{\circ}$

(d) $40^{\circ}$

(MAT 2010)

~~ 18. $A B C D$ is a square. $P, Q, R$ and $S$ are points on the sides $A B$, $B C, C D$ and $D A$ respectively such that $A P=B Q=C R=D S$. What is $\angle S P Q$ equal to?

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $90^{\circ}$

~~ 19. The middle points of the parallel sides $A B$ and $C D$ of a parallelogram $A B C D$ are $P$ and $Q$ respectively. If $A Q$ and $C P$ divide the diagonal $B D$ into three parts $B X, X Y$ and $Y D$, then which one of the following is correct?

(a) $B X \neq X Y \neq Y D$

(b) $B X=Y D \neq X Y$

(d) $X Y=2 B X$

(CDS 2010)

~~ 20. Let $A B C D$ be a parallellogram. Let $m$ and $n$ be positive integers such that $n<m<2 n$. Let $A C=2 m n$ and

$B D=m^{2}-n^{2}$ and $A B=\frac{(m^{2}+n^{2})}{2}$

Statement I. $A C>B D$.

Statement II. $A B C D$ is a rhombus.

Which one of the following is correct in respect of the above statements?

(a) Both the statement I and II are true and statement II is the correct explanation of statement I.

(b) Both the statements I and II are true but statement II is not the correct explanation of statement I.

(d) Statement II is true but statement I is false.

~~ 21. In the given figure, $A B C D$ is a square in which $A O=A X$. What is $\angle X O B$ ?

(a) $22.5^{\circ}$

(b) $25^{\circ}$

(d) $45^{\circ}$

(CDS 2009)

~~ 22. If $P Q R S$ is trapezium such that $P Q>R S$ and $L, M$ are the mid-points of the diagonals $P R$ and $Q S$ respectively then what is $L M$ equal to?

(a) $\frac{P Q}{2}$

(b) $\frac{R S}{2}$

(d) $\frac{P Q-R S}{2}$

(CDS 2006)

~~ 23. In the given figure, $\angle A B C=\angle A E D=90^{\circ}$. Consider the following statements: I. $A B C$ and $A D E$ are similar triangles.

II. The four points $B, C, E$ and $D$ may lie on a circle.

(a) Only I

(b) Only II

(d) Neither I nor II

~~ 24. Let $X$ be any point within a square $A B C D$. On $A X$ a square $A X Y Z$ is described such that $D$ is within it. Which one of the following is correct?

(a) $A X=D Z$

(b) $\angle A D Z=\angle B A X$

(d) $B X=D Z$

(CDS 2012)

~~ 25. In the given figure, $Y Z$ is parallel to $M N, X Y$ is parallel is $L M$ and $X Z$ is parallel to $L N$. Then $M Y$ is

(a) Median of $\triangle L M N$

(b) The anguler bisector of $\angle L M N$

(d) Perpendicular bisector of $L N$.

~~ 26. The locus of a point in rhombus $A B C D$ which is equidistant from $A$ and $C$ is

(a) a fixed point on diagonal $B D$

(b) diagonal $B D$

(d) None of the above

~~ 27. $A B C D$ is a square. The diagonals $A C$ and $B D$ meet at $O$. Let $K, L$ be the points on $A B$ such that $A O=A K, B O=B L$. If $\theta=\angle L O K$, then what is the value of $\tan \theta$ ?

(a) $\frac{1}{\sqrt{3}}$

(b) $\sqrt{3}$

(d) $\frac{1}{2}$

(CDS 2008)

~~ 28. $A B C D$ is a trapezium in which $A B | D C$ and $A D=B C$. If $P$, $Q, R$ and $S$ be respectively the mid-points of $B A, B D, C D$ and $C A$, then $P Q R S$ is a

(a) Rhombus

(b) Rectangle

(d) Square

~~ 29. A rigid square plate $A B C D$ of unit side rotates in its own plane about the middle-point of $C D$ until the new position of $A$ coincides with the old position of $B$. How far is the new position of $B$ from the old position of $A$ ?

(a) 4 units

(b) $5 \sqrt{5}$ units

(d) $4 \sqrt{5}$ units

(RMO)

~~ 30. A trapezium $A B C D$ in which $A B | C D$ is inscribed in a circle with centre $O$. Suppose the diagonals $A C$ and $B D$ of the trapezium intersect at $M$ and $O M=2$. If $\angle A M B=60^{\circ}$, the difference between the lengths of the parallel sides is:

(a) 2

(b) $\sqrt{3}$

(d) $2 \sqrt{3}$

~~ 31. $A B C D$ is a trapezium with $A B$ and $C D$ as parallel sides. The diagonals intersect at $O$. The area of the triangle $A B O$ is $\boldsymbol{p}$ and that of triangle $C D O$ is $\boldsymbol{q}$. The area of the trapezium is:

(a) $\sqrt{p}+\sqrt{q}$

(b) $\frac{1}{3}(\sqrt{p}+\sqrt{q})^{3}$

(d) $(\sqrt{p}+\sqrt{q})^{2}$

~~ 32. $P Q R S$ is a rectangle in which $P Q=2 P S . T$ and $U$ are the mid-points of $P S$ and $P Q$ respectively. $Q T$ and $U S$ intersect at $V$. Find the ratio of the area of quadrilateral $Q R S V$ to the area of triangle $P Q T$.

(a) $4: 1$

(b) $8: 3$

(d) $7: 4$

~~ 33. A rhombus has sides of length 1 and area $\frac{1}{2}$. Find the angle between the two adjacent sides of the rhombus.

(a) $60^{\circ}$

(b) $75^{\circ}$

(d) $30^{\circ}$

~~ 34. In the given figure $A B C D$ is a quadrilateal whose diagonals intersect at $O . \angle A O B=30^{\circ}, A C=24$ and $B D=22$. The area of quadrilateral $A B C D$ is:

(a) 132

(b) 264

(d) 66

~~ 35. $A B C D$ is a square and $A O B$ is an equilateral triangle. What is the value of $\angle D O C$ ?

(a) $120^{\circ}$

(b) $150^{\circ}$

(d) $80^{\circ}$

~~ 36. In a trapezium $A B C D, A B$ is parallel to $C D, B D$ is perpendicular to $A D$. $A C$ is perpendicular to $B C$. If $A D=B C$ $=15 cm$ and $A B=25 cm$, then the area of the trapezium is

(a) $192 cm^{2}$

(b) $232 cm^{2}$

(d) $172 cm^{2}$

~~ 37. $A B C D$ is a rectangle with $B C=a, A B=b$ and $a>b$. If $M$ is a point on $A D$ such that $\angle B M A=\angle B M C$, then $M D$ is equal to:

(a) $\sqrt{a^{2}+b^{2}}$

(b) $\sqrt{a^{2}-b^{2}}$

(d) $a-b$

~~ 38. Let $A B C D$ be a square. $M, N, R$ are the points on $A B, B C$ and $C D$ respectively such that $A M=B N=C R$. If $\angle M N R$ is a right angle, then $\angle M R N$ is equal to

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $75^{\circ}$

~~ 39. If for a regular pentagon $A B C D E$, the lines $A D$ and $B E$ intersect at points $P$, then $\angle B A D$ and $\angle A P E$ respectively are

(a) $36^{\circ}$ and $72^{\circ}$

(b) $54^{\circ}$ and $108^{\circ}$

(d) $36^{\circ}$ and $108^{\circ}$

LEVEL-3

~~ 40. Let $A B C D$ be a parallelogram. $P$ is any point on the side $A B$. If $D P$ and $C P$ are joined in such a way that they bisect the angles $A D C$ and $B C D$ respectively, then $D C$ is equal to

(a) $C B$

(b) $2 C B$

(d) $4 C B$

~~ 41. The adjacent sides of a parallelogram are $2 a$ and $a$. If the angle between them is $60^{\circ}$, then one of the diagonals of the parallelogram is

(a) $3 a$

(b) $\sqrt{5} a$

(d) $\sqrt{3} a$

~~ 42. In the given figure, $O$ is the centre of the circle. The radius $O P$ bisects a rectangle $A B C D$ at right angles. $D M=N C=2 cm$ and $A R$ $=S B=1 cm, K S=4 cm$ and $O P$ $=5 cm$. What is the area of the rectangle?

(a) $8 cm^{2}$

(b) $10 cm^{2}$

(d) $16 cm^{2}$

~~ 43. In the given figure $O$ is the centre of the circle. $\angle A O D$ $=120^{\circ}$. If the radius of the circle be ’ $r$ ‘, then find the sum of the areas of quadrilateral $A O D P$ and $O B Q C$

(a) $\frac{\sqrt{3}}{2} r^{2}$

(b) $3 \sqrt{3} r^{2}$

(d) $2 \sqrt{3} r^{2}$

~~ 44. Let $A B C D$ be a quadrilateral with $\angle C B D=2 \angle A D B$,

$\angle A B D=2 \angle C D B$ and $A B=C B$. Then which of the following statement hold true.

I $A D=C D$

II $A D=B D$

III $B E D F$ is a parallelogram

IV $A B=C D$

(a) I and III only

(b) All of above

(d) I, III and IV

(Canadian Mathematical Olympiad 2000)

~~ 45. A square sheet of paper $A B C D$ is so folded that $B$ falls on the mid-point $M$ of $C D$. The crease will divide $B C$ in the ratio

(a) $7: 4$

(b) $5: 3$

(d) $4: 1$

~~ 46. $A B C D$ is a square. $P, Q, R, S$ are the mid-points of $A B, B C$, $C D$ and $D A$ respectively. By joining $A R, B S, C P, D Q$, we get a quadrilateral which is a

(a) trapezium (b) rectangle

(d) rhombus

~~ 47. Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to $a$ units. The distance between their feet being equal to $b$ units. The area of the rhombus is

(a) $\frac{\sqrt{a^{2}+b^{2}}}{2 \sqrt{b^{2}-a^{2}}}$

(b) $\frac{2 a b}{2 \sqrt{b^{2}-a^{2}}}$

(d) $\frac{2 a^{2} b^{2}}{2 \sqrt{b^{2}-a^{2}}}$

~~ 48. In a trapezoid $A B C D$, side $B C$ is parallel to side $A D$. Also, the lengths of the sides $A B, B C, C D$ and $A D$ are $8,2,8$ and 10 units respectively. Find the radius of the circle that passes through all four of the points $A, B, C$ and $D$ ?

(a) $2 \sqrt{3}$

(b) $2 \sqrt{5}$

(d) $2 \sqrt{7}$

~~ 49. The length of the midline of a trapezoid equals $4 cm$ and the base angles are $40^{\circ}$ and $50^{\circ}$. The length of the bases if the distance of their mid-points equals $1 cm$ is equal to

(a) $5 cm, 3 cm$

(b) $4 cm, 3 cm$

(d) $6 cm, 5 cm$

~~ 50. In the figure, find the value of

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F$.

(a) $120^{\circ}$

(b) $720^{\circ}$

(d) $540^{\circ}$

ANSWERS

1. (c)	2. $(b)$	3. (b)	4. (a)	5. $(b)$	6. (c)	7. (c)	8. (a)	9. $(d)$	10. $(d)$
11. (c)	12. (b)	13. (a)	14. $(b)$	15. (b)	16. (a)	17. $(b)$	18. $(d)$	19. (c)	20. $(b)$
21. (a)	22. (d)	23. $(c)$	24. $(d)$	25. (a)	26. (a)	27. (c)	28. (a)	29. (c)	. (d)
31. (d)	32. (b)	33. $(d)$	34. (a)	35. (b)	36. (a)	37. (b)	38. (b)	39. (c)	40. $(b)$
41. $(d)$	42. $(b)$	43. $(c)$	44. (a)	45. $(b)$	46. (c)	47. $(c)$	48. $(d)$	49. (a)	50. $(d)$

HINTS AND SOLUTIONS

~~ 3. In $\triangle A O B, \angle A O B=180^{\circ}-(\frac{1}{2} \angle A+\frac{1}{2} \angle B)$

(Angle sum property of a $\triangle, O A$ bisects $\angle A$ and $O B$ bisects $\angle B$ ) In quadrilateral $A B C D$,

$\angle A+\angle B+\angle C+\angle D=360^{\circ}$

$\Rightarrow \angle C+\angle D=360^{\circ}-(\angle A+\angle B)$

$\Rightarrow \frac{1}{2}(\angle C+\angle D)$

$=180^{\circ}-\frac{1}{2}(\angle A+\angle B)$

(ii)

$\therefore$ From $(i)$ and $(i i)$

$ \angle A O B=\frac{1}{2}(\angle C+\angle D) $

~~ 4. Let $l$ and $m$ be the two given parallel lines and $n$ the transversal intersecting them at points $P$ and $Q$ respectively. $P S, P R$ and $Q S, Q R$ are the bisectors of the interior angles at $P$ and $Q$ respectively.

Given, $\angle 1=\angle 2, \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8$

Also, $\angle 1+\angle 2=\angle 7+\angle 8$,

(alternateangles)

$\Rightarrow 2 \angle 2=2 \angle 7 \Rightarrow \angle 2=\angle 7$

Also, $\angle 3+\angle 4=\angle 5+\angle 6$,

(alternateangles)

$\Rightarrow 2 \angle 3=2 \angle 6 \Rightarrow \angle 3=\angle 6$

Adding (i) and (ii), we get $\angle 2+\angle 3=\angle 6+\angle 7$

$\Rightarrow \angle S P R=\angle S Q R$

Also, $\angle 1+\angle 2+\angle 3+\angle 4=180^{\circ}$

$\Rightarrow 2(\angle 2+\angle 3)=180^{\circ} \Rightarrow \angle 2+\angle 3=90^{\circ}$

$\Rightarrow \angle S P R=\angle S Q R=90^{\circ}$

Also, $\angle 1+\angle 2+\angle 5+\angle 6=180^{\circ}$

(co-int. $\angle s$ are supplementary)

$\Rightarrow 2(\angle 2+\angle 6)=180^{\circ} \Rightarrow \angle 2+\angle 6=90^{\circ}$

$\therefore$ In $\triangle P S Q, \angle P S Q=180^{\circ}-(\angle 2+\angle 6)$

$ =180^{\circ}-90^{\circ}=90^{\circ} $

Similarly, $\angle P R Q=90^{\circ}$.

As all four angles of $P R Q S$ are each $=90^{\circ} \Rightarrow P R Q S$ is a rectangle.

~~ 5. In a cyclic quadrilateral $A B C D$

$\Rightarrow \angle A+\angle C=180^{\circ} \Rightarrow \angle A=180^{\circ}-\angle C$ $\Rightarrow \angle B+\angle D=180^{\circ} \Rightarrow \angle B=180^{\circ}-\angle D$

Also, given that

$\angle A+\angle B=2(\angle C+\angle D)$

$\Rightarrow 180^{\circ}-\angle C+180^{\circ}-\angle D=2(\angle C+\angle D)$

$\Rightarrow 360^{\circ}=3(\angle C+\angle D)$

$\Rightarrow \angle C+\angle D=120^{\circ}$

Given $\angle C>30^{\circ} \Rightarrow \angle D=120^{\circ}-\angle C \Rightarrow \angle D<90^{\circ}$.

~~ 6. If $B M$ bisects $\angle B$, then $A M$ bisects $\angle A$ as diagonals of a parallelogram bisect each other and here $M$ is the point of intersection of the diagonals $A C$ and $B D$. Also, diagonals of a prallelogram bisect the angles at the vertices they join.

$\Rightarrow \angle A+\angle B=90^{\circ}$

(Sum of adjacent angles of a parallelogram)

$\Rightarrow \frac{1}{2}(\angle A+\angle B)=90^{\circ}$

$\Rightarrow \angle M A B+\angle M B A=90^{\circ}$

In $\triangle A M B, \angle M A B+\angle M B A+\angle A M B=180^{\circ}$

$\Rightarrow 90^{\circ}+\angle A M B=180^{\circ} \Rightarrow \angle A M B=\mathbf{9 0}^{\circ}$.

~~ 7. Diagonals of a rhombus bisect each other at right angles.

$\therefore$ In $\triangle A O B$,

$O A^{2}+O B^{2}=A B^{2}$

$(\because \angle A O B=90^{\circ})$

$\Rightarrow(\frac{1}{2} A C)^{2}+(\frac{1}{2} B D)^{2}=A B^{2}$

$\Rightarrow \frac{1}{4} A C^{2}+\frac{1}{4} B D^{2}=A B^{2}$

$\Rightarrow A C^{2}+B D^{2}=4 A B^{2}$.

~~ 8. In $\triangle A O B$ and $\triangle C O D$,

$\angle B A O=\angle D C O$

$\angle A B O=\angle C D O$

$\angle A O B=\angle C O D$

(vert. opp. $\angle s$ )

$\Rightarrow \triangle A O B \sim \triangle C O D$

$\therefore \frac{O A}{O B}=\frac{O C}{O D}$

(alternate angle)

$\Rightarrow \frac{O A}{O C}=\frac{\boldsymbol{O B}}{\boldsymbol{O D}}$.

~~ 9. If a trapezium is inscribed in a circle, i.e., it is a cyclic trapezium, then it is an isosceles trapezium, i.e., non-parallel sides are equal $\Rightarrow A B=C D$.

~~ 10. Since $X$ and $Y$ are the mid-points of $A B$ and $D C$ respectively. $A X=\frac{1}{2} A B$ and $C Y=\frac{1}{2} D C$

But $A B=D C$

$\Rightarrow \frac{1}{2} A B=\frac{1}{2} D C$

$\Rightarrow A X=C Y$

Also, $A B|D C \Rightarrow A X| C Y$

Since a pair of opposite sides, $A X$ and $Y C$ are equal and parallel, $A X C Y$ is a parallelogram. Similarly, we can show that $B X D Y$ is a parallelogram.

Now, $A X C Y$ is a parallelogram.

$\Rightarrow A Y | C X$

(opp. sides are parallel)

$\Rightarrow P Y | Q X$

Also, $B X D Y$ is a parallelogram $\Rightarrow B Y | X D$

$\Rightarrow Q Y | P X$

Thus, in quadrilateral $P Y Q X, P Y | Q X$ and $Q Y | P X$.

When both pair of opposite sides are parallel in a quadrilateral, it is a prallelogram.

~~ 11. Each side of the rhombus

$ =\frac{8 \sqrt{5}}{4}=2 \sqrt{5} $

Let $O D=a$ and $O C=b$ Then, in $\triangle C O D$,

$a^{2}+b^{2}=(2 \sqrt{5})^{2}=20$

Also, $2 a+2 b=12$

$\Rightarrow a+b=6$

$\therefore$ From $(i),(a+b)^{2}-2 a b=20$

$\Rightarrow 36-2 a b=20 \quad \Rightarrow \quad 2 a b=16 \quad \Rightarrow \quad a b=8$

Now from (ii) and (iii), we get $a=4, b=2$

$\therefore$ Diagonals are $8 cm$ and $4 cm$.

~~ 12. $A E | B C$ and $A E=B C \Rightarrow A E C B$ is a parallelogram.

(For a parallelogram one pair of opposite sides can be shown equal and parallel)

$\therefore \angle B C E=\angle B A E=102^{\circ}$

(Opposite angles of a parallelogram are equal)

$A B=E C \quad$ (Opposite sides of a parallelogram)

$E D=C D=E C$

$(\because A B=E D=C D)$

$\therefore \triangle E D C$ is equilateral $\Rightarrow \angle E C D=60^{\circ}$

$\therefore \angle B C D=\angle B A E+\angle E C D=102^{\circ}+60^{\circ}=162^{\circ}$.

~~ 13. $A B C D$ and $A B E F$ are parallelogram and rectangle respectively on the same base $A B$ having equal perpendicular height. Therefore, the area of both the figures is equal. Now, $A D>A F$ and $B C>B E$

(hypotenuse $>$ perpendicular)

$\therefore 2(A B+B C)>2(A B+B E)$

$\Rightarrow$ Perimeter of parallelogram $>$

Perimeter of rectangle

$\Rightarrow k>1$.

~~ 14. By mid-point theorem, in $\triangle s O D A, O C D, O B C, O B A$ respectively

$\frac{E F}{A D}=\frac{F G}{D C}=\frac{G H}{C B}=\frac{H E}{B A}=\frac{1}{2}$

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\ldots \ldots$.

then each ratio is equal to $\frac{a+c+e+\ldots . .}{b+d+f+\ldots}$.

$\Rightarrow \frac{(E F+F G+G H+H E)}{(A D+D C+C B+B A)}=\frac{\mathbf{1}}{\mathbf{2}}$.

~~ 15. In a cyclic parallelogram each angle is equal to $90^{\circ}$ as, sum of opposite angles is $180^{\circ}$ and opposite angles are equal. So it is definitely either a square or a rectangle. As the given cyclic parallelogram has unequal adjacent sides, it is a rectangle.

~~ 16. Given,

$A D^{2}=A B^{2}+B C^{2}+C D^{2}$

$=A C^{2}+C D^{2}$

(In $\triangle A B C, A B^{2}+B C^{2}=A C^{2}$ )

$\Rightarrow \angle A C D=\mathbf{9 0}^{\circ}$.

~~ 17. Given, $\angle P=2 \angle R$

Also, $\angle P+\angle R=180^{\circ}$

$ \angle Q+\angle S=180^{\circ} $

$\therefore$ From (i) and (ii)

$\Rightarrow \angle R=60^{\circ}$

$\Rightarrow \angle P=120^{\circ}$

Also, given $\angle Q-\angle S=\frac{1}{3} \angle P=\frac{1}{3} \times 120^{\circ}=40^{\circ}$

Solving (iii) and (iv) simultaneously, we get

$\angle Q=110^{\circ}, \angle S=70^{\circ}$

$\therefore$ Minimum difference between any two angles of the quadrilateral is $10^{\circ}$.

~~ 18. In $\triangle s S A P$ and $P B Q$,

$.\begin{matrix} A S=B P \\ A P=B Q\end{matrix} \rbrace$ Given

$\angle A=\angle B=90^{\circ}$

$\therefore \triangle S A P \cong \triangle P B Q$

$\Rightarrow S P=P Q$

$\angle S P A=\angle B Q P=x$ (say)

and $\angle P S A=\angle Q P B=y$ (say)

In $\triangle S A P$ or $\triangle P B Q$

$x+y=90^{\circ}$

$ \begin{aligned} \therefore \quad \angle S P Q & =180^{\circ}-(\angle S P A+\angle B P Q) \\ & =180^{\circ}-(x+y)=180^{\circ}-90^{\circ}=\mathbf{9 0}^{\circ} . \end{aligned} $

~~ 19. $A B C D$ is a parallelogram.

$\Rightarrow A B=C D$ and $A B | C D$

$\Rightarrow A P=Q C$ and $A P | Q C$

$\Rightarrow A P C Q$ is a parallelogram.

$\Rightarrow A Q | P C$

$\therefore$ By intercept theorem in $\triangle D X C$,

we have $\frac{D Y}{Y X}=\frac{D Q}{Q C}=1$

$\Rightarrow D Y=Y X$

Also using intercept theorem in $\triangle A B Y$,

$\frac{B X}{X Y}=\frac{B P}{P A}=\frac{1}{1} \Rightarrow B X=X Y$.

$\therefore B X=X Y=D Y$.

~~ 20. Given, $A C=2 mn$

$B D=m^{2}-n^{2}$

$A B=\frac{m^{2}+n^{2}}{2}$

By Apollonius theorem, we know that

$A C^{2}+B D^{2}=2(A B^{2}+B C^{2})$

$\Rightarrow(2 m n)^{2}+(m^{2}-n^{2})^{2}=2 \lbrace \frac{1}{4}(m^{2}+n^{2})^{2}+B C^{2} \rbrace$

$\Rightarrow 4 m^{2} n^{2}+m^{4}+n^{4}-2 m^{2} n^{2}=\frac{1}{2}(m^{2}+n^{2})^{2}+2 B C^{2}$

$\Rightarrow(m^{2}+n^{2})^{2}=\frac{1}{2}(m^{2}+n^{2})^{2}+2 B C^{2}$

$\Rightarrow 2 B C^{2}=\frac{1}{2}(m^{2}+n^{2})^{2} \Rightarrow B C^{2}=\frac{(m^{2}+n^{2})^{2}}{4}$

$\Rightarrow B C=\frac{m^{2}+n^{2}}{2}=A B$

$\Rightarrow A B C D$ is a rhombus as adjacent sides are equal.

Let $A C>B D \Rightarrow 2 m n>m^{2}-n^{2}$

$\Rightarrow(m+n)^{2}>2 m^{2}$, which is always true for every positive integer $m, n$ where $n<m<2 n$.

~~ 21. Let $\angle X O B=\theta$.

In $\triangle O X B$

$\angle X O B+\angle X B O+\angle O X B=180^{\circ}$

$\Rightarrow \theta+45^{\circ}+\angle O X B=180^{\circ}$

$\Rightarrow \angle O X B=180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta$.

Also, $\angle O X B+\angle O X A=180^{\circ}$

$\Rightarrow \angle O X A=180^{\circ}-\angle O X B=180^{\circ}-(135^{\circ}-\theta)=45^{\circ}+\theta$.

In $\triangle O A X, A O=O X$

$\Rightarrow \angle O X A=\angle A O X=45^{\circ}+\theta$

Also, we know that diagonals of a square intersect each other at right angles, so

$ \begin{aligned} & \Rightarrow \angle A O X+\angle X O B=90^{\circ} \\ & \Rightarrow 45^{\circ}+\theta+\theta=90^{\circ} \\ & \Rightarrow 2 \theta=45^{\circ} \Rightarrow \theta=\mathbf{2 2 . 5 ^ { \circ }} . \end{aligned} $

~~ 22. Produce $L M$ to $N$ which intersects side $R Q$ at $N$.

By mid-point theorem,

In $\triangle P Q R, L N | P Q$

and $L N=\frac{1}{2} P Q$ and

in $\triangle Q R S, M N | R S$

and $M N=\frac{1}{2} R S$

$\therefore L M=L N-M N=\frac{\mathbf{1}}{\mathbf{2}}(\boldsymbol{P Q}-\boldsymbol{R} \boldsymbol{S})$.

~~ 23. In $\triangle s A B C$ and $\triangle A E D$

$\angle A B C=\angle A E D=90^{\circ}$

$\angle A=\angle A$

$\therefore \angle A B C \sim \angle A E D$ (AA similarly)

Also, $\angle D E C=180^{\circ}-90^{\circ}=90^{\circ}$.

In quadrilateral $E D B C$

$\angle E+\angle D+\angle C+\angle B=360^{\circ} \Rightarrow \angle C+\angle D=180^{\circ}$

$\therefore$ Opposite angles are supplementary, $E D B C$ is cylic quadrilateral.

~~ 24. In $\triangle A B X$ and $\triangle A D Z$,

$ A B=A D $

(Sides of square $A B C D$ )

$A X=A Z$

(Sides of suqare $A X Y Z$ )

Let $\angle B A X=\theta$

$\therefore \angle X A D=90^{\circ}-\theta$

Also, $A X Y Z$ being a square, $\angle Z A X=90^{\circ}$

$\Rightarrow \angle Z A D+\angle X A D=90^{\circ}$

$\Rightarrow \angle Z A D=90^{\circ}-\angle X A D$

$=90^{\circ}-(90^{\circ}-\theta)=\theta$

$\Rightarrow \angle B A X=\angle Z A D$

$\therefore \triangle A B X \cong \triangle A D Z$

$\Rightarrow B X=D Z$.

~~ 25. Since $Z Y | M N$ and $Z X | Y N$, $X N Y Z$ is a parallelogram

$\therefore Z X=Y N$

Also, $Z X | Y N$ and $X Y | Z L$

$(\because X Y | Z M)$

$\Rightarrow X Y L Z$ is a parallelogram

$\therefore Z X=L Y$

From (i) and (ii), $Y N=L Y$ $\Rightarrow M Y$ is the median of $\triangle L M N$.

~~ 26. In a rhombus, diagonals bisect each other. Suppose in the given rhombus $A B C D$, diagonals bisect each other at point $O$. Then, the distances of $O$ from the four vertices $A, B, C$ and $D$ are equal. Thus, even if we take fixed point on diagonal $B D$ as $O_1, O_2, O_3, O_4, \ldots$, they all are equidistant from the vertices $A$ and $C$ (by property of congruent triangles). Hence the locus of a point in rhombus $A B C D$ which is

equidistant from $A$ and $C$ is a fixed point on diagonal $B D$.

Alternatively, since diagonals of a rhombus bisect each at rt. $\angle s$, therefore, $BD$ is the single bisector of $AC$, and therefore, all points equidistant from $A$ and $C$ lie on it.

~~ 27. Let each side of the square be $a$

Then, $A C=a \sqrt{2}$ and $A O=O C=\frac{a}{\sqrt{2}}$

$A M=\frac{a}{2}, A K=A O=\frac{a}{\sqrt{2}}$

$LM=\frac{a}{\sqrt{2}}-\frac{a}{2}$ and $O M=\frac{a}{2}$

In $\triangle O M L, \tan \frac{\theta}{2}=\frac{L M}{O L}$

$=\frac{\frac{a}{\sqrt{2}}-\frac{a}{2}}{\frac{a}{2}}=\frac{\frac{\sqrt{2}-1}{2}}{\frac{1}{2}}=\sqrt{2}-1$

Now, $\tan \theta=\frac{2 \tan \theta / 2}{1-\tan ^{2} \theta / 2} \quad(\because \tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x})$

$=\frac{2(\sqrt{2}-1)}{1-(\sqrt{2}-1)^{2}}=\frac{2(\sqrt{2}-1)}{1-3+2 \sqrt{2}}=\frac{2(\sqrt{2}-1)}{2 \sqrt{2}-2}$

$\Rightarrow \tan \theta=1$.

~~ 28. In $\triangle B D C, Q$ and $R$ are the mid-points of $B D$ and $C D$ respectively.

$\therefore Q R | B C$ and $Q R=\frac{1}{2} B C$

Similarly, in $\triangle A B C, P S | B C$ and $P S=\frac{1}{2} B C$

$\therefore P S=Q R$

Also in $\triangle A B D, P Q | A D$ and $P Q=\frac{1}{2} A D$

In $\triangle A D C, S R | A D$ and $S R=\frac{1}{2} A D$

$\therefore P Q=S R$

$(A D=B C \Rightarrow P S=Q R=P Q=S R \Rightarrow P Q R S$ is a rhombus.

~~ 29. Let $O$ be the mid-point of $C D$. Since the new position of $A$ coincides with the old position of $B$, the rotation is in the counter-clockwise direction about $O$, through the angle $A O B$.

Let $B^{\prime}, C^{\prime}, D^{\prime}$ be the new positions of $B, C, D$, respectively. Also, let $O B$ and

$A B^{\prime}$ intersect at point $P$. We can also see that $\angle A O B=\angle B O B^{\prime}$ Also, $O A=O B=O B^{\prime}$. This means that $O B$ is the angle bisector of $\angle A O B^{\prime}$ of isosceles triangle $A O B^{\prime}$.

This implies that $O P \perp A B^{\prime}$ and $O P$ bisects $A B^{\prime}$

$\Rightarrow A P=P B^{\prime}$

Area of $\triangle A O B=\frac{1}{2}$ (Area of square $A B C D$ )

$=\frac{1}{2}$ square units

But area of $\triangle A O B=\frac{1}{2} \times A P \times O B \quad$… (ii) $(\because A P \perp O P)$

$O B=\sqrt{O C^{2}+C B^{2}}=\sqrt{(\frac{1}{2})^{2}+1^{2}}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

$\therefore$ From $(i)$ and (ii),

$\frac{1}{2}=\frac{1}{2} \times A P \times \frac{\sqrt{5}}{2}$

$\Rightarrow A P=\frac{2}{\sqrt{5}} \Rightarrow A B^{\prime}=2 A P=2 \times \frac{2}{\sqrt{5}}=\frac{4}{\sqrt{5}}=\frac{\mathbf{4} \sqrt{5}}{\mathbf{5}}$.

~~ 30. Since the trap. $A B C D$ is cyclic trapezium, therefore, it is an isosceles trapezium.

It is given that,

$\angle A M B=60^{\circ}$

$\Rightarrow \angle C M D=60^{\circ}$ (vert. opp. $\angle s$ )

Since $A M=B M$ and $D M=C M$,

(Property of Isos. trap.)

$\triangle A M B$ and so $\triangle C M D$ are equilateral $\triangle s$.

Draw $O P \perp B D$. Now $O M$ bisects $\angle A M B$ and so $\angle O M P$ $=30^{\circ}$

In $\triangle O M P$,

$\sin 30^{\circ}=\frac{O P}{O M} \Rightarrow O P=\frac{O M}{2}=\frac{2}{2}=1$

$\therefore \angle O P M=90^{\circ}, \therefore P M=\sqrt{O M^{2}-O P^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$

$\therefore A B-C D=A M-C D=B M-M D$

$(\because A B=A M$ and $C D=M D)$

$=(B P+P M)-(P D-P M)=B P+P M-B P+P M$

$(\because$ Perpendicular drawn from the centre bisects the chord, i.e., $B P=P D$ )

$=2 P M=2 \times \sqrt{3}=\mathbf{2} \sqrt{\mathbf{3}}$.

~~ 31. $\angle B D C=\angle D B A$

(alternate angles)

$\angle A C D=\angle C A B$

(alternate angles)

$\angle D O C=\angle A O B$

(vert. opp. $\angle s$ )

$\Rightarrow \triangle D C O \sim \triangle B A O$

$\Rightarrow \frac{\text{ Area of } \triangle A B O}{\text{ Area of } \triangle D C O}=\frac{A B^{2}}{D C^{2}}$

(By property of similar triangles)

$\Rightarrow \frac{p}{q}=\frac{A B^{2}}{D C^{2}} \Rightarrow \frac{A B}{D C}=\frac{\sqrt{p}}{\sqrt{q}}$

$\Rightarrow A B=\sqrt{p} \cdot k$ and $D C=\sqrt{q} \cdot k$ for some constant $k$.

Area of $\triangle D O C=\frac{1}{2} \times D C \times O L$

$\Rightarrow q=\frac{1}{2} \times \sqrt{q} \cdot k \times O L \Rightarrow O L=\frac{2 q}{\sqrt{q} \cdot k}=\frac{2 \sqrt{q}}{k}$

Area of $\triangle A O B=\frac{1}{2} \times A B \times O M$

$\Rightarrow p=\frac{1}{2} \times \sqrt{p} \cdot k \times O M \Rightarrow O M=\frac{2 q}{\sqrt{p} \cdot k}=\frac{2 \sqrt{p}}{k}$

Area of a trapezium $=\frac{1}{2} \times$ height of trapezium

$\times$ sum of parallel sides

$=\frac{1}{2} \times(O L+O M) \times(D C+A B)$

$=(\frac{\frac{2 \sqrt{q}}{k}+\frac{2 \sqrt{p}}{k}}{2})(\sqrt{q} k+\sqrt{p} k)$

$=(\sqrt{p}+\sqrt{q})^{2}$.

~~ 32. $U$ and $T$ are mid-points of $P Q$ and $P S$ respectively.

$\Rightarrow S U$ and $Q T$ are medians of $\triangle P S Q$.

$\Rightarrow V$ is the centroid of $\triangle P S Q$.

$\therefore \frac{\text{ Area of quad. } Q R S V}{\text{ Area of } \triangle P Q T}=\frac{\text{ Area of } \triangle Q R S+\text{ Area of } \triangle Q S V}{\frac{\text{ Area of } \triangle P Q S}{2}}$

$=\frac{\text{ Area of } \triangle P Q S+\frac{\text{ Area of } \triangle P Q S}{3}}{\frac{\text{ Area of } \triangle P Q S}{2}}$

$\because Q S$ divides rectangle $P Q R S$ into two equal $\triangle s P Q S$ and $Q R S$.

Also,Area of $\triangle Q S V=$ Area of $.\frac{\Delta P Q S}{3}]$

~~ 33. Let acute the angle between adjacent sides of the rhombus be $\theta$ and let each side of the rhombus $=\boldsymbol{a}$ units. Then,

Area of rhombus $A B C D=a^{2} \sin \theta$

Given, area $=\frac{1}{2}$ and $a=1$

$\therefore \frac{1}{2}=1 \times \sin \theta$

$\Rightarrow \sin \theta=\frac{1}{2} \Rightarrow \theta=\mathbf{3 0}^{\circ}$.

~~ 34. Area of quad. $A B C D$

$=$ Area $(\triangle A O B)+Area(\triangle B O C)$ + Area $(\triangle C O D)+Area(\triangle D O A)$

$=\frac{1}{2} \cdot O A \cdot O B \sin 30^{\circ}+\frac{1}{2} O B \cdot O C \sin 150^{\circ}$

$+\frac{1}{2} O C \cdot O D \sin 30^{\circ}+\frac{1}{2} O D \cdot O A \sin 150^{\circ}$

$=\frac{1}{4} O A \cdot O B+\frac{1}{4} O B \cdot O C+\frac{1}{4} O C \cdot O D+\frac{1}{4} O D \cdot O A$

$(\because \sin 30^{\circ}=\sin 150^{\circ}=\frac{1}{2})$

$=\frac{1}{4}(O A+O C)(O B+O D)=\frac{1}{4} \times A C \times B D$

$=\frac{1}{4} \times 24 \times 22=\mathbf{1 3 2}$.

~~ 35. $\triangle A O B$ is equilateral

$\Rightarrow A O=O B=A B$ and

$\angle O A B=\angle O B A=\angle A O B=60^{\circ}$

Also, $A B C D$ being a square,

$A B=B C=C D=A D$

$\Rightarrow A O=A D$ and $B O=B C$

$\angle D A O=\angle C B O=30^{\circ} \quad(\because \angle O A B=\angle O B A=60^{\circ})$

In $\triangle A D O, A O=A D \quad \Rightarrow \quad \angle A D O=\angle A O D$

$=\frac{1}{2}(180^{\circ}-\angle D A O)=\frac{1}{2}(180^{\circ}-30^{\circ})=75^{\circ}$

Similarly, $\angle B O C=75^{\circ}$.

$\therefore \angle D O C=360^{\circ}-(\angle A O B+\angle A O D+\angle B O C)$

$ =360^{\circ}-(60^{\circ}+75^{\circ}+75^{\circ})=360^{\circ}-210^{\circ}=\mathbf{1 5 0}^{\circ} . $

~~ 36. By Pythagoras’ Theorem,

$A C=\sqrt{A B^{2}-B C^{2}}$ and $B D=\sqrt{A B^{2}-A D^{2}}$

$\Rightarrow A C=B D=\sqrt{625-225}=\sqrt{400}=2 cm$

(By prop.of isos. trap. $AC=BD$ )

Now, Area of $\triangle D A B=\frac{1}{2} \times A D \times B D$

Also, Area of $\triangle D A B=\frac{1}{2} \times D M \times A B$

$\therefore A D \times B D=D M \times A B$

$\Rightarrow D M=\frac{A D \times B D}{A B}=\frac{15 \times 20}{25}=12 cm$

Also, $C N=D M=12 cm$

$A M=\sqrt{A D^{2}-D M^{2}}=\sqrt{15^{2}-12^{2}}=\sqrt{225-144}$

$ =\sqrt{81}=9 cm $

Also, $B N=A M=9 cm$

$M N=A B-(A M+B N)=25-(9+9)=25-18=7 cm$

$\Rightarrow C D=M N=7 cm$

$\therefore$ Area of trapezium $A B C D=\frac{1}{2} \times D M \times(A B+C D)$

$=\frac{1}{2} \times 12 \times(25+7)=\frac{1}{2} \times 12 \times 32=\mathbf{1 9 2} \mathbf{c m}^{2}$.

~~ 37. Let $\angle B M A=\angle B M C=x$

Then $\angle C M D=\pi-2 x$

Let $M D=y$. Then $A M=(a-y)$

In $\triangle C M D$,

$\tan (\pi-2 x)=\frac{b}{y}$

In $\triangle B A M$,

$\tan x=\frac{b}{a-y}$

From $(i)$, we have $\tan (\pi-2 x)=-\tan 2 x=\frac{b}{y}$

$\Rightarrow-\frac{2 \tan x}{1-\tan ^{2} x}=\frac{b}{y}$

$\Rightarrow-\frac{2 b /(a-y)}{1-(b /(a-y))^{2}}=\frac{b}{y}$

$\Rightarrow-\frac{2 b(a-y)}{(a-y)^{2}-b^{2}}=\frac{b}{y}$

$\Rightarrow-\frac{2 b a-2 b y}{a^{2}+y^{2}-2 a y-b^{2}}=\frac{b}{y}$

$\Rightarrow-2 b y a+2 b y^{2}=b a^{2}+b y^{2}-2 a b y-b^{3}$

$\Rightarrow b y^{2}=b(a^{2}-b^{2})$

$\Rightarrow y=\sqrt{\boldsymbol{a}^{2}-\boldsymbol{b}^{2}}$.

~~ 38. Let each side of the square $=x$ units

Let $A M=B N=C R=y$ units

$\Rightarrow M B=C N=D R$

$=(x-y)$ units

In $\triangle s M B N$ and $N C R$

$M B=N C=(x-y)$

$B N=C R=y$

$\angle M B N=\angle N C R=90^{\circ}$

$\Rightarrow \triangle M B N \cong \triangle N C R$

$ \Rightarrow M N=N R $

Given, $\angle M N R=90^{\circ}$

$\Rightarrow \triangle M N R$ is an isosceles right angled triangle.

$\Rightarrow \angle M R N=\angle N M R=\mathbf{4 5}^{\circ}$.

~~ 39. Sum of the interior angles of a regular pentagon $=540^{\circ}$

$\Rightarrow$ Each interior angle $=\frac{540^{\circ}}{5}=108^{\circ}$

As $A D | B C$

$\angle B A D+\angle A B C=180^{\circ}$ (co-int. $.\angle s)$

$\Rightarrow \angle B A D=180^{\circ}-108^{\circ}=\mathbf{7 2}^{\circ}$

$\Rightarrow \angle P A E=108^{\circ}-72^{\circ}=36^{\circ}$

Also, applying $B E | D C, \angle B E D=72^{\circ} \Rightarrow \angle P E A=36^{\circ}$

$\therefore \angle A P E=180^{\circ}-(\angle P A E+\angle P E A)$

(Angle sum property of a $\triangle$ )

$=180^{\circ}-(36^{\circ}+36^{\circ})=180^{\circ}-72^{\circ}=\mathbf{1 0 8}^{\circ}$.

~~ 40. Given, $D P$ bisects $\angle C D A$

$\Rightarrow \angle C D P=\angle A D P$

$\Rightarrow \angle A P D=\angle C D P$

(alt. $\angle s C D | A B$ )

$\Rightarrow A D=A P$ (Isosceles $\triangle$ property)

Similarly, $\angle C P B=\angle P C D=\angle P C B$

$\Rightarrow B P=B C$

$D C=A B=A P+B P=A D+B C=\mathbf{2 B C}$.

~~ 41. In $\triangle D A E$,

$\cos 60^{\circ}=\frac{A E}{a}, \sin 60^{\circ}=\frac{D E}{a}$

$\Rightarrow A E=\frac{1}{2} a, D E=\frac{\sqrt{3}}{2} a$

$\therefore B E=B A-A E$

$=2 a-\frac{1}{2} a=\frac{3 a}{2}$

$\therefore$ In right angled $\triangle B D E$,

$B D=\sqrt{B E^{2}+D E^{2}}=\sqrt{(\frac{3 a}{2})^{2}+(\frac{\sqrt{3}}{2} a)^{2}}$

$=\sqrt{\frac{9 a^{2}}{4}+\frac{3 a^{2}}{4}}=\sqrt{\frac{12 a^{2}}{4}}=\sqrt{3} a$.

~~ 42. $O S=O P=O N=$ radii of the circle $=5 cm$

$ \begin{matrix} (O S)^{2} & =(O K)^{2}+(K S)^{2} \\ \Rightarrow & 25 & =(O K)^{2}+16 \\ \Rightarrow & (O K)^{2} & =25-16=9 \\ \Rightarrow & O K & =3 cm \\ & & L C & =K B \\ & L N+N C & =K S+S B \end{matrix} $

$\Rightarrow L N+2=(4+1) cm \Rightarrow L N=3 cm$

Similarly, $(O N)^{2}=(O L)^{2}+(L N)^{2}$

$\Rightarrow \quad 25=(O L)^{2}+9$ $\Rightarrow \quad(O L)^{2}=16 \Rightarrow O L=4 cm$

$\therefore \quad L K=O L-O K=(4-3) cm=1 cm$

Also, $\quad A B=D C=10 cm$

$\therefore$ Area of rectangle $A B C D=A B \times B C=10 cm \times 1 cm$

$=\mathbf{1 0} \mathbf{c m}^{2}$

~~ 43. Let $O Q=O B=O C=r$ (radius of the circle)

Given, $\angle A O D=\angle B O C=120^{\circ}$

$\therefore \angle B O Q=\angle C O Q=60^{\circ}$

$\Rightarrow \frac{S B}{O B}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\Rightarrow S B=\frac{r \sqrt{3}}{2}$

$\therefore B C=2 S B=r \sqrt{3}$

$\therefore$ Area of quadrilateral $B Q C O$

$=\frac{1}{2} \times$ product of diagonals

$=\frac{1}{2} \times B C \times O Q$

$=\frac{1}{2} \times r \sqrt{3} \times r=\frac{r^{2} \sqrt{3}}{2} cm^{2}$

$\therefore$ Area of both the quadrilaterals

$=2 \times \frac{r^{2} \sqrt{3}}{2}=r^{2} \sqrt{3} cm^{2}$.

~~ 44. Given, $A B C D$ is a quadrilateral in which,

(1) $A B=B C$

(2) $\angle A B D=2 \angle B D C$

(3) $\angle D B C=2 \angle A D B$

Let $B E$ and $B F$ be the bisectors of $\angle A B D$ and

$\angle C B D$ respectively.

Then, $\angle A B E=\angle E B D=\angle B D F=y$ and

$ \angle C B F=\angle F B D=\angle B D E=x $

Now, $\angle E D B=\angle F B D$ and $\angle E B D=\angle B D F$

$\Rightarrow$ alternate angles are equal $\Rightarrow E D | B F$ and $B E | F D$

$\Rightarrow B E D F$ is a parallelogram.

Let $E F$ and $B D$ cut each other at $P$ and $A C$ and $B D$ cut each other at $Q$

$\therefore$ diagonals of a parallelogram bisect each other,

$E P=P F$

Now by the bisector theoram

$\frac{A E}{E D}=\frac{A B}{B D}=\frac{B C}{B D}=\frac{C F}{F D}$

$(\because A B=B C)$

$\Rightarrow \frac{A E}{E D}=\frac{C F}{F D}$

$\Rightarrow$ By the converse of Basic proportionality theoram,

$E F | A C$

$\therefore E P=P F \quad \Rightarrow \quad A Q=Q C$

$\therefore \frac{A B}{A Q}=\frac{B C}{Q C} \Rightarrow B Q$ bisects $\angle A B C \Rightarrow x=y$

$\Rightarrow \triangle A B D \cong \triangle B C D \Rightarrow A D=C D$.

~~ 45. By physically folding the square sheet of paper as per the conditions stated in the question, we can see that $E F$ is the crease along which the fold is made. Also,

$M F=F B$ and $M B \perp E F$.

We need to find $B F: F C$.

Let each side of the square be $a$ units and $C F=x$ units Then, $B F=(a-x)$ units and $M F=(a-x)$ units.

In $\triangle M F C, M F^{2}=M C^{2}+F C^{2}$

$($ Since $M F=F B)$

$\Rightarrow F B^{2}=M C^{2}+F C^{2}$

$\Rightarrow(a-x)^{2}=(a / 2)^{2}+x^{2}$

$\Rightarrow a^{2}+x^{2}-2 a x=a^{2} / 4+x^{2}$

$\Rightarrow 2 a x=a^{2}-\frac{a^{2}}{4}=\frac{3 a^{2}}{4}$

$\Rightarrow x=\frac{3 a}{8} \Rightarrow a-x=a-\frac{3 a}{8}=\frac{5 a}{8}$

$\therefore B F: F C=(x-a): x$

$ =\frac{5 a}{8}: \frac{3 a}{8}=5: 3 $

~~ 46. The quadrilateral formed by joining $A R, B S, C P$ and $D Q$ is EFGH.

Now, $\triangle D C Q \cong \triangle P B C$ as,

$C D=B C \quad$ (sides of a square)

$\angle D C Q=\angle P B C=90^{\circ}$

$C Q=P B$

(Half of sides of a square)

$\therefore \angle C D Q=\angle B C P=a$ (say)

Then, $\angle D Q C=\angle E Q C=90^{\circ}-a$

$\therefore$ In $\triangle E Q C, \angle C E Q=180^{\circ}-(\angle Q C E+\angle C Q E)$

$=180^{\circ}-(a+90^{\circ}-a)=90^{\circ}$

$\Rightarrow \angle F E H=90^{\circ}$

Similarly, $\angle E F G=\angle F G H=\angle G H E=90^{\circ}$

$\Rightarrow$ Vertex angles of quadrilateral $E F G H$ are $90^{\circ}$ each $\Rightarrow E F G H$ is a rectangle.

Now, $\triangle D R F \cong \triangle C Q E \quad \Rightarrow \quad R F=E Q, D F=E C$

Following the same concept,

$B H=C E=A G=D F$ and

$R F=E Q=P H=G S$

$\therefore E F=F G$

(Adjacent sides of $E F G H$ are equal)

$\therefore$ A rectangle whose adjacent sides are equal is a square

$\Rightarrow E F G H$ is a square.

~~ 47. Let $A E$ and $C F$ be the perpendiculars drawn from vertex $A$ and vertex $C$ respectively on sides $C D$ and $A B$.

$\Rightarrow E$ and $F$ are the feet of the perpendiculars on the sides.

Given, $A E=C F=a$ and $E F=b$.

$ A F=E C=\sqrt{b^{2}-a^{2}} $

$\therefore$ Area of $A F C E=AE \times AF=a \times \sqrt{b^{2}-a^{2}}$

Let $F B=x$. Then, $C B=\sqrt{a^{2}+x^{2}}$

As $A B C D$ is a rhombus,

$ \begin{aligned} & C B=A B \\ \Rightarrow \quad & \sqrt{a^{2}+x^{2}}=\sqrt{b^{2}-a^{2}}+x \\ \Rightarrow \quad & a^{2}+x^{2}=b^{2}-a^{2}+x^{2}+2 x \sqrt{b^{2}-a^{2}} \\ \Rightarrow \quad & 2 a^{2}-b^{2}=2 x \sqrt{b^{2}-a^{2}} \\ \Rightarrow & x=\frac{2 a^{2}-b^{2}}{2 \sqrt{b^{2}-a^{2}}} \end{aligned} $

$\therefore$ Area of rhombus $A B C D=$ base $\times$ height $=A B \times A E$

$ \begin{aligned} & =A B \times a=(\sqrt{b^{2}-a^{2}}+\frac{2 a^{2}-b^{2}}{2 \sqrt{b^{2}-a^{2}}}) \times a \\ & =(\frac{2(b^{2}-a^{2})+2 a^{2}-b^{2}}{2 \sqrt{b^{2}-a^{2}}}) a=\frac{\boldsymbol{a} \boldsymbol{b}^{2}}{\mathbf{2} \sqrt{\boldsymbol{b}^{2}-\boldsymbol{a}^{2}}} \end{aligned} $

~~ 48. Let $A B C D$ be the given trapezoid. Let $E$ and $F$ be the mid-points of the sides $B C$ and $A D$ respectively of trapezoid $A B C D$. By symmetry, the centre of the circle passing through the points $A, B, C$ and $D$ lies on $E F$. Let the centre be $O$.

Then, $O B=O A=r$ (say) (radii of the circle)

Let $h$ be the height of the altitude $E F$ of the trapezoid $A B C D$. Then, as we can see in the diagram, in $\triangle C G D$,

$h^{2}+4^{2}=(8)^{2} \Rightarrow h^{2}=64-16=48 \quad \Rightarrow \quad h=4 \sqrt{3}$

Now let $E O=x$, so $O F=h-x=4 \sqrt{3}-x$

In $\triangle O B E, O B=r=\sqrt{O E^{2}+B E^{2}}=\sqrt{x^{2}+1}$

In $\triangle O A F, O A=r=\sqrt{O F^{2}+A F^{2}}=\sqrt{(4 \sqrt{3}-x)^{2}+5^{2}}$

$\therefore$ From $(i)$ and (ii)

$\sqrt{x^{2}+1}=\sqrt{(4 \sqrt{3}-x)^{2}+5^{2}}$

$\Rightarrow x^{2}+1=48-8 \sqrt{3} x+x^{2}+25$

$\Rightarrow 8 \sqrt{3} x=72 \Rightarrow x=\frac{72}{8 \sqrt{3}}=3 \sqrt{3}$ $\therefore r=\sqrt{x^{2}+1}=\sqrt{(3 \sqrt{3})^{2}+1}=\sqrt{28}$

$\Rightarrow r=2 \sqrt{7}$.

~~ 49. Let $A B C D$ be the given trapezoid, whose base angles $\angle A$ and $\angle B$ equal $40^{\circ}$ and $50^{\circ}$. Also mid-line $E F=4 cm$.

Extend $A D$ and $B C$ to meet at $P$.

Now in $\triangle A P B$,

$\angle A P B=180^{\circ}-(40^{\circ}+50^{\circ})=90^{\circ}$

Also, $\triangle A B P \sim \triangle D C P$

$\Rightarrow \frac{A B}{D C}=\frac{A P}{D P}=\frac{B P}{C P}$

Let $Q$ and $R$ be the mid-points of $D C$ and $A B$ respectively.

$\therefore D C=2 D Q$ and $A B=2 A R$

$\therefore$ From $(i)$

$\frac{A B}{D C}=\frac{A P}{D P}=\frac{B P}{C P} \Rightarrow \frac{2 A R}{2 D Q}=\frac{A P}{D P}=\frac{B P}{C P}$

$\Rightarrow \frac{A R}{D Q}=\frac{A P}{D P}=\frac{B P}{C P}$

$\Rightarrow \triangle P D Q$ and $\triangle P A R$ are similar $\Rightarrow P, Q, R$ are collinear.

Now, as $\triangle A P B$ and $\triangle D P C$ are right angled $\Delta s$

$A R=R B=P R$ and $D Q=Q C=P Q$.

[If $D$ is the mid-pt. of hyp. $A C$, then $B D=A D=C D]$

$\therefore \frac{A B}{2}-\frac{D C}{2}=A R-D Q=P R-P Q=Q R=1$

$\Rightarrow A B-C D=2$

Also, $\frac{A B+C D}{2}=4 \quad \Rightarrow \quad A B+C D=8$

Solving (i) and (ii) simultaneously, we get

$A B=\mathbf{5}$ and $C D=\mathbf{3}$.

~~ 50. Sum of the exterior angles of a polygon $=360^{\circ}$

Sum of all the angles marked ’ $x$ '

$=2 \times$ Sum of ext. $\angle s$ of the polygon

$=2 \times 360^{\circ}=720^{\circ}$

$\therefore \angle A+\angle B+\angle C+\angle D+\angle E+\angle F+\angle G$

$=$ Sum of angles of the 7 triangles $-720^{\circ}$

$=180^{\circ} \times 7-720^{\circ}=\mathbf{5 4 0}^{\circ}$.

SELF ASSESSMENT SHEET

Let $A B C$ be a right angled triangle with $A C$ as its hypotenuse. Then,

(a) $A C^{3}>A B^{3}+A C^{3}$

(b) $A C^{2}>A B^{2}+B C^{2}$

(d) None of these

~~ 2. There are two congruent triangles each with area $198 cm^{2}$. Triangle $D E F$ is placed over triangle $A B C$ in such a way that the centroid of both the triangles coincide with each other and $A B | D E$ as shown in the figure forming a star. What is the area of the common region $P Q R S T U$ ?

(a) $49 cm^{2}$

(b) $66 cm^{2}$

(d) $162 cm^{2}$

~~ 3. In the given figure, $D F | A B$, $B C \perp A B, B C=5, B G=4, B A$ $=12, D A=3$. Then $C E$ is:

(a) 5.44

(b) 0.54

(d) 0.42

~~ 4. If $A B C D$ is a rectangle with $A B=4$ and $B C=2 \sqrt{3}$ and $M$ is the mid-point of $C D$, find $b-a$.

(a) $15^{\circ}$

(b) $20^{\circ}$

(d) $30^{\circ}$

~~ 5. $O D, O E$ and $O F$ are the perpendicular bisectors to the three sides of the triangle $A B C$. What is the relationship between $m \angle B A C$ and $m \angle B O C$ ?

(a) $m \angle B A C=180^{\circ}-m \angle B O C$

(b) $m \angle B O C=2 m \angle B A C$

~~ 6. In a $\triangle A B C, C A=C B$. On $C B$ square $B C D E$ is constructed away from the triangle. If $x$ is the number of degrees in angle $D A B$, then

(a) $x$ depends upon $\triangle A B C$

(b) $x$ is independent of the triangle

(d) $x$ is greater than $45^{\circ}$ but less than $90^{\circ}$.

~~ 7. Given triangle $A B C$ with medians $A E, B F, C D ; F H$ parallel and equal in length to $A E ; B H$ and $H E$ are drawn; $F E$ extended to meet $B H$ in $G$. Which one of the following statements is not necessarily

correct?

(a) $H E=H G$

(b) $B H=D C$

(d) $F G$ is the median of $\triangle B F H$.

~~ 8. Side $A C$ of a right triangle $A B C$ is divided into 8 equal parts. Seven line segments parallel to $B C$ are drawn to $A B$ from the points of division. If $B C$ $=10$, then the sum of the lengths of the seven line segments is

(a) 33

(b) 34

(d) 45

~~ 9. Given $A B=A C=B D, B D \perp A C$ in the figure shown alongside, the sum of the measures of angles $C$ and $D$ is

(a) $120^{\circ}$

(b) $140^{\circ}$

(d) $135^{\circ}$

~~ 10. Constructed externally on the sides $A B, A C$ of $\triangle A B C$ are equilateral triangle $A B X$ and $A C Y$. If $P, Q, R$ are the midpoints of $A X, A Y$ and $B C$ respectively, then $\triangle P Q R$ is

(a) right angle

(b) equilateral

(d) None of these

ANSWERS

~~ 2. (c)

~~ 3. (c)

~~ 4. $(d)$

~~ 5. (b)

~~ 6. $(b)$

~~ 7. $(a)$

~~ 8. (c)

~~ 9. $(d)$

~~ 10. $(b)$

HINTS AND SOLUTIONS

In a right angled triangle, hypotenuse is the longest side, so in $\triangle A B C, A C>B C$ and $A C>A B$

$ \begin{aligned} & A C^{3}=A C^{2} \cdot A C \\ & =(A B^{2}+B C^{2}) \cdot A C \\ & =A B^{2} \cdot A C+B C^{2} \cdot A C \\ & >A B^{2} \cdot A B+B C^{2} \cdot B C \quad \because A C>B C, A C>A B \end{aligned} $

$\Rightarrow A C^{3}>A B^{3}+B C^{3}$.

~~ 2. There are 12 similar triangles in the figure forming a star, each with equal area. But a larger triangle $A B C$ (or $D E F$ ) has only 9 smaller triangles. Out of the 9 triangles only 6 triangles are common.

$\therefore$ Area of common region PQRSTU

$=\frac{6}{9} \times 198 cm^{2}=132 cm^{2}$.

~~ 3. By Pythagoras’ Theorem,

$ \begin{aligned} G C^{2} & =(B C)^{2}+(G B)^{2} \\ & =25+16=41 \\ \Rightarrow G C & =\sqrt{41} \\ A C^{2} & =(B C)^{2}+(A B)^{2} \\ & =24+144=169 \end{aligned} $

$\Rightarrow A C=13$

$\therefore D C=A C-A D=13-3=10$

$\triangle D E C \sim \triangle A G C$

$(A A$ similarly $\because D F | A B)$

$\therefore \frac{D C}{A C}=\frac{C E}{G C}$

$\Rightarrow \frac{10}{13}=\frac{C E}{\sqrt{41}} \Rightarrow C E=\frac{10 \sqrt{41}}{13}=4.9$ (approx)

~~ 4. $C D=A B=4 \Rightarrow D M=2$

$A D=B C=2 \sqrt{3}$

We know that for $\triangle A D M, \angle A D M=90^{\circ}, A D=2 \sqrt{3}$, $D M=2$

$\Rightarrow A M=\sqrt{A D^{2}+D M^{2}}=\sqrt{12+4}=\sqrt{16}=4$

Now By the $30^{\circ}-60^{\circ}-90^{\circ}$ triangle theorem, the side half the hypotenuse is opposite the $30^{\circ}$ angle and the side $\frac{\sqrt{3}}{2}$ times the hypotenuse is opposite the $60^{\circ}$ angle.

$\therefore \triangle A D B$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

$\Rightarrow b=60^{\circ}, a=30^{\circ} \Rightarrow b-a=60^{\circ}-30^{\circ}=\mathbf{3 0}$.

~~ 5. $O$ being the point of intersection of the perpendicular bisectors of the sides of $\triangle A B C$, it is the circumcentre of $\triangle A B C$.

$ \therefore \angle B O C=2 \angle B A C(\Big( \text{ Angle subtended by an arc at the } \\ \text{ centre of a circle is twice the angle } \\ \text{ subtended by it at any other point } \\ \text{ on the remaining part of the circumference } \Big) \\ $

~~ 6. $\triangle A B C, C A=C B \\ $

$ \Rightarrow \angle C B A=\angle C A B=\angle A \\ $

$ \therefore \angle A C B=180^{\circ}-2 \angle A \\ $

$ \therefore \angle A C D=180^{\circ}-2 \angle A+90^{\circ} \\ $

$ \quad=270^{\circ}-2 \angle A \\ $

$ \because A C=A B=C D \\ $

$ \Rightarrow A C=C D \\ $

$ \Rightarrow In \triangle A D, \angle C D A=\angle C A D \\ $

$=\frac{1}{2}(180^{\circ}-(270^{\circ}-2 \angle A))=\angle A-45^{\circ} . $

$ =\angle A-x \quad(\therefore \angle C A E=\angle C A B-\angle E A B=\angle A-x) $

$\Rightarrow x=45^{\circ}$

So $X$ is independent of the triangle.

~~ 7. $F H=A E$ and $F H | A E$

$\Rightarrow A E H F$ is a parallelogram.

$\Rightarrow E H | A C$, when extended meets $A B$ in $D$. In congruent $\triangle s A C D$ and $H D B, D C=B H$ (corresponding sides)

$\therefore(b)$ is true.

Now $F G=F E+E G$

$ \begin{aligned} & =\frac{1}{2} A B+\frac{1}{2} D B \quad \begin{matrix} \text{ (By mid-point theorem in } \triangle s A B C \\ \text{ and } H D B \text{ respectively) } \end{matrix} \\ & =\frac{1}{2} A B+\frac{1}{2} \times \frac{1}{2} A B \quad(\because D \text{ is the mid-point of } A B) \\ & =\frac{3}{4} A B \end{aligned} $

(d) $F E | A B$, when extended to $G \Rightarrow E G | A B$

$\Rightarrow G$ is the mid-point of $H B$ (By converse of mid-point

theorem)

$\therefore$ In $\triangle B F H, F G$ is the median, so $(d)$ is also true.

(e) Cannot be proved true with the given information.

~~ 8. All the triangles formed with $A$ as vertex of $A B_1 C_1, A B_2 C_2$, $A B_3 C_3, A B_4 C_4, A B_5 C_5, A B_6 C_6$, $A B_7 C_7$, and $A B C$ are similar.

Let $h_k$ be the length of the side parallel to $B C$ at a distance $k / 8$. $A C$ from $A$,

where $k=1,2, \ldots ., 7$, i.e.,

$ \begin{aligned} & B_1 C_1=h_1=\frac{A C}{8}, B_2 C_2=h_2=\frac{2}{8} A C, B_3 C_3=h_3=\frac{3}{8} A C \text{, } \\ & B_7 C_7=h_7=\frac{7}{8} A C \end{aligned} $

$\therefore \triangle A B_1 C_1 \sim \triangle A B C$, their sides are proportional

$\Rightarrow \frac{B_1 C_1}{A C_1}=\frac{B C}{A C} \Rightarrow \frac{h_1}{1 / 8 A C}=\frac{10}{A C}$

Similarly, $\triangle A B_2 C_2 \sim \triangle A B C, \triangle A B_3 C_3 \sim \triangle A B C$,

$ \triangle A B_7 C_7 \sim \triangle A B C $

$\therefore$ In general, $\frac{h_k}{k / 8 A C}=\frac{10}{A C} \Rightarrow h_k=\frac{10 k}{8}$

$\therefore$ Required sum $=h_1+h_2+h_3+\ldots \ldots \ldots .+h_7$

$ =\frac{10}{8}(1+2+3+\ldots \ldots \ldots+7)=\frac{10}{8} \times \frac{7}{2} \times(1+7)=35 $

$(\because.$ Sum of $n$ terms $.=\frac{n}{2}(a+l))$

~~ 9. Let $A C$ and $B D$ intersect at point $K$ and $\angle A B K=\beta$, $\angle K B C=\alpha, \angle A D K=\gamma$.

Then, In rt. $\triangle A K B, \angle B A K=90^{\circ}-\beta$

Also, in rt. $\triangle A K D, \angle D A K=90^{\circ}-\gamma$

$\therefore A B=B D$

$\therefore \angle B A D=\angle B D A$

$\Rightarrow(90^{\circ}-\beta)+(90^{\circ}-\gamma)=\gamma$

$\Rightarrow 180^{\circ}-\beta-\gamma=\gamma$

$\Rightarrow 2 \gamma=180^{\circ}-\beta \Rightarrow \gamma=90^{\circ}-\beta / 2$.

In rt. $\triangle B K C, \angle C=90^{\circ}-\alpha=\angle A B C=\alpha+\beta(\because A B=A C)$

$\therefore 90^{\circ}-\alpha=\alpha+\beta \Rightarrow 2 \alpha=90^{\circ}-\beta$

$\Rightarrow \alpha=45^{\circ}-\beta / 2$.

$\angle C+\angle D=(90^{\circ}-\alpha)+\gamma=90^{\circ}-(45^{\circ}-\beta / 2)+90^{\circ}-\beta / 2$

From (i) and (ii)

$=45^{\circ}+\beta / 2+90^{\circ}-\beta / 2=135^{\circ}$.

~~ 10. Let $E$ and $F$ be the mid-points of $A B$ and $A C$ respectively. Then, by mid-point theorem, In $\triangle B A C$,

$E R | A C$ and $E R=\frac{1}{2} A C=A F$

In $\triangle C A B$,

$F R | A B$ and $F R=\frac{1}{2} A B=E A _{B}$

$\therefore A E R F$ is a parallelogram. $\therefore \triangle A B X$ being an equilateral $\triangle$ and $A E R F$ a parallelogram, $P E=A E=R F$.

Similarly for $\triangle A C Y$ and $| gm A E R F$

$Q F=A F=E R$

Also, $\angle P E R=\angle P E A+\angle A E R$

$ =60^{\circ}+\angle A E R $

(opp. $\angle s$ of a prallelogram are equal)

$ =60^{\circ}+\angle A F R=\angle Q F R $

$\therefore P E=R F, Q F=E R$ and $\angle P E R=\angle Q F R$

$\Rightarrow \triangle E P R \cong \triangle Q R F$

$\Rightarrow P R=Q R$ and $\angle E P R=\angle Q R F$

Also, $\angle P R Q=\angle E R F-\angle E R P-\angle Q R F$

$=180^{\circ}-\angle A E R-\angle E R P-\angle Q R F$

$=180^{\circ}-\angle A E R-\angle E R P-\angle E P R$

(From (iv)

$=180^{\circ}-\angle A E R-(\angle E R P+\angle E P R)$

$=180^{\circ}-\angle A E R-(180^{\circ}-\angle P E R)$

$=\angle P E R-\angle A E R=\mathbf{6 0}$.

Also, $P R=Q R \quad \Rightarrow \angle R P Q=\angle R Q P=\frac{1}{2}(180^{\circ}-\angle P R Q)$

$=\frac{1}{2}(180^{\circ}-60^{\circ})=60^{\circ}$.

$\Rightarrow \triangle P Q R$ is equilateral.

Permutations and Combinations

KEY FACTS

I. Fundamental Principle of Counting

(a) Multiplication: Let us understand this basic principle with the help of the following examples:

Ex. 1 . Suppose you have 3 shirts and 4 pairs of pants. In how many possible ways can you dress up by wearing a shirt and a pair of pants?

In the above case, you can wear any of the 3 shirts and after wearing one of these shirts any of these pairs of pants with it. If we label the shirts as $S_1, S_2, S_3$ and the pants as $P_1, P_2, P_3$ and $P_4$, then the different ways of dressing up can be as under:

$S_1 P_1$	$S_2 P_1$	$S_3 P_1$
$S_1 P_2$	$S_2 P_2$	$S_3 P_2$
$S_1 P_3$	$S_2 P_3$	$S_3 P_3$
$S_1 P_4$	$S_2 P_4$	$S_3 P_4$

Total number of ways $=12=\mathbf{3} \times \mathbf{4}$.

In this illustration, we multiply the number of ways in which you can wear a shirt and the number of ways in which you can wear a pair of pants.

Ex. 2 . There are 4 different routes between cities $A$ and $B$, and 3 different routes between cities $B$ and $C$. How many different routes are there from city $\boldsymbol{A}$ to city $\boldsymbol{C}$ by way of city $\boldsymbol{B}$ ?

Obviously, one can go from city $A$ to city $B$ by any of the 4 routes, i.e., in 4 ways. After having gone to $B$ by any of the different 4 routes, one can go to city $C$ by any of the three routes.

Thus, corresponding to one route taken, from $A$ to $B$, he has 3 choices from $B$ to $C$. Therefore, corresponding to 4 routes, there are 12 choices in all.

Therefore, he can go from $A$ to $C$ via $B$, in $4 \times 3=12$ ways as depicted in the given tree diagram.

The possible routes taken are :

$R_1 r_1$	$R_2 r_1$	$R_3 r_1$	$R_4 r_1$
$R_1 r_2$	$R_2 r_2$	$R_3 r_2$	$R_4 r_2$
$R_1 r_3$	$R_2 r_3$	$R_3 r_3$	$R_4 r_3$

Total number of ways $=12$.

Thus, the examples discussed above illustrate the use of a general principle, called the product rule or the fundamental principle of counting, which is stated below.

Suppose that the first operation is performed in any one of the $m$ ways, the second operation can then be performed in $n$ ways and with the particular first operation, we can associate any one of the $n$ ways of performing the second operation. This means that if the first operation could have been performed only in this one way, there would have been $1 \times n$, i.e., $n$ ways of performing both the operations. But it is given that the first operation can be performed in $m$ ways and there are $n$ ways of performing the second operation for every one way of performing the first operation. Therefore, there are $\boldsymbol{m} \times \boldsymbol{n}$ ways of performing both the operations.

Generalisation. The above principle can be extended to the case in which the different operations can be performed in $m, n, p, \ldots .$. ways. In this case, the number of ways of performing all the operations together would be $\boldsymbol{m} \times \boldsymbol{n} \times \boldsymbol{p}$…….

Ex. 3 . There are 10 buses running between two towns $X$ and $Y$. In how many ways can a man go from $X$ to $Y$ and return by a different bus?

Sol. The man can go from $X$ to $Y$ in 10 ways and as he is not to return by the same bus that he took while going, corresponding to each of the 10 ways of going, there are 9 ways of returning. Hence the total number of ways in which he can go to $Y$ and be back is $10 \times 9=90$.

Ex. 4 . How many different numbers of three digits can be formed with the digits 1, 2, 3, 4, 5, no digit is being repeated?

Sol. The unit’s place can be filled with either of these 5 digits and so the unit’s place can be filled in 5 ways. The ten’s place can be filled in 4 ways corresponding to each way of filling up the unit’s place, for we can have any digit here except the one used in the unit’s place. Similarly, the hundredth’s place can be filled in 3 ways as here we have any of the remaining three digits. Therefore, there are $5 \times 4 \times 3$ $=60$ ways of forming a number of three digits with the five given digits.

(b) Addition: If there are two jobs which can be performed independently in $m$ and $n$ ways respectively, then either of the two jobs can be performed in $m+n$ ways.

For example: Suppose you have 3 full-sleeve and 4 half-sleeve shirts. Since you have the choice of wearing any of these shirts, you can wear one shirt in $3+4=7$ ways. If in addition, you have 5 T-shirts, then you can wear one of them in $3+4+5=12$ ways.

The above example illustrates one way of counting, which we may call the sum rule and applies when one event has to happen out of given disjoint events.

In this case, we use the word ‘or’ between various jobs and the meaning of ‘or’ is addition.

II. Permutations

Def. : Each of the different arrangements which can be made by taking some or all of a number of things at a time is called a permutation.

Notation: The number of permutations of $n$ things taken $r$ at a time is denoted by ${ }^n \boldsymbol{P} _{\boldsymbol{r}}$ or $\boldsymbol{P}(\boldsymbol{n}, \boldsymbol{r})$. The letter $P$ is an abbreviation of the word ‘permutation’.

Thus ${ }^{6} P_4$ denotes the number of permutations or arrangements of 6 things taken 4 at a time.

The value of ${ }^{n} \boldsymbol{P}_r$

To find the number of permutations of $\boldsymbol{n}$ different things, taken $r$ at a time or to determine ${ }^{n} \boldsymbol{P} _{\boldsymbol{r}}$.

The number of permutations of $n$ things taken $r$ at a time will be the same as the number of ways in which $r$ blank places can be filled up with $n$ given things.

As the first place can be filled in by any one of the $n$ things so there are $n$ ways of filling up the first place.

After having filled in the first place by any one of the $n$ things, there are $(n-1)$ things left. Hence the second place can be filled in $(n-1)$ ways. Now, as for every one way of filling up the first place, there are $(n-1)$ ways of filling up the second place, so the first two places can be filled in $n(n-1)$ ways.

After having filled in the first two places in any one of the above ways, there are $(n-2)$ things left and so the third place can be filled in $(n-2)$ ways. Now for every one way of filling up the first two places, there are $(n-2)$ ways of filling up the third place and so the first three places can be filled up in $n(n-1)(n-2)$ ways. It may be observed that

(a) At every stage the number of factors is equal to the number of places filled up.

(b) Every factor is by one less than its preceding factor.

Position of the object	$1 st$	$2 nd$	$\ldots$	$(r-1)$ th	$r$ th
Number of ways	$n$	$n-1$	$\ldots$	$n-(r-2)$	$n-(r-1)$

Hence the number of ways of filling up all the $r$ places, i.e., the number of permutations of $n$ different things taken $r$ at a time is $n(n-1)(n-2) \ldots . . . r$ factors

Hence

$ =n(n-1)(n-2) \ldots(n-\overline{r-1}) $

Thus, ${ }^{7} P_2=7 \times 6 ;{ }^{10} P_4=10 \times 9 \times 8 \times 7 ;{ }^{20} P_3=20 \times 19 \times 18$.

Cor. : The number of permutations of $n$ things taken all at a time is

$ { }^{n} P_n=n(n-1)(n-2) \ldots . . .3 .2 .1 . \quad \text{ [Putting } n \text{ for } r \text{ ] } $

Ex. 1 . In how many ways can 5 persons occupy 3 vacant seats?

Sol. Total number of ways $={ }^{5} P_3=5 \times 4 \times 3=60$.

Ex. 2 . If ${ }^{12} P_r=1320$, find $\boldsymbol{r}$.

Sol. ${ }^{12} P_r=12 \times 11 \times \ldots .$. to $r$ factors $\Rightarrow 1320=12 \times 11 \times 10 \quad \therefore r=3$.

III. Factorial Notation

The product of $\boldsymbol{n}$ natural numbers from 1 to $\boldsymbol{n}$ is denoted by $\boldsymbol{n}$ ! or $\lfloor n$ and is read as factorial $\boldsymbol{n}$.

Thus, $\quad n$ ! or $\lfloor n=1.2 .3 \ldots \ldots(n-1) . n$

$4 !=1 \times 2 \times 3 \times 4=24 ; 6 !=1 \times 2 \times 3 \times 4 \times 5 \times 6=720$

$(n-1) !=1 \times 2 \times 3 \ldots \ldots(n-1)$.

It is easily seen that $8 !=8 \times(7 !)$.

IV. Values of ${ }^{\boldsymbol{n}} \boldsymbol{P} _{\boldsymbol{r}}$ in terms of Factorial Notation

$ \begin{aligned} { }^{n} P_r & .=n(n-1)(n-2) \ldots(n-r+1)=\frac{n(n-1) \ldots(n-r+1)}{\lfloor(n-r)} \cdot \rvert,(n-r) \\ & =\frac{n(n-1)(n-2) \ldots(n-r+1) \cdot(n-r)(n-r-1) \ldots 3.2 .1}{\lfloor(n-r)} \\ & =\frac{n(n-1)(n-2) \ldots 3.2 .1}{\lfloor(n-r)}=\frac{\lfloor n}{\mid(n-r)} \\ \therefore \quad{ }^{n} P_r & =\frac{\underline{\underline{n}}}{\mid \boldsymbol{n}-\boldsymbol{r})} \\ \text{ Thus, } \quad{ }^{18} P_5 & =\frac{\mid 18}{\lfloor 13} \end{aligned} $

Cor. 1. Putting $r=0,{ }^{n} P_0=\frac{\underline{n}}{\underline{\underline{n}}}=1$

Cor. 2. Value of 0 !.

Putting $r=n,{ }^{n} P_n=\frac{n !}{(n-n)}=\frac{n !}{0 !} ;$ But ${ }^{n} P_n=n ! \quad \therefore n !=\frac{n !}{0 !}$

Note. In fact, $0 !$ is meaningless but in order to avoid contradiction in the results, we suppose that $0 !=1$.

Ex. Find the value of $n$ if ${ }^{n} P _{13}:{ }^{n+1} P _{12}=\frac{3}{4}$.

Sol. Here, ${ }^{n} P _{13}=\frac{n !}{(n-13) !}$ and ${ }^{n+1} P _{12}=\frac{(n+1) !}{(n+1-12) !}=\frac{(n+1) !}{(n-11) !}$

$[.$ Using $.{ }^{n} P_r=\frac{n !}{(n-r) !}]$

${ }^{n} P _{13}:{ }^{n+1} P _{12}=\frac{n !}{(n-13) !} \times \frac{(n-11) !}{(n+1) !}=\frac{3}{4} \Rightarrow \frac{n !}{(n-13) !} \times \frac{(n-11)(n-12) \cdot(n-13) !}{(n+1) \cdot n !}=\frac{3}{4}$

$\Rightarrow \frac{(n-11)(n-12)}{(n+1)}=\frac{3}{4} \Rightarrow 4 n^{2}-95 n+525=0$ or $(n-15)(4 n-35)=0$

$\therefore \quad n=15$ (Rejecting the fractional value of $n$ ).

V. Restricted Permutations

Type I.

Ex. 1 . In how many of the permutations of 10 things taken 4 at a time will

(i) one thing always occur, (ii) never occur?

Sol. (i) Keeping aside the particular thing which will always occur, the number of permutations of 9 things taken 3 at a time is ${ }^{9} P_3$. Now this particular thing can take up any one of the four places and so can be arranged in 4 ways. Hence the total number of permutations $={ }^{9} P_3 \times 4=9 \times 8 \times 7 \times 4=2016$.

(ii) Leaving aside the particular thing which has never to occur, the number of permutations of 9 things taken 4 at a time is ${ }^{9} P_4=9 \times 8 \times 7 \times 6=3024$.

Ex. 2 . In how many of the permutations of $\boldsymbol{n}$ things taken $r$ at a time will 5 things

(i) always occur, (ii) never occur?

Sol. (i) Keeping aside the 5 things, the number of permutations of $(r-5)$ things taken out of $(n-5)$ things is ${ }^{n-5} P_{r-5}$. Now these 5 things can be arranged in $r$ places in ${ }^{r} P_5$ ways. Hence, the total number of permutations is ${ }^{r} P_5 \times{ }^{n-5} P_{r-5}$.

(ii) Total number of permutations $={ }^{n-5} P_r=\frac{(n-5) !}{(n-r-5) !}$

Type II. When certain things are not to occur together.

Case I. When the number of things not occurring together is two.

Procedure

Find the total number of permutations - when no restriction is imposed on the manner of arrangement.
Then find the number of permutations when the two things occur together.
The difference of the two results gives the number of permutations in which the two things do not occur together.

Ex. 3 . Prove that the number of ways in which $\boldsymbol{n}$ books can be placed on a shelf when two particular books are never together is $(n-2) \times(n-1)$ !.

Sol. Regarding the two particular books as one book, there are $(n-1)$ books now which can be arranged in ${ }^{n-1} P _{n-1}$, i.e., $(n-1)$ ! ways. Now, these two books can be arranged amongst themselves in 2 ! ways. Hence the total number of permutations in which these two books are placed together is $2 !$. (n-1)!. The number of permutations of $n$ books without any restriction is $n$ !.

Therefore, the number of arrangements in which these two books never occur together

$=n !-2 ! .(n-1) !=n \cdot(n-1) !-2 .(n-1) !=(n-2) .(n-1) !$

Case II. When the number of things not occurring together is more than two.

Ex. 4 . In how many ways can 6 boys and 4 girls be arranged in a straight line so that no two girls are ever together?

Sol. The seating arrangement may be done as desired in two operations.

(i) First we fix the positions of 6 boys. Their positions are indicated by $B_1, B_2, \ldots, B_6$.

$\times B_1 \times B_2 \times B_3 \times B_4 \times B_5 \times B_6 \times$

This can be done in 6 ! ways.

(ii) Now if the positions of girls are fixed at places (including those at the two ends) shown by the crosses, the four girls will never come together. In any one of these arrangements there are 7 places for 4 girls and so the girls can sit in ${ }^{7} P_4$ ways.

Hence the required number of ways of seating 6 boys and 4 girls under the given condition $={ }^{7} P_4 \times 6 !=7 \times 6 \times 5 \times 4 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=\mathbf{6 0 4 8 0 0}$.

Type III. Formation of numbers with digits.

Ex. 5 . Suppose the six digits 1, 2, 4, 5, 6,7 are given to us and we have to find the total number of numbers with no repetition of digits which can be formed under different conditions.

Sol. 1. There is no restriction. The number of 6-digit numbers.

$={ }^{6} P_6=6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$.

Numbers in which a particular digit occupies a particular place.

Suppose we have to form numbers in which 5 always occurs in the ten’s place. In this case, the ten’s place is fixed and the remaining five places can be filled in by the remaining 5 digits in ${ }^{5} P_5$, i.e., 5 ! $=120$ ways.

The number of numbers in which 5 occurs in the ten’s place $=120$.

Numbers divisible by a particular number. Suppose we have to form numbers which may be divisible by 2 . These numbers will have 2 or 4 or 6 in the unit’s place. Thus the unit’s place can be filled in 3 ways. After having filled up the unit’s place in any one of the above ways, the remaining five places can be filled in ${ }^{5} P_5=5 !=120$ ways.

$\therefore \quad$ The total number of numbers divisible by $2=120 \times 3=360$.

Numbers having particular digits in the beginning and the end. Suppose we have to form numbers which begin with 1 and end with 5 . Here, the first and the last places are fixed and the remaining four places can be filled in 4 !, i.e., 24 ways by the remaining four digits.

Therefore, the total number of numbers beginning with 1 and ending with $5=24$.

Note. If the numbers could have 1 or 5 in the beginning or the end, the number would have been 2 ! . 4 !, i.e., 48 .

Numbers which are smaller than or greater than a particular number. Suppose we have to form numbers which are greater than $4,00,000$. In these numbers, there will be 4 or a digit greater than 4 , i.e. 5, 6 or 7 in the lac’s place. Thus this place can be filled in 4 ways. The remaining 5 places can then be filled in $5 !=120$ ways.

$\therefore \quad$ The total number of numbers $=4 \times 120=480$.

Type IV. Word Building

The following cases may arise:

No letter may be repeated.
Some letters may be repeated.
There may be a particular letter in the beginning or the end.
Some letters may occur together.

Illustration. Suppose the word ‘PENCIL’ is given to us and we have to form words with the letters of this word.

Sol. 1. There is no restriction on the arrangement of the letters.

The six different letters can be arranged in ${ }^{6} P_6=6 !=720$ ways.

Hence the total number of words formed $=\mathbf{7 2 0}$.

All words begin with a particular letter.

Suppose all words begin with $E$. The remaining 5 places can be filled with remaining 5 letters in $5 !=\mathbf{1 2 0}$ ways.

All words begin and end with particular letters.

Suppose all words begin with $L$, and end with $P$. The remaining 4 places can then be filled in 4 ! ways.

$\therefore$ The total number of words formed $=4 !=\mathbf{2 4}$.

Note. If the words were to begin or end with $E$ or $L$, these two positions could have been filled in ${ }^{2} P_2=2$ ways. Hence the number of words in this case would have been $=2 \times 24=48$.

$N$ is always next to $E$.

$P$	$E N$	$C$	$I$	$L$

Since $N$ is always next to $E$, therefore ’ $E N$ ’ is considered to be one letter.

$\therefore \quad$ Required no. of permutations $=5$ ! $=5 \times 4 \times 3 \times 2 \times 1=\mathbf{1 2 0}$.

Vowels occur together: The vowels are $E$ and $I$. Regarding them as one letter, the 5 letters can be arranged in $5 !=120$ ways. These two vowels can be arranged amongst themselves in $2 !=2$ ways.

$\therefore \quad$ The total number of words $=2 \times 120=\mathbf{2 4 0}$.

Consonants occur together: Regarding these consonants as one letter the three letters $E, I,(P N C L)$ can be arranged in 3 !, i.e., 6 ways. The letters $P N C L$ can be arranged among themselves in $4 !=24$ ways.

$\therefore \quad$ The number of words in which consonants occur together $=6 \times 24=\mathbf{1 4 4}$.

Vowels occupy even places.

$\times$	$\times$	$\times$	$\times$	$\times$	$\times$
1	2	3	4	5	6

There are 6 letters and 3 even places. $E$ can be placed in any of of the three even places in ${ }^{3} P_1$, i.e., 3 ways. Having placed $E$ in any one of these places, $I$ can be placed in any one of the remaining 2 places in ${ }^{2} P_1$, i.e., 2 ways. Thus, the vowels can occupy even places in $2 \times 3=6$ ways. After the vowels have been placed the remaining 4 letters can take up their positions in ${ }^{4} P_4$, i.e., 24 ways.

$\therefore$ The total number of words $=6 \times 24=\mathbf{1 4 4}$.

VI. Permutations of Alike Things

The number of permutations of $\boldsymbol{n}$ things taken all at a time where $p$ of the things are alike and of one kind, $q$ others are alike and of another kind, $r$ others are alike and of another kind, and so on is

$ x=\frac{n !}{p ! q ! r ! . .} $

Illustration: In how many ways can the letters of the word ‘INDIA’ be arranged ?

Sol. The word contains 5 letters of which 2 are ‘I’s.

The number of words possible $=\frac{5 !}{2 !}=\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}=\mathbf{6 0}$.

VII. Permutations of Repeated Things

The number of permutations of $n$ different things taken $r$ at a time, when each thing may occur any number

of times is $\boldsymbol{n}^{r}$.Suppose $r$ places are to be filled with $n$ things. The first place can be filled in $n$ ways and when this has been filled up in any one of these ways, the second place can also be filled in $n$ ways for the thing occupying the first place may occupy the second place also. Thus the first two places can be filled in $n \times n=n^{2}$ ways. Similarly the third place can also be filled in $n$ ways.

Arguing in the same manner, we conclude that the $r$ places can be filled in $n \times n \times n \ldots . . r$ times, i.e. $n^{r}$ ways.

Illustration: In how many ways can 3 prizes be distributed among 4 boys, when

(i) no boy gets more than one prize;

(ii) a boy may get any number of prizes;

(iii) no boy gets all the prizes.

Sol. (i) The first prize can be given to any of the four boys. Then, the second prize can be given to any of the three boys. Lastly, the third prize can be given to any one of the remaining 2 boys.

$\therefore$ The number of ways in which all the 3 prizes can be given $=4 \times 3 \times 2=\mathbf{2 4}$.

(ii) In this case, each of the three prizes can be given in 4 ways since a boy can receive any number of prizes.

$\therefore$ The number of ways in which all the prizes can be given $=4 \times 4 \times 4=4^{3}=\mathbf{6 4}$.

(iii) Since anyone of the 4 boys can get all the prizes, therefore, the number of ways in which a boy gets all the 3 prizes $=4$.

$\therefore$ Number of ways in which a boy does not get all the prizes $=64-4=\mathbf{6 0}$.

VIII. Circular Permutations

Method I. If we have to arrange the four letters $A, B, C, D$, two of the arrangements would be $A B C D, D A B C$ which are two distinct arrangements. Now, if these arrangements are written along the circumference of a circle (Fig. 1 next page), the two arrangements are one and the same. Thus, we conclude that circular permutations are different only when the relative order of the objects is changed otherwise they are the same. Thus the arrangements in Fig. 2 are different.

As the number of circular permutations depends on the relative positions of objects, we fix the position of one object and then arrange the remaining ( $n-1$ ) objects in all possible ways. This can be done in $(n-1)$ ! ways.

Illustration: 20 persons were invited for a party. In how many ways can they and the host be seated around a circular table? In how many of these ways will two particular persons be seated on either side of the host?

Sol. There is 1 host and 20 guests, they are to be seated around a circular table.

(i) Let us fix the seat of one person, say the host, the 20 guests will be seated around the circular table in 20 ! ways, [or, $(n-1) !=(21-1) !=20 !]$

(ii) The two particular persons can be seated on either side of the host in 2 ways and for each way of their taking seats, the remaining 18 persons can be seated around the circular table in 18 ! ways Hence the number of ways of seating two particular persons on either side of the host $=2 \times 18$ !. [or, 2 ! $\times(19-1)$ !, considering the host and two particular persons as one entity.]

IX. Clockwise and Counter-Clockwise Permutations

We have two types of circular permutations:

(i) Those in which counter-clockwise and clockwise are distinguishable. Thus while seating 4 persons $A, B$, $C, D$ around a table, the following permutations are considered different. (Fig. 1)

(ii) Those in which counter-clockwise and anti-clockwise are not distinguishable. (Fig. 2)

Thus, (a) while forming a garland of roses or jasmine, the following arrangements are not disturbed if we turn the garland over.

Fig. 1

Fig. 2

(b) The distinction between clockwise and anti-clockwise is ignored when a number of people have to be seated around a table so as not to have the same neighbours.

Points to remember (Circular permutations)

We can summarise the above discussion as under:

Number of circular arrangements (permutations) of $n$ different things $=(n-1)$ !
Number of circular arrangements of $n$ different things when clockwise and anti-clockwise arrangements are not different $=\frac{1}{2}(n-1)$ !
Number of circular permutations of $n$ different things taken $r$ at a time when clockwise and anticlockwise arrangements are taken as different is $\frac{{ }^{n} P_r}{r}$.
Number of circular permutations of $n$ different things, taken $r$ at a time, when clockwise and anticlockwise arrangements are not different $=\frac{{ }^{n} P_r}{2 r}$.

Ex. Find the number of ways in which (i) $\boldsymbol{n}$ different beads, (ii) 10 different beads can be arranged to form a necklace.

Sol. Fixing the position of one bead, the remaining beads can be arranged in $(n-1)$ ! ways. As there is no distriction between the clockwise and anti-clockwise arrangements, the total number of ways in which 10 different beads can be arranged $=\frac{(10-1) !}{2}=\frac{1}{2}(9 !)$

X. Combinations

Def.: Each of the different groups or selections which can be made by taking some or all of a number of things at a time (irrespective of the order) is called a combination.

By the number of combinations of $n$ things taken $r$ at a time is meant the number of groups of $r$ things which can be formed from the $n$ things. The same is denoted by the symbol ${ }^{n} \boldsymbol{C} _{\boldsymbol{r}}$ or $\boldsymbol{C}(\boldsymbol{n}, \boldsymbol{r})$ or $(\begin{matrix} \boldsymbol{n} \\ \boldsymbol{r}\end{matrix} )$.

Value of ${ }^{n} C_r$

Each combination consists of $r$ different things which can be arranged among themselves in $r$ ! ways. Therefore, the number of arrangements for all the ${ }^{n} C_r$ combinations is ${ }^{n} C_r \times r !$. This is equal to the permutations of $n$ different things taken $r$ at a time.

$ \therefore \quad{ }^{n} C_r \times r !={ }^{n} P_r $

$ \begin{aligned} \therefore \quad{ }^{n} C_r & =\frac{{ }^{n} P_r}{r !}=\frac{n(n-1)(n-2) \ldots(n-r+1)}{1.2 .3 \ldots r} \\ { }^{n} P_r & =\frac{n !}{(n-r)} \quad \therefore{ }^{n} C_r=\frac{n !}{r !(n-r) !} \end{aligned} $

Some Important Properties of Combinations

${ }^{n} C_r=\frac{{ }^{n} P_r}{r !}$
${ }^{n} C_0={ }^{n} C_n=1$
${ }^{n} C_r={ }^{n} C _{n-r} \quad(0 \leq r \leq n)$
${ }^{n} C_x={ }^{n} C_y \Rightarrow x+y=n$ or $x=y(x, y \in W)$
${ }^{n} C _{r-1}+{ }^{n} C_r={ }^{n+1} C_r$
If $n$ is even, the greatest value of ${ }^{n} C_r={ }^{n} C_m$ where $m=\frac{n}{2}$
If $n$ is odd, the greatest value of ${ }^{n} C_r={ }^{n} C_m$, where, $m=\frac{n-1}{2}$ or $m=\frac{n+1}{2}$
${ }^{n} C_0+{ }^{n} C_1+{ }^{n} C_2+{ }^{n} C_3+\ldots+{ }^{n} C_r=2^{n}$
${ }^{n} C_1+{ }^{n} C_2+\ldots+{ }^{n} C_n=2^{n}-1$
${ }^{n} C_0+{ }^{n} C_2+{ }^{n} C_4+\ldots=2^{2 n}$

The proof of Formula 5 above is given below:

${ }^n \boldsymbol{C}_{\boldsymbol{r}} + { }^{n+1}C_r (1 \le r \le n) \quad \text{(Pascal’s Rule)} $

${ }^{n} C_r+{ }^{n} C _{r-1} =\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !} \\ $

$=n ! \lbrace \frac{(n-r+1)+r}{r ! \cdot(n-r+1)!}\rbrace =\frac{(n+1) n !}{r ! \cdot(n-r+1)}=\frac{(n+1)!}{r ! \cdot(n-r+1) !}={ }^{n+1} C_r $

Another form. $C(n, r)+C(n, r-1)=C(n+1, r)$,

i.e.,

$ C(n, r-1)=C(n+1, r)-C(n, r) $

Ex. Find the values of ${ }^{6} C_3$ and ${ }^{30} C _{28}$.

Sol. (i) ${ }^{6} C_3=\frac{6 \times 5 \times 4}{1 \times 2 \times 3}=20$.

(ii) ${ }^{30} C _{28}={ }^{30} C _{30-28}={ }^{30} C_2=\frac{30 \times 29}{1 \times 2}=435$.

XI. Miscellaneous Types

Type I. Total number of combinations.

To find the total number of combinations of $n$ dissimilar things taking any number of them at a time.

Case I. When all things are different.

Each thing may be disposed of in two ways. It may either be included or rejected.

$\therefore \quad$ The total number of ways of disposing of all the things $=2 \times 2 \times 2 \times \ldots . . n$ times $=2^{n}$

But this includes the case in which all the things are rejected.

Hence the total number of ways in which one or more things are taken $=2^{n}-1$.

Cor. $2^{n}-1$ is also the total number of the combinations of $n$ things taken $1,2,3, \ldots \ldots$ or $n$ at a time. Hence, ${ }^{n} C_1+{ }^{n} C_2+{ }^{n} C_3+\ldots+{ }^{n} C_n=2^{n}-1$.

Ex. There are 5 questions in a question paper. In how many ways can a boy solve one or more questions?

Sol. The boy can dispose of each question in two ways. He may either solve it or leave it. Thus the number of ways of disposing of all the questions $=2^{5}$.

But this includes the case in which he has left all the questions unsolved.

Hence the total number of ways of solving the paper $=2^{5}-1=31$.

Case II. When all things are not different.

Suppose, out of $(p+q+r+\ldots)$ things, $p$ are alike of one kind, $q$ are alike of a second kind, $r$ alike of a third kind, and the rest different.

Out of $p$ things we may take $0,1,2,3 \ldots \ldots$ or $p$. Hence they may be disposed of in $(p+1)$ ways.

Similarly, $q$ alike things may be disposed of in $(q+1)$ and $r$ alike things in $(r+1)$ ways. The $t$ different things may be disposed of in $2^{t}$ ways.

This includes that case in which all are rejected.

$\therefore \quad$ The total number of selections $=(p+1)(q+1)(r+1) 2^{t}-1$.

Ex. Prove that from the letters of the sentence, ‘Daddy did a deadly deed’, one or more letters can be selected in 1919 ways.

Sol. In the given sentence, there are $9 d s ; 3 a{ }^{\prime} ; 3 e s ; 2 y s ; 1 i$; and $1 l$.

$\therefore \quad$ The total number of selections $=(9+1)(3+1)(3+1)(2+1)(1+1)(1+1)-1$

$ =10 \times 4 \times 4 \times 3 \times 2 \times 2-1=1919 \text{. } $

Type II. Division into groups.

To find the number of ways in which $p+q$ things can be divided into two groups containing $p$ and $q$ things respectively.

Every time when a set of $p$ things is taken, a second set of $q$ things is left behind. Hence the required number of ways $=$ the number of combinations of $(p+q)$ things taken $p$ at a time

$ ={ }^{p+q} C_p=\frac{(p+q) !}{p ! q !} $

Cor. 1. Generalisation. The number of ways in which $p+q+r$ things can be divided into three groups containing $p, q$ and $r$ things respectively

$ ={ }^{p+q+r} C_p \times{ }^{q+r} C_q \times{ }^{r} C_r=\frac{(p+q+r) !}{p !(q+r) !} \times \frac{(q+r) !}{q ! r !} \times 1=\frac{(p+q+r) !}{p ! q ! r !} $

Similarly the result can be extended to the case of dividing a given number of things into more than three groups.

Cor. 2. The number of ways in which $3 p$ things can be divided equally into three distinct groups is

$\frac{(3 p) !}{(p !)^{3}} \cdot(q=p, r=p)$

Cor. 3. The number of ways in which $3 p$ things can be divided into three identical groups is $\frac{(3 p) !}{3 !(p !)^{3}}$.

Ex. 1 . In how many ways can 15 things be divided into 3 groups containing 8,4 and 3 things respectively?

Sol. The number of ways $=\frac{(15) !}{8 ! 4 ! 3 !}=225225$.

Ex. 2 . In how many ways can 18 different books be divided equally among 3 students?

Sol. The required number of ways $=\frac{(18) !}{(6 !)^{3}}$

SOLVED EXAMPLES

Ex. 1 . Each section in first year of plus two course has exactly 30 students. If there are 3 sections, in how many ways can a set of 3 student representatives be selected from each section?

Sol. 1 st representative can be selected from first section in 30 ways.

2nd representative can be selected from second section in 30 ways.

3rd representative can be selected from third section in 30 ways.

$\therefore$ Required number of ways $=30 \times 30 \times 30=\mathbf{2 7 0 0 0}$.

Ex. 2 . How many numbers are there between 100 and 1000 such that every digit is either 2 or 9 ?

Sol. Any number between 100 and 1000 is of 3 digits. The unit’s place can be filled by 2 or 9 in 2 ways.

Similarly ten’s place can be filled in 2 ways.

The hundred’s place can also be filled in 2 ways.

$\therefore$ Required no. of numbers $=2 \times 2 \times 2=\mathbf{8}$.

Ex. 3 . How many odd numbers less than 1000 can be formed using the digits $0,2,5,7$ repetition of digits are allowed?

Sol. Since the required numbers are less than 1000 therefore, they are 1-digit, 2-digit or 3-digit numbers.

One-digit numbers. Only two odd one-digit numbers are possible, namely, $\mathbf{5}$ and 7.

Two-digit numbers. For two-digit odd numbers the unit place can be filled up by 5 or 7 i.e., in two ways and ten’s place can be filled up by 2,5 or 7 (not 0 ) in 3 ways.

$\therefore \quad$ No. of possible 2-digit odd numbers $=2 \times 3=\mathbf{6}$.

Three-digit numbers. For three-digit odd numbers, the unit place can be filled up by 5 or 7 in 2 ways. The ten’s place can be filled up by any one of the digits $0,2,5,7$ in 4 ways. The hundred’s place can be filled up by 2,5 or 7 ( $not 0$ ) in 3 ways.

$\therefore \quad$ No. of possible 3-digit numbers $=2 \times 4 \times 3=24$

Hence total number of odd numbers $=2+6+24=\mathbf{3 2}$.

Ex. 4 . If $\frac{(2 n) !}{3 !(2 n-3) !}: \frac{n !}{2 !(n-2) !}=44: 3$, find $n$.

Sol. Given, $\frac{(2 n) !}{3 !(2 n-3) !}: \frac{n !}{2 !(n-2) !}=44: 3$

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{3 \times 2 ! \cdot(2 n-3) !} \times \frac{2 !(n-2) !}{n(n-1)(n-2) !}=\frac{44}{3}$

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{3} \times \frac{1}{n(n-1)}=\frac{44}{3}$

$\Rightarrow \frac{4(n)(2 n-1)(n-1)}{3} \times \frac{1}{n(n-1)}=\frac{44}{3}$

$\Rightarrow 4(2 n-1)=44 \Rightarrow 8 n-4=44 \Rightarrow 8 n=48 \Rightarrow n=\mathbf{6}$

Ex. 5 . How many numbers greater than 5000 can be formed with the digits $3,5,7,8,9$ no digit being repeated?

Sol. Obviously with the given 5-digits, numbers greater than 5000 are either 4-digit numbers having 5 or 7 or 8 or 9 at the thousand’s place or 5-digit numbers.

Since the thousand’s place in the required 4-digit numbers cannot take 3 as a value, we have 4 options for thousand’s place, the remaining 4 out of 5 (1 earning the one used up at thousand’s place) for hundred’s place, 3 for tens’ place and 2 for one’s place.

(Note: Repetition of digits is not allowed)

Four-Digit Numbers

Thousands Place $5,7,8$ or 9	Hundreds Place Any of the 4 remaining digits	Tens Place Any of the remaining 3 digits	Ones Place Any of the remaining 2 digits
4 ways	4 ways	3 ways	2 ways

$\therefore$ Number of 4-digit numbers greater than 5000

That can be formed with given digits $=4 \times 4 \times 3 \times 2=96$

For the 5-digit numbers, the various places can be filled up as shown:

$\therefore$ Number of 5-digit numbers $=5 \times 4 \times 3 \times 2 \times 1=120$.

$\therefore$ Total required numbers $=96+120=\mathbf{2 1 6}$.

Ex. 6 . How many 6-digit numbers can be formed from the digits 1, 2, 4, 5, 6, 7 (no digit being repeated) which are divisible by 5 ?

Sol. A number is divisible by 5 only if its unit’s digit is 5 or 0 .

So we fix the unit’s digit as 5 and fill up the remaining 5 places with the remaining 5 digits in the following way:

Note : Repetition of digits is not allowed

$\therefore$ Required number of numbers divisible by $5=5 \times 4 \times 3 \times 2 \times 1=\mathbf{1 2 0}$.

Ex. 7 . How many numbers can be formed by using any number of the digits $3,1,0,5,7,2,9$, no digit being repeated in any number?

Sol. The number of single digit numbers is ${ }^{7} P_1$.

The permutations of 7 digits taken 2 at a time are ${ }^{7} P_2$. But ${ }^{6} P_1$ of these have zero in the ten’s place and so reduce to one digit numbers.

Hence the number of two-digit numbers is ${ }^{7} P_2-{ }^{6} P_1$

Similarly the number of the three-digit numbers is ${ }^{7} P_3-{ }^{6} P_2$ and so on.

$\therefore$ The total number required

$ ={ }^{7} P_1+({ }^{7} P_2-{ }^{6} P_1)+({ }^{7} P_3-{ }^{6} P_2)+({ }^{7} P_4-{ }^{6} P_3)+({ }^{7} P_5-{ }^{6} P_4)+({ }^{7} P_6-{ }^{6} P_5)+({ }^{7} P_7-{ }^{6} P_6)=\mathbf{1 1 7 4 3 .} $

Ex. 8 . How many different numbers can be formed with the digits $1,3,5,7,9$, when taken all at a time, and what is their sum?

Sol. The total number of numbers $=5 !=120$. Suppose we have 9 in the unit’s place. We will have $4 !=24$ such numbers. The number of numbers in which we have $1,3,5$ or 7 in the unit’s place is also $4 !=24$ in each case.

Hence the sum of the digits in the unit’s place in all the 120 numbers

$=24(1+3+5+7+9)=600$.

The number of numbers when we have any one of the given digits in ten’s place is also 41=24 in each case. Hence the sum of the digits in the ten’s place

$=24(1+3+5+7+9)$ tens $=600$ tens $=600 \times 10$.

Proceeding similarly, the required sum

$=600$ units +600 tens +600 hundreds +600 thousands +600 ten thousands

$=600(1+10+100+1000+10000)=600 \times 11111=\mathbf{6 6 6 6 6 0 0}$.

Ex. 9 . Find the number of words that can be formed by taking all the letters of the word “COMBINE”, such that the vowels occupy odd places.

(WBJEE 2010)

Sol. There are 7 letters in the word COMBINE of which 4 are consonants and 3 are vowels $(O, I, E)$.

There are 4 odd places in a 7 letter word, so the number of ways 3 vowels can be arranged in 4 places $={ }^{4} P_3$ After arranging the 3 vowels, there are 4 places left (one at odd position and 3 at even positions). So the 4 consonants can be arranged in these 4 places in 4 ! ways.

$\therefore$ Required number of words $={ }^{4} P_3 \times 4 !=4 \times 4 \times 3 \times 2 \times 1=\mathbf{9 6}$.

Ex. 10 . How many ways are there to arrange the letter in the word GARDEN with the vowels in alphabetical order?

(AIEEE 2004)

Sol. The word GARDEN contains 6 letters - 4 consonants $(G, R, D, N)$ and 2 vowels $A, E$. The 4 consonants can be arranged in 6 places in ${ }^{6} P_4$ ways. In each of these arrangements two places will remain blank in which the first place will be filled by $A$ and the place following it by $E$ as vowels have to be in alphabetical order.

This can always be done in only 1 way, i.e., $E$ following $A$

$\therefore$ Required number of ways $={ }^{6} P_4 \times 1=\frac{6 !}{(6-4) !}=\frac{6 !}{2 !} \times 1=6 \times 5 \times 4 \times 3=\mathbf{3 6 0}$.

Ex. 11 . How many signals can be made by hoisting 2 blue, 2 red and 5 yellow flags on a pole at the same time?

Sol. The number of signals $=\frac{9 !}{2 ! 2 ! 5 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2 \times 5 \times 4 \times 3 \times 2 \times 1}=\mathbf{7 5 6}$.

Ex. 12 . A coin is tossed 6 times. In how many different ways can we obtain 4 heads and 2 tails?

Sol. Whether we toss a coin 6 times or toss 6 coins at a time, the number of arrangements will be the same.

$\therefore$ The number of arrangements of 4 heads and 2 tails out of 6 is $\frac{6 !}{4 ! 2 !}=15$.

Ex. 13 . How many numbers can be formed with digits $1,2,3,4,3,2$, 1, so that odd digits always occupy the odd places?

(IIT)

Sol. The odd digits having two 1 ’s alike and two 3 ’s alike can be arranged in four odd places in $\frac{4 !}{2 ! 2 !}=6$ ways.

The three even digits having two 2’s alike can be arranged in the three even places in $\frac{3 !}{2 !}=3$ ways.

$\therefore$ The number of numbers $=6 \times 3=\mathbf{1 8}$.

Ex. 14 . There are 3 copies each of 4 different books. Find the number of ways of arranging them on a shelf.

Sol. Total number of books $=3 \times 4=12$

Each of the 4 different titles has 3 copies each

$\therefore$ Required number of ways of arranging them on a shelf $=\frac{12 !}{3 ! 3 ! 3 ! 3 !}=\frac{12 !}{(3 !)^{4}}=\mathbf{3 6 9 6 0 0}$.

Ex. 15 . Find the number of arrangements of the letters of the word ‘BANANA’ in which the two $N$ ’s do not appear adjacently.

(IIT Prel. 2002)

Sol. Considering the two $N$ ’s as one letter, the number of letters to be arranged $=5$.

Therefore, the number of arrangements $=\frac{5 !}{3 !}=20$

$(\because A$ repeated 3 times)

Total number of arrangements if there were no restriction imposed $=\frac{6 !}{3 ! 2 !}=60$.

( $A$ repeated 3 times and $N$ repeated 2 times)

$\therefore$ Required number of arrangements $=60-20=\mathbf{4 0}$.

Ex. 16 . How many numbers greater than a million can be formed with the digits $2,3,0,3,4,2,3$ ?

Sol. A million is a 7-digit number. So any number greater than 1 million will contain all the seven digits. Since the digit 2 occurs twice and digit 3 occurs thrice and the rest are different, therefore, number of possible numbers which can be formed with the given seven digits $=\frac{7 !}{(2 !)(3 !)}=420$.

These possible numbers include those which have 0 at the millions place. Keeping 0 fixed at the millions place, the remaining 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are different can be arranged in $=\frac{6 !}{(2 !)(3 !)}=60$. ways.

$\therefore$ Number of numbers greater than 1 million made from the given digits $=420-60=\mathbf{3 6 0}$.

Ex. 17 . In how many ways can the letters of the word ‘ARRANGE’ be arranged such that the two $r$ ’s do not occur together?

Sol. There are two $a$ ’s, two $r$ ’s in the word ‘arrange’, therefore the number of arrangements $=\frac{7 !}{2 ! 2 !}=1260$.

The number of arrangements in which the two $r$ ’s occur together $=\frac{6 !}{2 !}=360$.

$\therefore \quad$ The number of arrangements in which two $r$ ’s do not occur together $=(1)-(2)=1260-360=\mathbf{9 0 0}$.

Ex. 18 . If the letters of the word ‘AGAIN’ be arranged in a dictionary, what is the fiftieth word?

Sol. In a dictionary, the words are arranged in an alphabetical order.

(i) Starting with $A$, the remaining 4 letters $G, A, I, N$ can be arranged in $4 !=24$ ways. These are the first

(ii) Then, starting with $G$, the remaining letters $A, A, I, N$ can be arranged in $\frac{4 !}{2 !}=12$ ways. Thus, there are 12 words starting will $G$.

(iii) Now, the words will start with $I$. Starting with $I$, the remaining letters $A, G, A, N$ can be arranged in $\frac{4 !}{2 !}=12$ ways. So, there are 12 words, which start with $I$.

(iv) Thus, so far, we have constructed $24+12+12$, i.e., 48 words. The 49 th word will start with $N$ and is NAAGI. Hence, the 50th word is NAAIG.

Ex. 19 . The letters of the word ‘RANDOM’ are written in all possible ways and these words are written out as in a dictionary. Find the rank of the word ‘RANDOM’.

Sol. The words in a dictionary are written in an alphabetical order, which, here is $A D M N O R$.

(i) Starting with $A$, the remaining letters $D, M, N, O, R$ can be arranged in $5 !=120$ ways. So, there are 120 words starting with $A$.

(ii) Similarly, number of words starting with $M=120$, starting with $N=120$, and starting with $O=120$. (iii) Now, the number of words starting with $R$ is also 120. Out of these words, one word is RANDOM. First, we find the words starting with $R A D$ and $R A M$.

Number of starting with $R A D=3 !=6$

Number of starting with $R A M=3 !=6$

(iv) Thus, so far, $120+120+120+120+120+6+6=612$ words have been constructed.

(v) Now, the words starting with $R A N$ will appear. Their number is also $=3 !=6$. One of these words is the word RANDOM itself.

The first word beginning with RAN is RANDMO. It is 613th word and so the next word is RANDOM. $R A N D O M$ is the 614 th word.

Hence, rank of $R A N D O M=\mathbf{6 1 4}$.

Ex. 20 . How many numbers of 3-digits can be formed with the digits $1,2,3,4,5$ when digits may be repeated?

Sol. Since repetition is allowed, each of the 3 places in a 3 -digit number can be filled in 5 ways.

$\mathbf{H}$	$\mathbf{T}$	$\mathbf{O}$
$\mathbf{5}$	$\mathbf{5}$	$\mathbf{5}$
ways	ways	ways

$\therefore$ Required number of 3-digit numbers $=5 \times 5 \times 5=5^{3}=125$

Ex. 21 . How many numbers each containing four digits can be formed, when a digit may be repeated any number of times?

Sol. There are in all 10 digits, including zero. As the first digit of the number cannot be zero, so it can be chosen in 9 ways. Again, as a digit may occur any number of times in a number, the second, third and fourth digits of the numbers can be any one of the ten digits and so each of the remaining three places can be filled in 10 ways. Hence the total number of 4-digit numbers $=9 \times 10^{3}=9000$.

Verification. All the 4-digit numbers will be between 1000 and 9999 and so their number is $9999-999=9000$.

Ex. 22 . Eight different letters of an alphabet are given. Words of 4 letters from these are formed. Find the number of such words with at least one letter repeated.

(EAMCET 2006)

Sol. If any letter can be used any number of times, then the number of words of 4 Letters with 8 different letters is $8 \times 8 \times 8 \times 8=8^{4}=4096$

Number of words of 4 letters with at least one letter repetition not allowed $={ }^{8} P_4=8 \times 7 \times 6 \times 5=1680$

$\therefore \quad$ Number of 4 letter words with at least one letter repeated is $8^{4}-{ }^{8} P_4=4096-1680=\mathbf{2 4 1 6}$.

Ex. 23 . (a) In how many ways can a party of 4 boys and 4 girls be seated at a circular table so that no 2 boys are adjacent

(b) In how many ways can 10 boys and 5 girls sit around a circular table, so that no two girls sit together?

Sol. (a) Let the girls first take up their seats. They can sit in 3 ! ways. When they have been seated, then there remain 4 places for the boys each between two girls. Therefore the boys can sit in 4 ! ways. Therefore there are $3 ! \times 4$ !, i.e., 144 ways of seating the party.

(b) Let $B$, denote the position of the boys around the table. 10 boys can be seated around the table in 9 ! ways.

There are 10 spaces between the boys, which can be occupied by 5 girls in ${ }^{10} P_5$ ways. Hence,

Total number of ways $=9 ! \times{ }^{10} P_5=\frac{9 ! 10 !}{5 !}$.

Ex. 24 . A round table conference is to be held between delegates of 20 countries. In how many ways can they be seated if two particular delegates are

(i) always together, (ii) never together?

Sol. (i) Let $D_1$ and $D_2$ be the two particular delegates. Considering $D_1$ and $D_2$ as one delegate, we have 19 delegates in all. 19 delegates can be seated round a circular table in $(19-1) !=18$ ! ways.

But two particular delegates can seat themselves in 2 ! ( $D_1 D_2$ or $.D_2 D_1)$ ways.

Hence, the total number of ways $=18 ! \times 2 !=2(18 !)$

(ii) To find the number of ways in which two particular delegates never sit together, we subtract the number of ways in which they sit together from the total number of ways of seating 20 persons i.e., $(20-1) !=19$ ! ways.

Hence the total number of ways in this case $=19 !-2(18 !)=19(18 !)-2(18 !)=\mathbf{1 7}(\mathbf{1 8}$ !).

Ex. 25 . Find the number of ways in which 10 different flowers can be strung to form a garland so that 4 particular flowers are never separated.

Sol. Consider 4 particular flowers as one flower. Then, we have 7 flowers which can be strung to form a garland in $(7-1) !=6$ ! ways. But 4 particular flowers can be arranged in 4 ! ways.

Hence, the required number of ways $=\frac{1}{2}(6 ! \times 4 !)=\mathbf{8 6 4 0}$.

Ex. 26 . In how many ways can 7 persons sit around a table so that all shall not have the same neighbours in any two arrangements.

Sol. 7 persons can sit around a table in 6 ! ways but as each person will have the same neighbours in clockwise and anti-clockwise arrangements, the required number $=\frac{1}{2} 6 !=\mathbf{3 6 0}$.

Ex. 27 . If ${ }^{n} C_{r-1}=36,{ }^{n} C_r=84$ and ${ }^{n} C_{r+1}=126$, then find the value of $r . \quad$ (DCE 2000, MPPET 2009, IIT, Pb CET)

Sol. $\frac{{ }^{n} C_r}{{ }^{n} C_{r-1}}=\frac{84}{36} \Rightarrow \frac{n !}{r !(n-r) !} \times \frac{(r-1) !(n-r+1) !}{n !}=\frac{84}{36} \Rightarrow \frac{n-r+1}{r}=\frac{7}{3}$

Also, $\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r-1}}=\frac{126}{84} \Rightarrow \frac{n !}{(r=1) !(n-r-1)} \times \frac{r !(n-r) !}{n !}=\frac{3}{2} \Rightarrow \frac{n-r}{r+1}=\frac{3}{2}$

From (ii) $2 n-2 \underline{r}=3 r+3 \Rightarrow n=\frac{5 r+3}{2 !}$

Substituting in $(i)$, we get $\frac{\frac{5 r+3}{2}-r+1}{r}=\frac{7}{3} \Rightarrow \boldsymbol{r}=\mathbf{3}$.

Ex. 28 . In how many ways can 4 persons be selected from amongst 9 persons? How many times will a particular person be always selected?

Sol. The number of ways in which 4 persons can be selected from amongst 9 persons

$={ }^{9} C_4=\frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4}=126$.

The number of ways in which a particular person is always to be selected

$={ }^{8} C_3=\frac{8 \times 7 \times 6}{1 \times 2 \times 3}=\mathbf{5 6}$.

Ex. 29 . Find the number of diagonals that can be drawn by joining the angular points of a

(i) heptagon

(ii) a polygon of 20 sides.

(J&K CET 2009)

Sol. (i) A heptagon has seven angular points and seven sides.

The join of two angular points is either a side or a diagonal.

The number of lines joining the angular points $={ }^{7} C_2=\frac{7 \times 6}{1 \times 2}=21$.

But the number of sides $=7 \quad \therefore$ The number of diagonals $=21-7=\mathbf{1 4}$.

(ii) Number of diagonals in a polygon of $n$ sides $={ }^{n} C_2-n$. Here, $n=20$

$\therefore$ Required number of diagonals $={ }^{20} C_2-20=\frac{20 \times 19}{2 \times 1}-20=\mathbf{1 7 0}$.

Ex. 30 . A committee of 4 is to be selected from amongst 5 boys and 6 girls. In how many ways can this be done so as to include (i) exactly one girl, (ii) at least one girl?

Sol. (i) In this case we have to select one girl out of 6 and 3 boys out of 5 .

The number of ways of selecting 3 boys $={ }^{5} C_3={ }^{5} C_2=10$.

The number of ways of selecting one girl $={ }^{6} C_1=6$.

$\therefore$ The required committee can be formed in $6 \times 10=60$ ways.

(ii) The committee can be formed with

(a) one boy and three girls, or (b) 2 boys and 2 girls, or (c) 3 boys and one girl, or (d) 4 girls alone.

The committee can be formed in $(a){ }^{5} C_1 \times{ }^{6} C_3$ ways ;

The committee can be formed in (b) ${ }^{5} C_2 \times{ }^{6} C_2$ ways;

The committee can be formed in (c) ${ }^{5} C_3 \times{ }^{6} C_1$ ways ;

The committee can be formed in $(d){ }^{6} C_4$ ways.

Hence the required number of ways of forming the committee

$={ }^{5} C_1 \times{ }^{6} C_3+{ }^{5} C_2 \times{ }^{6} C_2+{ }^{5} C_3 \times{ }^{6} C_1+{ }^{6} C_4=100+150+60+15=\mathbf{3 2 5}$ ways.

(ii) Method II. Required ways $=($ Committees of 4 out of 11 without any restriction $)-($ Committees in which no girl is included) $={ }^{11} C_4-{ }^{5} C_4=\mathbf{3 2 5}$.

Ex. 31 . A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. Find the number of choices available to him.

(AIEEE 2003)

Sol. Two cases are possible:

(i) Selecting 4 out of first five questions and 6 out of remaining 8 questions

$\therefore$ Number of choices in this case $={ }^{5} C_4 \times{ }^{8} C_6={ }^{5} C_1 \times{ }^{8} C_2=\frac{5 \times 8 \times 7}{1 \times 2}=140$

(ii) Selecting 5 out of first five questions and 5 out of remaining 8 questions.

$\Rightarrow$ Number of choices $={ }^{5} C_5 \times{ }^{8} C_5=1 \times{ }^{8} C_3=\frac{8 \times 7 \times 6}{1 \times 2 \times 3}=56$.

$\therefore$ Total number of choices $=140+56=\mathbf{1 9 6}$.

Ex. 32 . How many different words, each containing 2 vowels and 3 consonants, can be formed with 5 vowels and 17 consonants?

Sol. Two vowels can be selected in ${ }^{5} C_2$ ways.

Three consonants can be selected in ${ }^{17} C_3$ ways.

$\therefore \quad 2$ vowels and 3 consonants can be selected in ${ }^{5} C_2 \times{ }^{17} C_3$ ways.

Now, each group of 2 vowels and 3 consonants can be arranged in 5 ! ways.

$\therefore$ The total number of words $={ }^{5} C_2 \times{ }^{17} C_3 \times 5 !=\mathbf{8 1 6 0 0 0}$.

PRACTICE SHEET

How many words (with or without meaning) can be formed from three distinct letters of the English alphabet?

(a) 17576

(b) 15600

(d) None of these

~~ 2. Two persons enter a railway compartment where there are 6 vacant seats. In how many different ways can they seat themselves?

(a) 36

(b) 11

(d) 12

~~ 3. How many integers of four digits can be formed with the digits $0,1,3,5,6$ (assuming no repetitions)?

(a) 96

(b) 120

(d) 625

~~ 4. There are 12 true-false questions in an examination. How many sequences of answers are possible?

(a) 12 !

(b) 120

(d) $2^{12}-2$

~~ 5. How many seven-digit phone numbers are possible if 0 and 1 cannot be used as the first digit and the first three digits cannot be 555,411 or 936 ?

(a) 412560

(b) 7970000

(d) 362880

~~ 6. There are 5 roads leading to a town from a village. The number of different ways in which a villager can go to the town and return back is:

(a) 5

(b) 10

(d) 20

(AMU 2002)

~~ 7. Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.

(a) 120

(b) 20

(d) 240

~~ 8. How many numbers are there between 100 and 1000 such that at least one of their digits is 7 ?

(a) 900

(b) 648

(d) 252

~~ 9. How many numbers divisible by 5 and lying between 4000 and 5000 can be formed from the digits $4,5,6,7$ and 8 ?

(a) 625

(b) 125

(d) 20

~~ 10. How many three-digit odd numbers can be formed by using the digits 1, 2, 3, 4, 5, 6 if repetition of digits is not allowed?

(a) 72

(b) 108

(d) 60

~~ 11. The number of all 3-digit numbers in each of which the sum of the digits is even is

(a) 450

(b) 375

(d) 250

~~ 12. Find the total number of ways in which 20 balls can be put in 5 boxes so that the first box contains just one ball.

(a) $20^{5}$

(b) $5^{20}$

(d) $4^{20}$

~~ 13. Find the number of ways in which five large books, four medium sized books and three small books can be placed on a shelf so that all books of the same size together

(a) $5 \times 4 \times 3$

(b) $5 ! \times 4 ! \times 3$ !

(d) 3 ! $\times 5$ ! $\times 4$ ! $\times 3$ !

(VITEE 2006)

~~ 14. If $P(5, r)=2 P(6, r-1)$, find $r$.

(a) 2

(b) 4

(d) Cannot be determined

~~ 15. If ${ }^{2 n+1} P_n:{ }^{2 n-1} P_n=3: 5$, then $n$ equals

(a) 2

(b) 3

(d) 4

(Kerala PET 2009)

~~ 16. 6 different letters of an alphabet are given. Words with four letters are formed from these given letters. Determine the number of words which have at least one letter repeated.

(a) 1296

(b) 996

(d) 360

~~ 17. How many eight-distinct letter words can be formed with the letters of the word “COURTESY” beginning with $C$ and ending with $Y$ ?

(a) 576

(b) 640

(d) 720

~~ 18. How many different words can be formed from with the letters of the word “RAINBOW” so that the vowels occupy odd places.

(a) 676

(b) 336

(d) 144

~~ 19. How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together?

(a) 2880

(b) 4320

(d) 3200

~~ 20. How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together?

(a) 480

(b) 360

(d) 720

~~ 21. There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together?

(a) 36000

(b) 25600

(d) 15600

~~ 22. Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together?

(a) 10080

(b) 120960

(d) 36500

~~ 23. All the words that can be formed using the letters A, H, L, $U, R$ are written as in a dictionary (no alphabet is repeated). Then the rank of the word RAHUL is

(a) 71

(b) 72

(d) 74

(Kerala PET 2008)

~~ 24. The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. The number of words that appear before the word COCHIN is

(a) 48

(b) 96

(d) 360

(IIT 2007)

~~ 25. Four friends have 7 shirts, 6 pants and 8 ties. In how many ways can they wear them?

(a) 336000

(b) 33600

(d) 508032000

~~ 26. If all the L’s occur together and also all I’s occur together, when the letters of the word ‘HALLUCINATION’ are permuted, then the number of such arrangements of letters is:

(a) $\frac{7 !}{2 ! 2 ! 2 ! 2 !}$

(b) $\frac{11 !}{2 ! 2 !}$

(d) $\frac{11 !}{2 ! 2 ! 2 ! 2 !}$

~~ 27. In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places?

(a) 176

(b) 144

(d) 288

~~ 28. How many different numbers greater than 60000 can be formed with the digits $0,2,2,6,8$ ?

(a) 144

(b) 48

(d) 288

~~ 29. There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles are drawn, determine the number of different arrangements.

(a) 20772

(b) 27270

(d) 27720

~~ 30. How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, 4 if repetition is allowed?

(a) 123

(b) 113

(d) 313

~~ 31. In how many ways can 8 people sit around a circular Table?

(a) 5040

(b) 40320

(d) 2520

~~ 32. A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together.

(a) $\frac{10 !}{2 !}$

(b) $\frac{9 !}{2 !}$

(d) $\frac{11 !}{2 !}$

~~ 33. In how many ways can 7 men and 7 women sit on a round table such that no two women sit together?

(a) $2(7 !)$

(b) $(6 !)^{2}$

(d) $(7 !)^{2}$

(BITSAT 2007)

~~ 34. In how many ways can 6 gentlemen and 3 ladies be seated round a table so that every gentleman may have a lady by his side?

(a) 360

(b) 1440

(d) 1260

~~ 35. There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls sit adjacent to each other. How many such arrangements are possible?

(a) 44

(b) 72

(d) 12

~~ 36. Find the number of ways in which 10 different flowers can be strung to form a garland so that three particular flowers are always together.

(a) $\frac{9 ! \times 3 !}{2}$

(b) $\frac{7 ! \times 3 !}{2}$

(d) $7 ! \times 2 !$

~~ 37. If $C(2 n, 3): C(n, 2)=12: 1$ find $n$.

(a) 2

(b) 5

(d) 3

~~ 38. If ${ }^{n} C_r:{ }^{n} C_{r+1}=1: 2$ and ${ }^{n} C_{r+1}:{ }^{n} C_{r+2}=2: 3$ determine the values of $n$ and $r$.

(a) $n=10, r=6$

(b) $n=12, r=6$

(d) $n=14, r=4$

~~ 39. In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time?

(a) 35

(b) 42

(d) 56

~~ 40. In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only?

(a) 112

(b) 132

(d) 182

~~ 41. 15 men in a room shake hands with each other, then the total number of handshakes is

(a) 90

(b) 105

(d) 115

(Rajasthan PET 2002)

~~ 42. A man has 6 friends. Number of different ways, he can invite 2 or more for dinner is

(a) 28

(b) 57

(d) 96

(Orissa JEE 2011)

~~ 43. There are 5 professors and 6 students out of whom a committee of 2 professors and 3 students is to be formed such that a particular student is excluded.

(a) 100

(b) 150

(d) 125

~~ 44. Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included?

(a) 15

(b) 25

(d) 10

~~ 45. A committee of 5 persons is to be formed out of 6 gents and 4 ladies. In how many ways can this be done if at most two ladies are included?

(a) 168

(b) 156

(d) 165

~~ 46. At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for at least one candidate, then the number of ways he can vote is

(a) 385

(b) 1110

(d) 6210

(AIEEE 2006)

~~ 47. In a chess tournament, where participants were to play one game with one another, two players fell ill, having played 6 games each without playing among themselves. If the total number of games played was 117, then the number of participants at the beginning was:

(a) 15

(b) 16

(d) 18

(AMU 2005)

~~ 48. 7 relatives of a man comprise 4 ladies and 3 gentlemen; his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite in a dinner party, 3 ladies and 3 gentlemen so that there are 3 of man’s relatives and 3 of wife’s relatives?

(a) 265

(b) 375

(d) 485

(IIT 1985)

~~ 49. A person invites 15 guests for dinner and wishes to arrange them at two round tables that can accommodate 8 persons and 7 persons respectively. In how many ways can he arrange the guests?

(a) $7 ! \times 6$ !

(b) $2 \times 8 ! \times 7 !$

(d) $15 C_2 \times \frac{13 !}{56}$

~~ 50. How many diagonals can be drawn in a polygon of 15 sides?

(a) 16

(b) 60

(d) 90

(J&K CET 2009)

~~ 51. There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. The number of triangles formed by these points is:

(a) 805

(b) 806

(d) 816

(Rajasthan PET 2007)

~~ 52. In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game?

(a) 840

(b) 240

(d) None of these

~~ 53. In how many way can a committee of 4 women and 5 men be chosen from 9 women and $7 men$, if $Mr$. A refuse to serve on the committee if Ms. B is a member?

(a) 1608

(b) 1860

(d) 1806

~~ 54. If $m$ parallel lines in a plane are intersected by a family of $n$ parallel lines, find the number of parallelograms formed?

(a) $m^{n}$

(b) $(m+1)(n+1)$

(d) $\frac{m n(n-1)(m-1)}{4}$

~~ 55. In how many ways can a pack of 52 cards be divided equally among four players in order?

(a) $(52 !)^{4}$

(b) $4 \times(13$ !)

(d) None of these

~~ 56. In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards?

(a) $16 ! \times 52$ !

(b) $\frac{52 !}{(16 !)^{3}}$

(d) $\frac{52 !}{(16 !)^{3} \times(3 !)}$

~~ 57. 7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. The number of ways in which the 7 persons can leave the elevator, if each one of them can leave it at any of the ten floors is

(a) 10 !

(b) $10^{7}$

(d) $7^{10}$

~~ 58. Eighteen quests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three on the other side. Determine the number of ways in which the seating arrangement can be made?

(a) ${ }^{18} C_4 \times{ }^{14} C_3 \times 9 ! \times 2$

(b) ${ }^{11} C_5 \times{ }^{6} C_6 \times 9 ! \times 2 !$

(d) ${ }^{18} C_4 \times{ }^{14} C_3 \times 9 ! \times 2$ !

~~ 59. A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be selected so that there are at least two balls of each colour.

(a) 452

(b) 524

(d) 254

~~ 60. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways can we place all the balls so that no box remains empty?

(a) 6

(b) 36

(d) 150

(IIT 1981, 2012)

ANSWERS

1. (b)	2. (c)	3. (a)	4. (c)	5. (b)	6. (c)	7. (c)	8. (d)	9. (c)	10. (d)
11. (a)	12. (c)	13. (d)	14. (c)	15. (d)	16. (c)	17. (d)	18. (c)	19. (b)	20. (a)
21. (c)	22. (b)	23. (c)	24. (b)	25. (d)	26. (b)	27. (d)	28. (c)	29. (d)	30. (d)
31. (a)	32. (c)	33. (c)	34. (b)	35. (b)	36. (b)	37. (b)	38. (d)	39. (a)	40. (b)
41. (b)	42. (b)	43. (a)	44. (b)	45. (c)	46. (a)	47. (c)	48. (d)	49. (c)	50. (d)
51. (b)	52. $(a)$	53. (d)	54. $(d)$	55. (c)	56. $(d)$	57. $(b)$	58. (c)	59. (c)	60. $(d)$

HINTS AND SOLUTIONS

There are 26 letters in the English alphabet, so for words with three distinct letters, the first place has 26 choices of letters, the next has remaining 25 choices and the third place has 24 choices.

$\therefore \quad$ Total number of words $=26 \times 25 \times 24$

$ =15600 $

~~ 2. The first person can sit on any of the six vacant seats.

$\therefore \quad$ Number of ways to seat first person $=6$

Now the second person can sit on any of the remaining five seats.

So number of ways to seat second person $=5$

$\therefore$ Total number of ways in which both can be seated

$=6 \times 5=30$.

~~ 3. The thousands’ place cannot be occupied by 0 , as the number will then become three digit, so there are 4 choices for thousands’ place. As repetition is not allowed, so leaving the digit

Th	$H$	$T$
4	4	3	2

(Excluding 0) occupying thousands’ place, there are again 4 choices for hundreds’ place (including 0 )

Similarly for Tens’ place there 3 choices, for units’ place 2 choices.

$\therefore \quad$ Total number of 4-digit numbers without repetition of digits $=4 \times 4 \times 3 \times 2=96$.

~~ 4. For each of the 12 questions, there are 2 ways of answering ’true’ or ‘false’. Hence, total number of sequences of answers possible

$=\underbrace{2 \times 2 \times 2 \times \ldots \ldots \times 2} _{12 \text{ times }}=\mathbf{2}^{12}$.

~~ 5. For the first digit there are 8 choices (out of 10 digits) as 0 and 1 cannot be used. Since repetition can be done, the 2nd digit and the 3rd digit have 10 choices each. So, the first three digits can be filled in $(8 \times 10 \times 10-3)$ ways (We need to exclude the numbers 555, 411 and 936 also from first three digits)

The last four digits of the telephone number can be filled in $(10 \times 10 \times 10 \times 10)$ ways.

$\therefore$ Total number of seven digit phone numbers

$=(8 \times 10 \times 10-3) \times 10 \times 10 \times 10 \times 10=\mathbf{7 9 7 0 0 0 0}$.

~~ 6. Number of ways of going from the village to the town $=$ Number of roads between them $=5$

Since all the 5 roads can be used for going back from town to the village, so number of ways of returning $=5$.

$\therefore$ Total number of ways of going to the town and coming back to the village $=5 \times 5=\mathbf{2 5}$.

~~ 7. At least 2 flags in order means that a signal may consist of either 2 flags, 3 flags, 4 flags in succession one below the other.

Total number of signals $=$ Number of 2 flag signals

Number of 3 flag signals

Number of 4 flag signals

Number of 5 flag signals

$=5 \times 4+5 \times 4 \times 3+5 \times 4 \times 3 \times 2+5 \times 4 \times 3 \times 2 \times 1$

$=20+60+120+120=\mathbf{3 2 0}$.

~~ 8. A number between 100 and 1000 has 3-digits

$\therefore$ Total number of 3-digit numbers having at least one of their digits as $7=$ (Total number of 3-digit numbers) - (Total number of 3-digit numbers in which 7 does not appear at all)

(a) Total number of 3-digit numbers:

Clearly repetition of digits is allowed.

The hundred’s place can be filled in 9 ways, i.e., using any of the digits 1 to 9 excluding zero.

The ten’s and the unit’s place can be filled in 10 ways using all the digits from $0-9$ in each place.

$\therefore$ Total number of 3-digit numbers $=9 \times 10 \times 10=900$

(b) Total number of 3-digit numbers in which 7 does not appear at all:

Here, the digits to be used are $0,1,2,3,4,5,6,8,9$ i.e., 9 in number.

Now, the hundred’s place can be filled in 8 ways (excluding $0)$, the tens’ and ones’ place can be filled in 9 ways each.

$\therefore$ Total number of 3-digit numbers in which 7 does not appear, $=8 \times 9 \times 9=648$ $\therefore$ From $(a)$ and $(b)$, we have, total number of 3-digit numbers having at least one of their digits as 7 $=\mathbf{9 0 0}-\mathbf{6 4 8}=\mathbf{2 5 2}$.

~~ 9. A number between 4000 and 5000 will have a 4 in the thousand’s place and since the number has to be divisible by 5 , it will have 5 at unit’s place. The hundreds’ and tens’ place

Th	$H$	$T$	$O$
1	5	5	1

each can be filled with any of the five digits. So,

Total number of required numbers $=1 \times 5 \times 5 \times 1=\mathbf{2 5}$.

~~ 10. For a number to be odd, we should have 1 or 3 or 5 at the unit’s place. So there are 3 ways of filling the unit’s place.

As repetition of digits is not allowed, the ten’s place can be filled in 5 ways with any of the remaining 5 -digts and the hundred’s place can be filled in 4 ways by the remaining 4-digits. So,

Required number of three-digit odd numbers $=3 \times 5 \times 4$

$ =60 \text{. } $

~~ 11. Out of the 10 digits $0,1,2,3, \ldots, 9$, five digits, i.e., $0,2,4$, 6,8 are even and five digits, i.e., $1,3,5,7$ and 9 are odd.

The sum of the three digits $D_1, D_2, D_3$ of the number $D_1$ $D_2 D_3$ will be even if :

(i) All the three digits are even:

$\therefore$ Number of such numbers $=4 \times 5 \times 5=100$

(Here the position $D_1$ cannot be occupied by 0 )

(ii) One of the digits is even and the rest two are odd:

(a) $D_1$ is even, $D_2$ is odd, $D_3$ is odd

$\therefore$ Number of such numbers $=4 \times 5 \times 5=100$

(Again since $D_1 \neq 0$, so only 4 choices for $D_1$ )

(b) $D_1$ is odd, $D_2$ is even, $D_3$ is odd

$\therefore$ Number of such numbers $=5 \times 5 \times 5=125$

$\therefore$ Total number of required numbers

$=100+100+125+125=\mathbf{4 5 0}$.

~~ 12. One ball can be put in first box in 20 ways because we can put any one of the twenty balls in the first box. Now, remaining 19 balls are to be put into remaining 4 boxes. This can be done in $4^{19}$ ways, because there are 4 choices for each ball. Hence, the required number of ways $=\mathbf{2 0} \times \mathbf{4}^{\mathbf{1 9}}$.

~~ 13. Let us consider all the books of same size as one entity. Now there are three different entities which have to be arranged on the shelf and this can be done is 3 ! ways.

Also the five large books can be arranged amongst themselves in 5 ! ways, four medium books can be arranged amongst themselves in 4! ways and the three small books can be arranged amongst themselves in 3 ! ways. So, Required number of ways of arranging the books so that all same sized books are together $=3 ! \times 5 ! \times 4 ! \times 3$ !

~~ 14. $P(5, r)=2 \cdot P(6, r-1)$

$ \begin{aligned} & \Rightarrow \quad{ }^{5} P_r=2 \times{ }^{6} P _{r-1} \\ & \Rightarrow \quad \frac{\underline{5}}{\underline{5-r}}=2 \times \frac{\underline{6}}{\underline{6-(r-1)}} \\ & \Rightarrow \quad \frac{\underline{5}}{\underline{5-r}}=\frac{2 \times 6 \times \underline{5}}{17-r} \\ & \Rightarrow \quad \frac{\underline{5}}{\underline{5-r}}=\frac{12 \times \underline{5}}{(7-r)(6-r) \underline{5-r}} \\ & \Rightarrow \quad 1=\frac{12}{(7-r)(6-r)} \Rightarrow(7-r)(6-r)=12 \\ & \Rightarrow \quad(7-r)(6-r)=4 \times 3 \\ & \Rightarrow \quad 7-r=4 \text{ and } 6-r=3 \Rightarrow r=3 . \end{aligned} $

~~ 15. ${ }^{2 n+1} P _{n-1}:{ }^{2 n-1} P_n=3: 5$

$ \Rightarrow \quad \frac{(2 n+1) !}{(2 n+1-(n-1)) !}: \frac{(2 n-1) !}{(2 n-1-n) !}=3: 5 $

$\Rightarrow \quad \frac{(2 n+1) !}{(n+2) !} \times \frac{(n-1) !}{(2 n-1) !}=\frac{3}{5}$

$\Rightarrow \quad \frac{(2 n+1)(2 n)(2 n-1) !}{(n+2)(n+1)(n)(n-1)} \times \frac{(n-1) !}{(2 n-1) !}=\frac{3}{5}$

$\Rightarrow \quad \frac{4 n^{2}+2 n}{n(n^{2}+3 n+2)}=\frac{3}{5}$

$\Rightarrow \quad(4 n^{2}+2 n) 5=3 \times(n^{3}+3 n^{2}+2 n)$

$\Rightarrow \quad 20 n^{2}+10 n=3 n^{3}+9 n^{2}+6 n$

$\Rightarrow \quad 3 n^{3}-11 n^{2}-4 n=0$

$\Rightarrow \quad n(3 n^{2}-11 n-4)=0$

$\Rightarrow \quad n(3 n^{2}-12 n+n-4)=0$

$\Rightarrow \quad n(3 n(n-4)+1(n-4))=0$

$\Rightarrow \quad n=0$ or 4 or $-\frac{1}{3}$

Ignoring $n=0,-\frac{1}{3}$ as these are in admissible values, we have $\boldsymbol{n}=\mathbf{4}$.

~~ 16. The number of 4-letter words which can be formed from 6 letters when one or more of the letters is repeated $=6 \times 6 \times 6 \times 6=1296$

Number of 4-letter words which can be formed from the given 6 letters when none of the letters is repeated

$=$ Number of arrangements of 6 letters taken 4 at a time

$={ }^{6} P_4=\frac{6 !}{(6-4) !}=\frac{6 !}{2 !}=6 \times 5 \times 4 \times 3=360$

$\therefore \quad$ Number of 4 letter words which have at least one of their letters repeated $=1296-360=\mathbf{9 3 6}$.

~~ 17. The first place will always be filled by $C$ and the last place will always be filled with $Y$. The remaining six places can be filled by the remaining 6 letters in ${ }^{6} P_6$ ways.

$\therefore$ Total number of words beginning with $C$ and ending with $Y=1 \times 1 \times{ }^{6} P_6=6 !=\mathbf{7 2 0}$.

~~ 18. There are 7 letters in the word RAINBOW out of which 4 are consonants and 3 are vowels.

$ \begin{matrix} \underline{1} & 2 & \underline{3} & 4 & \underline{5} & 6 & \underline{7} \end{matrix} $

In order that the vowels may occupy odd places, we first of all arrange any 3 consonants in even places in ${ }^{4} P_3$ ways and then the odd places can be filled by 3 vowels and the remaining 1 consonant in ${ }^{4} P_4$ ways. So,

Required number of words $={ }^{4} P_3 \times{ }^{4} P_4=24 \times 24=\mathbf{5 7 6}$.

~~ 19. There are eight letters in the word “DAUGHTER” including three vowels (A, U, E) and 5 consonants (D, G, H, T, R)

If the vowels are to be together, we consider them as one letter, so the 6 letters now ( 5 consonants and 1 vowels entity) can be arranged in ${ }^{6} P_6=6$ ! ways. Also corresponding to each of these arrangements, the 3 vowels can be arranged amongst themselves in 3 ! ways.

$\therefore$ Required number of words $=6 ! \times 3$ !

$=6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2=\mathbf{4 3 2 0}$.

~~ 20. Total number of words in which vowels are never together

$=$ Total number of words that can be formed with the letters of the word SUNDAY - Number of words in which vowels are always together

There are 6 letters in the word SUNDAY. So,

Total number of words formed by using all 6 letters $={ }^{6} P_6=6 !=720$.

Considering the vowels $U$ and $A$ as one unit, we have 5 letters that can be arranged in ${ }^{5} P_5$ ways and also corresponding to each of these arrangement of 5 letters the vowels can be arranged amongst themselves in 2 ! ways.

So, number of words in which vowels are always together $=5 ! \times 2 !=\mathbf{2 4 0}$.

Hence, total number of words in which vowels are never together $=720-240=\mathbf{4 8 0}$.

~~ 21. First of all we arrange the 5 boys in 5 ! ways.

Then we arrange the 3 girls in the remaining 6 places between the 5 boys and on the extreme in ${ }^{6} P_3$ ways.

$\times B \times B \times B \times B \times B \times$

$\therefore$ Required number of ways $=5 ! \times{ }^{6} P_3$

$ =120 \times 6 \times 5 \times 4=\mathbf{1 4 4 0 0 0} \text{. } $

~~ 22. There are 11 letters in the word “MATHEMATICS” out of which 4 are vowels and the rest 7 are consonants.

Let the four vowels be written together. A A E I M, T, H, M, T, C, S

Consider the four vowels as one as unit, then these 8 letters (7 consonants and the vowel unit) can be permuted in $\frac{8 !}{2 ! 2 !}=10080$ ways. (There are two pairs of same letters

$AA$ and $MM$ )

Corresponding to each of these permutations, the 4 vowels can be arranged among themselves in $\frac{4 !}{2 !}=12$ ways.

$\therefore$ Required number of word in which vowels occur together

$=\frac{8 !}{2 ! 2 !} \times \frac{4 !}{2 !}=10080 \times 12=\mathbf{1 2 0 9 6 0}$.

~~ 23. The words coming before RAHUL in the dictionary will have $A, H, L$ or $R$ as their first letters.

I. When $A$ is the first letter, the rest of the 4 letters $H, L, R$, $U$ can fill the next 4 places in 4 !

$\therefore \quad$ Number of words beginning with $A=24$

II. When $H$ is the first letter, the rest of the 4 letters $A, L$, $R, U$ can fill the next 4 places in 4 !

Number of words beginning with $H=24$

Similarly, the number of words beginning with $L=24$

Now among the words having $R$ as their first letter, there is only one word which comes before RAHUL and that is RAHLU

Thus there are $(24 \times 3)+1=73$ words before RAHUL.

~~ 24. To determine the rank of the word COCHIN when the letters of the word ‘COCHIN’ are permuted and all permutations arranged in alphabetical order, we see that all words begin with $C$. So we have the following cases when permuted words are in alphabetical.

I. First two letters are CC

Then the number of ways of arranging the remaining 4 letters $0, H, I, N$ in remaining 4 places 4 !

$\therefore \quad$ Number of word having $C C$ has first two letters $=24$

II. First two letters are $\mathbf{C H}$ (next to $CC$ in alphabetical order)

Here also, as in case I,

Number of words with $CH$ as first two letters $=4 !=24$

Similarly, the number of words with $CI$ and $CN$ as their first two letters will be 24 each.

The first word with $CO$ as first two letters is COCHIN

$\therefore 24+24+24+24=96$ words appear before COCHIN.

~~ 25. 7 shirts can be worn by 4 friends in ${ }^{7} P_4$ ways.

Similarly, 6 pants and 8 ties can be worn by 4 friends in ${ }^{6} P_4$ and ${ }^{8} P_4$ ways respectively.

$\therefore \quad$ Total number of ways in which 7 shirts, 6 pants and 8 ties can be worn by 4 friends $={ }^{7} P_4 \times{ }^{6} P_4 \times{ }^{8} P_4$

$=\frac{7 !}{3 !} \times \frac{6 !}{2 !} \times \frac{8 !}{4 !}$ $=(7 \times 6 \times 5 \times 4) \times(6 \times 5 \times 4 \times 3) \times(8 \times 7 \times 6 \times 5)$

$=840 \times 360 \times 1680=\mathbf{5 0 8 0 3 2 0 0 0}$.

~~ 26. In the word ‘HALLUCINATION’ there are total 13 letters out of which 7 are consonants and 6 are vowels. Also there are $2 L$ ’s, $2 N$ ’s, $2 A$ ’s and $2 I$ ’s.

If L’s occur together and I’s occur together, then we consider them as one unit each. So, the arrangement can be written as: L, L, I, I , H, A, A, U, C, N, N, T, O

Now these 11 letters (9 letters and 2 units of $L$ and I) can be arranged in $\frac{11 !}{2 ! 2 !}$ ways. $(\because$ There are 2 A’s and $2 N$ ’s $)$

~~ 27. There are 8 letters in the word AFLATOON out of which 4 are vowels, i.e., A, A, O, O and 4 are consonants, i.e., $F, L$, $T, N$.

There are 2 ways in which the 4 consonants and 4 vowels occupy alternate places:

$VC VC VC VC$ or $CV CV CV CV$

where $V$-vowel, $C$-consonant.

The four vowels can be arranged in 4 places in $\frac{4 !}{2 ! 2 !}$ ways $=6$ ways

The four consonants can be arranged in their 4 places in 4 ! ways $=24$ ways.

Also, there are 2 cases, so,

$\therefore$ Required number of permutations $=6 \times 24 \times 2=\mathbf{2 8 8}$.

~~ 28. Numbers greater than 60000 will have either 6 or 8 in the TTh place and will consist of 5-digits.

If the digit 6 occupies the TTh place, the remaining 4 places can be occupied in $\frac{4 !}{2 !}$ ways. ( $\because$ There are two 2 ’s $)$

Number of numbers beginning with $6=\frac{4 !}{2 !}=12$

Similarly, number of numbers beginning with $8=\frac{4 !}{2 !}=12$

$\therefore$ Number of different numbers greater than 60000 formed with digits $0,2,2,6,8$

$=12+12=\mathbf{2 4}$

~~ 29. There are $5+4+3=12$ marbles of which 5 are red (alike), 4 are white (alike) and 3 are blue (alike).

$\therefore$ Required number of arrangements $=\frac{12 !}{5 ! \times 4 ! \times 3 !}$

$=\frac{\not 2 \times 11 \times 10 \times 9 \times 8^{4} \times 7 \times \not 6 \times 5 !}{5 ! \times 4 \times 5 \times 22 \times 1 \times 3 \times 2 \times 1}$

$=\mathbf{2 7 7 2 0}$.

~~ 30. As there are 4 digits 1, 2, 3, 4 and repetition of digits is allowed.

Total number of 1-digit numbers $=4$

Total number of 2-digit numbers $=4 \times 4=16$

Total number of 3-digit numbers $=4 \times 4 \times 4=64$ Number of 4-digit numbers beginning with $1=4 \times 4 \times 4$ $=64 \quad(\because$ The first place is occupied by 1$)$ Number of 4-digit numbers beginning with $2=4 \times 4 \times 4=64$ Number of 4-digit numbers beginning with $3=4 \times 4 \times 4=64$ Number of 4-digit numbers beginning with $41=4 \times 4=16$ Number of 4-digit numbers beginning with $42=4 \times 4=16$ Number of 4-digit numbers beginning with $431=4$

Number of 4-digit numbers beginning with $432=1$ (4321 only)

$\therefore$ Total number of 4-digit numbers $=64+64+64+16$

$+16+4+1=229$

$\therefore$ Total number of natural numbers not exceeding 4321 $=4+16+64+229=313$.

~~ 31. As is known in case of circular permutations, we keep one place fixed, so 8 people can sit around a circular table in $(8-1)$ ! ways $=7$ ! ways $=5040$ ways.

~~ 32. Considering the president and secretary as one member, we have now 10 members in all. These 10 members can be seated round the circular table in $(10-1) !=9$ ! ways.

Also the president and secretary can seat themselves in 2 ! ways PS or SP

$\therefore$ Required number of ways of seating $=9 ! \times 2$.

~~ 33. 7 men can seat themselves in a round table in $(7-1) !=6$ ! ways.

Now there are 7 places vacant between these 7 men.

$\therefore 7$ women can seat themselves in these 7 places in 7 ! ways.

$\therefore$ Total number of required arrangement where no two women sit together $=6 ! \times 7$.

~~ 34. There are 9 people, 6 gentlemen and 3 ladies. If each gentleman has to have a lady by his side, the seating arrangement can be done as shown below:

This can be done in 5 ! (Gentlemen) $\times 3$ ! (Ladies) $=720$ ways.

The arrangements can also be made in opposite direction. So,

Total number of required arrangements $=2 \times 720=\mathbf{1 4 4 0}$.

~~ 35. Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3 ! ways. 3 girls can be arrenged in 3 ! ways

Also, the girls can be seated before the boys.

Total number of required ways $=3 ! \times 3 !+3 ! \times 3$ !

$=2 \times(3 !)^{2}$

~~ 36. Consider the three particular flowers as one flower. Then we have $(10-3)+1=8$ flowers which can be strung in the garland.

Thus the garland can be formed in $(8-1)$ !, i.e., 7 ! ways

But the 3 particular flowers can be arranged amongst themselves in 3 ! ways.

$\therefore$ Required number of ways $=\frac{1}{2}(7 ! \times 3 !)$

~~ 37. Given ${ }^{2 n} C_3:{ }^{n} C_2=12: 1$

$\Rightarrow \frac{{ }^{2 n} C_3}{{ }^{n} C_2}=\frac{12}{1} \Rightarrow \frac{\frac{(2 n) !}{(2 n-3) ! 3 !}}{\frac{n !}{(n-2) ! 2 !}}=\frac{12}{1}$

$\Rightarrow \frac{(2 n) !}{n !} \times \frac{(n-2) ! 2 !}{(2 n-3) ! 3 !}=\frac{12}{1}$

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{n(n-1)(n-2) !} \times \frac{(n-2) ! \times 2 !}{(2 n-3) ! \times 3 \times 2 !}=\frac{12}{1}$

$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{3 n(n-1)}=\frac{12}{1}$

$\Rightarrow \frac{4(\not h)(2 n-1)(n-1)}{3 \not h(n-1)}=\frac{12}{1}$

$\Rightarrow 8 n-4=36 \Rightarrow 8 n=40 \Rightarrow \boldsymbol{n}=\mathbf{5}$.

~~ 38. Given, ${ }^{n} C_r:{ }^{n} C_{r+1}=1: 2$ and ${ }^{n} C_{r+1}:{ }^{n} C_{r+2}=2: 3$

$\Rightarrow \frac{n !}{r !(n-r) !}: \frac{n !}{(r+1) !(n-r-1) !}=1: 2$ and

$\frac{n !}{(r+1) !(n-r-1) !}: \frac{n !}{(r+2) !(n-r-2) !}=2: 3$

$\Rightarrow \frac{n !}{r !(n-r)(n-r-1) !} \times \frac{(r+1) r !(n-r-1) !}{n !}=\frac{1}{2}$

and $\frac{n !}{(r+1) !(n-r-1)(n-r-2) !}$

$ \times \frac{(r+2)(r+1) !(n-r-2) !}{n !}=\frac{2}{3} $

$\Rightarrow \frac{(r+1)}{n-r}=\frac{1}{2}$ and $\frac{r+2}{n-r-1}=\frac{2}{3}$

$\Rightarrow 2 r+2=n-r$ and $3 r+6=2 n-2 r-2$

$\Rightarrow n-3 r=2$ and $2 n-5 r=8$

Solving the two simultaneous equations, we get

$n=14, r=4$.

~~ 39. When a particular member is taken every time, then we need to choose the remaining 4 members from 7 members in ${ }^{7} C_4$ ways.

$\therefore$ Required number of ways $={ }^{7} C_4=\frac{7 !}{4 ! 3 !}=\frac{7 \times 6 \times 5}{3 \times 2}=\mathbf{3 5}$.

~~ 40. As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members.

So the number of ways in which a team of 11 can be selected

$={ }^{12} C _{10} \times{ }^{2} C_1=\frac{12 !}{10 ! \times 2 !} \times 2$

$=\frac{12 \times 11}{2} \times 2=\mathbf{1 3 2}$.

~~ 41. Since every person in the room shakes hand with every other person. So, total number of handshakes $=$ Number of ways of

selecting 2 men out of 15 men

$={ }^{15} C_2=\frac{15 !}{13 ! \times 2 !}=\frac{15 \times 14}{2}=\mathbf{1 0 5}$.

~~ 42. He can write 2 or more friends out of 6 friends in the given number of ways:

Invite (i) 2 friends out of 6 friends

or (ii) 3 friends out of 6 friends

or (iii) 4 friends out of 6 friends

or (iv) 5 friends out of 6 friends

or $(v) 6$ friends out of 6 friends.

$\therefore$ Total number of ways of inviting 2 or more friends

$={ }^{6} C_2+{ }^{6} C_3+{ }^{6} C_4+{ }^{6} C_5+{ }^{6} C_6$

$=\frac{6 !}{4 ! 2 !}+\frac{6 !}{3 ! 3 !}+\frac{6 !}{4 ! 2 !}+\frac{6 !}{5 ! 1 !}+\frac{6 !}{6 !}$

$=\frac{6 \times 5}{2}+\frac{6 \times 5 \times 4}{3 \times 2}+\frac{6 \times 5}{2}+6+1$

$=15+20+15+6+1=\mathbf{5 7}$.

~~ 43. If a particular student is excluded, then the committee has to be chosen as: 2 professors from 5 professors and 3 students from 5 students (as one student is excluded)

$\therefore$ Total number of ways of forming the committee

$={ }^{5} C_2 \times{ }^{5} C_3$

$=\frac{5 !}{2 ! 3 !} \times \frac{5 !}{3 ! 2 !}$

$=\frac{5 \times 4}{2} \times \frac{5 \times 4}{2}=\mathbf{1 0 0}$.

~~ 44. If at least one woman has to be included then the committee can be formed as:

1 woman and 2 men
2 women and 1 man

$\therefore$ Number of ways of forming the committee

$={ }^{2} C_1 \times{ }^{5} C_2+{ }^{2} C_2 \times{ }^{5} C_1$

$=2 \times 10+1 \times 5=20+5=\mathbf{2 5}$.

~~ 45. A committee of 5 persons, including at most two ladies can be constituted in the following ways :

I. Selecting 5 gents only out of 6 gents.

II. Selecting 4 gents out of 6 gents and 1 lady out of 4 ladies.

III. Selecting 3 gents out of 6 gents and 2 ladies out of 4 ladies.

$\therefore$ Total number of ways of forming the committee

$={ }^{6} C_5+{ }^{6} C_4 \times{ }^{4} C_1+{ }^{6} C_3 \times{ }^{4} C_2$

$=6+\frac{6 \times 5}{2} \times 4+\frac{6 \times 5 \times 4}{3 \times 2} \times \frac{4 \times 3}{2}$

$=6+60+120=\mathbf{1 8 6}$.

~~ 46. As there are 4 candidates to be elected and the voter votes for at least one candidate, the voter can vote for 1 or 2 or 3 or 4 candidates out of 10 candidates.

$\therefore$ Total number of ways in which the voter can vote

$={ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3+{ }^{4} C_4$

$=10+\frac{10 \times 9}{2 \times 1}+\frac{10 \times 9 \times 8}{3 \times 2 \times 1}+\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$

$=10+45+120+210=\mathbf{3 8 5}$.

~~ 47. Let the number of participants in the beginning be $\boldsymbol{n}$.

Total number of games played by the 2 players who fell ill $=6+6=12$

Number of remaining players is $(n-2)$

Number of games played by $(n-2)$ players with each other $={ }^{n-2} C_2$

$\therefore$ Total number of games played $={ }^{n-2} C_2+12$

Given ${ }^{n-2} C_2+12=117$

$\Rightarrow{ }^{n-2} C_2=105 \Rightarrow \frac{n-2 !}{(n-4) ! 2 !}=105$

$\Rightarrow \frac{(n-2)(n-3)}{2}=105 \Rightarrow n^{2}-5 n-204=0$

$\Rightarrow(n-17)(n+12) 0 \Rightarrow n=17 \quad$ (Neglecting - ve value)

~~ 48. The 6 people ( 3 ladies and 3 gentlemen) to be invited can be selected in the given ways:

(i) 3 ladies from man’s relatives and 3 gentlemen from wife’s relatives, i.e.,

No. of ways $={ }^{4} C_3 \times{ }^{3} C_0 \times{ }^{3} C_0 \times{ }^{4} C_3=16$

(ii) 2 ladies and 1 gentleman from man’s relatives and 1 lady and 2 gentlemen from wife’s relatives, i.e.,

No. of ways $={ }^{4} C_2 \times{ }^{3} C_1 \times{ }^{3} C_1 \times{ }^{4} C_2$

$=\frac{4 \times 3}{2} \times 3 \times 3 \times \frac{4 \times 3}{2}=324$

(iii) 1 lady and 2 gentlemen from man’s relatives and 2 ladies and 1 gentleman from wife’s relatives, i.e., No. of ways $={ }^{4} C_1 \times{ }^{3} C_2 \times{ }^{3} C_2 \times{ }^{4} C_1=4 \times 3 \times 4 \times 3$

$ =144 $

(iv) 3 gentlemen from man’s relatives and 3 ladies from wife’s relatives, i.e.,

No. of ways $={ }^{4} C_0 \times{ }^{3} C_3 \times{ }^{3} C_3 \times{ }^{4} C_0=1 \times 1=1$

$\therefore$ Total number of ways of selecting the relatives for dinner party $=16+324+144+1=\mathbf{4 8 5}$.

~~ 49. First the 15 guests have to be divided into two groups of 8 persons and 7 persons.

$\therefore \quad$ Number of ways of dividing the guests $=\frac{15 !}{8 ! 7 !}$

Now 8 persons can be seated in one round table in $(8-1)$ ! $=7$ ! ways

Also 7 persons can be seated in another round table in $(7-1) !=6$ ! ways $\therefore \quad$ Number of ways of arranging the guests $=\frac{15 !}{8 ! \not !} \times \not \subset ! \times 6 !$ $=\frac{15 !}{8 \times 7}=\frac{\mathbf{1 5} \text{ ! }}{\mathbf{5 6}}$.

~~ 50. Number of diagonals for a $n$-sided closed polygon

$ \begin{aligned} & ={ }^{n} C_2-n=\frac{n !}{(n-2) ! 2 !}-n \\ & =\frac{n(n-1)(n-2) !}{(n-2) ! \times 2}-n=\frac{n(n-1)-2 n}{2} \\ & =\frac{n^{2}-n-2 n}{2}=\frac{n-3 n}{2}=\frac{n(n-3)}{2} \end{aligned} $

$\therefore \quad$ Number of diagonals for a 15-sided polygon

$=\frac{15 \times(15-3)}{2}=\frac{15 \times 12}{2}=\mathbf{9 0}$.

~~ 51. Number of triangles that can be drawn with all 18 points in the plane $={ }^{18} C_3$

But 5 points are collinear

$\therefore$ Required number of triangles $={ }^{18} C_3-{ }^{5} C_3$

$=\frac{18 !}{15 ! 3 !}-\frac{5 !}{2 ! 3 !}=\frac{18 \times 17 \times 16}{3 \times 2}-\frac{5 \times 4}{2}$

$=816-10=\mathbf{8 0 6}$.

~~ 52. We can choose 2 men out of 8 men in ${ }^{8} C_2$ ways. Since no husband and wife are to play in the same game, two women out of the remaining 6 women can be chosen in ${ }^{6} C_2$ ways.

If $M_1, M_2, W_1, W_2$ are respectively the two men and two women chosen for the teams then the teams can be formed as $(M_1, W_1)(M_2, W_2)$ or $(M_1, W_2)(M_2, W_1)$, i.e., in 2 ways. Hence, total number of ways of arranging the game $={ }^{8} C_2 \times{ }^{6} C_2 \times 2$

$=\frac{8 \times 7}{2} \times \frac{6 \times 5}{2} \times 2=\mathbf{8 4 0}$.

~~ 53. The required committee can be chosen in the following ways:

I. When Ms. B is a member and Mr. A refuses to serve.

[Selection has to be made of 3 women from 8 women as Ms B is already there and 5 men from 6 men as Mr A is excluded]

II. When Ms B is not a member and Mr. A can serve.

(Selection is to made of 4 women from 8 women as Ms. B is excluded and 5 men from 7 men)

$\therefore$ Total number of ways in which the committee can be formed.

$={ }^{8} C_3 \times{ }^{6} C_5+{ }^{8} C_4 \times{ }^{7} C_5$

$=\frac{8 \times 7 \times 6}{3 \times 2} \times 6+\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \times \frac{7 \times 6}{2}$

$=56 \times 6+70 \times 21=336+1470=\mathbf{1 8 0 6}$.

~~ 54. A parallelogram is formed by choosing two straight lines from the set of $\boldsymbol{m}$ parallel lines and choosing two straight lines from the set of $\boldsymbol{n}$ parallel lines.

$\therefore$ Number of parallelograms formed $={ }^{m} C_2 \times{ }^{n} C_2$

$=\frac{m(m-1)}{2} \times \frac{n(n-1)}{2}=\frac{m n(m-1)(n-1)}{4}$

~~ 55. This means that first player gets 13 cards, then second player gets 13 cards, then third player gets 13 cards and then fourth gets 13 cards.

$\therefore$ First player gets 13 cards in ${ }^{52} C _{13}$ ways

Second player gets 13 cards in ${ }^{39} C _{13}$ ways

Third player gets 13 cards in ${ }^{26} C _{13}$ ways

Fourth player gets 13 cards in ${ }^{13} C _{13}$ ways

$\therefore$ Total number of ways $={ }^{52} C _{13} \times{ }^{39} C _{13} \times{ }^{26} C _{13} \times{ }^{13} C _{13}$

$=\frac{52 !}{39 ! 13 !} \times \frac{39 !}{26 ! 13 !} \times \frac{26 !}{13 ! 13 !} \times 1=\frac{52 !}{(13 !)^{4}}$

Alternatively, 52 cards can be divided equally among

4 players in $\frac{52 !}{(13 !)^{4} \times 4 \text{ ! }}$ ways.

But as here order is important, hence required number of ways

$=\frac{52 ! \times 4 !}{(13 !)^{4} \times 4 !}=\frac{52 !}{(13 !)^{4}}$

~~ 56. First player can get 16 cards in ${ }^{52} C _{16}$ ways

Second player can get 16 cards in ${ }^{36} C _{16}$ ways

Third player can get 16 cards in ${ }^{20} C _{16}$ ways

Fourth player can get 4 cards in ${ }^{4} C_4$ ways

But the first three sets can be interchanged in 3 ! ways

$\therefore$ Required number of ways

$={ }^{52} C _{16} \times{ }^{36} C _{16} \times{ }^{20} C _{16} \times{ }^{4} C_4 \times \frac{1}{3 !}$

$=\frac{52 !}{(16 !)^{3} \times 3 !}$

~~ 57. The first person can leave the elevator in any of the 10 floors $\therefore$ Number of ways 1 st person can leave the elevator $=10$ Similarly, for each of the remaining 6 persons, the number of ways each one can leave the elevator $=10$

$\therefore$ All the 7 persons can leave the elevator in

$=\underbrace{10 \times 10 \times 10 \times 10 \times 10 \times 10} _{7 \text{ terms }}=(\mathbf{1 0})^{7}$ ways.

~~ 58. As four particular guests want to sit on a particular side say $(S_1)$ and three others on the other side, say, $(S_2)$. So we are left with 11 guests out of which we choose 5 for side $S_1$ from remaining 11 in ${ }^{11} C_5$ ways and 6 from remaining 6 for side $S_2$ in ${ }^{6} C_6$ ways.

Also the 9 persons on each side can be arranged amongst themselves in 9 ! ways.

$\therefore$ Required number of ways of making the seating arrangement $={ }^{11} C_5 \times{ }^{6} C_6 \times 9 ! \times 9$ !

~~ 59. 6 balls consisting of at least two balls of each colour from 5 red and 6 white balls can be made in the following ways:

(a) Selecting 2 red balls out of 5 red balls and 4 white balls out of 6 , i.e.,

$ \begin{aligned} \text{ Number of ways } & ={ }^{5} C_2 \times{ }^{6} C_4=\frac{5 \times 4}{2} \times \frac{6 \times 5}{2} \\ & =10 \times 15=150 \end{aligned} $

(b) Selecting 3 red balls out of 5 red balls and 3 white balls out of 6 , i.e.,

Number of ways $={ }^{5} C_3 \times{ }^{6} C_3=\frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2}$

$ =10 \times 20=200 $

(c) Selecting 4 red balls out of 5 red balls and 2 white balls out of 6 , i.e., Number of ways $={ }^{5} C_3 \times{ }^{6} C_2=5 \times \frac{6 \times 5}{2}=5 \times 15=75$

$\therefore$ Total number of ways of selecting at least two balls of each colour $=150+200+75=\mathbf{4 2 5}$.

~~ 60. As no box has to be empty, each box should have at least 1 ball.

Let the boxes be labelled as $B_1, B_2$ and $B_3$. Then the distribution of the balls can be shown as:

$B_1$	$B_2$	$B_3$
1	1	3
1	2	2
1	3	1
2	1	2
2	2	1
3	1	1

$\therefore \quad$ Number of ways of placing the balls in different boxes so that no box remains empty

$ \begin{gathered} ={ }^{5} C_1 \times{ }^{4} C_1 \times{ }^{3} C_3+{ }^{5} C_1 \times{ }^{4} C_2 \times{ }^{2} C_2+{ }^{5} C_1 \times{ }^{4} C_3 \times{ }^{1} C_1+{ }^{5} C_2 \\ \times{ }^{3} C_1 \times{ }^{2} C_2+{ }^{5} C_2 \times{ }^{3} C_2 \times{ }^{1} C_1+{ }^{5} C_3 \times{ }^{2} C_1 \times{ }^{1} C_1 \\ =5 \times 4 \times 1+5 \times \frac{4 \times 3}{2} \times 1+5 \times 4 \times 1+\frac{5 \times 4}{2} \times 3 \times 1 \\ +\frac{5 \times 4}{2} \times 3 \times 1+\frac{5 \times 4}{2} \times 2 \times 1 \\ =20+30+20+30+30+20=\mathbf{1 5 0 .} \end{gathered} $

SELF ASSESSMENT SHEET

How many integers greater than 999 but not greater than 4000 can be formed with the digits $0,1,2,3$ and 4 if repetition of digits is allowed?

(a) 499

(b) 500

(d) 376

(CAT 2008)

~~ 2. Consider the word RADAR. Whichever way you read it, from left to right or right to left, you get the same word. Such a word is known as a palindrome. Find the maximum possible number of 5-letter palindromes (meaningful or non-meaningful)

(a) 676

(b) 78

(d) 130

~~ 3. How many arrangements can be formed out of the letters of the word EXAMINATION so that vowels always occupy odd places?

(a) 72000

(b) 86400

(d) 64000

(SNAP 2007)

~~ 4. A number lock consists of 3 rings each marked with 10 different numbers. In how many cases the lock cannot be opened?

(a) $3^{10}$

(b) $10^{3}$

(d) 999

(SNAP 2008)

~~ 5. A six digit number divisible by 3 is to be formed using the digits $0,1,2,3,4$ and 5 if repetition of digits is not allowed. The total number of ways this can be done is:

(a) 120

(b) 240

(d) 720

(Kerala PET)

~~ 6. How many different nine-digit numbers can be formed from the number 223355888 by rearranging its digits so that odd digits occupy even positions?

(a) 16

(b) 36

(d) 96

(IIT 2000)

~~ 7. In how many ways can the eight directors, the vice-chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the vice-chirman and the director?

(a) 9 ! $\times 2$

(b) $2 \times 8 !$

(d) None of these

(CAT)

~~ 8. A total of 28 handshakes was exchanged at the conclusion of a party. Assuming that each participants was equally polite toward all the others, the number of people present was

(a) 14

(b) 28

(d) 8

(FMS 2011)

~~ 9. Out of 6 ruling and 5 opposition party members, 4 are to be selected for a delegation. In how many ways can it be done so as to include at least one opposition member?

(a) 300

(b) 315

(d) 410

(JMET 2011)

~~ 10. A team of 8 players is to be chosen from a group of 12 players. Out of the eight players one is to be elected as captain and another as vice-captain. In how many ways can this be done?

(a) 27720

(b) 13860

(d) 495

(NDA/NA 2010)

~~ 11. There are 10 points on a line and 11 points on another line, which are parallel to each other. How many triangles can be drawn taking the vertices on any of the line?

(a) 1050

(b) 2550

(d) 1045

(CAT)

~~ 12. In how many ways is it possible to choose a white square and a black square on a chess board so that the squares must not lie in the same row or column?

(a) 56

(b) 896

(d) 768

(CAT 2002)

~~ 13. In a chess competition involving some boys and girls of a school, every student has to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

(a) 200

(b) 216

(d) 256

~~ 14. The value of $\sum _{r=1}^{n} \frac{{ }^{n} P_r}{r !}$ is

(CAT 2005)

(a) $2^{n}$

(b) $2^{n}-1$

(d) $2^{n}+1$

(IIFT 2007)

~~ 15. In how many ways can four letters of the word ‘SERIES’ be arranged?

(a) 24

(b) 42

(d) 102

(IIFT 2010)

ANSWERS

$($ d

~~ 2. (c)

~~ 3. (c)

~~ 4. $($ d)

~~ 5. $(c)$

~~ 6. $($ d $)$

~~ 7. (d)

~~ 8. (a)

~~ 9. $(b)$

~~ 10. (d)

~~ 11. (c)

~~ 12. $(b)$

~~ 13. (d)

~~ 14. (b)

~~ 15. (a)

HINTS AND SOLUTIONS

The smallest number in the required series is 1000 and the greatest is 4000 (The only number to start with 4)

$\therefore \quad$ The thousands’ place can be filled with any of the 3 -digits 1,2 and 3.

The next three places (hundred’s, ten’s and unit’s place) can take any of the five values 0 or 1 or 2 or 3 or 4 .

$\therefore \quad$ Number of numbers from 100 to 3999

$=3 \times 5 \times 5 \times 5=375$

As 4000 is also included, so required number $=\mathbf{3 7 6}$.

~~ 2. The first letter can be chosen in 26 ways as there are 26 letters in the English alphabet.

Having chosen the first letter, we have 26 choices for the second letter also.

Similarly for the third letter also, we have 26 choices.

The fourth letter has to be same as the second and the fifth letter has to be same as the first letter so 1 choice for each.

$\therefore$ Maximum possible number of 5-letter palindromes

$=26 \times 26 \times 26 \times 1 \times 1=\mathbf{1 7 5 7 6}$.

~~ 3. There are 11 letters in the word EXAMINATION of which there are 2A’s, 2I’s.

Also there are 5 consonants and 6 vowels.

The 6 vowels can be arranged in 6 odd places in $\frac{6 !}{2 ! 2 !}$ ways (2A’s, 2I’s)

The 5 consonants can be arranged in 5 even possible $=\frac{5 !}{2 !}$ ways ( $2 N$ ’s)

$\therefore$ Total number of arrangements possible $=\frac{6 !}{2 ! \times 2 !} \times \frac{5 !}{2 !}$

$=\frac{6 \times 5 \times 4 \times 3}{2} \times 5 \times 4 \times 3=\mathbf{1 0 8 0 0}$.

~~ 4. The first ring can be marked with any of the 10 numbers, i.e., Number of ways of marking the first ring $=10$

Similarly the second and third ring can also be marked with any of the 10 numbers.

$\therefore$ Number of number combinations that can appear on the 3 rings $=10 \times 10 \times 10=1000$

The lock can be opened with a single combination only.

$\therefore$ Number of cases in which the lock cannot be opened $=1000-1=999$.

~~ 5. As repetition of digits is not allowed and the six digit number has to be formed using all the 6-digits $0,1,2,3,4,5$ implies that all the six digits will be used in making the number.

Sum of the 6-digits $=0+1+2+3+4+5=15$

Hence any 6-digit number formed with these 6 digits will be divisible by 3 .

The first place from the left can be filled with any of the non zero numbers.

Now, as we can see in the diagram given

The second place with remaining 5 numbers (including 0 ), the third with remaining 4 , the fourth with remaining 3 and so on.

$\therefore \quad$ Number of ways in which all the six places can be filled $=$ Number of ways of forming the 6 -digit number divisible by 3

$=5 \times 5 \times 4 \times 3 \times 2 \times 1=\mathbf{6 0 0}$.

~~ 6. There are 9-digits in the given number of which there are 4 odd-digits (Two 3’s, Two 5’s) and 5 even digits (Two 2’s, Three 8’s).

In case of a 9-digit number there are 5 odd positions and 4 even positions

The 4 odd digts can occupy the 4 even positions in $\frac{4 !}{2 ! 2 !}$ ways (Two 3’s, Two 5’s)

$ -X-X-X-X- $

$ (\begin{matrix} \text{ X Odd } \\ \text{ - Even } \end{matrix} ) $

Also the 5 even digits can occupy the 5 odd positions in $\frac{5 !}{3 ! 2 !}$ ways (Three 8 ’s, Two 2 ’s)

$\therefore$ Required number of 9-digit numbers $=\frac{4 !}{2 ! 2 !} \times \frac{5 !}{3 ! 2 !}$

$=6 \times 10=\mathbf{6 0}$.

~~ 7. Let the vice-chairman and chairman form 1 unit along with 8 directors. So now we have 9 units to arrange in a circle which can be done in $(9-1)$ ! ways $=8$ ! ways.

$ \begin{matrix} D & C & & VC & D \\ & & \text{ or } & & \\ D & VC & & C & D \end{matrix} $

Also the chairman and vice chairman can be arranged amongst themselves in 2 different ways.

$\therefore$ Required number of ways $=\mathbf{2} \times \mathbf{8}$ !.

~~ 8. Let the number of people present be $n$. Then if $n$ people shake hands with one another, number of handshakes $={ }^{n} C_2$. $\Rightarrow{ }^{n} C_2=28$

$\Rightarrow \frac{n !}{(n-2) ! 2 !}=28 \Rightarrow \frac{n(n-1)}{2}=28$

$\Rightarrow n(n-1)=56 \Rightarrow n(n-1)=8 \times 7 \quad \Rightarrow n=8$.

~~ 9. 4 members of the delegation can be selected in the following ways: I. 1 opposition member and 3 Ruling party members, i.e., Number of ways of this selection $={ }^{5} C_1 \times{ }^{6} C_3$

II. 2 opposition members and 2 ruling party members, i.e., Number of ways of this selection $={ }^{5} C_2 \times{ }^{6} C_2$

III. 3 opposition members and 1 ruling party member, i.e., Number of ways of this selection $={ }^{5} C_3 \times{ }^{6} C_1$

IV. 4 opposition members, i.e.,

Number of ways of this selection $={ }^{5} C_4$.

$\therefore$ Total number of ways for required selection

$={ }^{5} C_1 \times{ }^{6} C_3+{ }^{5} C_2 \times{ }^{6} C_2+{ }^{5} C_3 \times{ }^{6} C_1+{ }^{5} C_4$

$ \begin{aligned} & =5 \times \frac{6 \times 5 \times 4}{3 \times 2}+\frac{5 \times 4}{2} \times \frac{6 \times 5}{2}+\frac{5 \times 4}{2} \times 6+5 \\ & =100+150+60+5=\mathbf{3 1 5 .} \end{aligned} $

~~ 10. Number of ways to choose 8 players from 12 players

$={ }^{12} C_8=\frac{12 !}{8 ! 4 !}=495$

Number of ways to choose a captain and a vice captain $={ }^{8} C_1 \times{ }^{7} C_1$

$=8 \times 7=56$

$\therefore$ Required number of ways $=495 \times 56=\mathbf{2 7 7 2 0}$.

~~ 11. For a triangle, two points on one line and one on the other has to be chosen.

No. of ways $=2$ points from 10 points and 1 point from 11 points or 1 point from 10 points and 2 points from 11 points

$={ }^{10} C_2 \times{ }^{11} C_1+{ }^{10} C_1 \times{ }^{11} C_2$

$=\frac{10 \times 9}{2} \times 11+10 \times \frac{11 \times 10}{2}=495+550=\mathbf{1 0 4 5}$.

~~ 12. There are 32 black and 32 white squares on a chess board. Then number of ways of choosing one white and one black square on the chose board

$={ }^{32} C_1 \times{ }^{32} C_1=32 \times 32=1024$

There are 8 rows and 8 columns on a chess board each containing 4 white squares and 4 black squares.

$\therefore \quad$ Number of ways to choose a white and a black square from the same row $={ }^{4} C_1 \times{ }^{4} C_1 \times 8=128$

Similarly, number of ways to choose a white and a black square from the same column $={ }^{4} C_1 \times{ }^{4} C_1 \times 8=128$

$\therefore \quad$ Number of ways of choosing a white square and a black square on the chess board so that squares do not lie in the same row or column

$=1024-(128+128)=1024-256=\mathbf{7 6 8}$.

~~ 13. Let the number of girls be $x$ and let the number of boys be $y$.

Then, as in 45 games 2 girls played against each other,

${ }^{x} C_2=45$

Similarly, ${ }^{y} C_2=190$

$ \begin{aligned} & { }^{x} C_2=45 \Rightarrow \frac{x !}{(x-2) ! 2 !}=45 \Rightarrow x(x-1)=90 \\ & \Rightarrow x(x-1)=10 \times 9 \Rightarrow x=10 \\ & { }^{y} C_2=190 \Rightarrow \frac{y(y-1)}{2}=190 \Rightarrow y(y-1)=380 \\ & \Rightarrow y(y-1)=20 \times 19 \Rightarrow y=\mathbf{2 0 .} \end{aligned} $

$\therefore \quad$ There are 10 girls and 20 boys.

Hence, the total number of games in which one player is a girl and other player is a boy $=10 \times 20=\mathbf{2 0 0}$.

~~ 14. $\sum_{r=1}^{n} \frac{{ }^{n} P_r}{r !}=\sum_{r=1}^{n} \frac{n !}{(n-r) ! r !}=\sum_{r=1}^{n}{ }^{n} C_r$

$ ={ }^{n} C_1+{ }^{n} C_2+{ }^{n} C_3+\ldots+{ }^{n} C _{n-1}+{ }^{n} C_n $

We know that,

$ \begin{aligned} & { }^{n} C_0+{ }^{n} C_1+{ }^{n} C_2+\ldots+{ }^{n} C _{n-1}+{ }^{n} C_n=2^{n} \\ & \therefore \quad \sum _{r=1}^{n}{ }^{n} C_r=2^{n}-{ }^{n} C_0=\mathbf{2}^{\boldsymbol{n}}-\mathbf{1} . \end{aligned} $

~~ 15. The given word SERIES has $2 S$ and $2 E$ and the rest are distinct. The number of ways of arranging the 4 letters of word are as follows: I. 4 letters are distinct, i.e., S, E, R, I

$\therefore \quad$ Number of ways of arranging these distinct 4 letters $=4 !=24$

II. 2 letters are same and 2 are distinct, i.e.

SSRI, SSRE, SSIE, EERI, EERS, EEIS

$\therefore \quad$ Number of ways of arranging letters in this way

$=\frac{4 !}{2 !} \times 6=72$

III. Two are same of one kind and two are same of other kind

$\therefore \quad$ Number of ways of above arrangement $=\frac{4 !}{2 ! 2 !}=6$

$\therefore \quad$ Total number of ways $=24+72+6=\mathbf{1 0 2}$.

Probability

KEY FACTS

Probability: It is the chance of happening of an event when measured quantitatively.
Trial: It is the performance of an experiment, such as throwing a dice or tossing a coin.
Random Experiment. A random experiment is an experiment in which

(i) all the outcomes of the experiment are known in advance and

(ii) the exact outcome of any specific performance of the experiment is unpredictable, i.e., not known in advance.

For example,

(i) Tossing a coin is a random experiment as :

(a) Assuming that the coin does not land on the edge, there are only two possible outcomes of tossing- a head or a tail.

(b) But on preforming the experiment we are not sure whether we shall get a head or a tail.

(ii) Drawing a card from a well shuffled pack of 52 cards is a random experiment as:

(a) Total number of outcomes is 52 since there are 52 cards.

(b) We do not know in advance which card is drawn in a specific draw.

Event: It is something that happens, whose probability we may want to measure, as drawing a spade from a pack of 52 cards.
Occurrence of an Event: If a die* is thrown and the desired event is getting an even number on the top, then the event $A$ is said to have occurred if the number appearing on top of the die is either 2 or 4 or 6 . If 1 or 3 or 5 appears on the top, then event $\boldsymbol{A}$ has not occurred.
Sample Space is the set of all possible outcomes of a random experiment generally denoted by $\boldsymbol{S}$. For example, when a die is thrown, $S=\lbrace 1,2,3,4,5,6$, . Each element of a sample space is called a sample point.
Elementary or Simple Event. An event containing only a single sample point is called an elementary event. For example,

(i) If a coin is tossed, getting head or tail are two elementary events.

(ii) If a die is thrown, getting anyone of $1,2,3,4,5$ or 6 on the top are six elementary events.

Compound or Composite Event. An event containing more than one sample point is called a compound or composite event. It is a subset of the sample spaces.

For example,

(i) If a die is thrown, then the statement “getting an odd number on the top” is a compound event containing the sample points 1,3 and 5 .

Let $A$ denote the given event, then $A=\lbrace 1,3,5\rbrace $

Sample space $S=\lbrace 1,2,3,4,5,6\rbrace $

Clearly $A \subset S$^1 If two coins are tossed simultaneously then the sample space $S$ is given by :

$S=\lbrace (H H),(H T),(T H),(T T)\rbrace $

If $A$ be the event that “one head and one tail” turn up, then

$A=\lbrace (H T),(T H)\rbrace $

Sure Event: Let $S$ be the sample space of a random experiment since $S$ is a subset of itself, it also represents an event. Since all the outcomes of a random experiment belong to sample space $S$, the event $S$ represents a sure event.
Impossible Event: If we assume that $S$ is the sample space of a random experiment then $\phi$ (null set) is a subset of $S$.

As no outcome of the experiment belong to $\phi$, the event $\phi$ can never occur when the experiment is performed. Hence $\phi$ is an impossible event.

Probability of a sure event is 1 and the probability of an impossible event is 0 .

Probability is never less than 0 and greater than 1.

Equally Likely Events. When two or more events have an equal chance of happening, then they are called equally likely.

Ex. (i) In a toss of a coin, both head and tail are equally likely to occur.

(ii) In a throw of a die, all the six faces $(1,2,3,4,5,6)$ are equally likely to occur.

Complementary Event. Let $S$ be the sample space associated with a random experiment and let $E$ be an event. The event of non-occurrence of $E$ is called the complementary event of $E$. It is represented by $\bar{E}$ or $E^{\prime} . \bar{E}$ is also known as not $\boldsymbol{E}$.

Ex. If in a throw of a die, getting a prime number is a favourable event $A$, then the event of getting a number that is not prime is called the complementary event of $A$.

If $A=\lbrace 2,3,5\rbrace $, then $\bar{A}=\lbrace 1,4,6\rbrace $

Exhaustive Events. The total number of possible outcomes of a random experiment is called exhaustive events.

Ex. (i) In tossing a coin, exhaustive events are 2 (Head or tail)

(ii) In throwing a die, the exhaustive number of cases is 6 , since any of the faces marked with 1, 2, 3, 4, 5 or 6 may come uppermost.

(iii) In drawing 4 cards from a well shuffled pack of 52 cards, the exhaustive number of cases is $52 C_4$, since 4 cards can be drawn out of 52 cards in $52 C_4$ ways.

Mutually Exclusive Events. The events are said to be mutually exclusive if they cannot occur simultaneously in a single draw. Events such as tossing a head or a tail with a coin, drawing a queen or a jack from a pack of cards, throwing an even number or an odd number with a dice are all mutually exclusive events. Here, the occurrence of an event rules out the happening of the other event in the same experiment, i.e,.

If we toss a coin, we can never get a head and a tail in the same toss.

Ex : Suppose a card is drawn from a pack of cards and we define events as below :

$A:$ The card drawn is a spade;

$B:$ The card drawn is a heart;

$C$ : The card drawn is a king.

$A$ and $B$ are mutually exclusive, since if a card is a spade, it cannot be a heart, i.e, if $A$ occurs, $B$ cannot occur and vice versa. On the other hand, events $A$ and $C$ are not mutually exclusive as the card can be the king of spades. Similarly for $B$ and $C$, the card can be king of hearts.

Algebra of Events. Let $A, B$ and $C$ be any two events associated with a random experiment whose sample space is $S$. Then,

(i) $A \cup B$. (Union of $\boldsymbol{A}$ and $B$ ) is the event that occurs if $A$ occurs or $B$ occurs or both $A$ and $B$ occur.

(ii) $A \cap B$. (Intersection of $\boldsymbol{A}$ and $\boldsymbol{B}$ ). It is the event set which contains all sample points or outcomes which the two events $A$ and $B$ have in common.

Ex: (In a throw of a die), $A$ : Event of getting an odd number

$ \begin{aligned} & B: \text{ Event of getting a prime number } \\ & \Rightarrow A=\lbrace 1,3,5\rbrace , B=\lbrace 2,3,5\rbrace . \text{ Then, } A \cap B=\lbrace 3,5\rbrace . \end{aligned} $

(iii) $\bar{A}$, (Complement of an event $\boldsymbol{A})$. It is the set of all sample points of sample spaces that are not contained in $A$.

In the toss of a coin, if $A$ is getting a tail, then $\bar{A}$ is getting a head.

(iv) $(\boldsymbol{A} \cup \boldsymbol{B} \cup \boldsymbol{C}$ ) is the event that occurs when at least one of the events $A, B$ or $C$ occurs

(v) $(\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C}$ ) is the event that occurs when all the three events $A, B$ and $C$ occur.

(vi) As mutually exclusive events cannot occur together, if events $A$ and $B$ are mutually exclusive, then $\boldsymbol{A} \cap \boldsymbol{B}=\phi$ since $A$ and $B$ have nothing in common.

(vii) Mutually exclusive and exhaustive events.

Let $S$ be the sample space associated with a random experiment. If $E_1, E_2, \ldots, E_n$ are mutually exclusive elementary events associated with the random experiment, then $E i \cap E j=\phi$ for all $i \neq j$ and $E_1 \cup E_2 \cup E_3 \ldots \cup E_n=S$

(Since exhaustive means the total number of possible outcomes) Therefore, an event and its complementary event are both mutually exclusive and exhaustive since:

$A \cap \bar{A}=\phi$ and $A \cup \bar{A}=S$.

Ex. Let 1 ball be drawn from a bag containing 12 balls of which 4 balls are white, 4 are red and 4 are green.

Let $A:$ Event-ball drawn is white

$B:$ Event-ball drawn is red

$C:$ Event-ball drawn is green.

It is obvious that one of the three events must occur as the ball drawn is either white or red or green.

This means that $A, B$ and $C$ form a mutually exclusive and exhaustive set of events.

$\therefore A \cap B=\phi, B \cap C=\phi, A \cap C=\phi$ and $A \cup B \cup C=S$.

17. Generally probability examples involve coins, dice and a pack of cards.

We give below the sample spaces $(S$ ) associated with some common events.

(i) Toss of a coin:

$S=\lbrace H, T\rbrace \Rightarrow n(S)=2 \quad$ where $\boldsymbol{n}($ Event) is number of outcomes.

(ii) Toss of two coins simultaneously:

$S=\lbrace H H, H T, T H, T T\rbrace \Rightarrow n(S)=4=(2 \times 2)$

(iii) Toss of three coins simultaneously :

$S=\lbrace H H H, H H T, H T T, T T T, T H H, T T H, T H T, H T H\rbrace $

$\Rightarrow n(S)=8=(2 \times 2 \times 2)$

(iv) Throw of a single die:

$S=\lbrace 1,2,3,4,5,6\rbrace \Rightarrow n(S)=6$

(v) Throw of two dice together :

$ \boldsymbol{S}=\lbrace \begin{matrix} (1,1) & (1,2) & \ldots & (1,6) \\ (2,1) & (2,2) & \ldots & (2,6) \\ (3,1) & (3,2) & \ldots & (3,6) \\ (4,1) & (4,2) & \ldots & (4,6) \\ (5,1) & (5,2) & \ldots & (5,6) \\ (6,1) & (6,2) & \ldots & (6,6) \end{matrix} \rbrace \Rightarrow n(S)=36=(6 \times 6) $

(vi) Drawing a card from a normal pack of 52 cards :

Here $S=$ All 52 cards. $\Rightarrow n(S)=52$.

Note : A pack of cards consists of 52 cards, divided in $\mathbf{4}$ suits, $\mathbf{2}$ red (Hearts and Diamonds) and $\mathbf{2}$ black (Spades and Clubs).

Each of the suits has 13 cards bearing the values 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, Queen, King and Ace. The Jack, Queen and King are called picture or face cards. Therefore there are 12 face cards in all.

Definition of Probability: If in a random experiment there are $n$ mutually exclusive and equally likely elementary events and $m$ of them are favourable to an event $A$, then the probability $P$ of happening of $A$ denoted by $P(A)$ is defined as the ratio $\frac{m}{n}$.

i.e, $\quad P(A)=\frac{\text{ Number of favourable outcomes }}{\text{ Total number of possible outcomes }}=\frac{n(A)}{n(S)}$,

where $\quad \boldsymbol{n}(\boldsymbol{A})=$ number of sample points in event $A$

$\boldsymbol{n}(\boldsymbol{S})=$ number of sample points in sample space $S$.

If $\quad \boldsymbol{P}(\boldsymbol{A})=$ probability of occurrence of $A$, then

$\boldsymbol{P}(\overline{\boldsymbol{A}})=$ probability of failure of $A$ or non-occurrence of $A=\mathbf{1}-\boldsymbol{P}(\boldsymbol{A})$ as $\boldsymbol{P}(A)+\boldsymbol{P}(\overline{\boldsymbol{A}})=\mathbf{1}$

Note : The probabilities of mutually exclusive and exhaustive events always adds up to 1 .

Odds in favour of an event and Odds against an event.

Probabilities are often expressed in terms of “odds”. Suppose an outcome $\boldsymbol{A}$ occurs $\boldsymbol{m}$ times out of $\boldsymbol{n}$ equally outcomes in a random experiment, then $\boldsymbol{A}$ does not occur $(\boldsymbol{n}-\boldsymbol{m})$ times. Thus,

Odds in favour of $A=\frac{\text{ Number of times } A \text{ occurs }}{\text{ Number of times } A \text{ does not occur }}$

$ =\frac{m}{n-m}=m:(n-m) $

Odds against $A=\frac{\text{ Number of times } A \text{ does not occur }}{\text{ Number of times } A \text{ occurs }}$

$ =\frac{n-m}{m}=(n-m): m $

Thus, $\bullet$ Odds in favour of an event $=$ Probability (success) : Probability (failure)

Odds against an event = Probability (failure) : Probability (success)

Ex : What are the odds in favour of and against the event $A$ : “getting a 5 in a single throw of die”.

Here, $A=\lbrace 5\rbrace $ and sample space $S=\lbrace 1,2,3,4,5,6\rbrace $

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{1}{6} \Rightarrow P(not A)=1-\frac{1}{6}=\frac{5}{6}$

$\therefore$ Odds in favour of $A=P(A): P(\bar{A})=1 / 6: 5 / 6=1: 5$

Odds against $A=P(\bar{A}): P(A)=5 / 6: 1 / 6=5: 1$

20. Addition Theorem of Probability

(a) For Two Events. If $A$ and $B$ are two events associated with a random experiment, then $\boldsymbol{P}(\boldsymbol{A} \cup \boldsymbol{B})=\boldsymbol{P}(\boldsymbol{A})+\boldsymbol{P}(\boldsymbol{B})-\boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{B})$

$\Rightarrow \quad P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$

Corollary 1: If $A$ and $B$ are mutually exclusive events, then,

$ \begin{matrix} P(A \cap B)=0, \text{ therefore } \\ \boldsymbol{P}(\boldsymbol{A} \cup \boldsymbol{B})=\boldsymbol{P}(\boldsymbol{A})+\boldsymbol{P}(\boldsymbol{B}) \end{matrix} $

Corollary 2: $\boldsymbol{P}(\boldsymbol{A}$ or $\boldsymbol{B}) \leq P(\underline{A})+P(B)$

Given, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Since, $P(A \cap B)$ is greater than or equal to 0 ,

$ P(A \text{ or } B)<P(A)+P(B) $

Equality in the above result holds when $A$ and $B$ are mutually exclusive as $P(A \cap B)=0$.

(b) For three events

If $A, B$ and $C$ be any three events associated with a random experiment, then

$\boldsymbol{P}(\boldsymbol{A} \cup B \cup C)=\boldsymbol{P}(\boldsymbol{A})+\boldsymbol{P}(B)+\boldsymbol{P}(\boldsymbol{C})-\boldsymbol{P}(\boldsymbol{A} \cap B)-\boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{C})-\boldsymbol{P}(\boldsymbol{B} \cap \boldsymbol{C})+\boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C})$

Corollary 1: If $A, B$ and $C$ are mutually exclusive events, then

$P(A \cap B)=P(B \cap C)=P(A \cap C)=P(A \cap B \cap C)=0$

$\therefore \quad \boldsymbol{P}(\boldsymbol{A} \cup B \cup \boldsymbol{C})=\boldsymbol{P}(\boldsymbol{A})+\boldsymbol{P}(B)+\boldsymbol{P}(\boldsymbol{C})$

Note : (i) If $A$ and $B$ are two events such that $A \subseteq B$, then $P(A) \leq P(B)$

(ii) If $E$ is an event associated with a random experiment, then $0 P(E) \leq 1$

21. Independent and Dependent Events.

An event $E$ is said to be independent of another event $F$, when the actual happening of $F$ does not influence in any degree the probability of the happening of $E$. If the probability of the happening of $E$ is dependent or influenced by the previous happening then $E$ is said to be dependent on $F$.

Illustration :

Imagine a bag containing eight balls - 5 of which are red and 3 of other colours. If you pick out two balls from the bag at random one after the other, what is the probability that both will be red.

The probability will be different depending on whether or not you put the first ball back in the bag before you pick the second ball.

Case I : If the first ball picked is replaced back in the bag, then the second event is independent of first as first draw of a red ball has no effect on the second draw.

Case II : If the first ball picked is red and you do not put it back, there will be one less ball to pick from. In this way the probability of picking a second red ball depends on whether the first ball was red, hence a case of dependent events.

22. Probability of Independent Events - Multiplication Rule

Study the following situations.

Umesh tosses a coin and Preeti rolls a die. Umesh knows that the probability of his coin landing “head” is $\frac{1}{2}$. The probability of Preeti’s die showing a six is $\frac{1}{6}$.

What is the probability of getting a head and a six?

If we draw the sample space diagram, we see that there are 12 possible outcomes, and

$ \begin{aligned} P(H \text{ and } 6) & =\frac{1}{12} \\ P(H) & =\frac{1}{2}, P(6)=\frac{1}{6} \\ P(H \text{ and } 6) & =\frac{1}{2} \times \frac{1}{6}=\frac{1}{12} \end{aligned} $

Coin | | 1 | 2 | 3 | 4 | 5 | 6 | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $H$ | | | | | | $H 6$ | | $T$ | | | | | | |

Whether or not the coin lands “head” has no effect or whether the die shows a ‘six’ or any score has no effect. $P(H$ and 6$)=P(H) \times P(6)$

If two events, $A$ and $B$, are independent $P(A$ and $B)=P(A) \times P(B)$

The two events, getting a “head” and scoring a ‘six’, are independent.

The above rule is called the multiplication or the ‘AND’ rule and is used to find the probabilities of joint occurrence of independent events.

In the situation described in point 22 , the tree diagram to show the probability of picking two red balls, is as under:

Independent events - first ball replaced before second ball is picked

Here the probability of two red balls being chosen is:

$\frac{5}{8} \times \frac{5}{8}=\frac{25}{64}$

or:

$P($ red $) \times P($ red $)=P($ red and red $)$ Dependent events - first ball not replaced before second ball is picked

If a red ball is picked first and not replaced there are only seven balls left to choose from. Only four of these are red, so for the second ball:

$P($ red $)=\frac{4}{7}$

and $P($ not red $)=\frac{3}{7}$

If a ’not red’ ball is picked first and not replaced there are still seven balls left to choose from, but now the five left are red, so for the second ball:

$P($ red $)=\frac{5}{7}$ and $P($ not red $)=\frac{2}{7}$

So when the first ball is not replaced, the probability of two reds being picked is:

$\frac{5}{8} \times \frac{4}{7}=\frac{20}{56}=\frac{5}{14}$

23. Multiplication Theorem on Probability

Statement I. If two events $A$ and $B$ are independent, then probability that they will both occur is equal to the product of their individual probabilities.

i.e. $\quad \boldsymbol{P}(\boldsymbol{A}$ and $B)=\boldsymbol{P}(\boldsymbol{A}) \times \boldsymbol{P}(\boldsymbol{B})$

[‘AND’ rule]

In set notation, $\quad \boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{B})=\boldsymbol{P}(\boldsymbol{A}) \times \boldsymbol{P}(\boldsymbol{B})$

Note. If $A, B, C$ are independent events, then

$ P(A \text{ and } B \text{ and } C)=P(A) \times P(B) \times P(C) $

In set notation, $\boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C})=\boldsymbol{P}(\boldsymbol{A}) \cdot \boldsymbol{P}(\boldsymbol{B}) \cdot \boldsymbol{P}(\boldsymbol{C})$.

In general, if $A_1, A_2, \ldots . A_n$ are $n$ independent events, then

$P(A_1.$ and $A_2$ and $A_3$ and $.\ldots . . A_n)=P(A_1) \times P(A_2) \times P(A_3) \times \ldots . \times P(A_n)$

In set notation, $P(A_1 \cap A_2 \cap A_3 \ldots . \cap A_n)=P(A_1) \times P(A_2) \ldots . . \times P(A_n)$

Finding probabilities of simultaneous occurrence of two independent events.

Method. Use the relation $P(A \cap B)=P(A) . P(B)$.

Ex. 1 . Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other.

Sol. Let $A$ : Getting an odd number on first die; $B$ : Getting a multiple of 3 on second die, then

$ A=\lbrace 1,3,5\rbrace , B=\lbrace 3,6\rbrace \quad \therefore \quad P(A)=\frac{3}{6}=\frac{1}{2}, P(B)=\frac{2}{6}=\frac{1}{3} $

The events $A$ and $B$ are independent

$\therefore \quad$ Required probability $=P(A) . P(B)=\frac{1}{2} \times \frac{1}{3}=\frac{\mathbf{1}}{\mathbf{6}}$.

Ex. 2 . A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. What is the probability of selecting a fair complexioned rich girl?

Sol. Let $P(A)=$ Probability of selecting a fair complexioned person. Then $P(A)=\frac{20}{80}=\frac{1}{4}$

Let $P(B)=$ Probability of selecting a rich person. Then $P(B)=\frac{10}{80}=\frac{1}{8}$

Let $P(C)=$ Probability of selecting a girl. Then $P(C)=\frac{25}{80}=\frac{5}{16}$

Since the events $A, B, C$ are independent, so probability of selecting a fair complexioned rich girl

3. Formal Definition of Independence

$ =P(A) . P(B) . P(C)=\frac{1}{4} \times \frac{1}{8} \times \frac{5}{16}=\frac{5}{512}=\mathbf{0 . 0 0 9} . $

Events $A$ and $B$ are independent if $P(A \cap B)=P(A) . P(B)$, otherwise they are dependent.

Caution. Any physical description of the events will not tell us if they are independent or not.

Proving the independence or dependence of events.

Ex. 1 . A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events $E$ and $F$ are independent or not?

$E:$ the number of heads is odd.

$F$ : the number of tails is odd.

Sol. When a coin is tossed three times, the sample space is given by

$S=[H H H$, HHT, HTH, THT, THH, HTT, TTH, TTT $]$

$E=\lbrace H H H, H T T, T H T, T T H\rbrace , F=\lbrace T T T, H T H, T H H, H H T\rbrace $

$ \begin{aligned} E \cap F & =\phi \\ P(E) & =\frac{4}{8}=\frac{1}{2}, P(F)=\frac{4}{8}=\frac{1}{2}, P(E \cap F)=\phi \\ P(E) . P(F) & =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq P(E \cap F) \end{aligned} $

$\therefore \quad E$ and $F$ are not independent events.

Ex. 2 . A fair coin is tossed three times. Let $A, B$ and $C$ be defined as follows:

$A=\lbrace $ first toss is heads $\rbrace , B=\lbrace $ second toss is heads $\rbrace $ and

$C=\lbrace $ exactly two heads are tossed in a row $\rbrace $

Check the independence of (i) $A$ and $B$, (ii) $B$ and $C$ and (iii) $C$ and $A$.

Sol. The sample space is $S=\lbrace H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\rbrace $

$ A=\lbrace H H H, H H T, H T H, H T T\rbrace , B=\lbrace H H H, H H T, T H H, T H T\rbrace \text{ and } C=\lbrace H H T, T H H\rbrace $

Also,

$ A \cap B=\lbrace H H H, H H T\rbrace , B \cap C=\lbrace H H T, T H H\rbrace , C \cap A=\lbrace H H T\rbrace $

$ P(A)=\frac{n(A)}{n(S)}=\frac{4}{8}=\frac{1}{2}, P(B)=\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}, P(C)=\frac{n(C)}{n(S)}=\frac{2}{8}=\frac{1}{4}, $

$ P(A \cap B)=\frac{n[(A \cap B)]}{n(S)}=\frac{2}{8}=\frac{1}{4} \text{. Similarly, } P(B \cap C)=\frac{2}{8}=\frac{1}{4} \text{ and } P(C \cap A)=\frac{1}{8} \text{. } $

Now, $\quad P(A) \cdot P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}=P(A \cap B)$, therefore, $A$ and $B$ are independent events.

$P(B) . P(C)=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8} \neq P(B \cap C)$ which is $\frac{1}{4}$. Therefore, $B$ and $C$ are not independent events.

$P(C) . P(A)=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}=P(C \cap A)$. Therefore, $C$ and $A$ are independent events.

Three events $A, B$ and $C$ are independent if:

(i) $P(A \cap B)=P(A) \cdot P(B), P(A \cap C)=P(A) . P(C)$ and $P(B \cap C)=P(B) . P(C)$, i.e., if the events are pairwise independent and

(ii) $P(A \cap B \cap C)=P(A) . P(B) . P(C)$.

The example given below shows that condition (ii) does not follow from condition (i). In other words, three events may be pairwise independent but not independent themselves.

Ex. 3 . Let a pair of fair coins be tossed. Here $S=\lbrace H H, H T, T H, T T\rbrace $. Consider the events $A=\lbrace $ heads on the first coin $\rbrace =\lbrace H H, H T\rbrace , B=\lbrace $ heads on the second coin $\rbrace =\lbrace H H, T H\rbrace , C=\lbrace $ heads on exactly one coin $\rbrace $ $=\lbrace \boldsymbol{H T}, \boldsymbol{T H}\rbrace $

Sol. Then

$ \begin{gathered} P(A)=P(B)=P(C)=\frac{2}{4}=\frac{1}{2} \text{ and } \\ P(A \cap B)=P(\lbrace H H\rbrace )=\frac{1}{4}, P(A \cap C)=P(\lbrace H T\rbrace )=\frac{1}{4} \\ P(B \cap C)=P(\lbrace T H\rbrace )=\frac{1}{4},(A \cap B \cap C)=\phi \end{gathered} $

$ \therefore \quad P(A \cap B \cap C)=P(\phi)=0 \neq P(A) \cdot P(B) . P(C) $

Thus condition $(i)$ is satisfied, i.e., the events are pairwise independent. But condition (ii) is not satisfied and so the three events are not independent.

Ex. 4 . Let $A$ and $B$ be two possible events of an experiment such that $P(A)=0.4, P(A \cup B)=0.7$ and $P(B)=p$. (i) For what value of $p$ are $A$ and $B$ mutually exclusive ? (ii) For what value of $p$ are $A$ and $B$ independent?

(AP 1996)

Sol. (i) Since $A$ and $B$ are mutually exclusive, therefore, $A \cap B=\phi$.

Substituting in $P(A \cup B)=P(A)+P(B)-P(A \cap B)$, we have $0.7=0.4+p-0 \Rightarrow p=0.3$

(ii) Since $A$ and $B$ are independent, therefore, $P(A \cap B)=P(A) . P(B)$

i.e., $P(A)+P(B)-P(A \cup B)=P(A) . P(B)$

Substituting the given values we have $0.4+p-0.7=0.4 \times p \Rightarrow \boldsymbol{p}=\mathbf{0 . 5}$.

Also, we have the following :

If $p_1, p_2, p_3 \ldots \ldots . . p_n$ are the probabilities that certain events happen, the probability of failing of all these events is given by

$ P=(1-p_1)(1-p_2) \ldots(1-p_n) $

Probability of the occurrence of AT LEAST ONE of the several independent events of a random experiment. If $p_1, p_2, p_3, \ldots . . p_n$ are the probabilities that certain events happen, then the probability that at least one of these events must happen is

$ 1-(1-p_1)(1-p_2)(1-p_3) \ldots \ldots(1-p_n) . $

Probability of the occurrence of exactly one of the two events.

Let $A$ and $B$ be two events associated with a random experiment and let $p_1$, and $p_2$ be the probabilities of their occurrence respectively, then

$ \begin{aligned} & P(A)=p_1, P(B)=p_2 \quad \therefore P(\bar{A})=1-P(A)=1-p_1=q_1 \text{ (say) } \\ & P(\bar{B})=P(\text{ non-occurrence of } B)=1-P(B)=1-p_2=q_2 \text{ (say) } \end{aligned} $

Since only one of the two events $A$ and $B$ is to occur, therefore, there are two possibilities (i) $A$ occurs and $B$ does not (ii) $B$ occurs and $A$ does not

Since these are mutually exclusive events, therefore,

Reqd. probability $=P(A \bar{B}) + P(\bar{A} B)=P(A) \cdot P(\bar{B})+P(B) \cdot P(\bar{A})=\boldsymbol{p} _{\mathbf{1}} \boldsymbol{q} _{\mathbf{2}} + \boldsymbol{p} _{\mathbf{2}} \boldsymbol{q} _{\mathbf{1}}$.

Ex. 1 . A husband and a wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is $\frac{1}{7}$ and that of wife’s selection is $\frac{1}{5}$. What is the probability that

(i) Both of them will be selected

(iii) None of them will be selected (ii) Only one of them will be selected

(iv) At least one of them will be selected.

Sol. Let $A$ : Event of husband being selected

$B$ : Event of wife being selected

Then $P(A)=\frac{1}{7}, \quad P(B)=\frac{1}{5}$

$P(\bar{A})=1-\frac{1}{7}=\frac{6}{7}, P(\bar{B})=1-\frac{1}{5}=\frac{4}{5}$

(i) $P($ Both are selected $)=P(A) \times P(B)=\frac{1}{7} \times \frac{1}{5}=\frac{1}{35}$

(ii) $P($ only one is selected $)=P(A$ selected $) \times P(B$ not selected $)+P(A$ not selected $) \times P(B$ selected $)$

$ =P(A) \times P(\bar{B})+P(\bar{A}) \times P(B)=\frac{1}{7} \times \frac{4}{5}+\frac{6}{7} \times \frac{1}{5}=\frac{4}{35}+\frac{6}{35}=\frac{10}{35}=\frac{\mathbf{2}}{7} $

(iii) $P($ none selected $)=P(\bar{A}) \times P(\bar{B})=\frac{6}{7} \times \frac{4}{5}=\frac{24}{35}$

(iv) $P($ at least one selected $)=1-P($ none selected $)=1-\frac{24}{35}=\frac{\mathbf{1 1}}{\mathbf{3 5}}$

Addition theorem for independents events:

If $A_1, A_2, \ldots ., A_n$ are $n$ independent events associated with a random experiment, then

$\boldsymbol{P}(A_1 \cup A_2 \cup \ldots . . A_n)=1-\boldsymbol{P}(\overline{A_1}) \boldsymbol{P}(\overline{A_2}) \ldots . \boldsymbol{P}(\overline{A_n})$

5. Repeated events

If the probability that an event will happen is $p$, the chance that it will happen in any succession of $r$ trials is $p^{r}$.

The probability of non-occurrence is $q=1-p$. Therefore, for $\boldsymbol{n}$ repeated non-occurrence of the event we have the probability $\boldsymbol{q}^{r}=(\mathbf{1}-\boldsymbol{p})^{r}$.

Ex. 2 . A machine manufactured by a firm consists of two parts $A$ and $B$. But of $100 A$ ’s manufactured, 9 are likely to be defective and out of $100 B$ ’s manufactured 5 are likely to be defective. Find the probability that a machine manufactured by the firm is free of any defect. Give your answer, rounded off to two places of decimal.

Sol. Let $E=\lbrace $ part $A$ is defective $\rbrace , F=\lbrace $ part $B$ is defective $\rbrace $.

Then, $P(E)=\frac{9}{100}, P(F)=\frac{5}{100} \therefore P(\bar{E})=1-\frac{9}{100}=\frac{91}{100}$ and $P(\bar{F})=1-\frac{5}{100}=\frac{95}{100}$

Since $E$ and $F$ are independent, therefore, $\bar{E}$ and $\bar{F}$ are also independent.

Now $P($ none is defective $)=P(\bar{E}) \cdot P(\bar{F})=\frac{91}{100} \times \frac{95}{100}=0.91 \times 0.95=0.8645=\mathbf{0 . 8 6}$ to 2 d.p.

Finding probability of occurrence of at least one event for independent events.

Ex. 1 . The probability that Dimpu gets scholarship is 0.9 and Pintu will get is 0.8 . What is the probability that at least one of them gets the scholarship.

(Pb 1995 C)

Sol. Let $A$ be the event that Dimpu gets scholarship and $B$ the event that Pintu gets scholarship. It is given that $P$ $(A)=0.9, P(B)=0.8$.

The probability that none of them gets the scholarship $=(1-0.9)(1-0.8)=0.1 \times 0.2=0.02$

$\therefore \quad$ The probability that at least one of them gets the scholarship $=1-0.02=\mathbf{0 . 9 8}$.

Ex. 2 . A can solve $90 %$ of the problems given in a book and B can solve $\mathbf{7 0 \%}$. What is the probability that at least one of them will solve a problem selected at random from the book ?

Sol. Let $E$ be the event that $A$ solve the problem and $F$ the event that $B$ solves the problem.

Then $P(E)=\frac{90}{100}=\frac{9}{10}, P(F)=\frac{70}{100}=\frac{7}{10}, P(\bar{E})=1-\frac{9}{10}=\frac{1}{10}, P(\bar{F})=1-\frac{7}{10}=\frac{3}{10}$ $\therefore \quad$ Probability that no one solves the problem $P(\bar{E}) \cdot P(\bar{F})=\frac{1}{10} \times \frac{3}{10}=\frac{3}{100}$.

Hence, probability that at least one of them will solve a problem $=1-\frac{3}{100}=\frac{97}{100}=\mathbf{0 . 9 7}$.

Ex. 3 . A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white.

Sol. Let $A, B, C, D$ denote the events of not getting a white ball in first, second, third and fourth draw respectively. Since the balls are drawn with replacement, therefore, $A, B, C, D$ are independent events such that $P(A)$ $=P(B)=P(C)=P(D)$.

Since out of 16 balls, 11 are not white, therefore, $P(A)=11 / 16$.

$\therefore \quad$ Required probability $=P(A) . P(B) . P(C) . P(D)$

$ =\frac{11}{16} \times \frac{11}{16} \times \frac{11}{16} \times \frac{11}{16}=(\frac{\mathbf{1 1}}{\mathbf{1 6}})^{4} . $

Ex. 4 . A Police man fires six bullets on a decoit. The probability that the decoit will be killed by one bullet is 0.6 . What is the probability that the decoit is still alive?

Sol. Let $A_i$ be the event that the decoit is killed by the $i$ th bullet $(1 \leq i \leq 6)$. Then $\overline{A_i}$ is the event that the decoit is not killed, $\therefore P(A_i)=0.6$ and $P(\overline{A_i})=1-0.6=0.4$

$\therefore \quad$ Probability of the decoit being still alive

$ =P(\overline{A_1}) P(\overline{A_2}) P(\overline{A_3}) P(\overline{A_4}) P(\overline{A_5}) P(\overline{A_6})=0.4 \times 0.4 \times 0.4 \times 0.4 \times 0.4 \times 0.4=\mathbf{0 . 0 0 4 0 9 6} $

Note. Using simpler notation, we may write success : Bullet will hit the decoit

$ p=0.6 \Rightarrow q=1-p=1-0.6=0.4 $

Required probability $.=q q q q q q=(q)^{6}=\mathbf{( 0 . 4})^{\mathbf{6}}$.

Problems solved by using both the addition and multiplication theorems of probability.

Ex. 1 . A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.

Sol. Let $W_1$ and $W_2$ denote the events of drawing a white ball from the first and one from the second bag respectively. Let $B_1$ and $B_2$ denote the events of drawing black balls from the two bags in the same order. Then

$P($ both balls are white $)=P$ [white ball from (I bag) and white ball from II bag]

$ =P(W_1) \cdot P(W_2)=\frac{3}{9} \times \frac{4}{9}=\frac{12}{81} $

[‘AND’ Theorem]

Similarly, $P($ both balls are black $)=\frac{6}{9} \times \frac{5}{9}=\frac{30}{81}$.

[‘AND’ Theorem]

$\therefore \quad$ The probability of getting two balls of the same colour

[‘OR’ Theorem]

$=\frac{12}{81}+\frac{30}{81}=\frac{42}{81}=\frac{\mathbf{1 4}}{\mathbf{2 7}}$. (By addition theorem for mutually exclusive events.

Ex. 2 . Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls, one by one without replacement. Find the probability that both the balls are of different colours.

Sol. Given, 3 white ( $3 W$ ), 3 red ( $3 R$ ), 4 green $(4 G), 4$ black ( $4 B$ ) balls

Total no. of balls $=3+3+4+4=14$

Two balls are to be drawn, one by one without replacement.

There are 4 possibilities.

First Ball

Case I

Case II

Case III

Case IV White

Red

Green

Black

Second Ball

Not white

Not red

Not green

Not black

Since all cases are mutually exclusive, therefore

$ \text{ Reqd. probability }=\frac{3}{14} \times \frac{11}{13}+\frac{3}{14} \times \frac{11}{13}+\frac{4}{14} \times \frac{10}{13}+\frac{4}{14} \times \frac{10}{13}=\frac{33+33+40+40}{14 \times 13}=\frac{146}{182}=\frac{\mathbf{7 3}}{\mathbf{9 1}} \text{. } $

Ex. 3 . The probability of student A passing examination is $\frac{3}{7}$ and of student $B$ passing is $\frac{5}{7}$. Assuming the two events “A passes”, “B passses” as independent, find the probability of (i) only A passing the examination (ii) only one of them passing the examination.

Sol. $p_1=P(A)=\frac{3}{7}, p_2=P(B)=\frac{5}{7} \therefore q_1=P(\bar{A})=1-P(A)=1-\frac{3}{7}=\frac{4}{7}$.

$q_2=P(\bar{B})=1-P(B)=1-\frac{5}{7}=\frac{2}{7}$

(i) $P($ only $A$ passes $)=p_1 q_2=\frac{3}{7} \times \frac{2}{7}=\frac{\mathbf{6}}{\mathbf{4 9}} \cdot P($ only $B$ passes $)=q_1 p_2=\frac{4}{7} \times \frac{5}{7}=\frac{20}{49}$

(ii) $P($ only one of them passes $)=p_1 q_2+q_1 p_2=\frac{3}{7} \times \frac{2}{7}+\frac{4}{7} \times \frac{5}{7}=\frac{\mathbf{2 6}}{\mathbf{4 9}}$.

Ex. 4 . Two cards are drawn from a well shuffled pack of 52 cards one after another without replacement. Find the probability that one of these is an ace and the other is a queen of the opposite shade.

Sol. Probability of drawing an ace in the first draw $=\frac{4}{52}$.

Probability of drawing a queen of opposite shade in the second draw $=\frac{2}{51}$.

Probability of drawing a queen in the first draw $=\frac{4}{52}$.

Probability of drawing an ace of opposite shade in the second draw $=\frac{2}{51}$.

$\therefore \quad$ Required probability $=\frac{4}{52} \times \frac{2}{51}+\frac{4}{52} \times \frac{2}{51}=\frac{\mathbf{4}}{\mathbf{6 6 3}}$.

[‘AND’ and ‘OR’ Theorems]

Ex. 5 . $A$ can hit a target three times in five shots, $B$ two times in five shots and $C$ three times in four shots. They fire a volley. What is the probability that two shots hit the target?

Sol. $P(A)=\frac{3}{5}, P(B)=\frac{2}{5}$ and $P(C)=\frac{3}{4}$. In order that two shots may hit the target the following three mutually exclusive events are possible,

(i) $A$ and $B$ hit the target and not $C$. (ii) $B$ and $C$ hit the target and not $A$.

(iii) $A$ and $C$ hit the target and $not B$.

If $p_1, p_2, p_3$ are the probabilities of these three events then

$ \begin{aligned} & p_1=P(A) \cdot P(B) . P(not C)=\frac{3}{5} \times \frac{2}{5} \times(1-\frac{3}{4})=\frac{3}{5} \times \frac{2}{5} \times \frac{1}{4}=\frac{6}{100} \\ & p_2=P(B) \times P(C) \times P(not A)=\frac{2}{5} \times \frac{3}{4} \times(1-\frac{3}{5})=\frac{2}{5} \times \frac{3}{4} \times \frac{2}{5}=\frac{12}{100} \end{aligned} $

$ p_3=P(A) \times P(C) \times P(not B)=\frac{3}{5} \times \frac{3}{4} \times(1-\frac{2}{5})=\frac{3}{5} \times \frac{3}{4} \times \frac{3}{5}=\frac{27}{100} $

$\therefore \quad$ Required probability $=p_1+p_2+p_3=\frac{6}{100}+\frac{12}{100}+\frac{27}{100}=\frac{45}{100}=\frac{\mathbf{9}}{\mathbf{2 0}}$.

Ex. 6 . The probability of $A, B, C$ solving a problem are $\frac{1}{3}, \frac{2}{7}$ and $\frac{3}{8}$ respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.

Sol. Let $E_1, E_2, E_3$ be the events that the problem is solved by $A, B$ respectively and let $p_1, p_2, p_3$ be corresponding probabilities. Then,

$ \begin{aligned} & p_1=P(E_1)=\frac{1}{3}, p_2=P(E_2)=\frac{2}{7}, p_3=P(E_3)=\frac{3}{8}, q_1=P(\overline{E_1})=1-\frac{1}{3}=\frac{2}{3}, \\ & q_2=P(\overline{E_2})=1-\frac{2}{7}=\frac{5}{7}, q_3=P(\overline{E_3})=1-\frac{3}{8}=\frac{5}{8} . \end{aligned} $

The problem will be solved by exactly one of them if it happens in the following mutually exclusive ways:

(1) $A$ solves and $B$, and $C$ do not solve;

(2) $B$ solves and $A$, and $C$ do not solve;

(3) $C$ solves and $A$, and $B$ do not solve;

Required probability $=p_1 q_2 q_3+q_1 p_2 q_3+q_1 q_2 p_3$

$ =\frac{1}{3} \times \frac{5}{7} \times \frac{5}{8}+\frac{2}{3} \times \frac{2}{7} \times \frac{5}{8}+\frac{2}{3} \times \frac{5}{7} \times \frac{3}{8}=\frac{25}{168}+\frac{5}{42}+\frac{5}{28}=\frac{\mathbf{2 5}}{\mathbf{5 6}} . $

MORE SOLVED EXAMPLES

Ex. 1 . If two coins are tossed once, what is the probability of getting at least one head ?

Sol. When two coins are tossed once, there are four possible outcomes, i.e., $S=\lbrace H H, H T, T H, T T\rbrace $

$\therefore$ Total number of outcomes $=n(S)=4$

Let $A$ : Event of getting at least one head

$\Rightarrow A=\lbrace H H, H T, T H\rbrace \Rightarrow n(A)=3$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{\mathbf{3}}{\mathbf{4}}$.

Ex. 2 . Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ?

Sol. When two unbiased dice are rolled, the possible out comes are

$ \begin{aligned} & S= \begin{bmatrix} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{bmatrix} \\ & \therefore n(S)=36 \end{aligned} $

Let $A$ : getting a multiple of 2 on one die and a multiple of 3 on the other die.

$\Rightarrow A=\lbrace (2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2),(3,4),(3,6),(6,2),(6,4)\rbrace $

$\Rightarrow n(A)=11$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{\mathbf{1 1}}{\mathbf{3 6}}$.

Ex. 3 . In an examination there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all the three questions. What is the probability that the student will not answer all the 3 questions correctly?

Sol. Probability of selecting a correct choice for a question $=\frac{1}{4} \quad(\because$ Out of 4 choices only one is correct $)$

$\therefore$ Probability of answering all the three questions correctly $=\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{64}$

$\Rightarrow$ Probability of not answering all the three questions correctly $=1-\frac{1}{64}=\frac{\mathbf{6 3}}{\mathbf{6 4}}$.

Ex. 4 . What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non-leap year ?

(Orissa JEE 2010)

Sol. A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week.

Possible outcomes $n(S)=7=$ \lbrace Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday $\rbrace $.

$\therefore \quad$ Number of possible outcomes $n(S)=7$

(As there are seven days in a week)

Let $A$ : Getting the extra day as Sunday or Tuesday or Thursday

$\Rightarrow A=\lbrace $ Sunday, Tuesday, Thursday $\rbrace $

$\Rightarrow n(A)=3$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{3}{7}$.

Ex. 5 . An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ?

Sol. As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, $n(S)=200$

Let $A$ : Event of integer chosen from 1 to 200 being divisible by 6

$\Rightarrow n(A)=33 \quad(\frac{200}{6}=33 \frac{1}{3})$

Let $B$ : Event of integer chosen from 1 to 200 being divisible by 8

$\Rightarrow n(B)=\frac{200}{8}=25$

$A \cap B$ : Event of integer chosen from 1 to 200 being divisible by both 6 and 8 , i.e, divisible by 24

$(\because$ LCM of 6 and $8=24)$

$\Rightarrow \quad n(A \cap B)=8(\frac{200}{24}=8.$ (approx) $)$

$\therefore \quad P(A \cup B)=P($ Integer chosen is divisible by 6 or 8$)$

$=P(A)+P(B)-P(A \cap B)=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)}=\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{50}{200}=\frac{\mathbf{1}}{\mathbf{4}}$.

Ex. 6 . There are three events $E_1, E_2$ and $E_3$, one of which is must and only one can happen. The odds are 7 to 4 against $E_1$ and 5 to 3 against $E_2$. Find the odds against $E_3$ ?

Sol. Since one and only one of the three events $E_1, E_2$ and $E_3$ can happen, i.e, they are mutually exclusive.

Therefore, $P(E_1)+P(E_2)+P(E_3)=1$

Odds against $E_1$ are $7: 4 \Rightarrow$ Odds in favour of $E_1$ are $4: 7$

$\Rightarrow \quad P(E_1)=\frac{4}{4+7}=\frac{4}{11}$

Odds against $E_2$ are $5: 3 \Rightarrow$ Odds in favour of $E_2$ are $3: 5$

$\Rightarrow \quad P(E_2)=\frac{3}{3+5}=\frac{3}{8}$ $\therefore$ From (i), (ii) and (iii)

$ \begin{aligned} & \frac{4}{11}+\frac{3}{8}+P(E_3)=1 \Rightarrow P(E_3)=1-(\frac{4}{11}+\frac{3}{8})=1-(\frac{32+33}{88})=1-\frac{65}{88}=\frac{23}{88} \\ & \therefore \text{ Odds against } E_3 \text{ are } \frac{1-P(E_3)}{P(E_3)}=\frac{1-\frac{23}{88}}{\frac{23}{88}}=\frac{65 / 88}{23 / 88}=\mathbf{6 5 : 2 3 .} \end{aligned} $

Ex. 7 . Find the probability that the three cards drawn from a pack of 52 cards are all black ?

Sol. Number of ways in which three cards can be drawn from a pack of 52 cards $n(S)={ }^{52} C_3$.

Let $A$ : Event of drawing all the three cards as black

Then, $n(A)={ }^{26} C_3$

( $\because$ There are 26 black cards)

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{{ }^{26} C_3}{{ }^{52} C_3}=\frac{26 \times 25 \times 24}{52 \times 51 \times 50}=\frac{\mathbf{2}}{\mathbf{1 7}}$.

Ex. 8 . Four cards are drawn from a full pack of cards. Find the probability that :

(a) there is one card of each suit

(b) all the four are spades, and one of them is a king

Sol. 4 cards can be drawn from a pack of cards in ${ }^{52} C_4$ ways

$\therefore$ Exhaustive number of cases $=n(S)={ }^{52} C_4$

(a) There are 4 suits, each containing 13 cards.

Let $A$ : Event of drawing one card from each suit

$\Rightarrow$ Favourable number of cases $=n(A)=13 C_1 \times 13 C_1 \times 13 C_1 \times 13 C_1$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{{ }^{13} C_1 \times{ }^{13} C_1 \times{ }^{13} C_1 \times{ }^{13} C_1}{{ }^{52} C_4}=\frac{13 \times 13 \times 13 \times 13}{\underline{52 \times 51 \times 50 \times 49}} \quad \because[n _{C_r}=\frac{\underline{n}}{\underline{n-r} \underline{r}}]$ $=\frac{\mathbf{2 1 9 7}}{\mathbf{2 0 8 2 5}}$

$4 \times 3 \times 2 \times 1$

(b) Let $A$ : Drawing 4 spade cards of which one is king of spades.

Then, Favourable number of cases $=n(A)={ }^{12} C_3 \times 1$

( $\because$ There is only one king of spades and the rest of the three spades we draw from remaining 12 spade cards)

$ n(S)={ }^{52} C_4 $

$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{{ }^{12} C_3 \times 1}{{ }^{52} C_4}=\frac{\frac{12 \times 11 \times 10}{3 \times 2 \times 1}}{\frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}}=\frac{12 \times 11 \times 10 \times 4}{52 \times 51 \times 50 \times 49}=\frac{\mathbf{4 4}}{\mathbf{5 4 1 4 5}}$

Now since there are 4 aces in the pack of 52 cards, therefore, the number of ways of drawing 4 cards so that no card is an ace $={ }^{48} C_4$

$\therefore$ Probability of drawing four cards so that none is an ace

$ P(\bar{A})=\frac{{ }^{48} C_4}{{ }^{52} C_4}=\frac{48 \times 47 \times 46 \times 45}{52 \times 51 \times 50 \times 49}=\frac{38916}{54145} $

[Here $\bar{A}$ denotes the complement of event $A$, i.e, non-happening of event $A$ ]

$ \therefore \quad P(A)=1-P(\bar{A})=1-\frac{38916}{54145}=\frac{15229}{54145} \quad(\because P(\text{ Event })+P(\text{ complement of event })=1) $

Ex. 9 . Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers?

(UPSEE 2002)

Sol. The chosen consists of players $(6+5)$.

$\therefore$ Number of ways of selecting 11 players out of 15 players $=n(S)={ }^{15} C _{11}$

Let $A$ : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers

Then, $n(A)={ }^{8} C_6 \times{ }^{7} C_5$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{{ }^{8} C_6 \times{ }^{7} C_5}{{ }^{15} C _{11}}=\frac{\frac{8 \times 7}{2} \times \frac{7 \times 6}{2}}{\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}}=\frac{8 \times 7}{2} \times \frac{7 \times 6}{2} \times \frac{4 \times 3 \times 2 \times 1}{15 \times 14 \times 13 \times 12}=\frac{\mathbf{2 8}}{\mathbf{6 5}}$.

Ex. 10 . A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black. ?

Sol. Let $A$ : Event of getting at least 3 black balls

Then $n(A)={ }^{5} C_3 \times{ }^{11} C_1+{ }^{5} C_4$

( $\because$ Besides 5 black balls, there are 11 other balls)

(3 black + others) ( 4 black)

$ =\frac{5 \times 4}{2} \times 11+5=115 $

Total numbers of ways in which 4 balls can be drawn from $(7+5+4)=16$ balls

$ \begin{aligned} & n(S)={ }^{16} C_4=\frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1}=1820 \\ \therefore \quad & P(A)=\frac{n(A)}{n(S)}=\frac{115}{1820}=\frac{\mathbf{2 3}}{\mathbf{3 6 4}} . \end{aligned} $

Ex. 11 . Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ?

Sol. Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) $=n(S)=7$ !

Let $A$ : Event in which the 4 boys and 3 girls occupy alternate position.

This is possible when the 4 boys occupy the 4 odd places standing form position 1 and the girls fill in the three even places between then as :

alt text

$\therefore 4$ boys can be arranged in a queue in 4 ! ways

3 girls can fill the gaps in the queue in 3 ! ways.

$ \begin{matrix} \therefore & n(A)=4 ! \times 3 ! \\ \therefore & P(A)=\frac{n(A)}{n(S)}=\frac{4 ! \times 3 !}{7 !}=\frac{4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}=\frac{1}{35} . \end{matrix} $

Ex. 12 . A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order. Find the probability that the third number is 20.

Sol. Total number of ways in which 5 tickets can be drawn $=n(S)={ }^{30} C_5$.

The tickets are arranged in the form $T_1, T_2, T_3(=20), T_4, T_5$

Where $T_1, T_2 \in\lbrace 1,2,3, \ldots, 19\rbrace $ and $T_4, T_5 \in\lbrace 21,22, \ldots, 30\rbrace $

$\therefore$ Number of favourable cases $={ }^{19} C_2 \times 1 \times{ }^{10} C_2$

$\therefore$ Required probability $=\frac{{ }^{19} C_2 \times{ }^{10} C_2}{{ }^{30} C_5}=\frac{19 \times 18}{2} \times \frac{10 \times 9}{2} \times \frac{5 \times 4 \times 3 \times 2 \times 1}{30 \times 29 \times 28 \times 27 \times 26}=\frac{\mathbf{2 8 5}}{\mathbf{5 2 7 8}}$.

Ex. 13 . Find the probability that a two digit number formed by the digit $1,2,3,4$ and 5 is divisible by 4 . (UPSEE)

Sol. The two digit numbers can be formed by putting any of 5 digits at the one ’s place and also one of the 5 digits at ten’s place. So,

Total number of 2-digit numbers that can be formed using these 5 -digits $=5 \times 5=25$

The 2-digit numbers formed by 1, 2, 3, 4 and 5 that are divisible by 4 are $\lbrace 12,24,32,44,52\rbrace $, i.e, 5 in number.

$\therefore$ Required probability $=\frac{5}{25}=\frac{\mathbf{1}}{\mathbf{5}}$.

Ex. 14 . A five digit number is formed by the digits $0,1,2,3,4$ (without repetition). Find the probability that the number formed is divisible by 4 ?

Sol. Without repetition, a five -digit number can be formed using the five digits in 5 ! ways ( $5 \times 4 \times 3 \times 2 \times 1)$

Out of these 5! numbers, 4! numbers will be starting with digit 0 . ( $\mathbf{0}$ (fixed) $4 \times 3 \times 2 \times 1$ )

$\therefore$ Total number of 5-digit numbers that can be formed using the digits $0,1,2,3,4=5 !-4 !=120-24=\mathbf{9 6}$.

$\Rightarrow$ No. of exhaustive cases $=n(S)=96$

Now, only these numbers are divisible by 4 in which the numbers formed by the last two digits is divisible by 4 .

Thus, the numbers ending in $04,12,20,24,32$ and 40 will be divisible by 4 .

(i) If the numbers end in 04 , the remaining three numbers $1,2,3$, can be arranged in 3 ! ways $=6$ ways

(ii) If the numbers end in 20, the remaining three numbers, i.e., 1, 4, 3 can be arranged in 3 ! ways $=6$ ways

(iii) If the numbers end in 40 , the remaining three numbers, i.e., $1,2,3$ can be arranged in 3 ! ways $=6$ ways

(iv) If the numbers end in 12, the remaining three numbers, i.e., 0, 3, 4 can be arranged 3 ! ways, but these cases in which 0 is the extreme left digit one to be discorded. The number of such cases is 2 !.

$\therefore$ Number of numbers ending in $12=3 !-2 !=4$

(v) Similarly the numbers ending in 24 and 32 are 4 each.

$\therefore$ Total number of favourable cases, i.e., number of 5 -digit numbers divisible by 4

$\therefore$ Required probability $=\frac{30}{96}=\frac{\mathbf{5}}{\mathbf{1 6}}$.

$=6+6+6+4+4+4=30$

Ex. 15 . There are 10 persons who are to be seated around a circular table. Find the probability that two particular persons will always sit together.

Sol. Total number of ways in which 10 person can sit around a circular table $=9$ !

( $\because$ We shall keep one place fixed and the rest of the 9 places will be filled in $(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)$ ways as there is no repetition in such case. The first variable place can be filled by any of the remaining 9 person, second place by any of the remaining 8 persons and so on.)

If two particular persons (say $A$ and $B$ ) sit together, then the number of ways in which 9 person (considering the pair of $A$ and $B$ as one unit) can sit round the table $=2 ! \times 8 !$. (Here $A$ and $B$ can interchange places amongst themselves in 2 ! ways)

$\therefore$ Favourable number of cases $=2 \times 8$ !

$\therefore$ Required probability $=\frac{2 \times 8 !}{9 !}=\frac{2 \times 8 !}{9 \times 8 !}=\frac{\mathbf{2}}{\mathbf{9}}$.

Ex. 16 . The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together?

Sol. There are 7 letters in the word SOCIETY.

$\therefore$ Total number of ways of arranging all the 7 letters $=n(S)=7$ !. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the number of ways in which 5 letters (SCTY-4, IEO -1 ) can be arranged $=5$ !

Also the 3 vowels can be arranged amongst themselves in 3 ! ways

$\therefore$ Total number of favourable cases $=5 ! \times 3$ !

$\therefore$ Required probability $=\frac{5 ! \times 3 !}{7 !}=\frac{1}{7}$

Ex. 17 . There are $n$ letters and $n$ addressed envelopes. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelope?

Sol. Total number of ways of placing $n$ letters in $n$ envelopes $=n$ !

All the letters can be placed correctly in only 1 way

$\therefore$ Probability of placing all the letters in the right envelopes $=\frac{1}{n !}$

$\therefore$ Probability that all the letters are not placed in the right envelope $=1-\frac{1}{n !}$.

Ex. 18 . In a group there are 3 women and 3 men. 4 people are selected at random from this group. Find the probability that 3 women and 1 man or 1 woman and 3 men may be selected ?

Sol. $A$ : Selected 3 women and 1 man

$B$ : Selected 1 women and 3 men

$S$ : Selected 4 people from 6 people $(3+3)$

Then $n(A)={ }^{3} C_3 \times{ }^{3} C_1, \quad n(B)={ }^{3} C_1 \times{ }^{3} C_3, \quad n(S)={ }^{6} C_4$

$\therefore$ Required probability $=P(A)+P(B)=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}=\frac{{ }^{3} C_3 \times{ }^{3} C_1}{{ }^{6} C_4}+\frac{{ }^{3} C_1 \times{ }^{3} C_3}{{ }^{6} C_4}=\frac{2 \times 1 \times 3}{15}=\frac{\mathbf{2}}{\mathbf{5}}$.

Ex. 19 . Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings?

Sol. Let $S$ : Drawing 2 cards out of 52 card

$A$ : Drawing 2 red cards

$B$ : Drawing 2 kings

$A \cup B:$ Drawing 2 red cards or 2 kings

$\therefore n(S)={ }^{52} C_2$

$ \begin{matrix} n(A)={ }^{26} C_2 \\ n(B)={ }^{4} C_2 & (\because \text{ There are } 26 \text{ red cards }) \\ \end{matrix} $

But there are 2 red kings, so

$A \cap B:$ Drawing 2 red kings

$\Rightarrow \quad n(A \cap B)={ }^{2} C_2$.

$\therefore$ Required probability $=P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$ \begin{aligned} & =\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)}=\frac{{ }^{26} C_2}{{ }^{52} C_2}+\frac{{ }^{4} C_2}{{ }^{52} C_2}-\frac{{ }^{2} C_2}{{ }^{52} C_2} \\ & =\frac{26 \times 25}{52 \times 51}+\frac{4 \times 3}{52 \times 51}-\frac{2}{52 \times 51}=\frac{660}{2652}=\frac{\mathbf{5 5}}{\mathbf{2 2 1}} . \end{aligned} $

Ex. 20 . One bag contains 3 black and 4 white balls and the other bag contains 4 black and 3 white balls. A die is rolled. If two or five comes up, a ball is taken from the first bag, otherwise a ball is drawn from the second bag. Find the probability of choosing a white ball?

(NMOC)

Sol. Let $A$ : Getting 2 or 5

$B$ : Getting white ball from first bag

$C$ : Getting white ball from second bag.

$\therefore P(A)=\frac{2}{6}=\frac{1}{3} \Rightarrow P(\bar{A})=1-P(A)=1-\frac{1}{3}=\frac{2}{3}$ $\therefore$ Required probability $=$ First bag white ball + Second bag white ball

$=P(A) P(B)+P(\bar{A}) P(C)=\frac{1}{3} \times \frac{4}{7}+\frac{2}{3} \times \frac{3}{7}=\frac{4+6}{21}=\frac{\mathbf{1 0}}{\mathbf{2 1}}$.

Ex. 21 . An urn contains 3 white and 5 blue balls and a second urn contains 4 white and 4 blue balls. If one ball is drawn from each urn, what is the probability that they will be of the same colour?

Sol. Let $E$ : Event of drawing both the balls of same colour from the two urns

$E_1$ : Getting 1 white ball from the first urn and 1 white ball from the second urn

$E_2$ : Getting 1 blue ball from the first urn and 1 blue ball from the second urn

Then, $P(E)=P(E)+P(E_2) \quad(\because.$ The two events $E_1$ and $E_2$ are mutually exclusive)

$=\lbrace P(1$ white from 1 st urn $) \times P(1$ white from 2 nd urn $)\rbrace +\lbrace P(1$ blue from 1 st urn $)$

$=\frac{3}{8} \times \frac{4}{8}+\frac{5}{8} \times \frac{4}{8} \quad(\because$ Event of drawing a ball from one urn is independent of drawing

$=\frac{12}{64}+\frac{20}{64}=\frac{32}{64}=\frac{1}{2}$.

a ball from other urn)

Ex. 22 . A speaks truth in $75 %$ and $B$ in $80 %$ of the cases. In what per cent of the cases are they likely to contradict each other in narrating the same incident?

(AIEEE 2004, IIT)

Sol. Let $E_1$ : Event that $A$ speak the truth

$E_2:$ Event that $B$ speaks the truth

Then, $\bar E_1$ : Event that $A$ tells a lie

$\bar E_2:$ Event that $B$ tells a lie

$E_1, E_2$ and so $\bar E_1, \bar E_2$ are independent events.

$\therefore \quad A$ and $B$ will contradict each other in the following two mutually exclusive ways

(i) $A$ speaks the truth, $B$ tells a lie

(ii) $A$ tells a lie, $B$ speaks the truth

Now, $P(E_1)=\frac{75}{100}=\frac{3}{4} \Rightarrow P(\bar E_1)=1-\frac{3}{4}=\frac{1}{4}$

$P(E_2)=\frac{80}{100}=\frac{4}{5} \Rightarrow P(\bar E_2)=1-\frac{4}{5}=\frac{1}{5}$

$\therefore$ Required probability $=P(E_1) \times P(\bar E_2)+P(\bar E_1) \times P(E_2)=\frac{3}{4} \times \frac{1}{5}+\frac{1}{4} \times \frac{4}{5}=\frac{7}{\mathbf{2 0}}$.

$\therefore \quad %$ of cases in which $A$ and $B$ contradict each other $=(\frac{7}{20} \times 100) \%=\mathbf{3 5 \%}$.

Ex. 23 . A problem in mathematics is given to four students $A, B, C$ and $D$. Their chances of solving the problem are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively. What is the probability that the problem will be solved?

Sol. Given, $P(A)=\frac{1}{2} \quad \Rightarrow P(\bar{A})=P(A$ not solving the problem $)=1-\frac{1}{2}=\frac{1}{2}$

$ \begin{aligned} & P(B)=\frac{1}{3} \Rightarrow P(\bar{B})=1-\frac{1}{3}=\frac{2}{3} \\ & P(C)=\frac{1}{4} \Rightarrow P(\bar{C})=1-\frac{1}{4}=\frac{3}{4} \\ & P(D)=\frac{1}{5} \Rightarrow P(\bar{D})=1-\frac{1}{5}=\frac{4}{5} \end{aligned} $

$\therefore \quad P($ problem will not be solved $)=P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) \times P(\bar{D})=\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}=\frac{1}{5}$

$\therefore \quad P($ problem will be solved $)=1-\frac{1}{5}=\frac{\mathbf{4}}{\mathbf{5}}$.

Ex. 24 . $A$ and $B$ throw a coin alternately till one of them gets a ‘head’ and wins the game. Find their respective probabilities of winning .

Sol. Let $\boldsymbol{A}$ : Event of $A$ getting a head

$\Rightarrow \bar{A}:$ Event of $A$ not getting a head

$\therefore P(A)=\frac{1}{2}$ and $P(\bar{A})=1-\frac{1}{2}=\frac{1}{2}$

Similarly, $\boldsymbol{B}$ : Event of $B$ getting a head

$\overline{\boldsymbol{B}}:$ Event of $B$ not getting a head.

So $P(B)=\frac{1}{2}$ and $P(\bar{B})=1-\frac{1}{2}=\frac{1}{2}$.

Let $A$ start the game. $A$ ’s wins if he throws a head in the 1st throw or 3rd throw or 5th throw or throw and so $\begin{aligned} & \text{ On. } \\ & \text{ Probability of } A \text{ ’s winning in } 1 \text{ st throw }\end{aligned} P(A)=\frac{1}{2}$

Probability of $A$ ’s winning in 3rd throw $=P(not A) \cdot P(not B) . P(A)=P(\bar{A}) \times P(\bar{B}) \times P(A)$

$ =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=(\frac{1}{2})^{3} $

Similarly, probability of $A$ ’s winning in 5th throw $=P(\bar{A}) \times P(\bar{B}) \times P(\bar{A}) \times P(\bar{B}) \times P(A)$

Since all these events are mutually exclusive,

$ =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=(\frac{1}{2})^{5} \text{ and so on. } $

$p(A$ winning the game first $)=\frac{1}{2}+(\frac{1}{2})^{3}+(\frac{1}{2})^{5}+(\frac{1}{2})^{7} \ldots=\frac{1}{2}[1+(\frac{1}{2})^{2}+(\frac{1}{2})^{4}+(\frac{1}{2})^{6}+\ldots]$

$ =\frac{1}{2}[\frac{1}{1-(\frac{1}{2})^{2}}]=\frac{1}{2} \times \frac{1}{3 / 4}=\frac{1}{2} \times \frac{4}{3}=\frac{2}{\mathbf{3}} . $

$ (\because \text{ For an infinite } G P, S _{\infty}=\frac{\text{ First term }}{1-\text{ common ratio }}) $

Since $A$ and $B$ are mutually exclusive events, as either of them will win, $P(B$ winning the game first $)=1-\frac{2}{3}=\frac{\mathbf{1}}{\mathbf{3}}$.

PRACTICE SHEET

What is the probability that a number selected at random from the set of numbers $\lbrace 1,2,3, \ldots, 100\rbrace $ is a perfect cube?

(a) $\frac{1}{25}$

(b) $\frac{2}{25}$

(d) $\frac{4}{25}$

~~ 2. Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails.

(a) $3 / 8$

(b) $7 / 8$

(d) $1 / 2$

~~ 3. Three identical dice are rolled. The probability that same number will appear on each of them is

(a) $\frac{1}{6}$

(b) $\frac{1}{12}$

(d) $\frac{1}{36}$

(DCE 2001)

~~ 4. Two dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is

(a) $\frac{15}{36}$

(b) $\frac{25}{52}$

(d) $\frac{5}{6}$

~~ 5. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is red and a king.

(a) $\frac{1}{13}$

(b) $\frac{1}{52}$

(d) $\frac{5}{6}$

~~ 6. If two dice are rolled simultaneously, the probability that the product of the numbers thrown will be greater than 16 is

(a) $\frac{1}{6}$

(b) $\frac{1}{9}$

(d) $\frac{5}{36}$

~~ 7. Counters marked 1,2,3 are placed in a bag and one is drawn and replaced. The operation is repeated three times. The chance of obtaining a total of 6 is

(a) $\frac{2}{9}$

(b) $\frac{7}{27}$

(d) $\frac{14}{27}$

(AMU 2000)

~~ 8. A bag contains 7 red and 5 green balls. The probability of drawing all four balls as red balls, when four balls are drawn at random is

(a) $\frac{14}{99}$

(b) $\frac{7}{99}$

(d) $\frac{4}{12}=\frac{1}{3}$

~~ 9. From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected.

(a) $\frac{1}{5}$

(b) $\frac{7}{10}$

(d) $\frac{5}{6}$

~~ 10. Five horses are in a race. $Mr A$. Selects two of the horses at random and bets on them. The probability that $Mr A$ selected the winning horse is

(a) $\frac{1}{5}$

(b) $\frac{2}{5}$

(d) $\frac{4}{5}$

~~ 11. Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7.

(a) $\frac{5}{12}$

(b) $\frac{1}{4}$

(d) $\frac{5}{9}$

~~ 12. A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ?

(a) $\frac{11}{12}$

(b) $\frac{11}{36}$

(d) $\frac{1}{6}$

~~ 13. Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ?

(a) $\frac{55}{112}$

(b) $\frac{55}{221}$

(d) $\frac{33}{112}$

~~ 14. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack?

(a) $\frac{10}{13}$

(b) $\frac{15}{26}$

(d) $\frac{7}{13}$

~~ 15. A natural number is chosen at random from amongst the first 300 . What is the probability that the number chosen is a multiple of 2 or 3 or 5 ?

(a) $\frac{1}{10}$

(b) $\frac{11}{15}$

(d) $\frac{17}{30}$

~~ 16. A drawer contains 5 brown and 4 blue socks well mixed. A man reaches the drawer and pulls out 2 socks at random. What is the probability that they match?

(a) $\frac{2}{9}$

(b) $\frac{1}{3}$

(d) $\frac{5}{9}$

(J&K CET 2009)

~~ 17. If three natural numbers from 1 to 100 are selected randomly, then the probability that all are divisible by both 2 and 3 is

(a) $\frac{2}{105}$

(b) $\frac{3}{767}$

(d) $\frac{5}{1224}$

(IIT 2004)

~~ 18. 6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is

(a) $\frac{1}{36}$

(b) $\frac{1}{462}$

(d) $\frac{1}{231}$

~~ 19. A four digit number is formed by the digits $1,2,3,4$ with no repetition. The probability that the number is odd is

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(d) $\frac{1}{6}$

(UPSEE 2005)

~~ 20. Probability of all 4-digit numbers having all the digits same is

(a) $\frac{1}{4}$

(b) $\frac{9}{1000}$

(d) $\frac{1}{100}$

~~ 21. The letters of the word ‘NATIONAL’ are arranged at random. What is the probability that the last letter will be $T$ ?

(a) $\frac{1}{4}$

(b) $\frac{5}{8}$

(d) $\frac{3}{8}$

~~ 22. The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’ the two I’s do not come together is

(a) $\frac{1}{5}$

(b) $\frac{4}{5}$

(d) $\frac{9}{10}$

(UPSEE 2001)

~~ 23. An urn contains nine balls, of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

(a) $\frac{1}{3}$

(b) $\frac{2}{7}$

(d) $\frac{2}{23}$

(AIEEE 2010)

~~ 24. The probability of India winning a test match against Westindies is $\frac{1}{2}$. Assuming independence from match to match, the probability that in a match series, India’s second win occurs at the third test is

(a) $\frac{1}{2}$

(b) $\frac{2}{3}$

(d) $\frac{2}{5}$

(IIT 1993)

~~ 25. A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two cards are drawn at random. The probability that at least one them is an ace is

(a) $\frac{1}{5}$

(b) $\frac{1}{9}$

(d) $\frac{9}{20}$

(Rajasthan PET 2005)

~~ 26. A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour?

(a) $\frac{5}{11}$

(b) $\frac{7}{22}$

(d) $\frac{31}{66}$

~~ 27. Two players $A$ and $B$ play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. The first one to draw a queen wins the game. If $A$ begins the game, then the probability of his winning the game is

(a) $\frac{13}{25}$

(b) $\frac{7}{26}$

(d) $\frac{15}{52}$

~~ 28. 6 boys and 6 girls are seated in a row. Probability that all the boys sit together is

(a) $\frac{1}{102}$

(b) $\frac{1}{112}$

(d) $\frac{1}{132}$

~~ 29. The probability that $A$ hits a target is $\frac{1}{3}$ and the probability that $B$ hits it is $\frac{2}{5}$. What is the probability that the target will be hit, if each one of $A$ and $B$ shoots the target ?

(a) $\frac{5}{6}$

(b) $\frac{3}{5}$

(d) $\frac{1}{6}$

~~ 30. A coin and six faced die, both unbiased are thrown simultaneously. The probability of getting a tail on the coin and an even number on the die is

(a) $\frac{1}{2}$

(b) $\frac{3}{4}$

(d) $\frac{2}{3}$

~~ 31. An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at $1 st, 2 nd, 3 rd$ and 4 th shots are $0.4,0.3,0.2$ and 0.1 respectively. What is the probability that at least one shot hits the plane?

(a) 0.0024

(b) 0.3024

(d) 0.9976

(IIT 1981)

~~ 32. A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

(a) $\frac{23}{42}$

(b) $\frac{19}{42}$

(d) $\frac{16}{39}$

~~ 33. If from each of the three boxes containing 3 blue and 1 red balls, 2 blue and 2 red balls, 1 blue and 3 red balls, one ball is drawn at random, then the probability that 2 blue and 1 red ball will be drawn is :

(a) $\frac{13}{32}$

(b) $\frac{27}{32}$

(d) None of these

~~ 34. A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. What is the probability that they are of different colours?

(a) $\frac{5}{32}$

(b) $\frac{1}{10}$

(d) $\frac{1}{7}$

~~ 35. A management institute has six senior professors and four junior professors. Three professors are selected at random for a government project. The probability that at least one of the junior professors would get selected is

(a) $5 / 6$

(b) $2 / 3$

(d) $1 / 6$ (e) None of these

(XAT 2007)

~~ 36. A card is drawn at random from a well shuffled pack of 52 cards

$X$ : The card drawn is black or a king

$Y:$ The card drawn is a club or heart or a jack

$Z:$ The card drawn is an ace or a diamond or a queen

Then, which of the following is correct.

(a) $P(X)>P(Y)>P(Z)$

(b) $P(X) \geq P(Y)=P(Z)$

(d) $P(X)=P(Y)=P(Z)$

(IIFT 2009)

~~ 37. The game of “chuck-a-luck” is played at carnivals in some parts of Europe. Its rules are as follows : You pick a number from 1 to 6 and the operator rolls three dice. If the number you picked comes up an all the three dice, the operator pays you ₹ 3, if it comes up on two dice, you are paid ₹ 2; and if it comes on just one die, you are paid ₹ 1 . Only if the number you picked does not come up at all, you pay the operator $₹ 1$. The probability that you will win money playing in this game is :

(a) 0.52

(b) 0.753

(d) None of these

(IIFT 2008)

~~ 38. The odds against certain event are $5: 2$ and the odds in favour of another in dependent event are 6:5. The probability that at least one of the event will happen is

(a) $\frac{12}{77}$

(b) $\frac{25}{77}$

(d) $\frac{65}{77}$

(IIFT 2005)

~~ 39. If the chance that a vessel arrives safely at a port is $\frac{9}{10}$, then what is the chance that out of 5 vessels expected at least 4 will arrive safely?

(a) $\frac{(14 \times 9^{4})}{10^{5}}$

(b) $\frac{(15 \times 9^{5})}{10^{4}}$

(d) $\frac{(14 \times 9^{6})}{10^{5}}$ (JMET 2011)

~~ 40. A group of $2 n$ boys and $2 n$ girls is divided at random into two equal batches. The probability that each batch will have equal number of boys and girls is

(a) $\frac{1}{2} n$

(b) $\frac{1}{4} n$

(d) ${ }^{2 n} C_n$

(ATMA 2008)

~~ 41. A box contains 100 balls numbers from 1 to 100 . If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

(a) $\frac{3}{8}$

(b) $\frac{1}{2}$

(d) $\frac{1}{4}$

(ATMA 2005)

~~ 42. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals :

(a) $\frac{1}{6}$

(b) $\frac{1}{3}$

(d) None of these

~~ 43. A die is rolled three times. The probability of getting a larger number than the pervious number each time is:

(a) $\frac{5}{72}$

(b) $\frac{5}{54}$

(d) $\frac{1}{18}$

(MPPET 2011)

~~ 44. A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is

(a) $\frac{{ }^{20} C _{10} \times 5^{6}}{6^{20}}$

(b) $\frac{120 \times 5^{7}}{6^{10}}$

(d) None of these

(Odisha JEE 2013)

~~ 45. A bag contains $a$ white and $b$ black balls. Two players $A$ and $B$ alternately draw a ball from the bag replacing the ball each time after the draw till one of them draws a white ball and wins the game. $A$ begins the game. If the probability of $A$ winning the game is 3 times that of $B$, then the ratio $a: b$ is

(a) $1: 1$

(b) $1: 2$

(d) None of these

(BITSAT 2009)

~~ 46. A determinant of second order is made with the elements 0,1 . What is the probability that the determinant is positive?

(a) $\frac{7}{12}$

(b) $\frac{11}{12}$

(d) $\frac{15}{16}$

(AMU 2009)

~~ 47. Out of $3 n$ consecutive natural numbers 3 natural numbers are chosen at random without replacement. The probability that the sum of the chosen numbers in divisible by 3 is

(a) $\frac{n(3 n^{2}-3 n+2)}{2}$

(b) $\frac{(3 n^{2}-3 n+2)}{2(3 n-1)(3 n-2)}$

(d) $\frac{n(3 n-1)(3 n-2)}{3(n-1)}$

(DCE 2009)

~~ 48. A bag contains $2 n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $(n+1)$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

(a) 10

(b) 11

(d) 13

(EAMCET 2013)

~~ 49. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answer just by guessing is

(a) $\frac{17}{3^{5}}$

(b) $\frac{13}{3^{5}}$

(d) $\frac{10}{3^{5}}$

(IIT JEE Main 2013)

~~ 50. If the three independent events $E_1, E_2$ and $E_3$, the probability that only $E_1$ occurs is $\alpha, E_2$ occurs is $\beta$, only $E_3$ occurs is $\gamma$. Let the probability $p$ that none of the event $E_1, E_2$ or $E_3$ occurs satisfy the equations $(\alpha-2 \beta) p=\alpha \beta$ and $(\beta-3 \gamma) p$ $=2 \beta \gamma$. All the given probabilities are assumed to lie in the interval $(0,1)$.

Then, $\frac{\text{ Probability of occurrence of } E_1}{\text{ Probability of occurrence of } E_3}$ is equal to

(a) 3

(b) 5

(d) 8

(IIT JEE Advance 2013)

ANSWERS

1. (a)	2. $(d)$	3. (d)	4. (c)	5. $(c)$	6. $(c)$	7. $(b)$	8. (b)	9. $(b)$	10. $(b)$
11. (c)	12. (b)	13. $(b)$	14. $(d)$	15. (b)	16. $(c)$	17. $(c)$	18. (b)	19. (a)	20. (c)
21. (c)	22. (b)	23. (b)	24. (c)	25. $(d)$	26. $(d)$	27. (a)	28. $(d)$	29. (b)	30. (c)
31. (c)	32. $(b)$	33. (a)	34. (c)	35. (a)	36. (c)	37. (c)	38. $(c)$	39. (a)	40. (c)
41. $(d)$	42. $(c)$	43. $(b)$	44. (c)	45. (c)	46. (c)	47. $(c)$	48. (a)	49. $(c)$	50. $(c)$

HINTS AND SOLUTIONS

Let us assume $S$ as the sample space in all questions. $S$ means the set denoting the total number of outcomes possible. Let $S=\lbrace 1,2,3, \ldots, 100\rbrace $ be the sample space. Then, $n(S)=100$

Let $A$ : Event of selecting a cube from the given set $S$.

$\Rightarrow A=\lbrace 18,27,64\rbrace $

$\Rightarrow n(A)=4$

$\therefore$ Required probability $P(A)=\frac{n(A)}{n(S)}=\frac{4}{100}=\frac{1}{25}$

~~ 2. Let $S$ be the sample space. Then,

$ \begin{aligned} & S=\lbrace H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T\rbrace \\ & \Rightarrow n(S)=8 \end{aligned} $

Let $A$ : Event of getting more heads than number of tails.

Then, $A=\lbrace H H H, H H T, H T H, T H H\rbrace $

$\Rightarrow n(A)=4$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{4}{8}=\frac{1}{2}$.

~~ 3. Total number of outcomes when three identical dice are rolled, $n(S)=6 \times 6 \times 6=216$

Let $A$ : Event of rolling same number or each dice

$\Rightarrow A=\lbrace (1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5)$, $(6,6,6)\rbrace $

$\Rightarrow n(A)=6$

$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{6}{216}=\frac{\mathbf{1}}{\mathbf{3 6}}$.

~~ 4. Total number of outcomes when two identical dice are rolled, $n(S)=6 \times 6=36$

Let $A$ : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die

$\Rightarrow A=\lbrace (2,3),(2,6),(4,3),(4,6),(6,3),(6,6),(3,2)$, $(6,2),(3,4),(6,4),(3,6)\rbrace $

$\Rightarrow \quad n(A)=11$

$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{\mathbf{1 1}}{\mathbf{3 6}}$.

~~ 5. There is a total of 52 cards $n(S)=52$

Let $A$ : Event of drawing a red king

Since there are only two red kings in the pack, $n(A)=2$

$\therefore \quad P(A)=\frac{2}{52}=\frac{\mathbf{1}}{\mathbf{2 6}}$.

~~ 6. Let $S$ be the sample space of drawing a counter three times and replacing it each time.

Then, $n(S)=3 \times 3 \times 3=27$

Let $A$ : Event of obtaining a total of 6 in the three draws of counters.

Then, $A=\lbrace (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2)$, $(3,2,1),(2,2,2)\rbrace $

$\Rightarrow n(A)=7$

$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{7}{\mathbf{2 7}}$.

~~ 8. There are $(7+5)=12$ balls in the bag.

4 balls can be drawn at random from 12 balls in ${ }^{12} C_4$ ways.

$\therefore n(S)={ }^{12} C_4=\frac{\mid 12}{\underline{8} \underline{4} 4}=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}=495$

Let $A$ : Event of drawing all four balls as red.

4 red balls can be drawn from 7 red balls in ${ }^{7} C_4$ ways

$\therefore \quad n(A)={ }^{7} C_4=\frac{\lfloor 7}{\lfloor\lfloor 4}=\frac{7 \times 6 \times 5}{3 \times 2}=35$

$\therefore \quad$ Required probability $=\frac{n(A)}{n(S)}=\frac{35}{495}=\frac{7}{99}$

~~ 9. Total number of ways of selecting 2 persons at random out of 5 persons $={ }^{5} C_2$

$\therefore \quad n(S)={ }^{5} C_2=\frac{\lfloor 5}{\lfloor\underline{5} \underline{2}}=\frac{5 \times 4}{2 \times 1}=10$

Let $A$ : Event of selecting at least one woman. $\Rightarrow A=$ Selecting 1 woman +1 man or selecting 2 women

$\Rightarrow n(A)={ }^{2} C_1 \times{ }^{3} C_1+{ }^{2} C_2=2 \times 3+1=7$

$\therefore \quad$ Required probability $=\frac{n(A)}{n(S)}=\frac{7}{\mathbf{1 0}}$.

~~ 10. As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race $={ }^{5} C_1$

$\therefore \quad n(S)={ }^{5} C_1=\frac{\underline{5}}{\lfloor 4 \underline{1}}=5$

To find the chance that $Mr A$ selects the winning horses, it is essential that one of the two horses selected by him wins the race.

$E: Mr A$ selecting the winning horse.

$\Rightarrow n(E)={ }^{2} C_1=2$

$\therefore \quad$ Required probability $=\frac{n(E)}{n(S)}=\frac{\mathbf{2}}{\mathbf{5}}$.

~~ 11. Total number of ways in which 2 dice are rolled $=6 \times 6=36$ $\Rightarrow \quad n(S)=36$

Let $A$ : Event of rolling an even number of 1st dice $B$ : Event of rolling a total of 7

$\Rightarrow A=\lbrace (2,1),(2,2) \ldots,(2,6),(4,1),(4,2) \ldots,(4,6)$, $(6,1),(6,2), \ldots,(6,6)\rbrace $

and $B=\lbrace (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\rbrace $

$\Rightarrow n(A)=3 \times 6=18, n(B)=6$

$\Rightarrow A \cap B$ : Event of rolling an even number on 1st dice and a total of 7

$\Rightarrow A \cap B=\lbrace (2,5),(4,3),(6,1)\rbrace \Rightarrow n(A \cap B)=3$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{18}{36} ; P(B)=\frac{n(B)}{n(S)}=\frac{6}{36} ; P(A \cap B)=\frac{3}{36}$

$\Rightarrow \quad P(A \cup B)=P($ Event of rolling an even number 1 st dice or a total of 7)

$=P(A)+P(B)-P(A \cap B)$

$=\frac{18}{36}+\frac{6}{36}-\frac{3}{36}=\frac{21}{36}=\frac{7}{\mathbf{1 2}}$.

~~ 12. Let $S=$ total ways in which two dice can be rolled

$\Rightarrow n(S)=6 \times 6=36$

Let $A$ : Event of throwing 3 with 1 st dice,

$B$ : Event of throwing 3 with 2 nd dice.

Then, $A=\lbrace (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\rbrace $

$\Rightarrow n(A)=6$

$B=\lbrace (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)\rbrace \Rightarrow n(B)=6$

$A \cap B=\lbrace (3,3)\rbrace \Rightarrow n(A \cap B)=1$

$ \begin{gathered} \therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}, P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}, \\ P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{36} \end{gathered} $

$\therefore$ Required probability $=P$ (Throwing a 3 with at least one of the dice)

$ \begin{aligned} & =P(A)+P(B)-P(A \cap B) \\ & =\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{\mathbf{1 1}}{\mathbf{3 6}} . \end{aligned} $

~~ 13. $S$ : Drawing 2 cards out of 52 cards

$\Rightarrow \quad n(S)={ }^{52} C_2=\frac{\underline{52}}{\underline{50}\lfloor 2}=\frac{52 \times 51}{2}=1326$

$A$ : Event of drawing 2 black cards out of 26 black cards

$\Rightarrow n(A)={ }^{26} C_2=\frac{26 \times 25}{2}=325$

$B$ : Event of drawing 2 kings out of 4 kings

$\Rightarrow n(B)={ }^{4} C_2=\frac{\lfloor 4}{\lfloor 2}=6$

$\Rightarrow A \cap B$ : Event of drawing 2 black kings

$\Rightarrow n(A \cap B)={ }^{2} C_2=1$

$\therefore \quad P(A)=\frac{325}{1326}, P(B)=\frac{6}{1326}, P(A \cap B)=\frac{1}{1326}$

$\therefore \quad P$ (Both cards are black or both kings)

$=P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$=\frac{325}{1326}+\frac{6}{1326}-\frac{1}{1326}=\frac{330}{1326}=\frac{\mathbf{5 5}}{\mathbf{2 2 1}}$.

~~ 14. Here $n(S)=52$

Let $A, B, C$ be the events of getting a red card, a diamond and a jack respectively.

$\because \quad$ There are 26 red cards, 13 diamonds and 4 jacks,

$ n(A)=26, n(B)=13, n(C)=4 $

$\Rightarrow \quad n(A \cap B)=n($ red and diamond $)=13$,

$n(B \cap C)=n($ diamond and jack $)=1$,

( $\because$ There is only jack of diamonds)

$n(A \cap C)=n($ red and jack $)=2$

( $\because$ There are two red jacks)

$n(A \cap B \cap C)=n($ red, diamond, jack $)=1$

$\therefore P(A)=\frac{26}{52}, P(B)=\frac{13}{52}, P(C)=\frac{4}{52}, P(A \cap B)=\frac{13}{52}$,

$P(B \cap C)=\frac{1}{52}, P(A \cap C)=\frac{2}{52}, P(A \cap B \cap C)=\frac{1}{52}$

$\therefore \quad P($ a red card or a diamond or a jack $)=P(A \cup B \cup C)$

$=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(A \cap C)$

$+P(A \cap B \cap C)$

$=\frac{26}{52}+\frac{13}{52}+\frac{4}{52}-(\frac{13}{52}+\frac{1}{52}+\frac{2}{52})+\frac{1}{52}$

$=\frac{44}{52}-\frac{16}{52}=\frac{28}{52}=\frac{7}{13}$.

~~ 15. $n(S)=300$

Let $A$ : Event of getting a number divisible by 2

$B$ : Event of getting a number divisible by 3

$C$ : Event of getting a number divisible by 5

$\therefore A \cap B$ : Event of getting a number divisible by both 2 and 3 , i.e., 6

$B \cap C$ : Event of getting a number divisible by both 3 and 5, i.e, 15

$A \cap C$ : Event of getting a number divisible by both 2 and 5, i.e., 10

$A \cap B \cap C$ : Event of getting a number divisible by all 2, 3 and 5 , i.e., 30

Then, $n(A)=150, n(B)=100, n(C)=60$

$n(A \cap B)=50, n(B \cap C)=20, n(A \cap C)=30$,

$n(A \cap B \cap C)=10$

$\Rightarrow \quad P(A)=\frac{150}{300}, P(B)=\frac{100}{300}, P(C)=\frac{60}{300}$

$P(A \cap B)=\frac{50}{300} P(B \cap C)=\frac{20}{300}, P(A \cap C)=\frac{30}{300}$,

$P(A \cap B \cap C)=\frac{10}{300}$

$\therefore \quad P($ Number divisible by 2 or 3 or 5$)=P(A \cup B \cup C)$

$=P(A)+P(B)+P(C)-(P(A \cap B)-P(B \cap C)-P(A \cap C))$ $+P(A \cap B \cap C)$

$=\frac{150}{300}+\frac{100}{300}+\frac{60}{300}-(\frac{50}{300}+\frac{20}{300}+\frac{30}{300})+\frac{10}{300}$

$=\frac{320}{300}-\frac{100}{300}=\frac{220}{300}=\frac{\mathbf{1 1}}{\mathbf{1 5}}$.

~~ 16. $n(S)=$ Total number of ways in which 2 socks can be drawn out of 9 socks ( 5 brown and 4 blue socks)

$={ }^{9} C_2=\frac{\underline{9}}{\lfloor\underline{2}}=\frac{9 \times 8}{2}=\mathbf{3 6}$

Let $A$ : Event of drawing 2 socks of same colour

$\Rightarrow A=$ Drawing 2 brown socks out of 5 brown socks or Drawing 2 blue sock out of 4 blue socks

$\Rightarrow n(A)={ }^{5} C_2+{ }^{4} C_2=\frac{\underline{5}}{\lfloor\underline{3} \underline{2}}+\frac{\underline{4}}{\lfloor\underline{2} 2}$

$=\frac{5 \times 4}{2}+\frac{4 \times 3}{2}=10+6=16$

$\therefore \quad$ Required probability $P(A)=\frac{n(A)}{n(S)}=\frac{16}{36}=\frac{4}{\mathbf{9}}$.

~~ 17. Let $n(S)=$ Number of ways of selecting 3 numbers from 100 numbers

$={ }^{100} C_3$

Let $E$ : Event of selecting three numbers divisible by both 2 and 3 from numbers 1 to 100

$=$ Event of selecting three multiples of 6 from 1 to 100

Selecting 3 numbers from

$=\lbrace 6,12,18,24,30,36,42,48,54,60,66,72,78$, $84,90,96\rbrace $

$\Rightarrow n(E)={ }^{16} C_3$

$\therefore \quad P$ (Selecting 3 numbers exactly divisible by 6 between

1 and 100$)=\frac{n(E)}{n(S)}=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{16 \times 15 \times 14}{100 \times 99 \times 98}=\frac{\mathbf{4}}{\mathbf{1 1 5 5}}$.

~~ 18. Let $S$ be the sample space.

Then, $n(S)=$ Number of ways in which 6 boys and 6 girls can sit in a row $=12$ !

Let $E$ : Event of 6 girls and 6 boys sitting alternately.

Then, the 6 girls and 6 boys can be arranged in alternate position in two ways.

Ist way : We start with a boy. Then the arrangement is :

$B G \quad B G \quad B G \quad B G \quad B G \quad B G$

$\therefore \quad$ Number of ways of arranging 6 boys in 6 places $=6$ !

Number of ways of arranging 6 girls in 6 places $=6$ !

$\therefore \quad$ Number of ways of arranging 6 boys and 6 girls in alternate places $=6 ! \times 6 !$

Similarly IInd way : Here we start with a girl. Then the

$\therefore$ Number of ways of arranging 6 boys and 6 girls alternately this way

$=6 ! \times 6$ !

$\therefore \quad n(E)=6 ! \times 6 !+6 ! \times 6$ !

$=2 \times 6 ! \times 6 !$

$\therefore \quad n(E)=\frac{n(E)}{n(S)}=\frac{2 \times 6 ! \times 6 !}{12 !}$

$=\frac{2 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{12 \times 11 \times 10 \times 9 \times 8 \times 7}=\frac{1}{462}$.

~~ 19. Let $S$ be the sample space

Then $n(S)=$ Number of four digit numbers that can be formed with out repetition

1 of 4 digits	1 of 3 digits	1 of 2 digits	1 of 1 digit
$Th$	$H$	$T$	$O$

$=4 !=4 \times 3 \times 2 \times 1$ ways $=24$ ways.

Let $E$ : Event of forming odd 4 digit numbers with the digits $1,2,3,4$ without repetition.

Then, the ones place can be filled by the odd digits 1 or 3 , in 2 ways.

The thousands place can be filled with the remaining 3 digits in 3 ways

The hundreds place can be filled with the remaining 2 digits in 2 ways

The tens place can be filled with the remaining 1 digit in 1 way

$ \begin{matrix} \therefore & n(E)=2 \times 3 \times 2 \times 1=12 \\ \therefore & P(E)=\frac{n(E)}{n(S)}=\frac{12}{24}=\frac{1}{2} . \end{matrix} $

~~ 20. Let $S$ be the sample space where 4 digit numbers are formed using digits 0 to 9 repetition. Then,

$n(S)=$ Total number of 4 digit number formed

$=9 \times 10 \times 10 \times 10=9000$

( $\because$ The thousands place rest can only be accuried by the 9 non-zero digits, while the rest of the three places, i.e., hundreds, tens and ones can each be filled in 10 ways)

Let $E$ : Event of forming four digit numbers having all the digits same. Then $n(E)=9 \times 1 \times 1 \times 1=9$

( $\because$ For one place (thousands place) we choose the number from the non-zero 9 digits and for the rest of the places, the same digit is repeated)

$ \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{9}{9000}=\frac{\mathbf{1}}{\mathbf{1 0 0 0}} $

~~ 21. Let $S$ be the sample space.

Then $n(S)=$ Total number of ways in which the letters of word NATIONAL can be arranged

$=\frac{8 !}{2 ! 2 !} \quad(\because.$ There are $2 A$ ’s and $2 N^{\prime}$ s in 8 letters $)$

Let $E$ : Event of arranged letters of the word NATIONAL, so that the last position is occupied by letter ’ $T$ “.

Here position of $T$ is fixed, so

$n(E)=$ Number of ways in which the rest of the 7 letters can be arranged in 7 places.

$=\frac{7 !}{2 ! 2 !}$

( $\because$ There are $2 A^{\text{s }}$ s and $2 N^{\star}$ s)

$\therefore$ Required probability $=\frac{n(E)}{n(S)}=\frac{\frac{7 !}{2 ! 2 !}}{\frac{8 !}{2 ! 2 !}}=\frac{7 !}{8 !}=\frac{7 !}{8 \times 7 !}=\frac{\mathbf{1}}{\mathbf{8}}$.

~~ 22. Let $S$ be the sample space. Then,

$n(S)=$ Total number of ways in which the letters of the word ‘UNIVERSITY’ can be arranged

$=\frac{10 !}{2 !}$

$(\because.$ There are $.2 I^{\star} s)$

Let $E$ : Event of arranging the letters of the word ‘UNIVERSITY’ such that 2 I’s remain together.

Not considering the two I’s, the remaining 8 letters of ‘UNIVERSITY’ can be arranged in 8 ! ways. There seven places between these 8 letters and two on extreme ends which can be occupied by the two I’s, so that they are not together.

$-X-X-X-X-X-X-X-X-$

The places marked ‘-’ can be filled with I’s.

$\therefore$ Now 2I’s can be arranged in any of the 2 out of 9 places

in ${ }^{9} C_2$ ways $=\frac{\underline{9}}{\underline{7} \times \underline{2}}=\frac{9 \times 8}{2}=36$ ways.

$\therefore \quad n(E)=8 ! \times 36$

$\therefore \quad P(E)=\frac{8 ! \times 36}{10 ! / 2 !}=\frac{8 ! \times 36 \times 2 !}{10 \times 9 \times 8 !}=\frac{4}{\mathbf{5}}$.

~~ 23. Let $S$ be the sample space having 9 balls $(3 R+4 B+2 G)$

Then $n(S)=$ Total number of ways in which 3 balls can be drawn out of the 9 balls

$=\frac{9 \times 8 \times 7}{3 \times 2}=\mathbf{8 4}$

Let $A$ : Event of drawing three balls of different colours from the urn. $\Rightarrow A=$ Event of drawing 1 red ball out of 3 red balls, 1 blue ball out of 4 blue balls and 1 green ball out of 2 green balls

$\Rightarrow n(A)={ }^{3} C_1 \times{ }^{4} C_1 \times{ }^{2} C_1=3 \times 4 \times 2=24$.

$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{24}{84}=\frac{\mathbf{2}}{7}$.

~~ 24. Let $A=$ Event that India wins the match. Then,

$\bar{A}=$ Event that India loses the match.

$P(A)=\frac{1}{2}$ and $P(\bar{A})=1-\frac{1}{2}=\frac{1}{2} \quad(\because P(A)+P(\bar{A})=1)$

$P$ (Third match is India’s second winning match)

$=P(.$ India wins the $1^{\text{st }}$ match, loses $2^{\text{nd }}$ match, wins $3^{\text{rd }}$ match $)+P(.$ India loses $1^{\text{st }}$ match, wins $2^{\text{nd }}$ match, wins $3^{\text{rd }}$ match)

(or)

$\Rightarrow$ Required probability $=P(A \bar{A} A)+P(\bar{A} A A)$

$ \begin{aligned} & =P(A) \times P(\bar{A}) \times P(A)+P(\bar{A}) \times P(A) \times P(A) \\ & =\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{\mathbf{1}}{\mathbf{4}} . \end{aligned} $

~~ 25. Let $S$ be the sample space for drawing 2 cards out of 4 aces, 4 kings, 4 queens and 4 jacks i.e, 16 cards.

Then $n(S)={ }^{16} C_2$

$P($ Drawing at least one ace $)=1-P($ Drawing no ace $)$

Let $E$ : Event of drawing no aces in the 2 drawn cards

$\Rightarrow n(E)={ }^{12} C_2($ Cards leaving aces $=16-4-12)$

$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{{ }^{12} C_2}{{ }^{16} C_2}=\frac{12 \times 11}{16 \times 15}=\frac{11}{20}$

$\therefore \quad P($ drawing at least one ace $)=1-\frac{11}{20}=\frac{\mathbf{9}}{\mathbf{2 0}}$.

~~ 26. Total number of balls in the bag $=12$ ( 5 Green +7 Red $)$

Let $S$ be the sample space of drawing 2 balls out of 12 balls. Then

$n(S)={ }^{12} C_2=\frac{12 \times 11}{2}=66$

$\therefore$ Let $A$ : Event of drawing two red balls

$\Rightarrow n(A)={ }^{7} C_2=\frac{7 \times 6}{2}=21$

$B$ : Event of drawing two green balls

$\Rightarrow n(B)={ }^{5} C_2 \frac{5 \times 4}{2}=10$

$\therefore P($ Event of drawing 2 balls of same colour $)=P($ Drawing two red balls) or $P$ (drawing two green balls)

$=P(A)+P(B)=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}=\frac{21}{66}+\frac{10}{66}=\frac{\mathbf{3 1}}{\mathbf{6 6}}$.

~~ 27. Let $E$ : Event of drawing a queen in a single draw the pack of 52 cards.

As there are 4 queens in a pack of 52 cards,

$P(E)=\frac{4}{52}=\frac{1}{13}$ $P(\bar{E})=P($ not drawing a queen $)=1-P(E)=1-\frac{1}{13}=\frac{12}{13}$

$\therefore P(A$ wins $)=P(A$ draws a queen in the 1 st or 3 rd or 5 th or …draws with $B$ not drawing the queen in the 2 nd or 4 th or 6th or …draws)

$=P(E$ or $\bar{E} \bar{E} E$ or $\bar{E} \bar{E} \bar{E} \bar{E} E$ or $\ldots)$

$=P(E)+P(\bar{E}) \cdot P(\bar{E}) \cdot P(E)+P(\bar{E}) P(\bar{E}) P(\bar{E})$ . $P(\bar{E}) . P(E)+\ldots \infty$

$=\frac{1}{13}+(\frac{12}{13} \times \frac{12}{13} \times \frac{1}{13})+(\frac{12}{13} \times \frac{12}{13} \times \frac{12}{13} \times \frac{12}{13} \times \frac{1}{13})$

$=\frac{1}{13}[1+(\frac{12}{13})^{2}+(\frac{12}{13})^{4}+\ldots \infty]$

$=\frac{1}{13}[\frac{1}{1-(\frac{12}{13})^{2}}]=\frac{1}{13}[\frac{1}{\frac{(169-144)}{169}}]$

$=\frac{1}{13} \times \frac{169}{25}=\frac{\mathbf{1 3}}{\mathbf{2 5}}$.

[Sum of a G.P with infinite terms $=\frac{a}{1-r}$

where $a=1$ st term, $r=$ common ratio.]

~~ 28. Let $S$ be the sample space for arranging 6 boys and 6 girls in a row. Then, $n(S)=12$ !

If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the remaining, i.e., 6 girls can be arranged in a row in 6 ! ways. There are 5 places between the 6 girls and 2 on the extreme ends, where the entity of 6 boys can be placed, i.e., for the single entity of 6 boys, we have 7 places where they can be arranged in 7 ways and also amongst themselves they can be arranged in 6! ways.

$\therefore$ No. of ways of arranging 6 boys and 6 girls in a row where the 6 boys are together $=6 ! \times 7 \times 6$ !

$\therefore \quad$ Required probability

$=\frac{6 ! \times 7 \times 6 !}{12 !}=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{12 \times 11 \times 10 \times 9 \times 8 \times 7}=\frac{1}{\mathbf{1 3 2}}$.

~~ 29. Let $A$ : Event that $A$ hits the target

$B$ : Event that $B$ hits the target

Given, $P(A)=\frac{1}{3}$

$\Rightarrow P(A$ not hitting the target $)=P(\bar{A})=1-\frac{1}{3}=\frac{2}{3}$

$P(B)=\frac{2}{5} \Rightarrow P(B$ not hitting the target $)=P(\bar{B})=1-\frac{2}{5}=\frac{3}{5}$

$\therefore \quad P($ Target is hit $)=P(A$ hits $) \times P(B$ does not hit $)$

$+P(A$ does not hit $) \times P(B$ hits $)+P(A$ hits $) \times P(B$ hits $)$

$=P(A) \times P(\bar{A})+P(\bar{A}) \times P(B)+P(A) \times P(B)$

$=\frac{1}{3} \times \frac{3}{5}+\frac{2}{3} \times \frac{2}{5}+\frac{1}{3} \times \frac{2}{5}=\frac{3}{15}+\frac{4}{15}+\frac{2}{15}=\frac{9}{15}=\frac{\mathbf{3}}{\mathbf{5}}$.

~~ 30. Let $A$ : Event of getting a tail on the coin

$B:$ Event of getting an even number on the die.

Then, $P(A)=\frac{1}{2}$

$P(B)=\frac{3}{6}=\frac{1}{2}$ as $B=\lbrace 2,4,6\rbrace $

$A$ and $B$ being independent events,

$P($ Getting tail on the coin and even number on the die)

$=P(A \cap B)=P(A) \times P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{\mathbf{1}}{\mathbf{4}}$.

~~ 31. Let $A, B, C, D$ be the events that 1 st, $2 nd, 3 rd$ and 4 th shots hits the plane respectively. Then,

$P(A)=0.4, P(B)=0.3, P(C)=0.2, P(D)=0.1$

$\Rightarrow \quad P(not A)=1-P(A), P(not B)=1-P(B)$

$=1-0.4=0.6 \quad=1-0.3=0.7$

$P(not C)=1-P(C), \quad P(not D)=1-P(D)$

$=1-0.2=0.8$

$=1-0.1=0.9$

$\therefore \quad P($ None of the shots hits the plane)

$ \begin{aligned} & =P(not A) \times P(not B) \times P(not C) \times P(D) \\ & =0.6 \times 0.7 \times 0.8 \times 0.9=0.3024 \end{aligned} $

(As $A, B, C$ and $D$ are independent events)

$P$ (At least one shot hits the plane)

$=1-P($ None hits the plane $)$

$=1-0.3024=\mathbf{0 . 6 9 7 6}$.

~~ 32. A red ball can be selected in two mutually exclusive ways.

(i) Selecting bag I and then drawing a red ball from it

(ii) Selecting bag II and them drawing a red ball from it

$\therefore P($ red ball $)=P($ Selecting bag $I) \times P($ Red ball from bag I) $+($ selecting bag II $) \times P($ Red ball from bag II $)$

$=\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{2}{6}=\frac{2}{7}+\frac{1}{6}=\frac{\mathbf{1 9}}{\mathbf{4 2}}$.

~~ 33. The three boxes $B_1, B_2$ and $B_2$ contain the different coloured balls as follows :

	Blue	Red
Box 1	3	1
Box 2	2	2
Box 3	1	3

There can be three mutually exclusive cases of drawing 2

	Box 1	Box 2	Box 3
Case I	1 Blue	1 Blue	1 Red
Case II	1 Blue	1 Red	1 Blue
Case III	1 Red	1 Blue	1 Blue
$\therefore P($ Drawing 2 blue and 1 red ball $)$

blue balls and 1 red balls in the ways as given : $=\frac{3}{4} \times \frac{2}{4} \times \frac{3}{4}+\frac{3}{4} \times \frac{2}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{2}{4} \times \frac{1}{4}$

$=\frac{18}{64}+\frac{6}{64}+\frac{2}{64}=\frac{26}{64}=\frac{\mathbf{1 3}}{\mathbf{3 2}}$.

~~ 34. $P$ (Drawing of two balls of different colours from one of the bags)

$=P($ choosing the 1 st bag $) \times P($ Drawing 1 green out 5 green and 1 out of 11 blue balls) $+P$ (choosing the 2 nd bag) $\times P($ Drawing 1 green out of 3 green balls and 1 out of 7 blue balls)

$P($ choosing 1 st bag or 2 nd bag $)=\frac{1}{2}(\because$ There are 2 bags $)$

$P$ (choosing 1 green and 1 blue ball from bag 1 )

$=\frac{{ }^{5} C_1 \times{ }^{11} C_1}{{ }^{16} C_2}=\frac{5 \times 11 \times 2}{16 \times 15}=\frac{11}{24}$

$P($ choosing 1 green and 1 blue ball from bag 2 )

$=\frac{{ }^{3} C_1 \times{ }^{7} C_1}{{ }^{10} C_2}=\frac{3 \times 7 \times 2}{10 \times 9}=\frac{7}{15}$

$\therefore$ Required probability $=\frac{1}{2} \times \frac{11}{24}+\frac{1}{2} \times \frac{7}{15}$

$ =\frac{11}{48}+\frac{7}{30}=\frac{55+56}{240}=\frac{\mathbf{1 1 1}}{\mathbf{2 4 0}} . $

~~ 35. $P$ (At least one junior professor is selected)

$=P($ Selecting 1 Junior $) \times P($ Selecting 2 Seniors $)$

$+P($ Selecting 2 Junior $) \times P($ Selecting 1 Senior $)$

$+P($ Selecting all 3 Juniors $)$

$\therefore$ Required probability $=\frac{{ }^{4} C_1 \times{ }^{6} C_2}{{ }^{10} C_3}+\frac{{ }^{4} C_2 \times{ }^{6} C_1}{{ }^{10} C_3}+\frac{{ }^{4} C_3}{{ }^{10} C_3}$

$=\frac{1}{2}+\frac{3}{10}+\frac{1}{30}=\frac{15+9+1}{30}=\frac{25}{30}=\frac{\mathbf{5}}{\mathbf{6}}$.

~~ 36. $P(X)=\frac{26}{52}+\frac{4}{52}-\frac{2}{52}=\frac{28}{52}$

( $\because$ There are 26 black cards, 4 kings and 2 black kings)

$P(Y)=\frac{13}{52}+\frac{13}{52}+\frac{4}{52}-\frac{1}{52}-\frac{1}{52}=\frac{28}{52}$

( $\because$ There are 13 clubs, 13 hearts and 4 jacks, 1 jack of clubs and 1 jack of hearts)

$P(Z)=\frac{4}{52}+\frac{13}{52}+\frac{4}{52}-\frac{1}{52}-\frac{1}{52}=\frac{19}{52}$

( $\because$ There are 4 aces, 13 diamonds, 4 queens, 1 ace of diamond, 1 queen of diamond)

$\therefore \quad \boldsymbol{P}(X)=\boldsymbol{P}(\boldsymbol{Y})>\boldsymbol{P}(Z)$.

~~ 37. $P($ Particular number comes on the dice $)=\frac{1}{6}$

( $\because$ there are in all 6 numbers)

$P($ particular number does not come on the dice $)=1-\frac{1}{6}=\frac{5}{6}$

As there are 3 dices, so,

$P($ Picked number does not come in any of dice $)=(\frac{5}{6})^{3}$

$P($ You lose money $)=(\frac{5}{6})^{3}=\frac{125}{216} \approx 0.58$

$\therefore \quad P($ Winning $)=1-0.58=\mathbf{0 . 4 2}$

~~ 38. Given, odds against Event $1=5: 2$

$\Rightarrow \quad P($ Event 1 not happening $)=\frac{5}{5+2}=\frac{5}{7}$

Odds in favour of Event $2=6: 5$

$\Rightarrow \quad P($ Event 2 happens $)=\frac{6}{6+5}=\frac{6}{11}$

$\Rightarrow \quad P($ Event 2 not happening $)=1-\frac{6}{11}=\frac{5}{11}$

$\therefore \quad P($ None of the events happen $)=\frac{5}{7} \times \frac{5}{11}=\frac{25}{77}$

( $\because$ Both event are independent)

$\Rightarrow \quad P($ At least one event happens $)=1-\frac{25}{77}=\frac{\mathbf{5 2}}{77}$.

~~ 39. $P$ (At least 4 vessels arrive safely)

$=P($ exactly four vessels arrive safely $)$

$+P$ (All five vessels arrive safely)

$={ }^{5} C_4 \times(\frac{9}{10})^{4} \times(\frac{1}{10})+(\frac{9}{10})^{5}$

$=\frac{5}{10} \times(\frac{9}{10})^{4}+(\frac{9}{10})^{5}=(\frac{\mathbf{9}}{\mathbf{1 0}})^{4}(\frac{\mathbf{5}}{\mathbf{1 0}}+\frac{\mathbf{9}}{\mathbf{1 0}})=\frac{\mathbf{1 4} \times \mathbf{9}^{4}}{\mathbf{1 0}^{5}}$.

~~ 40. Total number of boys and girls $=2 n+2 n=4 n$

Since, there are two equal batches, each batch has $2 n$ members

$\therefore$ Let $S$ (Sample space): Selecting one batch out of 2

$\Rightarrow S$ : Selecting $2 n$ members out of $4 n$ members.

$\Rightarrow n(S)={ }^{4 n} C _{2 n}$

If each batch has to have equal number of boys and girls, each batch should have $n$ boys and $n$ girls.

Let $E$ : Event that each batch has ’ $n$ ’ boys and ’ $n$ ’ girls

$\Rightarrow n(E)={ }^{2 n} C_n \times{ }^{2 n} C_n=({ }^{2 n} C_n)^{2}$

$\therefore \quad$ Required probability $=\frac{n(E)}{n(S)}=\frac{({ }^{2 n} C_n)^{2}}{{ }^{4 n} C _{2 n}}$.

~~ 41. The box contains 100 balls numbered from 1 to 100. Therefore, there are 50 even and 50 odd numbered balls. The sum of the three numbers drawn will be odd, if all three are odd or one is even and 2 are odd.

$\therefore$ Required probability $=P($ odd $) \times P($ odd $) \times P($ odd $)$

$+P($ even $) \times P($ odd $) \times P($ odd $)$

$ \begin{aligned} & =\frac{{ }^{50} C_1}{{ }^{100} C_1} \times \frac{{ }^{50} C_1}{{ }^{100} C_1} \times \frac{{ }^{50} C_1}{{ }^{100} C_1}+\frac{{ }^{50} C_1}{{ }^{100} C_1} \times \frac{{ }^{50} C_1}{{ }^{100} C_1}+\frac{{ }^{50} C_1}{{ }^{100} C_1} \\ & =2 \times \frac{{ }^{50} C_1 \times{ }^{50} C_1 \times{ }^{50} C_1}{{ }^{100} C_1 \times{ }^{100} C_1 \times{ }^{100} C_1} \\ & =2 \times \frac{50 \times 50 \times 50}{100 \times 100 \times 100}=\frac{\mathbf{1}}{\mathbf{4}} . \end{aligned} $

~~ 42. Let $S$ be the sample space.

Then $n(S)=$ Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon

$={ }^{6} C_3=\frac{6 \times 5 \times 4}{3 \times 2}=20$.

Let $A$ : Event that the selected three vertices form an equilateral triangle.

Then $n(A)=2$

(As only two equilateral triangles are formed from the vertices of a regular hexagon)

$\therefore \quad$ Required probability $=\frac{n(A)}{n(S)}=\frac{2}{20}=\frac{\mathbf{1}}{\mathbf{1 0}}$.

~~ 43. Total number of ways three die can be rolled

$ \begin{aligned} & =6 \times 6 \times 6 \\ & =216 \end{aligned} $

A larger number than the previous number can be got in the three throws as

$(1,2,3),(1,2,4),(1,2,5)(1,2,6),(1,3,4),(1,3,5)$

$(1,3,6),(1,4,5)(1,4,6)(1,5,6)(2,3,4),(2,3,5),(2,3,6)$

$(2,4,5),(2,4,6),(2,5,6),(3,4,5),(3,4,6)(3,5,6),(4,5,6)$.

$\therefore$ Total number of favourable cases $=20$

$\therefore$ Required probability $=\frac{20}{216}=\frac{\mathbf{5}}{\mathbf{5 4}}$

~~ 44. In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn’t matter.

$\therefore$ Required probability

$ ={ }^{9} C_3(\frac{1}{6})^{3} \times(\frac{5}{6})^{6} \times \frac{1}{6} \times \underbrace{1 \times 1 \times 1 \ldots . . \times 1} _{10 \text{ times }}=\frac{\mathbf{8 4} \times \mathbf{5}^{\mathbf{6}}}{\mathbf{6}^{\mathbf{1 0}}} $

~~ 45. Let $W$ denote the event of drawing a white ball at any draw and $B$ that of drawing a black ball.

Then, $\quad P(W)=\frac{a}{a+b}, P(B)=\frac{b}{a+b}$

$\therefore \quad P(A$ wins the game $)=P(W$ or $B B W$ or $B B B B W$ or ……)

$ \begin{aligned} & =P(W)+P(B) \cdot P(B) \cdot P(W)+P(B) \cdot P(B) \cdot P(B) . \\ & =P(W)(1+(P(B))^{2}+(P(B))^{4}+\ldots) . . . P \\ & =\frac{P(W)}{1-(P(B))^{2}}=\frac{\frac{a}{a+b}}{1-\frac{b^{2}}{(a+b)^{2}}}=\frac{a(a+b)}{a^{2}+2 a b}=\frac{a+b}{a+2 b} \end{aligned} $

$\therefore \quad P(B$ wins the game $)=1-P(A$ wins the game $)$ $=1-\frac{(a+b)}{(a+2 b)}=\frac{b}{a+2 b}$

According to the given condition,

$ \frac{a+b}{a+2 b}=3 \cdot \frac{b}{a+2 b} \Rightarrow a=2 b \Rightarrow a: b=\mathbf{2}: \mathbf{1} \text{. } $

~~ 46. Total number of determinants that can be formed using 0 and $1=16(4 \times 4)$ The positive determinants are $ \begin{vmatrix} 1 & 0 \\ 1 & 1\end{vmatrix} , \begin{vmatrix} 1 & 0 \\ 0 & 1\end{vmatrix} , \begin{vmatrix} 1 & 1 \\ 0 & 1\end{vmatrix} $, i.e.,

$\therefore$ Required probability $=\frac{\mathbf{3}}{\mathbf{1 6}}$.

~~ 47. In $3 n$ consecutive natural numbers,

(i) $n$ numbers are of the form $3 p$

(ii) $n$ numbers are of the form $3 p+1$

(iii) $n$ numbers are of the form $3 p+2$

For the sum of the chosen 3 numbers to be divisible by 3 , we can proceed in two ways:

(i) Either we can select all the three numbers from any one of the set or

(ii) We can select one number from each set.

$\therefore \quad$ Favourable number of cases $={ }^{n} C_3+{ }^{n} C_3+{ }^{n} C_3+({ }^{n} C_1.$

$\times{ }^{n} C_1 \times{ }^{n} C_1$ )

$=3 \times(\frac{\underline{n}}{\underline{\underline{n-3} \underline{3}}})+n^{3}=3 \times(\frac{n(n-1)(n-2)}{6})+n^{3}$

Total number of selections

$ \begin{aligned} & ={ }^{3 n} C_3=\frac{\underline{3 n}}{|3 n-3| 3}=\frac{3 n(3 n-1)(3 n-2)}{6} \\ & \therefore \quad \text{ Required probability }=\frac{\frac{3 n(n-1)(n-2)}{6}+n^{3}}{\frac{3 n(3 n-1)(3 n-2)}{6}} \\ & =\frac{\frac{3 n[(n-1)(n-2)+2 n^{2}]}{6}}{\frac{3 n(3 n-1)(3 n-2)}{6}} \\ & =\frac{n^{2}-3 n+2+2 n^{2}}{(3 n-1)(3 n-2)}=\frac{\mathbf{3} n^{2}-\mathbf{3}+\mathbf{2}}{(\mathbf{3 n}-\mathbf{1})(\mathbf{3} \boldsymbol{n}-\mathbf{2})} . \end{aligned} $

~~ 48. As $(n+1)$ coins are fair

$ P(\text{ Tossing a tail })=\frac{\frac{n+1}{2}}{2 n+1}=\frac{n+1}{2(2 n+1)} $

$\therefore \quad P($ Tossing a head $)=1-\frac{n+1}{2(2 n+1)}$

Given, $\frac{3 n+1}{4 n+2}=\frac{31}{42}$

$ =\frac{4 n+2-n-1}{2(2 n+1)}=\frac{3 n+1}{4 n+2} $

$\Rightarrow 126 n+42=124 n+62 \Rightarrow 2 n=20 \Rightarrow n=\mathbf{1 0}$.

~~ 49. Probability of guessing a correct answer $=\frac{1}{3}$

Probability of guessing an incorrect answer $=\frac{2}{3}$

$\therefore \quad$ Probability of guessing 4 or more correct answers

$ ={ }^{5} C_4(\frac{1}{3})^{4}(\frac{2}{3})+{ }^{5} C_5(\frac{1}{3})^{5}=5 \times \frac{2}{3^{5}}+\frac{1}{3^{5}}=\frac{\mathbf{1 1}}{\mathbf{3}^{5}} . $

~~ 50. Let $x, y, z$ be the probabilities of happening of events $E_1, E_2$ and $E_3$ respectively. Then,

$\alpha=P(.$ occurrence of $E_1$ only $)=x(1-y)(1-z)$

$\beta=P(.$ occurrence of $E_2$ only $)=(1-x) y(1-z)$

$\gamma=P(.$ occurrence of $E_3$ only $)=(1-x)(1-y) z$

$p=P(.$ not occurrence of $.E_1, E_2, E_3)=(1-x)(1-y)(1-z)$.

$\therefore \quad(\alpha-2 \beta) p=\alpha \beta$

$ \begin{aligned} & \Rightarrow \quad[x(1-y)(1-z)-2(1-x) y(1-z)](1 - x)(1-y)(1-z) \\ & \quad=x(1-y)(1-z)(1-x) y(1-z) \\ & \Rightarrow \quad(1-z)[x(1-y)-2 y(1-x)]=x y(1-z) \\ & \Rightarrow \quad x-x y-2 y+2 x y=x y \\ & \Rightarrow \quad x=2 y \end{aligned} $

Also, $(\beta-3 \gamma) p=2 \beta \gamma$

$ \begin{aligned} & \Rightarrow \quad y=3 z \\ & \therefore \quad x=6 z(\text{ From }(i) \text{ and }(i i)) \\ & \Rightarrow \frac{x}{z}=6 \Rightarrow \frac{P(E_1)}{P(E_3)}=\mathbf{6} \text{. } \end{aligned} $

SELF ASSESSMENT SHEET

Thirty days are in September, April, June and November. Some months are of thirty one days. A month is chosen at random. Then its probability of having exactly three days less than a maximum of 31 day is

(a) $\frac{15}{16}$

(b) 1

(d) None of these

(SNAP 2007)

~~ 2. There are four hotels in a town. If 3 men check into the hotels in a day, then what is the probability that each checks into a different hotel?

(a) $\frac{6}{7}$

(b) $\frac{1}{8}$

(d) $\frac{5}{9}$

(JMET 2011)

~~ 3. A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that either 2 are blue and 2 are red or 2 are green and 2 are black ?

(a) $\frac{3}{7}$

(b) $\frac{3}{91}$

(d) $\frac{36}{1001}$

(MBA CET Mah. 2007)

~~ 4. There are three events $A, B$ and $C$, one of which must and can only happen. If the odds one $8: 3$ against $A, 5: 2$ against $B$, the odds against $C$ must be

(a) $13: 7$

(b) $3: 2$

(d) $43: 77$

(ATMA 2008)

~~ 5. Three letters are randomly selected from the 26 capital letters of the English alphabet. What is the probability that letters $A$ will not be included in the choice ?

(a) $\frac{1}{12}$

(b) $\frac{23}{26}$

(d) $\frac{25}{26}$

(NDA/NA 2009)

~~ 6. $A$ and $B$ are two mutually exclusive and exhaustive events with $P(B)=3 P(A)$. What is the value of $P(\bar{B})$ ?

(a) $\frac{3}{4}$

(b) $\frac{1}{4}$

(d) $\frac{2}{3}$

(NDA/NA 2009)

~~ 7. Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. Then, the probability that the player gets all distinct cards is

(a) ${ }^{52} C _{26} \mid{ }^{104} C _{26}$

(b) $2 \times{ }^{52} C _{26} \mid{ }^{104} C _{26}$

(d) $2^{26} \times{ }^{52} C _{26} \mid{ }^{104} C _{26}$.

(WBJEE 2012)

~~ 8. An article manufactured by a company consists of two parts $X$ and $Y$. In the process of manufacture of the part $X, 9$ out of 100 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part $Y$. Calculate the probability that the assembled product will not be defective.

(a) 0.6485

(b) 0.6565

(d) None of these

~~ 9. If $n$ integers taken at random are multiplied together, then the probability that the last digit of the product is $1,3,7$ or 9 is

(a) $\frac{2^{n}}{5^{n}}$

(b) $\frac{4^{n}-2^{n}}{5^{n}}$

(d) None of these (UPSEE 2013)

~~ 10. A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is

(a) $\frac{1}{45}$

(b) $\frac{13}{90}$

(d) None of these

(AMU 2007)

ANSWERS

$(d)$

~~ 2. (c)

~~ 3. $(d)$

~~ 4. $(c)$

~~ 5. (b)

~~ 6. (b)

~~ 7. $(d)$

~~ 8. $(c)$

~~ 9. (a)

~~ 10. (c)

HINTS AND SOLUTIONS

The month having 3 days less than 31 days has 28 days, i.e, it is the month of February.

$P($ Choosing February $)=\frac{\mathbf{1}}{\mathbf{1 2}}$

~~ 2. Total number of ways of checking in the 4 hotels by 3 men $=4 \times 4 \times 4=\mathbf{4}^{3}$.

Number of ways in which each man checks into a different hotel $=4 \times 3 \times 2$

(As for the 1st person, there are 4 choices, for 2nd remaining 3 , and for the 3rd remaining 2).

$\therefore P($ Each person checks into a different hotel $)$

$=\frac{4 \times 3 \times 2}{4 \times 4 \times 4}=\frac{\mathbf{3}}{\mathbf{8}}$.

~~ 3. Required probability $=\frac{P(2 \text{ blue balls }) \times P(2 \text{ red balls })}{P(4 \text{ balls out of } 14 \text{ ball })}$

$ \begin{aligned} & +\frac{P(2 \text{ green balls }) \times P(2 \text{ black balls })}{P(4 \text{ out of } 14 \text{ balls })} \\ = & \frac{{ }^{2} C_2 \times{ }^{4} C_2}{{ }^{14} C_4}+\frac{{ }^{3} C_2 \times{ }^{5} C_2}{{ }^{14} C_4} \\ = & \frac{\frac{4 \times 3}{2}}{\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}}+\frac{\frac{3 \times 5 \times 4}{2}}{\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2}} \\ = & \frac{6}{1001}+\frac{30}{1001}=\frac{\mathbf{3 6}}{\mathbf{1 0 0 1}} \end{aligned} $

~~ 4. Given, odds against $A=8: 3$

$\Rightarrow P(not A)=\frac{8}{8+3}=\frac{8}{11} \Rightarrow P(A$ happens $)=\frac{3}{11}$

Odds against $B=5: 2$

$\Rightarrow \quad P(not B)=\frac{5}{5+2}=\frac{5}{7}$

$\Rightarrow \quad P(B$ happens $)=\frac{2}{7}$

As out of $A, B$ and $C$, one and only one can happen, so

$P(A)+P(B)+P(C)=1$

$P(C)=1-(P(A)+P(B))$

$=1-(\frac{3}{11}+\frac{2}{7})=1-(\frac{21+22}{77})=1-\frac{43}{77}=\frac{34}{77}$

$P(not C)=1-\frac{34}{77}=\frac{43}{77}$

odds against $C=\frac{P(\text{ not } c)}{P(c)}=\frac{43 / 77}{34 / 77}=\mathbf{4 3}: \mathbf{3 4}$.

~~ 5. Total number of ways in which 3 letters can be selected from 26 letters

$={ }^{26} C_3$.

If $A$ is not to be included in the choice, there are 25 letters left, so number of ways in which 3 letters can be selected without including

$A={ }^{25} C_3$.

$\therefore$ Required probability $=\frac{{ }^{25} C_3}{{ }^{26} C_3}=\frac{25 \times 24 \times 23}{26 \times 25 \times 24}=\frac{\mathbf{2 3}}{\mathbf{2 6}}$.

~~ 6. Since $A$ and $B$ are mutually exclusive and exhaustive events, therefore,

$P(A \cap B)=0, P(A \cup B)=1$

We know that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\Rightarrow 1=P(A)+3 P(A)$

(Given $P(B)=3 P(A)$ )

$\Rightarrow 4 P(A)=1 \Rightarrow P(A)=\frac{1}{4}$

$\therefore \quad P(B)=1-\frac{1}{4}=\frac{3}{4} \quad(\because P(A)+P(B)=1)$

$\Rightarrow \quad P(\bar{B})=1-P(B)=1-\frac{3}{4}=\frac{\mathbf{1}}{\mathbf{4}}$.

~~ 7. Since there are 52 distinct cards in a deck and each distinct card is 2 in number.

$\therefore 2$ decks will also contain only 52 distinct cards, two each.

$\therefore$ Probability that the player gets all distinct cards

$ =\frac{{ }^{52} C _{26} \times 2^{26}}{{ }^{104} C _{26}} . $

~~ 8. Required probability $=P(X$ not defective and $Y$ not defective $)$

$=P(\bar{X}) \times P(\bar{Y})$

$=(1-P(X))(1-P(Y))$

$=(1-\frac{9}{100})(1-\frac{5}{100})$

$=\frac{91}{100} \times \frac{95}{100}=\frac{8645}{10000}=\mathbf{0 . 8 6 4 5}$.

~~ 9. In any number the last digits can $0,1,2,3,4,5,6,7,8$ and 9. Therefore, the last digit of each number can be chosen in 10 ways.

$\therefore \quad$ Exhaustive number of ways $=10^{n}$. If the last digit be $1,3,7$ or 9 , none of the numbers can be even or end in 0 or 5. That is, there is a choice of only 4 digits for the last digit of each of these $n$ numbers.

So, favourable number of ways $=4^{n}$

$\therefore$ Required probability $=\frac{4^{n}}{10^{n}}=\frac{\mathbf{2}^{n}}{\mathbf{5}^{n}}$.

~~ 10. In the words ‘ASSISTANT’ and STATISTICS’, ’ $N$ ’ and ’ $C$ ’ are the uncommon letters.

The same letters are $A, I, S$ and $T$ whose numbers in both the words are as follows:

STATISTICS $arrow 1 \quad 2 \quad 3 \quad 3$

Probability of choosing ’ $A$ ’ $=\frac{{ }^{2} C_1}{{ }^{9} C_1} \times \frac{{ }^{1} C_1}{{ }^{10} C_1}=\frac{2}{9} \times \frac{1}{10}=\frac{1}{45}$

Probability of choosing ’ $I$ ’ $=\frac{1}{{ }^{9} C_1} \times \frac{{ }^{2} C_1}{{ }^{10} C_1}=\frac{1}{9} \times \frac{2}{10}=\frac{1}{45}$

Probability of choosing ’ $S$ ’ $=\frac{{ }^{3} C_1}{{ }^{9} C_1} \times \frac{{ }^{3} C_1}{{ }^{10} C_1}=\frac{3}{9} \times \frac{3}{10}=\frac{1}{10}$

Probability of choosing ’ $T$ ’ $=\frac{{ }^{2} C_1}{{ }^{9} C_1} \times \frac{{ }^{3} C_1}{{ }^{10} C_1}=\frac{2}{9} \times \frac{3}{10}=\frac{1}{15}$

$\therefore \quad$ Required probability $=\frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}=\frac{\mathbf{1 9}}{\mathbf{9 0}}$.

10 Trigonometry

KEY FACTS

I. The common systems of measuring angles are:

(a) Sexagesimal System: Here, one complete revolution is divided into 360 equal parts, each called a degree, a degree is divided into sixty equal parts, each called a second. Thus,

One complete revolution $=360^{\circ}$

$ \begin{aligned} & 1^{\circ}=60^{\prime}(\text{ minutes }) \\ & 1^{\prime}=60^{\prime \prime}(\text{ seconds }) \end{aligned} $

(b) Circular System: The unit of measuring an angle in this system is radian. A radian is the measure of the central angle subtended by an arc equal in length to the radius of the circle. $\mathbf{1}$ radian is written as $\mathbf{1}^{c}$.

II. $\pi$ radians $=180$ degrees

(a) To convert an angle in radians to its equivalent in degrees, multiply the number of radians by $\frac{180^{\circ}}{\pi}$.

$\therefore \frac{\pi}{3}$ radians $=\frac{\pi}{3} \times \frac{180^{\circ}}{\pi}=60^{\circ}$

(b) To convert an angle in degrees to its equivalent in radians, multiply the number of degrees by $\frac{\pi}{180^{\circ}}$. $\therefore 45^{\circ}=\frac{\pi}{180^{\circ}} \times 45^{\circ}=\frac{\pi}{4} rad$.

III. Trigonometrical functions (or Ratios)

In a right angled triangle $O M P$, where

$\angle O M P=90^{\circ}, \angle P O M=\alpha$, base $O M=x$, perpendicular $P M=y$ and hypotenuse $O P=r$,

(a) $\boldsymbol{\operatorname { s i n }} \boldsymbol{\alpha}=\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}=\frac{\boldsymbol{y}}{\boldsymbol{r}}$

(b) $\boldsymbol{\operatorname { c o s } \alpha} \alpha=\frac{\text{ Base }}{\text{ Hypotenuse }}=\frac{\boldsymbol{x}}{\boldsymbol{r}}$

(c) $\boldsymbol{\operatorname { t a n }} \boldsymbol{\alpha}=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{\boldsymbol{y}}{\boldsymbol{x}}$

(d) $\boldsymbol{\operatorname { c o s e c }} \alpha=\frac{1}{\sin \alpha}=\frac{\text{ Hypotenuse }}{\text{ Pr ependicular }}=\frac{\boldsymbol{r}}{\boldsymbol{y}}$

(e) $\sec \boldsymbol{\alpha}=\frac{1}{\cos \alpha}=\frac{\text{ Hypotenuse }}{\text{ Base }}=\frac{\boldsymbol{r}}{\boldsymbol{y}}$

(f) $\cot \alpha=\frac{1}{\tan \alpha}=\frac{\text{ Base }}{\text{ Perpendicular }}=\frac{\boldsymbol{x}}{\boldsymbol{y}}$

Note:

In the right $\angle d \triangle A B C$, it can be easily seen that,

(a) $\sin A=\frac{a}{c}=\cos B$

(b) $\cos A=\frac{b}{c}=\sin B$

(d) $\tan B=\frac{b}{a}=\cot A$

~~ 2. $(\sin \theta)^{2}=\sin ^{2} \theta,(\cos \theta)^{3}=\cos ^{3} \theta$ read as sine square $\theta$ and cos cube $\theta$ respectively.

IV. Fundamental relations between trigonometric functions

1. Reciprocal relations

(i) $cosec \theta=\frac{1}{\sin \theta} \Rightarrow cosec \theta \times \sin \theta=1$ (ii) $\sec \theta=\frac{1}{\cos \theta} \Rightarrow \sec \theta \times \cos \theta=1$ (iii) $\cot \theta=\frac{1}{\tan \theta} \Rightarrow \cot \theta \times \tan \theta=1$

2. Quotient relations

(i) $\tan \theta=\frac{\sin \theta}{\cos \theta}$ (ii) $\cot \theta=\frac{\cos \theta}{\sin \theta}$

~~ 3. Square relations (Pythagorean Identities)

(i) $\sin ^{2} \theta+\cos ^{2} \theta=1$

(ii) $\sec ^{2} \theta=1+\tan ^{2} \theta$ or $\sec ^{2} \theta-\tan ^{2} \theta=1$ or $\sec ^{2} \theta-1=\tan ^{2} \theta$

(iii) $cosec^{2} \theta=1+\cot ^{2} \theta$ or $cosec^{2} \theta-\cot ^{2} \theta=1$ or $cosec^{2} \theta-1=\cot ^{2} \theta$

V. Signs of trigonometric functions

The sign of a particular $t$-function is any quadrant can be remembered by

the phrase. “All $-\sin -\tan -\operatorname{cos”}$ (Add Sugar To Coffee).

quadrant along with its reciprocal is positive in that quadrant and the rest are negative.

VI. Values of trigonometric ratios of some standard angles

$\boldsymbol{t}$-function	$\mathbf{0}^{\circ}$	$\mathbf{3 0}^{\circ}$ or $\frac{\boldsymbol{\pi}}{\mathbf{3}}$	$\mathbf{4 5}^{\circ}$ or $\frac{\boldsymbol{\pi}}{\mathbf{4}}$	$\mathbf{6 0}^{\circ}$ or $\frac{\boldsymbol{\pi}}{\mathbf{6}}$	$\mathbf{9 0}^{\circ}$ or $\frac{\boldsymbol{\pi}}{\mathbf{2}}$	$\mathbf{1 8 0}^{\circ}$ or $\pi$
$\boldsymbol{\operatorname { s i n }} \theta$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1	0
$\boldsymbol{\operatorname { c o s }} \theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0	-1
$\boldsymbol{\operatorname { t a n }} \theta$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	undefined	0
$cosec \theta$	undefined	2	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	1	undefined
$\sec \theta$	1	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	2	undefined	-1
$\cot \theta$	undefined	$\sqrt{3}$	1	$\frac{1}{\sqrt{3}}$	0	undefined

VII. Trigonometric functions of angles of any magnitude

The following formulae are helpful in reducing functions of any larger angle to function of smaller angles.

	$\mathbf{- \theta}$	$\mathbf{9 0}^{\circ}-\boldsymbol{\theta}$	$\mathbf{9 0}^{\circ}+\boldsymbol{\theta}$	$\mathbf{1 8 0}^{\circ}-\boldsymbol{\theta}$	$\mathbf{1 8 0}^{\circ}+\boldsymbol{\theta}$	$\mathbf{2 7 0}^{\circ}-\boldsymbol{\theta}$	$\mathbf{2 7 0}^{\circ}+\boldsymbol{\theta}$	$\mathbf{3 6 0}^{\circ}-\boldsymbol{\theta}$	$\mathbf{3 6 0}^{\circ}+\boldsymbol{\theta}$
$\sin$	$-\sin$	$\cos \theta$	$\cos \theta$	$\sin \theta$	$-\sin \theta$	$-\cos \theta$	$-\cos \theta$	$-\sin \theta$	$\sin \theta$
$\cos$	$\cos \theta$	$\sin \theta$	$-\sin \theta$	$-\cos \theta$	$-\cos \theta$	$-\sin \theta$	$\sin \theta$	$\cos \theta$	$\cos \theta$
$\tan$	$-\tan \theta$	$\cot \theta$	$-\cot \theta$	$-\tan \theta$	$\tan \theta$	$\cot \theta$	$-\cot \theta$	$-\tan \theta$	$\tan \theta$
$cosec$	$-cosec \theta$	$\sec \theta$	$\sec \theta$	$cosec \theta$	$-cosec \theta$	$-\sec \theta$	$-\sec \theta$	$-cosec \theta$	$cosec \theta$
$\sec$	$\sec \theta$	$cosec \theta$	$-cosec \theta$	$-\sec \theta$	$-\sec \theta$	$-cosec \theta$	$cosec \theta$	$\sec \theta$	$\sec \theta$
$\cot$	$-\cot \theta$	$\tan \theta$	$-\tan \theta$	$-\cot \theta$	$\cot \theta$	$\tan \theta$	$-\tan \theta$	$\cot \theta$	$\cot \theta$

	$2 n \pi-\theta$ or $n .360^{\circ}-\theta$	$2 n \pi+\theta$ or $n \cdot 360^{\circ}+\boldsymbol{\theta}$
$\sin$	$-\sin \theta$	$\sin \theta$
$\cos$	$\cos \theta$	$\cos \theta$
$\tan$	$-\tan \theta$	$\tan \theta$

VIII. Some important formulae

(a) Sum Formulae (i) $\sin (A+B)=\sin A \cos B+\cos A \sin B$ (ii) $\cos (A+B)=\cos A \cos B+\sin A \sin B$ (iii) $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$ (iv) $\cot (A+B)=\frac{\cot A \cot B-1}{\cot B+\cot A}$

(v) $\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}$

(b) Difference formulae (i) $\sin (A-B)=\sin A \cos B-\cos A \sin B$ (ii) $\cos (A-B)=\cos A \cos B+\sin A \sin B$ (iii) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$ (iv) $\cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A}$

$\cos (A+B) \cos (A-B)=\cos ^{2} A-\sin ^{2} B$

(d) $\sin C+\sin D=2 \sin (\frac{C+D}{2}) \cos (\frac{C-D}{2})$

$\sin C-\sin D=2 \cos (\frac{C+D}{2}) \sin (\frac{C-D}{2})$

$\cos C+\cos D=2 \cos (\frac{C+D}{2}) \cos (\frac{C-D}{2})$

$\cos C-\cos D=2 \sin (\frac{C+D}{2}) \sin (\frac{D-C}{2})$

(e) Product formulae

$2 \sin A \cos B=\sin (A+B)+\sin (A-B)$

$2 \cos A \sin B=\sin (A+B)+\sin (A-B)$

$2 \cos A \cos B=\cos (A+B)+\cos (A-B)$

$2 \sin A \sin B=\cos (A-B)-\cos (A+B)$ (f) Double angle formulae

$\sin 2 A=2 \sin A \cos A$ $\sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}$
$\cos 2 A=\cos ^{2} A-\sin ^{2} A$ $\cos 2 A=1-2 \sin ^{2} A$ $\cos 2 A=2 \cos ^{2} A-1$ $\cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$ $2 \sin ^{2} A=1-\cos 2 A$
$2 \cos ^{2} A=1+\cos 2 A$
$\tan ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}=\frac{1-\cos 2 A}{1+\cos 2 A}$ $\Rightarrow \sin ^{2} A=\frac{1-\cos 2 A}{2}$ $\Rightarrow \cos ^{2} A=\frac{1+\cos 2 A}{2}$
$\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$

(g) Half angle formulae

$\sin A=2 \sin A / 2 \cos A / 2$
$\cos A=\cos ^{2} A / 2-\sin ^{2} A / 2 \quad-\tan A=\frac{2 \tan A / 2}{1-\tan ^{2} A / 2}$

$\sin A=\frac{2 \tan A / 2}{1-\tan ^{2} A / 2}$

$\cos A=1-2 \sin ^{2} A / 2$

$\cos A=2 \cos ^{2} A / 2-1$

$\cos A=\frac{1-\tan ^{2} A / 2}{1+\tan ^{2} A / 2}$

$1-\cos A$

$\sin A / 2= \pm \sqrt{\frac{1-\cos A}{2}}$

$\cos A / 2= \pm \sqrt{\frac{1-\cos A}{2}}$

$\tan A / 2= \pm \sqrt{\frac{1-\cos A}{1+\cos A}}$

(h) Triple-angle formulae

$\sin 3 A=3 \sin A-4 \sin ^{3} A \quad \bullet \cos 3 A=4 \cos ^{3} A-3 \cos A \quad \bullet \tan 3 A=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$
$\cot 3 A=\frac{\cot ^{3} A-3 \cot A}{3 \cot ^{2} A-1}$

SOLVED EXAMPLES

Ex. 1 . If $\alpha$ is an acute angle and $\sin \alpha=\sqrt{\frac{x-1}{2 x}}$, then what is $\tan \alpha$ equal to?

(CDS 2010)

Sol. In $\triangle A B C$ where $\angle B=90^{\circ}, \sin \alpha=\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}=\frac{\sqrt{x-1}}{\sqrt{2 x}}=\frac{A B}{A C}$

$\therefore B C^{2}=A C^{2}-A B^{2}=2 x-(x-1)=x+1$

$\Rightarrow B C=\sqrt{x+1}$

$\therefore \quad \tan \alpha=\frac{\text{ Prependicular }}{\text{ Base }}=\frac{A B}{B C}=\frac{\sqrt{\boldsymbol{x}-\mathbf{1}}}{\sqrt{\boldsymbol{x}+\mathbf{1}}}$.

Ex. 2 . What is the value of the expression

$3 \tan ^{2} \frac{\pi}{6}+\frac{4}{3} \cos ^{2} \frac{\pi}{6}-\frac{1}{2} \cot ^{3} \frac{\pi}{4}-\frac{2}{3} \sin ^{2} \frac{\pi}{3}+\frac{1}{8} \sec ^{4} \frac{\pi}{3} ?$

Sol. $3 \tan ^{2} \frac{\pi}{6}+\frac{4}{3} \cos ^{2} \frac{\pi}{6}-\frac{1}{2} \cot ^{3} \frac{\pi}{4}-\frac{2}{3} \sin ^{2} \frac{\pi}{3}+\frac{1}{8} \sec ^{4} \frac{\pi}{3}$

$=3 \tan ^{2} 30^{\circ}+\frac{4}{3} \cos ^{2} 30^{\circ}-\frac{1}{2} \cot ^{3} 45^{\circ}-\frac{2}{3} \sin ^{2} 60^{\circ}+\frac{1}{8} \sec ^{4} 60^{\circ}$

$=3 \times(\frac{1}{\sqrt{3}})^{2}+\frac{4}{3} \times(\frac{\sqrt{3}}{2})^{2}-\frac{1}{2} \times(1)^{3}-\frac{2}{3} \times(\frac{\sqrt{3}}{2})^{2}+\frac{1}{8} \times(2)^{4}$

$=3 \times \frac{1}{3}+\frac{4}{3} \times \frac{3}{4}-\frac{1}{2}-\frac{2}{3} \times \frac{3}{4}+\frac{1}{8} \times 16=1+1-\frac{1}{2}-\frac{1}{2}+2=3$.

Ex. 3 . If $A+B=90^{\circ}$, then what is the value of $\sqrt{\sin A \sec B-\sin A \cos B}$ ?

Sol. Given $A+B=90^{\circ}$

$ \begin{aligned} \therefore \quad \sqrt{\sin A \sec B-\sin A \cos B} & =\sqrt{\sin A \sec (90^{\circ}-A)-\sin A \cos (90^{\circ}-A)} \\ & =\sqrt{\sin A \cdot cosec A-\sin A \cdot \sin A}=\sqrt{1-\sin ^{2} A}=\sqrt{\cos ^{2} A}=\cos A \end{aligned} $

Ex. 4 . If $\sin (A+B+C)=\mathbf{1}, \tan (A-B)=\frac{1}{\sqrt{3}}$ and $\sec (A+C)=\mathbf{2}$, then find the values of the angles $A, B$ and $C$ in degrees.

Sol. $\sin (A+B+C)=1 \Rightarrow \sin (A+B+C)=\sin 90^{\circ} \Rightarrow A+B+C=90^{\circ}$

$\tan (A-B)=\frac{1}{\sqrt{3}} \Rightarrow \tan (A-B)=\tan 30^{\circ} \Rightarrow A-B=30^{\circ}$

$\sec (A+C)=2 \Rightarrow \sec (A+C)=\sec 60^{\circ} \Rightarrow A+C=60^{\circ}$

$Eq(i)-Eq(i i i) \Rightarrow B=30^{\circ}$

$\therefore$ From eqn (ii), $A-30^{\circ}=30^{\circ} \Rightarrow A=60^{\circ}$

$\therefore$ From eqn $(i), 60^{\circ}+30^{\circ}+C=90^{\circ} \Rightarrow C=0^{\circ} \Rightarrow A=\mathbf{6 0}^{\circ}, B=\mathbf{3 0}^{\circ}, C=\mathbf{0}^{\circ}$.

Ex. 5 . If $\sec \theta=\sqrt{2}$ and $\frac{3 \pi}{2}<\theta<2 \pi$, find the value of $\frac{1+\tan \theta+cosec \theta}{1+\cot \theta-cosec \theta}$.

Sol. $\sec \theta=\sqrt{2} \Rightarrow \cos \theta=\frac{1}{\sqrt{2}}$

$\therefore \sin \theta= \pm \sqrt{1-\cos ^{2} \theta}= \pm \sqrt{1-\frac{1}{2}}= \pm \frac{1}{\sqrt{2}}$

Since $\theta$ lies in the fourth quadrant, so $\sin \theta$ is $-v e$ and $\cos \theta$ is $+v e$.

$\therefore \sin \theta=-\frac{1}{\sqrt{2}}, cosec \theta=-\sqrt{2}$

$\tan \theta=\frac{\sin \theta}{\cos \theta}=-\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{1}=-1 \Rightarrow \cot \theta=-1$

$\therefore \frac{1+\tan \theta+cosec \theta}{1+\cot \theta-cosec \theta}=\frac{1-1-\sqrt{2}}{1-1+\sqrt{2}}=\mathbf{- 1}$.

Ex. 6 . What is the value of $\cot (-870^{\circ})$ ?

Sol. $\cot (-870^{\circ})=-\cot 870^{\circ}$

$ =-\cot (720^{\circ}+150^{\circ})=-\cot (2 \times 360^{\circ}+150^{\circ}) $

$ \begin{aligned} & =-\cot 150^{\circ}=-\cot (90^{\circ}+60^{\circ}) \\ & =-(-\tan 60^{\circ}) \\ & =\tan 60^{\circ}=\sqrt{3} . \end{aligned} $

$(\because \cot (n .360^{\circ}+\theta)=\cot \theta, n \in N)$

$(\because \cot (90^{\circ}+\theta)=-\tan \theta)$

Ex. 7 . What is the value of $\cos 480^{\circ} \sin 150^{\circ}+\sin 600^{\circ} \cos 390^{\circ}$ ?

(Kerala PET 2006)

Sol. $\cos 480^{\circ} \sin 150^{\circ}+\sin 600^{\circ} \cos 390^{\circ}$

$ \begin{aligned} & =\cos (360^{\circ}+120^{\circ}) \sin (180^{\circ}-30^{\circ})+\sin (2 \times 360^{\circ}-120^{\circ}) \cos (360^{\circ}+30^{\circ}) \\ & =\cos 120^{\circ} \sin 30^{\circ}-\sin (120^{\circ}) \cos 30^{\circ} \\ & =-\cos 60^{\circ} \sin 30^{\circ}-\sin (180^{\circ}-60^{\circ}) \cos 30^{\circ} \\ & =-\cos 60^{\circ} \sin 30^{\circ}-\sin 60^{\circ} \cos 30^{\circ} \\ & =-(\sin 30^{\circ} \cos 60^{\circ}+\sin 60^{\circ} \cos 30^{\circ}) \\ & =-\sin (30^{\circ}+60^{\circ})(\because \sin (A+B)=\sin A \cos B+\cos A \sin B) \\ & =-\sin 90^{\circ}=-1 . \end{aligned} \quad \begin{bmatrix} \because \cos (360^{\circ}+\theta)=\cos \theta \\ \sin (n .360^{\circ}-\theta)=-\sin \theta, n \in N \\ \sin (180^{\circ}-\theta)=\sin \theta \\ \sin (180^{\circ}+\theta)=-\sin \theta \\ \cos 120^{\circ}=\cos (180^{\circ}-60^{\circ})=-\cos 60^{\circ} \end{bmatrix} $

Ex. 8 . If $x=y \cos (\frac{2 \pi}{3})=z \cos (\frac{4 \pi}{3})$, then what is $x y+y z+z x$ equal to?

(NDA/NA 2011)

Sol. $x=y \cos 120^{\circ}=z \cos 240^{\circ}$

$ \begin{aligned} & \Rightarrow x=y \cos (180^{\circ}-60^{\circ})=z \cos (180^{\circ}+60^{\circ}) \\ & \Rightarrow x=-y \cos 60^{\circ}=-z \cos 60^{\circ}(\because \cos (180^{\circ}-\theta)=-\cos \theta=\cos (180^{\circ}+\theta)) \\ & \Rightarrow x=-\frac{1}{2} y=-\frac{z}{2} \Rightarrow 2 x=-y=-z \\ & \Rightarrow \frac{x}{\frac{1}{2}}=\frac{y}{(-1)}=\frac{z}{(-1)}=k \Rightarrow x=\frac{k}{2}, y=-k, z=-k \\ & \therefore x y+y z+z x=(\frac{k}{2})(-k)+(-k)(-k)+(-k)(\frac{k}{2})=\frac{-k^{2}}{2}+k^{2}-\frac{k^{2}}{2}=\mathbf{0} . \end{aligned} $

Ex. 9 . If $A$ lies in the third quadrant and $3 \tan A-4=0$, then what is the value of $5 \sin 2 A+3 \sin A+4 \cos A$ ?

Sol. $3 \tan A-4=0$

$ \begin{aligned} & \Rightarrow 3 \tan A=4 \Rightarrow \tan A=\frac{4}{3} \\ & \Rightarrow \tan A=\frac{4}{3} \Rightarrow \frac{B C}{A B}=\frac{4}{3} \Rightarrow A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{9+16}=\sqrt{25}=\mathbf{5 .} \end{aligned} $

$\therefore$ A being in the third quadrant, $\sin A$ and $\cos A$ are negative

So, $\sin A=-\frac{4}{5}$ and $\cos A=-\frac{3}{5}$.

$\therefore$ Given expression $5 \sin 2 A+3 \sin A+4 \cos A$

$ \begin{aligned} & =5 \times 2 \sin A \cos A+3 \sin A+4 \cos A \\ & =10 \times(\frac{-4}{5}) \times(\frac{-3}{5})+3 \times(\frac{-4}{5})+4 \times(\frac{-3}{5}) \\ & =\frac{24}{5}-\frac{12}{5}-\frac{12}{5}=0 . \end{aligned} $

Ex. 10 . Show that $\cos (\frac{3 \pi}{2}+x) \cos (2 \pi+x)\lbrace \cot (\frac{3 \pi}{2}-x)+\cot (2 \pi+x)\rbrace =1$.

Sol. $\cos (\frac{3 \pi}{2}+x) \cos (2 \pi+x)\lbrace \cot (\frac{3 \pi}{2}-x)+\cot (2 \pi+x)\rbrace $

$=\cos (270^{\circ}+x) \cos (360^{\circ}+x)\lbrace \cot (270^{\circ}-x)+\cot (360^{\circ}+x)\rbrace $

$=\sin x \cos x\lbrace \tan x+\cot x\rbrace $

$=\sin x \cos x[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}]=\sin x \cos x[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}]=\mathbf{1}$.

$ \begin{bmatrix} \because \cos (360^{\circ}+\theta)=\cos \theta \\ \cot (360^{\circ}+\theta)=\cot \theta \\ \cos (270^{\circ}+\theta)=+\sin \theta \\ \cot (270^{\circ}-\theta)=\tan \theta \end{bmatrix} $

Ex. 11 . What is the value of $\frac{\cos 15^{\circ}+\cos 45^{\circ}}{\cos ^{3} 15^{\circ}+\cos ^{3} 45^{\circ}}$ ?

(NDA/NA 2013)

Sol. $\quad \frac{\cos 15^{\circ}+\cos 45^{\circ}}{\cos ^{3} 15^{\circ}+\cos ^{3} 45^{\circ}}=\frac{\cos 15^{\circ}+\cos 45^{\circ}}{(\cos 15^{\circ}+\cos 45^{\circ})(\cos ^{2} 15^{\circ}+\cos ^{2} 45^{\circ}-\cos 45^{\circ} \cos 15^{\circ})} \quad (\because a^{3}+b^{3}=(a+b)(a^{2}-a b+b^{2}))$

$=\frac{1}{(\cos ^{2} 15^{\circ}+\cos ^{2} 45^{\circ}-\cos 45^{\circ} \cos 15^{\circ})} \\ $

$ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \quad $ alt text

Ex. 12 . What is the value of $\sqrt{3} cosec 20^{\circ}-\sec 20^{\circ}$ ?

Sol. $\sqrt{3} cosec 20^{\circ}-\sec 20^{\circ}$

$=\frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}=\frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$ $=\frac{4}{2 \sin 20^{\circ} \cos 20^{\circ}}(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ})$

$=\frac{4}{\sin 40^{\circ}}(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ})$

(Multiplying numerator and denominator by 4)

$\begin{matrix} =\frac{4}{\sin 40^{\circ}}(\sin (60^{\circ}-20^{\circ})) \\ = & \frac{4}{\sin 40^{\circ}} \cdot \sin 40^{\circ}=4 .\end{matrix} \quad(\because \sin (A-B)=\sin A \cos B-\cos A \sin B)$

Ex. 13 . Given, $3 \sin \theta+4 \cos \theta=5$, then what is $3 \cos \theta-4 \sin \theta$ equal to?

Sol. $3 \sin \theta+4 \cos \theta=5$

$ \begin{aligned} & \Rightarrow(3 \sin \theta+4 \cos \theta)^{2}=25 \Rightarrow 9 \sin ^{2} \theta+16 \cos ^{2} \theta+24 \sin \theta \cos \theta=25 \\ & \Rightarrow 9(1-\cos ^{2} \theta)+16(1-\sin ^{2} \theta)+24 \sin \theta \cos \theta=25 \\ & \Rightarrow 9-9 \cos ^{2} \theta+16-16 \sin ^{2} \theta+24 \sin \theta \cos \theta=25 \\ & \Rightarrow 9 \cos ^{2} \theta+16 \sin ^{2} \theta-24 \sin \theta \cos \theta=0 \\ & \Rightarrow(3 \cos \theta-4 \sin \theta)^{2}=0 \Rightarrow 3 \cos \theta-4 \sin \theta=\mathbf{0} . \end{aligned} \quad(\because \sin ^{2} \theta+\cos ^{2} \theta=1) $

Ex. 14 . If $\tan x=b / a$, then what is the value of $a \cos 2 x+b \sin 2 x$ ?

(UPSEAT 2007, AMU 2002)

Sol. Given $\tan x=b / a$

$ \begin{aligned} a \cos 2 x+b \sin 2 x & =a(\frac{1-\tan ^{2} x}{1+\tan ^{2} x})+b(\frac{2 \tan x}{1+\tan ^{2} x})=\frac{a(1-\tan ^{2} x)+b(2 \tan x)}{(1+\tan ^{2} x)} \\ & =\frac{a(1-b^{2} / a^{2})+b(2 \cdot b / a)}{1+b^{2} / a^{2}}=\frac{a(\frac{a^{2}-b^{2}}{a^{2}})+\frac{2 b^{2}}{a}}{\frac{a^{2}+b^{2}}{a^{2}}} \\ & =\frac{\frac{a^{2}-b^{2}}{a}+\frac{2 b^{2}}{a}}{\frac{a^{2}+b^{2}}{a^{2}}}=\frac{\frac{a^{2}-b^{2}+2 b^{2}}{a}}{\frac{a^{2}+b^{2}}{a^{2}}}=\frac{a^{2}+b^{2}}{a} \cdot \frac{a^{2}}{a^{2}+b^{2}}=\boldsymbol{a} . \end{aligned} $

Ex. 15 . If $\theta$ and $\phi$ are angles in the first quadrant such that $\tan \theta=\frac{1}{7}$ and $\sin \phi=\frac{1}{\sqrt{10}}$,

then show that $\theta+2 \phi=45^{\circ}$.

Sol. Since $\theta$ and $\phi$ lie in the Ist quadrant, $\sin \theta, \sin \phi ; \cos \theta, \cos \phi$ and $\tan \theta, \tan \phi$ are all positive.

$ \begin{aligned} & \Rightarrow \sin \phi=\frac{1}{\sqrt{10}} \Rightarrow \cos \phi=\sqrt{1-\sin ^{2} \phi}=\sqrt{1-\frac{1}{10}}=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}} \\ & \Rightarrow \tan \phi=\frac{\sin \phi}{\cos \phi}=\frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}}=\frac{1}{3} \quad \therefore \quad \tan 2 \phi=\frac{2 \tan \phi}{1-\tan ^{2} \phi}=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{8}}{\frac{8}{9}}=\frac{3}{4} \end{aligned} $

Also, given $\tan \theta=\frac{1}{2}$

$ \begin{aligned} \therefore \tan (\theta+2 \phi) & =\frac{\tan \theta+\tan 2 \phi}{1-\tan \theta \tan 2 \phi} \\ & =\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \cdot \frac{3}{4}}=\frac{\frac{4+21}{28}}{1-\frac{3}{28}}=\frac{\frac{25}{28}}{\frac{25}{28}}=1 \\ \Rightarrow \tan (\theta+2 \phi) & =\tan 45^{\circ} \Rightarrow \theta+2 \phi=45^{\circ} . \end{aligned} $

Ex. 16 . If $A+B+C=\frac{\pi}{2}$, then find $\sin 2 A+\sin 2 B+\sin 2 C$.

Sol. $\sin 2 A+\sin 2 B+\sin 2 C$

$=(\sin 2 A+\sin 2 B)+2 \sin C \cos C$

$=2 \sin (\frac{2 A+2 B}{2}) \cos (\frac{2 A-2 B}{2})+2 \sin C \cos C \quad(.$ Using $.\sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2})$

$=2 \sin (A+B) \cos (A-B)+2 \sin C \cos C$

$=2 \sin (\frac{\pi}{2}-C) \cos (A-B)+2 \sin C \cos C$ $(.$ Given $.A+B+C=\frac{\pi}{2} \Rightarrow A+B=\frac{\pi}{2}-C)$

$=2 \cos C \cos (A-B)+2 \sin C \cos C$

(Using $.\sin (90^{\circ}-\theta)=\cos \theta)$

$=2 \cos C(\cos (A-B)+\sin C)$

$=2 \cos C(\cos (A-B)+\sin \lbrace \frac{\pi}{2}-(A+B)\rbrace .$

$=2 \cos C(\cos (A-B)+\cos (A+B))$

$=2 \cos C(2 \cos (\frac{A-B+A+B}{2}) \cos (\frac{A-B-A-B}{2})) \quad(.$ Using $.\cos C+\cos D=2 \cos \frac{C+D}{2} \cos \frac{C-D}{2})$

$=2 \cos C(2 \cos A \cdot \cos (-B))=4 \cos A \cos B \cos C$

(Using $\cos (-\theta)=\cos \theta$ )

Ex. 17 . If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$, then show that $\tan \alpha=\tan \beta=2 \tan \gamma$.

(IIT 2003)

Sol. $\alpha+\beta=\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{2}-\beta \Rightarrow \tan \alpha=\tan (\frac{\pi}{2}-\beta)$

$\Rightarrow \tan \alpha=\cot \beta \Rightarrow \tan \alpha \tan \beta=1$.

Now, $\beta+\gamma=\alpha \Rightarrow \gamma=\alpha-\beta$

$\Rightarrow \tan \gamma=\tan (\alpha-\beta) \Rightarrow \tan \gamma=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}$

$\Rightarrow \tan \gamma=\frac{\tan \alpha-\tan \beta}{1+1}=\frac{\tan \alpha-\tan \beta}{2}$

$(\because \tan \alpha \tan \beta=1)$

$\Rightarrow 2 \tan \gamma=\tan \alpha-\tan \beta \Rightarrow \tan \alpha=\tan \beta+2 \tan \gamma$.

Ex. 18 . If $\cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}$, then show that $\tan A, \tan B$ and $\tan C$ are in G.P.

Sol. $\cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}$

$\Rightarrow \frac{1-\tan ^{2} B}{1+\tan ^{2} B}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C}$

$\Rightarrow \frac{1-\tan ^{2} B}{1+\tan ^{2} B}=\frac{1-\tan A \tan C}{1+\tan A \tan C} \quad$ (On dividing the numerator and denominator of RHS by $\cos A \cos C$ )

$\Rightarrow 1+\tan A \tan C-\tan ^{2} B-\tan A \tan ^{2} B \tan C=1+\tan ^{2} B-\tan A \tan C-\tan A \tan ^{2} B \tan C$

$\Rightarrow 2 \tan A \tan C=2 \tan ^{2} B \Rightarrow \tan A . \tan C=\tan ^{2} B \Rightarrow \tan A, \tan B, \tan C$ are in G.P.

Ex. 19 . If $\cos x=\frac{2 \cos y-1}{2-\cos y}$, where $x \in(0, \pi)$, then show that $\tan \frac{x}{2} \cot \frac{y}{2}=\sqrt{3}$. (Manipal Engineering 2010)

Sol. $\cos x=\frac{2 \cos y-1}{2-\cos y}$

$ \begin{aligned} & \Rightarrow \frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}=\frac{2[\frac{1-\tan ^{2} y / 2}{1+\tan ^{2} y / 2}]-1}{2-[\frac{1-\tan ^{2} y / 2}{1+\tan ^{2} y / 2}]} \Rightarrow \frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}=\frac{2(1-\tan ^{2} y / 2)-(1+\tan ^{2} y / 2)}{2(1+\tan ^{2} y / 2)-(1-\tan ^{2} y / 2)} \\ & \Rightarrow \frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}=\frac{1-3 \tan ^{2} y / 2}{1+3 \tan ^{2} y / 2} \\ & \Rightarrow 1+3 \tan ^{2} y / 2-\tan ^{2} x / 2-3 \tan ^{2} x / 2 \tan ^{2} y / 2=1-3 \tan ^{2} y / 2+\tan ^{2} x / 2-3 \tan ^{2} x / 2 \tan ^{2} y / 2 \\ & \Rightarrow 6 \tan ^{2} y / 2=2 \tan ^{2} x / 2 \Rightarrow \tan ^{2} x / 2 \cdot \frac{1}{\tan ^{2} y / 2}=3 \Rightarrow \tan x / 2 \cdot \cot y / 2=\sqrt{3} . \end{aligned} $

Ex. 20 . If $A, B$ and $C$ are the angles of a triangle such that $\sec (A-B), \sec A$ and $\sec (A+B)$ are in arithmetic progression then show that $2 \sec ^{2} A=\sec ^{2} \frac{B}{2}$.

(UPSEEE 2011)

Sol. Since $\sec (A-B), \sec A, \sec (A+B)$ are in A.P., therefore

$ \begin{aligned} & \sec A=\frac{\sec (A-B)+\sec (A+B)}{2}=\frac{1}{2}[\frac{1}{\cos (A-B)}+\frac{1}{\cos (A+B)}] \\ & =\frac{\cos (A+B)+\cos (A-B)}{2 \cos (A-B) \cos (A+B)}=\frac{2 \cos A \cos B}{2[(\cos A \cos B+\sin A \sin B)(\cos A \cos B-\sin A \sin B)]} \\ & =\frac{\cos A \cos B}{(\cos ^{2} A \cos ^{2} B-\sin ^{2} A \sin ^{2} B)}=\frac{\cos A \cos B}{(\cos ^{2} A \cos ^{2} B)-(1-\cos ^{2} A)(1-\cos ^{2} B)} \\ & \Rightarrow \sec A=\frac{\cos A \cos B}{[\cos ^{2} A \cos ^{2} B-1+\cos ^{2} A+\cos ^{2} B-\cos ^{2} A \cos ^{2} B]} \\ & \Rightarrow \cos ^{2} A+\cos ^{2} B-1=\cos ^{2} A \cos B \Rightarrow \cos ^{2} A(1-\cos B)=1-\cos ^{2} B \\ & \Rightarrow \cos ^{2} A=1+\cos B \Rightarrow \cos ^{2} A=2 \cos ^{2} \frac{B}{2} \Rightarrow \sec ^{2} A=\frac{1}{2} \sec ^{2} \frac{B}{2} \Rightarrow \mathbf{2} \sec ^{2} A=\mathbf{s e c}^{2} \frac{\boldsymbol{B}}{\mathbf{2}} . \end{aligned} $

Ex. 21 . If $\tan \theta+\sec \theta=4$, then what is the value of $\sin \theta$ ?

(NDA/NA 2012)

Sol. Given, $\tan \theta+\sec \theta=4$

$\Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=4 \Rightarrow \frac{1+\sin \theta}{\cos \theta}=4$

$\Rightarrow \frac{\sin ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+2 \sin \theta / 2 \cos \theta / 2}{(\cos ^{2} \theta / 2-\sin ^{2} \theta / 2)}=4$

(Using the formulas $\sin ^{2} \theta+\cos ^{2} \theta=1, \sin 2 \theta=2 \sin \theta \cos \theta, \cos 2 \theta$ $=\cos ^{2}-\sin ^{2} \theta$ )

$\Rightarrow \frac{(\sin \frac{\theta}{2}+\cos \frac{\theta}{2})^{2}}{(\cos \frac{\theta}{2}+\sin \frac{\theta}{2})(\cos \frac{\theta}{2}-\sin \frac{\theta}{2})}=4$

$ \begin{aligned} & \text{ Alternatively, } \\ & \text{ Given, } \tan \theta+\sec \theta=4 \\ & \sec ^{2} \theta=1+\tan ^{2} \theta \\ & \Rightarrow \sec ^{2} \theta-\tan ^{2} \theta=1 \\ & \text{ eq. (ii) } \div \text{ eq. (i) } \\ & \Rightarrow \quad \sec \theta-\tan \theta=\frac{1}{4} \\ & \text{ eq. (i) }+ \text{ eq. (iii) } \\ & \Rightarrow 2 \sec \theta=\frac{17}{4} \Rightarrow \sec \theta=\frac{17}{8} \\ & \Rightarrow \cos \theta=\frac{8}{17} \end{aligned} $

Now, use $\sin \theta=\sqrt{1-\cos ^{2} \theta}$

$ \begin{matrix} \Rightarrow \quad \frac{(\sin \theta / 2+\cos \theta / 2)}{\cos \theta / 2-\sin \theta / 2}=4 & \Rightarrow \frac{1+\tan \theta / 2}{1-\tan \theta / 2}=4 \\ \Rightarrow 1+\tan \theta / 2=4-4 \tan \theta / 2 & \Rightarrow 5 \tan \theta / 2=3 \Rightarrow \tan \theta / 2=3 / 5 \\ \therefore & \sin \theta=\frac{2 \tan \theta / 2}{1+\tan ^{2} \theta / 2}=\frac{2 \times 3 / 5}{1+9 / 25}=\frac{6 / 5}{34 / 25}=\frac{30}{34}=\frac{\mathbf{1 5}}{\mathbf{1 7}} \end{matrix} $

Ex. 22 . Find the value of $\tan (7 \frac{1^{0}}{2})$.

$ =\frac{1-\cos 15^{\circ}}{\sin 15^{\circ}} \quad(\text{ Using } 1-\cos 2 \theta=2 \sin ^{2} \theta, \sin 2 \theta=2 \sin \theta \cos \theta) $

$ \begin{aligned} & =\frac{1-\cos (45^{\circ}-30^{\circ})}{\sin (45^{\circ}-30^{\circ})}=\frac{1-(\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ})}{\sin 45^{\circ} \cos 30^{\circ}-\sin 30^{\circ} \sin 45^{\circ}} \\ & =\frac{1-(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2})}{\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{\sqrt{2}}}=\frac{1-(\frac{\sqrt{3}+1}{2 \sqrt{2}})}{\frac{\sqrt{3}-1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1} \\ & =\frac{2 \sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2 \sqrt{6}-3-\sqrt{3}+2 \sqrt{2}-\sqrt{3}-1}{3-1} \\ & =\frac{2 \sqrt{6}-2 \sqrt{3}-4-2 \sqrt{2}}{2}=\sqrt{6}-\sqrt{3}-\sqrt{2}-2 . \end{aligned} $

Ex. 23 . If $\tan (\frac{\pi}{4}+\theta)+\tan (\frac{\pi}{4}-\theta)=p \sec 2 \theta$, then find the value of $p$.

Sol. $\tan (\frac{\pi}{4}+\theta)+\tan (\frac{\pi}{4}-\theta)=p \sec 2 \theta$

$\Rightarrow \frac{\tan \pi / 4+\tan \theta}{1-\tan \pi / 4 \cdot \tan \theta}+\frac{\tan \pi / 4-\tan \theta}{1+\tan \pi / 4 \cdot \tan \theta}=p \sec 2 \theta$

$\Rightarrow \frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta}=p \sec \theta$

$\Rightarrow \frac{(1+\tan \theta)^{2}+(1-\tan \theta)^{2}}{1-\tan ^{2} \theta}=p \sec ^{2} \theta$

$\Rightarrow \frac{1+2 \tan \theta+\tan ^{2} \theta+1-2 \tan \theta+\tan \theta}{1-\tan ^{2} \theta}=p \sec 2 \theta$

$\Rightarrow \frac{2(1+\tan ^{2} \theta)}{1-\tan ^{2} \theta}=p \sec 2 \theta \Rightarrow \frac{2}{\cos 2 \theta}=p \sec 2 \theta$

$\Rightarrow 2 \sec 2 \theta=p \sec 2 \theta \Rightarrow \boldsymbol{p}=\mathbf{2}$. $(\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B})$ $(\because \tan \frac{\pi}{4}=\tan 45^{\circ}=1)$

$(.$ Using $.\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta})$

Ex. 24 . Prove that $\frac{\sin A+\sin 3 A+\sin 5 A+\sin 7 A}{\cos A+\cos 3 A+\cos 5 A+\cos 7 A}=\tan 4 A$.

Sol. LHS $=\frac{(\sin 7 A+\sin A)+(\sin 5 A+\sin 3 A)}{(\cos 7 A+\cos A)+(\cos 5 A+\cos 3 A)}$

$=\frac{2 \sin (\frac{7 A+A}{2}) \cos (\frac{7 A-A}{2})+2 \sin (\frac{5 A+3 A}{2}) \cos (\frac{5 A-3 A}{2})}{2 \cos (\frac{7 A+A}{2}) \cos (\frac{7 A-A}{2})+2 \cos (\frac{5 A+3 A}{2}) \cos (\frac{5 A-3 A}{2})}$

$(Using \sin C+\sin D=2 \sin (\frac{C+D}{2}) \cos (\frac{C-D}{2}) ; \cos C+\cos D=2 \cos (\frac{C+D}{2}) \cos (\frac{C-D}{2}))$

$=\frac{2 \sin 4 A \cos 3 A+2 \sin 4 A \cos A}{2 \cos 4 A \cos 3 A+2 \cos 4 A \cos A}=\frac{2 \sin 4 A(\cos 3 A+\cos A)}{2 \cos 4 A(\cos 3 A+\cos A)}=\tan 4 A$.

Ex. 25 . If $\cos \alpha+\cos \beta+\cos \gamma=0$, then prove that $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=12 \cos \alpha \cos \beta \cos \gamma$.

Sol. $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=(4 \cos ^{3} \alpha-3 \cos ^{3} \alpha)+(4 \cos ^{3} \beta-3 \cos \beta)+(4 \cos ^{3} \gamma-3 \cos \gamma)$

$=4(\cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma)-3(\cos \alpha+\cos \beta+\cos \gamma)$

$=4(\cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma)-3 \times 0=4(\cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma)$

$=4 \times 3 \cos \alpha \cos \beta \cos \gamma$

$(\because a+b+c=0 \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c)$

$=12 \cos \alpha \cos \beta \cos \gamma$.

Ex. 26 . If $A=\frac{41 \pi}{12}$, then what is the value of $\frac{1-3 \tan ^{2} A}{3 \tan A-\tan ^{3} A}$.

Sol. $\because \tan 3 A=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$

$ \begin{aligned} \therefore \quad \frac{1-3 \tan ^{2} A}{3 \tan A-\tan ^{3} A} & =\frac{1}{\tan 3 A}=\frac{1}{\tan 3 \times \frac{41 \pi}{12}}=\frac{1}{\tan \frac{41 \pi}{4}} \\ & =\frac{1}{\tan (10 \pi+\frac{\pi}{4})}=\frac{1}{\tan \frac{\pi}{4}}=1 . \end{aligned} $

[Using $\tan (2 n \pi+\theta)=\tan \theta]$

Ex. 27 . If sec $\alpha$ and $cosec \alpha$ are the roots of the equation $x^{2}-p x+q=0$, then prove that $p^{2}=q(q+2)$.

(Kerala PET 2007)

Sol. Sum of roots $=\sec \alpha+cosec \alpha=p$

$\Rightarrow \frac{1}{\cos \alpha}+\frac{1}{\sin \alpha}=p \Rightarrow \frac{\sin \alpha+\cos \alpha}{\sin \alpha \cos \alpha}=p$

Product of roots $=\sec \alpha \cdot cosec \alpha=q \Rightarrow \frac{1}{\sin \alpha \cos \alpha}=q$

Now $p^{2}=(\frac{\sin \alpha+\cos \alpha}{\sin \alpha \cos \alpha})^{2}=\frac{\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha}{\sin \alpha^{2} \cos ^{2} \alpha}$

$\Rightarrow p^{2}=\frac{1+2 \sin \alpha \cos \alpha}{\sin ^{2} \alpha \cos ^{2} \alpha}=\frac{1}{\sin ^{2} \alpha \cos ^{2} \alpha}+\frac{2}{\sin \alpha \cos \alpha}=q^{2}+2 q=\boldsymbol{q}(\boldsymbol{q}+\mathbf{2})$.

Ex. 28 . If $\theta$ and $\phi$ are acute angles such that $\sin \theta=\frac{1}{2}$ and $\cos \phi=\frac{1}{3}$, than $\theta+\phi$ lies in

(a) $] \frac{\pi}{3}, \frac{\pi}{2}[$

(b) $] \frac{2 \pi}{3}, \frac{5 \pi}{3}[$

(d) $] \frac{5 \pi}{6}, \pi[$

(IIT 2004)

Sol. $\sin \theta=\frac{1}{2} \Rightarrow \sin \theta=\sin \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \quad(\because \theta$ and $\phi$ are acute angles lying in the first quadrant $)$

Now $\cos \phi=\frac{1}{3} \Rightarrow 0<\cos \phi<\frac{1}{2} \Rightarrow \cos \frac{\pi}{2}<\cos \phi<\cos \frac{\pi}{3} \Rightarrow \frac{\pi}{2}<\phi<\frac{\pi}{3}$

$\therefore$ From (i) and (ii) $\frac{\pi}{2}+\frac{\pi}{6}<\theta+\phi<\frac{\pi}{3}+\frac{\pi}{6}$

$\Rightarrow \frac{\pi}{2}<\theta+\phi<\frac{2 \pi}{3} \Rightarrow \theta+\phi$ lies in the open interval $] \frac{\pi}{2}, \frac{2 \pi}{3}[$.

Hence (c) is the correct option.

Ex. 29 . The angle $A$ lies in the third quadrant and it satisfies the equation $4(\sin ^{2} x+\cos x)=1$. What is the measure of angle $A$ ?

(NDA/NA 2010)

Sol. $4 \sin ^{2} x+4 \cos x=1$

$\Rightarrow 4 \sin ^{2} x+4 \cos x-1=0 \Rightarrow 4(1-\cos ^{2} x)+4 \cos x-1=0$

$\Rightarrow-4 \cos ^{2} x+4 \cos x+3=0 \Rightarrow 4 \cos ^{2} x-4 \cos x-3=0$

$\Rightarrow 4 \cos ^{2} x-6 \cos x+2 \cos x-3=0 \Rightarrow 2 \cos x(2 \cos x-3)+1(2 \cos x-3)=0$

$\Rightarrow(2 \cos x+1)(2 \cos x-3)=0$

$\Rightarrow \cos x=-\frac{1}{2}$ and $\cos x=\frac{3}{2}$ (not possible)

Now $\cos x=-\frac{1}{2}$

Since $A$ lies in the third quadrant and $\cos A=-\frac{1}{2}$, therefore,

$\cos A=\cos (180^{\circ}+60^{\circ})=\cos 240^{\circ} \Rightarrow A=\mathbf{2 4 0 ^ { \circ }}$.

Ex. 30 . Find the value of $\cos ^{2} 76+\cos ^{2} 16-\cos 76^{\circ} \cos 16^{\circ}$.

(EAMCET)

Sol. $\cos ^{2} 76^{\circ}+\cos ^{2} 16^{\circ}-\cos ^{2} 76^{\circ} \cos 16^{\circ}$

$ \begin{aligned} & =\frac{1+\cos (76^{\circ} \times 2)}{2}+\frac{1+\cos (16^{\circ} \times 2)}{2}-\frac{1}{2}[\cos (76^{\circ}-16^{\circ})+\cos (76^{\circ}-16^{\circ})] \\ & =\frac{1}{2}[1+\cos 152^{\circ}+1+\cos 32^{\circ}-\cos 92^{\circ}+\cos 60^{\circ}] \\ & =\frac{1}{2}[(2-\frac{1}{2})+\cos 152^{\circ}+\cos 32^{\circ}-\cos 92^{\circ}][\because \cos 60^{\circ}=\frac{1}{2}] \\ & =\frac{1}{2}[\frac{3}{2}+2 \cos (\frac{152^{\circ}+32^{\circ}}{2}) \cos (\frac{152^{\circ}-32^{\circ}}{2})-\cos 92^{\circ}] \\ & =\frac{1}{2}[\frac{3}{2}+2 \cos 92^{\circ} \cos 60^{\circ}-\cos 92^{\circ}](\because \cos C+\cos D=2 \cos (\frac{C+D}{2}) \cos (\frac{C-D}{2})) \quad(\because \cos 60^{\circ}=\frac{1}{2}) \\ & =\frac{1}{2}[\frac{3}{2}+\cos 92^{\circ}-\cos 92^{\circ}]=\frac{\mathbf{3}}{\mathbf{4}} . \end{aligned} $

PRACTICE SHEET

If $\cot \theta=\frac{2 x y}{x^{2}-y^{2}}$, then what is $\cos \theta$ equal to?

(a) $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

(b) $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}$

(d) $\frac{2 x y}{\sqrt{x^{2}+y^{2}}}$

(CDS 2009)

~~ 2. If $\sec \theta=\frac{13}{5}$, then what is the value of $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}$ ?

(a) 1

(b) 3

(d) 4

(CDS 2007)

~~ 3. If $cosec A=2$, then the value of $\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$ is

(a) $\sqrt{2}-1$

(b) $\sqrt{3}+2$

(d) 2 ~~ 4. The value of $4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\cos ^{2} 90^{\circ}$ is

(a) $\frac{1}{2}$

(b) $\frac{3}{4}$

(d) $\frac{11}{16}$

~~ 5. Evaluate $\frac{1}{4}(\cot ^{4} 30^{\circ}-cosec^{4} 60^{\circ})$

$ +\frac{3}{2}(\sec ^{2} 45^{\circ}-\tan ^{2} 30^{\circ})-5 \cos ^{2} 60^{\circ} . $

(a) $\frac{55}{18}$

(b) $\frac{11}{6}$

(d) 0

~~ 6. If $2 x^{2} \cos 60^{\circ}-4 \cot ^{2} 45^{\circ}-2 \tan 60^{\circ}=0$, what is the value of $x$ ?

(a) 2

(b) 3

(d) $\sqrt{3}+1$

(CDS 2007)

~~ 7. The cotangent of the angles $\pi / 3, \pi / 4$ and $\pi / 6$ are in

(a) A.P

(b) G.P.

(d) None of these

(AMU)

~~ 8. The value of $\cos 1^{\circ}, \cos 2^{\circ} \ldots \cos 100^{\circ}$ is

(a) -1

(b) 0

(d) None of these

(AIEEE 2002)

~~ 9. If $\pi=22 / 7$, then a unit radian is approximately equal to

(a) $57^{\circ} 16^{\prime} 22^{\prime \prime}$

(b) $57^{\circ} 15^{\prime} 22^{\prime \prime}$

(d) $57^{\circ} 15^{\prime} 20^{\prime \prime}$

(CDS 2007)

~~ 10. What is the value of

$\frac{5 \sin 75^{\circ} \sin 77^{\circ}+2 \cos 13^{\circ} \cos 15^{\circ}}{\cos 15^{\circ} \sin 77^{\circ}}-\frac{7 \sin 81^{\circ}}{\cos 90} ?$

(a) -1

(b) 0

(d) 2

(CDS 2009)

~~ 11. What is the value of

$\sin ^{2} 15^{\circ}+\sin ^{2} 20^{\circ}+\sin ^{2} 25^{\circ}+\ldots \ldots . .+\sin ^{2} 75^{\circ} ?$

(a) $\tan ^{2} 15^{\circ}+\tan ^{2} 20^{\circ}+\tan ^{2} 25^{\circ}+$ $+\tan ^{2} 75^{\circ}$

(b) $\cos ^{2} 15^{\circ}+\cos ^{2} 20^{\circ}+\cos ^{2} 25^{\circ}+$ $+\cos ^{2} 75^{\circ}$

(d) $\sec ^{2} 15^{\circ}+\sec ^{2} 20^{\circ}+\sec ^{2} 25^{\circ}+$ $+\sec ^{2} 75^{\circ}$.

~~ 12. If $\tan (x^{2}-8 x+60)^{\circ}=\cot (6 x-5)^{\circ}$, what is one of the values of $x$ ?

(a) 7

(b) 8

(d) 10 .

(CDS 2005)

~~ 13. What is the value of $\sec (90-\theta)^{\circ} . \sin \theta \sec 45^{\circ}$ ?

(a) 1

(b) $\frac{\sqrt{3}}{2}$

(d) $\sqrt{3}$

(CDS 2012)

~~ 14. If $x+y=90^{\circ}$, then what is the value of $(1+\frac{\tan x}{\tan y}) \sin ^{2} y$ ?

(a) 0

(b) $1 / 2$

(d) 2

~~ 15. If $\frac{\tan 26^{\circ}+\tan 19^{\circ}}{x(1-\tan 26^{\circ} \tan 19^{\circ})}=\cos 60^{\circ}$, then the value of $x$ is

(a) 1

(b) $\sqrt{2}$

(d) $\sqrt{3}$

~~ 16. The value of $\cot 105^{\circ}$ is

(a) $\sqrt{3}-2$

(b) $2-\sqrt{3}$

(d) $\sqrt{3}+2$

(Odisha JEE)

~~ 17. $\sin 120^{\circ} \cos 150^{\circ}-\cos 240^{\circ} \sin 330^{\circ}$ is equal to

(a) $-(\frac{\sqrt{3}+1}{4})$

(b) -1

(d) $\frac{2}{3}$

(EAMCET 2006)

~~ 18. If $A, B, C, D$ are the successive angles of a cyclic quadrilateral, then what is $\cos A+\cos B+\cos C+\cos D$ equal to:

(a) 4

(b) 2

(d) 0

~~ 19. What is the value of $\tan (-1575^{\circ})$ ?

(CDS 2011)

(a) 1

(b) $\frac{1}{2}$

(d) -1

(NDA/NA 2009)

~~ 20. The value of $\frac{\sin 300^{\circ} \tan 330^{\circ} \sec 420^{\circ}}{\tan 135^{\circ} \sin 210^{\circ} \sec 315^{\circ}}$ is equal to

(a) $\frac{1}{\sqrt{3}}$

(b) $\frac{1}{\sqrt{2}}$

(d) $\sqrt{3}$

~~ 21. $\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots \ldots . .+\cos 180^{\circ}$ is equal to

(a) -1

(b) 0

(d) 2

(KCET 2003)

~~ 22. What is $\frac{\sin ^{6} \theta-\cos ^{6} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}$ equal to?

(a) $\sin ^{4} \theta-\cos ^{4} \theta$

(b) $1-\sin ^{2} \theta \cos ^{2} \theta$

(d) $1-3 \sin ^{2} \theta \cos ^{2} \theta$

(CDS 2011)

~~ 23. If $\frac{\cos x}{1+cosec x}+\frac{\cos x}{cosec x-1}=2$, which one of the following is one of the values of $x$ ?

(a) $\pi / 2$

(b) $\pi / 3$

(d) $\pi / 6$

(CDS 2009)

~~ 24. If $x=a(1+\cos \theta \cos \phi), y=b(1+\cos \theta \sin \phi)$ and $z=c(1+\sin \theta)$, then which of the following is correct?

(a) $(\frac{x-a}{a})^{2}+(\frac{y-b}{b})^{2}+(\frac{z-c}{c})^{2}=1$

(b) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$

(d) $\frac{(x-a)^{2}}{a}+\frac{(y-b)^{2}}{b}+\frac{(z-c)^{2}}{c}=1$

(CDS 2005)

~~ 25. If $\sin ^{4} x+\sin ^{2} x=1$ then what is 1 are the value of $\cot ^{4} x+\cot ^{2} x$ ?

(a) $\cos ^{2} x$

(b) $\sin ^{2} x$

(d) 1

(CDS 2006)

~~ 26. If $a \sin \theta+b \cos \theta=c$, what is/are the values of $(a \cos \theta-b \sin \theta)$ ?

(a) $c-a+b$

(b) $c-b+a$

(d) $\pm \sqrt{c^{2}+b^{2}-a^{2}}$

(CDS 2005)

~~ 27. What is the value of $\sin A \cos A \tan A+\cos A \sin A \cot A$ ?

(a) $\sin ^{2} A+\cos A$

(b) $\sin ^{2} A+\tan ^{2} A$

(d) $cosec^{2} A-\cot ^{2} A$

~~ 28. $\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)+\cos \gamma \sin (\alpha-\beta)$ is equal to

(a) 0

(b) $1 / 2$

(d) $4 \cos \alpha \cos \beta \cos \gamma$

(EAMCET)

~~ 29. If $\sin \theta, \cos \theta$ and $\tan \theta$ are in geometric progression, then $\cot ^{6} \theta-\cot ^{2} \theta$ is equal to

(a) $\frac{1}{2}$

(b) 1

(d) 3

~~ 30. If $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$, then the value of $\cos (\theta-\frac{\pi}{4})$ is

(a) $\frac{1}{2 \sqrt{2}}$

(b) $\frac{1}{\sqrt{2}}$

(d) $\frac{1}{4}$

(EAMCET 2013)

~~ 31. If $3 \sin x+4 \cos x=5$, then $6 \tan \frac{x}{2}-9 \tan ^{2} \frac{x}{2}$ is equal to

(a) 0

(b) 1

(d) 4

(EAMCET 2012)

~~ 32. $\cos ^{2} \alpha+\cos ^{2}(\alpha-120^{\circ})+\cos ^{2}(\alpha+120^{\circ})$ is equal to

(a) 0

(b) $\frac{1}{2}$

(d) $\frac{3}{2}$

(MPPET)

~~ 33. The expression $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ can be written as

(a) $\sin A \cos A+1$

(b) $\sec A cosec A+1$

(d) $\sec A+cosec A$

(IIT 2013)

~~ 34. The value of $\tan 5 \theta$ is

(a) $\frac{\tan ^{5} \theta+10 \tan ^{3} \theta-5 \tan \theta}{5 \tan ^{4} \theta-10 \tan ^{2} \theta+1}$

(b) $\frac{5 \tan \theta+10 \tan ^{3} \theta-\tan ^{5} \theta}{1+10 \tan ^{2} \theta-5 \tan ^{4} \theta}$

(d) $\frac{\tan ^{5} \theta-10 \tan ^{3} \theta+5 \tan \theta}{1-10 \tan ^{2} \theta+5 \tan ^{4} \theta}$

~~ 35. If $\alpha$ and $\beta$ are such that $\tan \alpha=2 \tan \beta$, then what is $\sin (\alpha+\beta)$ equal to?

(a) 1

(b) $2 \sin (\alpha-\beta)$

(d) $3 \sin (\alpha-\beta)$

(NDA/NA 2007)

~~ 36. If $\sin x+cosec x=2$, then $\sin ^{n} x+cosec^{n} x$ is equal to

(a) 2

(b) $2^{n}$

(d) $2^{n-2}$

(AMU 2008)

~~ 37. The value of $\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \frac{16 \pi}{15}$ is equal to:

(a) $\frac{1}{16}$

(b) $\frac{1}{32}$

(d) $\frac{1}{8}$

(Manipal Engineering 2010)

~~ 38. If $\alpha+\beta+\gamma=\pi$, then the value of $\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma$ is equal to

(a) $2 \sin \alpha$

(b) $2 \sin \alpha \cos \beta \sin \gamma$

(d) $2 \sin \alpha \sin \beta \sin \gamma$

(Manipal Engineering 2012)

~~ 39. What is the value of $\cos 15^{\circ}$ ?

(a) $\frac{1}{2}(\sqrt{2-\sqrt{3}})$

(b) $\frac{1}{2}(\sqrt{2+\sqrt{3}})$

(d) $\sqrt{2}-\sqrt{3}$

~~ 40. What is $\sqrt{2+\sqrt{2+\sqrt{2+\cos 4 A}}}$ equal to?

(a) $\cos A$

(b) $\cos (2 A)$

(d) $\sqrt{2 \cos A}$

(NDA/NA 2008)

~~ 41. If $\cos x \neq 1$, then what is $\frac{\sin x}{1+\cos x}$ equal to?

(a) $-\cot \frac{x}{2}$

(b) $\cot \frac{x}{2}$

(d) $-\tan \frac{x}{2}$

(NDA/NA 2010)

~~ 42. The value of $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}$ is equal to

(a) $\frac{1}{2}$

(b) 0

(d) $\frac{\sqrt{3}}{2}$

~~ 43. If $\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$, then $\tan \alpha$ is equal to

(a) $\frac{\tan \beta}{\sqrt{2}}$

(b) $\tan \beta$

(d) $\sqrt{2} \tan \beta$

(Odisha JEE 2006)

~~ 44. $\frac{\sin 5 \theta}{\sin \theta}$ is equal to

(a) $16 \cos ^{4} \theta-12 \cos ^{2} \theta-1$

(b) $16 \cos ^{4} \theta-12 \cos ^{2} \theta+1$

(d) $16 \cos ^{4} \theta+12 \cos ^{2} \theta+1$

(EAMCET 2001)

~~ 45. If $\sin A+\sin B=a, \cos A-\cos B=b$, then the value of $\cos (A-B)$ is

(a) $\frac{2 a b}{a^{2}+b^{2}}$

(b) $\frac{2 a b}{a^{2}-b^{2}}$

(d) $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

~~ 46. The value of $\sin 16^{\circ}+\cos 16^{\circ}$ is

(a) $\frac{1}{\sqrt{3}}(\sqrt{2} \cos 1^{\circ}+\sin 1^{\circ})$

(b) $\frac{1}{\sqrt{2}}(\cos 1^{\circ}+\sqrt{3} \sin 1^{\circ})$

(d) $\frac{1}{\sqrt{2}}(\sqrt{3} \cos 1^{\circ}+\sin 1^{\circ})$

~~ 47. $[1+\cos \frac{\pi}{8}][1+\cos \frac{3 \pi}{8}][1+\cos \frac{5 \pi}{8}][1+\cos \frac{7 \pi}{8}]$ is equal to

(a) $\frac{1}{8}$

(b) $\frac{1}{2}$

(d) $\cos \frac{\pi}{8}$

(AIEEE 2002, DCE 2003)

~~ 48. Evaluate:

$\cos 2(\theta+\phi)+4 \cos (\theta+\phi) \sin \theta \sin \phi+2 \sin ^{2} \phi$

(a) $\sin 2 \theta$

(b) $\cos 2 \theta$

(d) $\cos 3 \theta$

(Odisha CET 2004)

~~ 49. If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then what is $\cot (A-B)$ equal to?

(a) $\frac{1}{y}-\frac{1}{x}$

(b) $\frac{1}{x}-\frac{1}{y}$

(d) $-\frac{1}{x}-\frac{1}{y}$

(NDA/NA 2011)

~~ 50. $\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha$ is equal to

(a) 0

(b) $2 \tan \alpha$

(d) $\tan 16 \alpha$

(IIT)

ANSWERS

1. $(c)$	2. $(b)$	3. $(d)$	4. $(b)$	5. $(a)$	6. $(d)$	7. $(b)$	8. $(b)$	9. $(a)$
10. $(b)$	11. $(b)$	12. $(a)$	13. $(c)$	14. $(c)$	15. $(c)$	16. $(a)$	17. $(b)$	18. $(d)$
19. $(a)$	20. $(c)$	21. $(a)$	22. $(b)$	23. $(c)$	24. $(a)$	25. $(d)$	26. $(c)$	27. $(d)$
28. $(a)$	29. $(b)$	30. $(a)$	31. $(b)$	32. $(d)$	33. $(b)$	34. $(d)$	35. $(d)$	36. $(a)$
37. $(a)$	38. $(c)$	39. $(b)$	40. $(c)$	41. $(c)$	42. $(c)$	43. $(d)$	44. $(b)$	45. $(c)$
46. $(d)$	47. $(a)$	48. $(b)$	49. $(c)$	50. $(c)$

HINTS AND SOLUTIONS

$\cot \theta=\frac{\text{ Base }}{\text{ Perpendicular }}=\frac{B C}{A B}($ where $\angle A C B=\theta)$

$ \begin{aligned} \therefore \quad & B C=(2 x y) k \\ & A B=(x^{2}-y^{2}) k \\ \Rightarrow & A C^{2}=A B^{2}+B C^{2} \\ & =k^{2}((x^{2}+y^{2})^{2}+(2 x y)^{2}) \\ & =k^{2}[x^{4}+y^{4}-2 x^{2} y^{2}+4 x^{2} y^{2}] \\ & =k^{2}[x^{4}+y^{4}+2 x^{2} y^{2}]=k^{2}(x^{2}+y^{2})^{2} \\ \Rightarrow & A C=k(x^{2}+y^{2}) \end{aligned} $

$\therefore \quad \cos \theta=\frac{\text{ Base }}{\text{ Hyp. }}=\frac{B C}{A C}=\frac{\mathbf{2} \boldsymbol{x y}}{\boldsymbol{x}^{\mathbf{2}}+\boldsymbol{y}^{\mathbf{2}}}$.

~~ 2. Given, $\sec \theta=\frac{13}{5} \Rightarrow \sec ^{2} \theta=\frac{169}{25}$

$ \begin{aligned} & \Rightarrow 1+\tan ^{2} \theta=\frac{169}{25} \quad(\because \sec ^{2} \theta-\tan ^{2} \theta=1) \\ & \Rightarrow \tan ^{2} \theta=\frac{169}{25}-1=\frac{144}{25} \Rightarrow \tan \theta=\frac{12}{5} \end{aligned} $

$ \therefore \frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}=\frac{\frac{2 \sin \theta}{\cos \theta}-3}{\frac{4 \sin \theta}{\cos \theta}-9} $

(On dividing each term of numerator and denominator by $\cos \theta$ )

$ \begin{aligned} & =\frac{2 \tan \theta-3}{4 \tan \theta-9}=\frac{2 \times \frac{12}{5}-3}{4 \times \frac{12}{5}-9} \\ & =\frac{24-15}{48-45}=\frac{9}{3}=\mathbf{3} \end{aligned} $

~~ 3. $cosec A=\frac{\text{ Hypotenuse }}{\text{ Perpendicular }}=\frac{2}{1}$

$\Rightarrow A C=2 k, B C=k$

$\Rightarrow A B^{2}=\sqrt{A C^{2}-B C^{2}}=\sqrt{4 k^{2}-k^{2}}=\sqrt{3 k^{2}}=k \sqrt{3}$

$\therefore \quad \tan A=\frac{\text{ Perp. }}{\text{ Base }}=\frac{B C}{A B}=\frac{k}{k \sqrt{3}}=\frac{1}{\sqrt{3}}$

$\sin A=\frac{1}{cosec A}=\frac{1}{2}$

$\cos A=\frac{\text{ Base }}{\text{ Hypotenuse }}=\frac{A B}{A C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$

$\therefore \quad \frac{1}{\tan A}+\frac{\sin A}{1+\cos A}=\frac{1}{1 / \sqrt{3}}+\frac{1 / 2}{1+\sqrt{3} / 2}$

$=\frac{\sqrt{3}}{1}+\frac{1 / 2}{\frac{2+\sqrt{3}}{2}}=\frac{\sqrt{3}}{1}+\frac{1}{2+\sqrt{3}}$

$=\frac{2 \sqrt{3}+3+1}{2+\sqrt{3}}=\frac{2 \sqrt{3}+4}{2+\sqrt{3}}=\frac{2(\sqrt{3}+2)}{2+\sqrt{3}}=\mathbf{2}$.

~~ 4. $4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\cos ^{2} 90^{\circ}$

$=4(\cot 45^{\circ})^{2}-(\sec 60^{\circ})^{2}+(\sin 60^{\circ})^{2}+(\cos 90^{\circ})^{2}$

$=4 \times 1^{2}-(2)^{2}+(\frac{\sqrt{3}}{2})^{2}+(0)^{2}$

$=4-4+\frac{3}{4}+0=\frac{\mathbf{3}}{\mathbf{4}}$.

~~ 5. $\frac{1}{4}(\cot ^{4} 30^{\circ}-cosec^{4} 60^{\circ})$

$ +\frac{3}{2}(\sec ^{2} 45^{\circ}-\tan ^{2} 30^{\circ})-5 \cos ^{2} 60^{\circ} $

$=\frac{1}{4}[(\cot 30^{\circ})^{4}-(cosec 60^{\circ})^{4}]$

$ +\frac{3}{2}[(\sec 45^{\circ})^{2}-(\tan 30^{\circ})^{2}]-5(\cos 60^{\circ})^{2} $

$=\frac{1}{4}[(\sqrt{3})^{4}-(\frac{2}{\sqrt{3}})^{4}]+\frac{3}{2}[(\sqrt{2})^{2}-(\frac{1}{\sqrt{3}})^{2}]-5(\frac{1}{2})^{2}$ $=\frac{1}{4}(9-\frac{16}{9})+\frac{3}{2}(2-\frac{1}{3})-5 \times \frac{1}{4}$

$=\frac{1}{4}[\frac{81-16}{9}]+\frac{3}{2}[\frac{6-1}{3}]-\frac{5}{4}$

$=\frac{1}{4} \times \frac{65}{9}+\frac{3}{2} \times \frac{5}{3}-\frac{5}{4}=\frac{65}{36}+\frac{5}{2}-\frac{5}{4}$

$=\frac{65+90-45}{36}=\frac{110}{36}=\frac{\mathbf{5 5}}{\mathbf{1 8}}$.

~~ 6. $2 x^{2} \cos 60^{\circ}-4 \cot ^{2} 45^{\circ}-2 \tan 60^{\circ}=0$

$\Rightarrow 2 x^{2} \times \frac{1}{2}-4 \times(1)^{2}-2 \times \sqrt{3}=0$

$\Rightarrow x^{2}-4-2 \sqrt{3}=0 \Rightarrow x^{2}=4+2 \sqrt{3}$

$\Rightarrow x^{2}=3+1+2 \sqrt{3} \Rightarrow x^{2}=(\sqrt{3})^{2}+(1)^{2}+2 \sqrt{3}$

$\Rightarrow x^{2}=(\sqrt{3}+1)^{2} \Rightarrow x=\sqrt{\mathbf{3}}+\mathbf{1}$.

~~ 7. $\cot \frac{\pi}{3}=\cot 60^{\circ}=\frac{1}{\sqrt{3}},+\cot \frac{\pi}{4}=\cot 45^{\circ}=1$,

$\cot \frac{\pi}{6}=\cot 30^{\circ}=\sqrt{3}$

We can see that

$(\cot \frac{\pi}{3}) \cdot(\cot (\frac{\pi}{6}))=\frac{1}{\sqrt{3}} \cdot \sqrt{3}=1=(\cot \frac{\pi}{4})^{2}$

$\therefore \cot \pi / 3, \cot \pi / 4, \cot \pi / 6$ are in G.P

~~ 8. $\because \cos 90^{\circ}=0$

$\therefore \cos 1^{\circ} \cdot \cot 2^{\circ}$. $\cos 90^{\circ} \cdot \cos 91^{\circ}$ $\cos 100^{\circ}=\mathbf{0}$.

~~ 9. $\pi$ radians $=180^{\circ}$

$ \begin{aligned} \Rightarrow 1 \text{ radian } & =\frac{180}{\pi} \text{ degree }=\frac{180}{22} \times 7 \text{ degree } \\ & =\frac{630}{11} \text{ degree }=57 \frac{3}{11} \text{ degree } \\ & =57^{\circ}+\frac{3}{11} \times 60 min=57^{\circ}+\frac{180}{11} \min \\ & =57^{\circ}+16 \frac{4}{11} min \\ & =57^{\circ}+16^{\prime}+\frac{4}{11} \times 60 s \\ & =57^{\circ}+16^{\prime}+21.8^{\prime \prime} \\ & =\mathbf{5 7}^{\circ} \mathbf{1 6}^{\prime} \mathbf{2 2}^{\prime \prime} \text{ (approx). } \end{aligned} $

~~ 10. $\frac{5 \sin 75^{\circ} \sin 77^{\circ}+2 \cos 13^{\circ} \cos 15^{\circ}}{\cos 15^{\circ} \sin 77^{\circ}}-\frac{7 \sin 81^{\circ}}{\cos 9^{\circ}}$

$ \begin{matrix} \frac{5 \sin (90^{\circ}-15^{\circ}) \sin 77^{\circ}+2 \cos (90^{\circ}-77^{\circ}) \cos 15^{\circ}}{\cos 15^{\circ} \sin 77^{\circ}} \\ -\frac{7 \sin (90^{\circ}-9^{\circ})}{\cos 9^{\circ}} \end{matrix} $

$=\frac{5 \cos 15^{\circ} \sin 77^{\circ}+2 \sin 77^{\circ} \cos 15^{\circ}}{\cos 15^{\circ} \sin 77^{\circ}}-\frac{7 \sin 9^{\circ}}{\cos 9^{\circ}}$

$=\frac{7 \cos 15^{\circ} \sin 77^{\circ}}{\cos 15^{\circ} \sin 77^{\circ}}-\frac{7 \cos 9^{\circ}}{\cos 9^{\circ}}=7-7=\mathbf{0}$.

~~ 11. $\sin ^{2} 15^{\circ}+\sin ^{2} 20^{\circ}+\sin ^{2} 25^{\circ}+$ $+\sin ^{2} 75^{\circ}$

$=\sin ^{2}(90^{\circ}-75^{\circ})+\sin ^{2}(90^{\circ}-70^{\circ})+\ldots+\sin ^{2}(90^{\circ}-15^{\circ})$

$=\cos ^{2} 75^{\circ}+\cos ^{2} 70^{\circ}+\ldots \ldots . .+\cos ^{2} \mathbf{1 5}^{\circ}$

~~ 12. $\tan (x^{2}-8 x+60)^{\circ}=\cot (6 x-5)^{\circ}$

$\Rightarrow \tan (x^{2}-8 x+60)^{\circ}=\tan [90-(6 x-5)]^{\circ}$

$\Rightarrow(x^{2}-8 x+60)^{\circ}=90^{\circ}-(6 x-5)^{\circ}$

$\Rightarrow x^{2}-8 x+60=90-6 x+5 \Rightarrow x^{2}-2 x-35=0$

$\Rightarrow(x-7)(x+5)=0 \Rightarrow x=7,-5 \Rightarrow x=7$.

~~ 13. $\sec (90-\theta)^{\circ} \sin \theta \sec 45^{\circ}$

$=cosec \theta \sin \theta .(\sqrt{2})=\frac{1}{\sin \theta} \cdot \sin \theta \cdot \sqrt{2}=\sqrt{2}$.

~~ 14. $(1+\frac{\tan x}{\tan y}) \cdot \sin ^{2} y$

$=(1+\frac{\tan (90^{\circ}-y)}{\tan y}) \cdot \sin ^{2} y \quad(\because x+y=90^{\circ})$

$=(1+\cot y \cdot \cot y) \cdot \sin ^{2} y$

$=(1+\cot ^{2} y) \cdot \sin ^{2} y=cosec^{2} y \cdot \sin ^{2} y=\mathbf{1}$.

~~ 15. $\frac{\tan 26^{\circ}+\tan 19^{\circ}}{x(1-\tan 26^{\circ} \tan 19^{\circ})}=\cos 60^{\circ}$

$=\frac{\tan 26^{\circ}+\tan 19^{\circ}}{1-\tan 26^{\circ} \tan 19^{\circ}}=x \cos 60^{\circ}$

$=\tan (26^{\circ}+19^{\circ})=x \times \frac{1}{2}$

$[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}]$

$=\tan 45^{\circ}=x / 2 \Rightarrow x / 2=1 \Rightarrow x=\mathbf{2}$.

~~ 16. $\cot 105^{\circ}=\cot (90^{\circ}+15^{\circ})=-\tan 15^{\circ}$

$=-\tan (45^{\circ}-30^{\circ})=-[\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}]$

$(\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B})$

$=-[\frac{1-\frac{1}{\sqrt{3}}}{1+1 \cdot \frac{1}{\sqrt{3}}}]=-\frac{\sqrt{3}-1}{\sqrt{3}+1}$

$=-\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=-\frac{4-2 \sqrt{3}}{2}=\sqrt{3}-2$.

~~ 17. $\sin 120^{\circ} \cos 150^{\circ}-\cos 240^{\circ} \sin 330^{\circ}$

$=\sin (180^{\circ}-60^{\circ}) \cos (180^{\circ}-30^{\circ})$

$-\cos (180^{\circ}+60^{\circ}) \sin (360^{\circ}-30^{\circ})$ $=(\sin 60^{\circ}) \cdot(-\cos 30^{\circ})-(-\cos 60^{\circ})(-\sin 30^{\circ})$

$=\frac{\sqrt{3}}{2} \times-\frac{\sqrt{3}}{2}-(\frac{-1}{2})(\frac{-1}{2})=-\frac{3}{4}-\frac{1}{4}=-\frac{4}{4}=-\mathbf{1}$.

~~ 18. In a cyclic quadrilateral, the sum of the opposite angles is $180^{\circ}$.

$\Rightarrow A+C=180^{\circ}$ and $B+D=180^{\circ}$

$\therefore \cos A+\cos B+\cos C+\cos D$

$=\cos A+\cos B+\cos (180^{\circ}-C)+\cos (180^{\circ}-B)$

$=\cos A+\cos B-\cos A-\cos B$

$\mathbf{0} \mathbf{0}$.

~~ 19. $\tan (-1575^{\circ})$

$ \begin{aligned} & =-\tan (1575^{\circ}) \\ & =-\tan (4 \times 360^{\circ}+135^{\circ}) \\ & =-\tan (135^{\circ}) \\ & =-\tan (90^{\circ}+45^{\circ}) \\ & =-(-\cot 45^{\circ})=\cot 45^{\circ}=1 . \end{aligned} \quad(\because \tan (-\theta)=-\tan \theta) $

~~ 20. $\frac{\sin 300^{\circ} \tan 330^{\circ} \sec 420^{\circ}}{\tan 135^{\circ} \sin 210^{\circ} \sec 315^{\circ}}$

$=\frac{\sin (360^{\circ}-60^{\circ}) \tan (360^{\circ}-30^{\circ}) \sec (360^{\circ}+60^{\circ})}{\tan (180^{\circ}-45^{\circ}) \sin (180^{\circ}+30^{\circ}) \sec (360^{\circ}-45^{\circ})}$

$=\frac{(-\sin 60^{\circ})(-\tan 30^{\circ})(\sec 60^{\circ})}{(-\tan 45^{\circ})(-\sin 30^{\circ})(\sec 45^{\circ})}$

$=\frac{\sin 60^{\circ} \times \tan 30^{\circ} \times \sec 60^{\circ}}{\tan 45^{\circ} \times \sin 30^{\circ} \times \sec 45^{\circ}}$

$=\frac{\sqrt{3} / 2 \times 1 / \sqrt{3} \times 2}{1 \times 1 / 2 \times \sqrt{2}}=\sqrt{2}$.

~~ 21. $\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+$ $+\cos 180^{\circ}$

$=(\cos 1^{\circ}+\cos 179^{\circ})+(\cos 2^{\circ}+\cos 178^{\circ})+$ $+(\cos 89^{\circ}+\cos 91^{\circ})+\cos 90^{\circ}+\cos 180^{\circ}$ $=[\cos 1^{\circ}+\cos (180^{\circ}-1^{\circ})]+[\cos 2^{\circ}+\cos (180^{\circ}-2^{\circ})]+$ $+[\cos 89^{\circ}+\cos (180^{\circ}-91^{\circ})]+\cos 90^{\circ}+\cos 180^{\circ}$ $=(\cos 1^{\circ}-\cos 1^{\circ})+(\cos 2^{\circ}-\cos 2^{\circ})+$

$ +(\cos 89^{\circ}-\cos +89^{\circ})+0+(-1)=-\mathbf{1} $

$(\because \cos (180^{\circ}-\theta)=-\cos \theta)$

~~ 22. $\frac{\sin ^{6} \theta-\cos ^{6} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}=\frac{(\sin ^{2} \theta)^{3}-(\cos ^{2} \theta)^{3}}{\sin ^{2} \theta-\cos ^{2} \theta}$

$=\frac{(\sin ^{2} \theta-\cos ^{2} \theta)(\sin ^{4} \theta+\cos ^{4} \theta+\sin ^{2} \theta \cos ^{2} \theta)}{(\sin ^{2} \theta-\cos ^{2} \theta)}$

(Using $.a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+a b))$

$=\sin ^{4} \theta+\cos ^{4} \theta+2 \sin ^{2} \theta \cos ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta$

$=(\sin ^{2} \theta+\cos ^{2} \theta)^{2}-\sin ^{2} \theta \cos ^{2} \theta$

$=1-\sin ^{2} \boldsymbol{\theta} \cos ^{2} \boldsymbol{\theta} \quad(\because \sin ^{2} \theta+\cos ^{2} \theta=1)$

~~ 23. $\frac{\cos x}{1+cosec x}+\frac{\cos x}{cosec x-1}=2$

$\Rightarrow \frac{\cos x(cosec x-1)+\cos x(1+cosec x)}{cosec^{2} x-1}=2$

$\Rightarrow \frac{2 \cos x cosec x}{\cot ^{2} x}=2$

$\Rightarrow \frac{\cos x}{\sin x} \times \frac{\sin ^{2} x}{\cos ^{2} x}=1$

$\Rightarrow \tan x=1 \Rightarrow \tan x=\tan \frac{\pi}{4} \Rightarrow x=\frac{\pi}{4}$.

~~ 24. Given, $x=a(1+\cos \theta \cos \phi)$

$\Rightarrow \frac{x}{a}=1+\cos \theta \cos \phi \Rightarrow \frac{x}{a}-1=\cos \theta \cos \phi$

$\Rightarrow \frac{x-a}{a}=\cos \theta \cos \phi$

Similarly, $y=b(1+\cos \theta \sin \phi)$

$\Rightarrow \frac{y-b}{b}=\cos \theta \sin \phi$

$z=c(1+\sin \theta) \Rightarrow \frac{z-c}{c}=\sin \theta$

Squaring eqns (i), (ii) and (iii) and adding, we get

$(\frac{x-a}{a})^{2}+(\frac{y-b}{b})^{2}+(\frac{z-c}{c})^{2}$

$=\cos ^{2} \theta \cos ^{2} \phi+\cos ^{2} \theta \sin ^{2} \phi+\sin ^{2} \theta$

$=\cos ^{2} \theta(\cos ^{2} \phi+\sin ^{2} \phi)+\sin ^{2} \theta$

$=\cos ^{2} \theta+\sin ^{2} \theta=\mathbf{1}$.

~~ 25. $\sin ^{4} x+\sin ^{2} x=1$

$\Rightarrow \sin ^{4} x=1-\sin ^{2} x$

$\Rightarrow \sin ^{4} x=\cos ^{2} x$

$\therefore \cot ^{4} x+\cot ^{2} x$

$=\cot ^{2} x(1+\cot ^{2} x)=\cot ^{2} x \cdot cosec^{2} x$

$=\frac{\cos ^{2} x}{\sin ^{2} x \cdot \sin ^{2} x}=\frac{\cos ^{2} x}{\sin ^{4} x}$

$=\frac{\cos ^{2} x}{\cos ^{2} x}=1$.

[Using $(i)$ ]

~~ 26. Given, $a \sin \theta-b \cos \theta=c$

$(a \cos \theta-b \sin \theta)^{2}$

$=a^{2} \cos ^{2} \theta-2 a b \cos \theta \sin \theta+b^{2} \sin ^{2} \theta$

$=a^{2}(1-\sin ^{2} \theta)-2(a \sin \theta)(b \cos \theta)+b^{2}(1-\cos ^{2} \theta)$

$=a^{2}-a^{2} \sin ^{2} \theta-2(a \sin \theta)(b \cos \theta)+b^{2}-b^{2} \cos ^{2} \theta$

$=a^{2}+b^{2}-[a^{2} \sin ^{2} \theta+2(a \sin \theta)(b \cos \theta)+b^{2} \cos ^{2} \theta]$

$=a^{2}+b^{2}-(a \sin \theta+b \cos \theta)^{2}=a^{2}+b^{2}-c^{2}$

$\Rightarrow(a \cos \theta-b \sin \theta)= \pm \sqrt{\boldsymbol{a}^{2}+\boldsymbol{b}^{2}-\boldsymbol{c}^{2}}$.

~~ 27. $\sin A \cos A \tan A+\cos A \sin A \cot A$

$=\sin A \cos A \cdot \frac{\sin A}{\cos A}+\cos A \sin A \frac{\cos A}{\sin A}$

$=\sin ^{2} A+\cos ^{2} A=1=cosec^{2} \boldsymbol{A}-\cot ^{2} \boldsymbol{A}$.

~~ 28. $\cos \alpha \sin (\beta-\gamma)+\cos \beta \sin (\gamma-\alpha)+\cos \gamma \sin (\alpha-\beta)$

$=\cos \alpha[\sin \beta \cos \gamma-\cos \beta \sin \gamma]$

$+\cos \beta[\sin \gamma \cos \alpha-\cos \alpha \sin \beta]$

$+\cos \gamma[\sin \alpha \cos \beta-\cos \alpha \sin \beta]$

$=\cos \alpha \sin \beta \cos \gamma-\cos \alpha \cos \beta \sin \gamma$

$+\cos \beta \sin \gamma \cos \alpha-\cos \beta \cos \alpha \sin \gamma$

$+\cos \gamma \sin \alpha \cos \beta-\cos \gamma \cos \alpha \sin \beta$

$=\mathbf{0}$.

~~ 29. $\sin \theta, \cos \theta$ and $\tan \theta$ are in G.P.

$\Rightarrow \sin \theta \times \tan \theta=\cos ^{2} \theta \Rightarrow \sin \theta \cdot \frac{\sin \theta}{\cos \theta}=\cos ^{2} \theta$

$\Rightarrow \sin ^{2} \theta=\cos ^{3} \theta$

Now, $\cot ^{6} \theta-\cot ^{2} \theta=\frac{\cos ^{6} \theta}{\sin ^{6} \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}$

$=\frac{(\cos ^{3} \theta)^{2}}{\sin ^{6} \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}$

$=\frac{\sin ^{4} \theta}{\sin ^{6} \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}$

(By $(i))$

$=\frac{1}{\sin ^{2} \theta}-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}$

$=\frac{1-\cos ^{2} \theta}{\sin ^{2} \theta}=\frac{\sin ^{2} \theta}{\sin ^{2} \theta}=\mathbf{1}$.

~~ 30. $\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$

$\Rightarrow \tan (\pi \cos \theta)=\tan (\pi / 2-\pi \sin \theta)$

$\Rightarrow \pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta$

$\Rightarrow \pi(\cos \theta-\sin \theta)=\frac{\pi}{2} \quad \Rightarrow \cos \theta+\sin \theta=\frac{1}{2}$

$\Rightarrow \frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta=\frac{1}{2 \sqrt{2}}$

(Multiplying both sides by $\frac{1}{\sqrt{2}}$ )

$\Rightarrow \cos \frac{\pi}{4} \cos \theta+\sin \frac{\pi}{4} \sin \theta=\frac{1}{2 \sqrt{2}}$

$\Rightarrow \cos (\pi / 4-\theta)=\frac{1}{2 \sqrt{2}}$

$ (\because \cos (A-B)=\cos A \cos B+\sin A \sin B) $

~~ 31. $3 \sin x+4 \cos x=5$

$ \begin{aligned} & \Rightarrow 3(\frac{2 \tan x / 2}{1+\tan ^{2} x / 2})+4(\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2})=5 \\ & (\because \sin x=\frac{2 \tan x / 2}{1+\tan ^{2} x / 2}, \cos x=\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}) \\ & \Rightarrow \frac{6 \tan x / 2+4-4 \tan ^{2} x / 2}{1+\tan ^{2} x / 2}=5 \\ & \Rightarrow 6 \tan x / 2+4-4 \tan ^{2} x / 2=5+5 \tan ^{2} x / 2 \\ & \Rightarrow 6 \tan x / 2-9 \tan ^{2} x / 2=1 . \end{aligned} $

~~ 32. $\cos ^{2} \alpha+\cos ^{2}(\alpha-120^{\circ})+\cos ^{2}(\alpha+120^{\circ})$ $=\cos ^{2} \alpha+\lbrace \cos \alpha \cos 120^{\circ}+\sin \alpha \sin 120^{\circ}\rbrace ^{2}$

$+\lbrace \cos \alpha \cos 120^{\circ}-\sin \alpha \sin 120^{\circ}\rbrace ^{2}$ $(\because \cos (A+B)=\cos A \cos B-\sin A \sin B)$

$=\cos ^{2} \alpha+\lbrace -\frac{1}{2} \cos \alpha+\frac{\sqrt{3}}{2} \sin \alpha\rbrace ^{2}$ $+\lbrace -\frac{1}{2} \cos \alpha-\frac{\sqrt{3}}{2} \sin \alpha\rbrace ^{2}$ $(\cos 120^{\circ}=\cos (180^{\circ}-60^{\circ})=-\cos 60^{\circ}=-\frac{1}{2})$ $=\cos ^{2} \alpha+\frac{1}{4} \cos ^{2} \alpha+\frac{3}{4} \sin ^{2} \alpha-\frac{\sqrt{3}}{2} \sin \alpha \cos \alpha$

$ +\frac{1}{4} \cos ^{2} \alpha+\frac{3}{4} \sin ^{2} \alpha+\frac{\sqrt{3}}{2} \sin \alpha \cos \alpha $

$=\cos ^{2} \alpha+\frac{1}{4} \cos ^{2} \alpha+\frac{1}{4} \cos ^{2} \alpha+\frac{3}{4} \sin ^{2} \alpha+\frac{3}{4} \sin ^{2} \alpha$

$=\frac{3}{2} \cos ^{2} \alpha+\frac{3}{2} \sin ^{2} \alpha=\frac{3}{2}(\cos ^{2} \alpha+\sin ^{2} \alpha)=\frac{\mathbf{3}}{\mathbf{2}}$.

~~ 33. $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$

$=\frac{\sin A}{\cos A} \times \frac{1}{1-\frac{\cos A}{\sin A}}+\frac{\cos A}{\sin A} \times \frac{1}{1-\frac{\sin A}{\cos A}}$

$=\frac{\sin A}{\cos A} \times \frac{\sin A}{\sin A-\cos A}+\frac{\cos A}{\sin A} \times \frac{\cos A}{\cos A-\sin A}$

$=\frac{\sin ^{2} A}{\cos A(\sin A-\cos A)}-\frac{\cos ^{2} A}{\sin A(\sin A-\cos A)}$

$=\frac{1}{(\sin A-\cos A)}[\frac{\sin ^{3} A-\cos ^{3} A}{\cos A \sin A}]$

$=\frac{(\sin A-\cos A)(\sin ^{2} A+\sin A \cos A+\cos ^{2} A)}{(\sin A-\cos A) \cos A \sin A}$

$=\frac{1+\sin A \cos A}{\cos A \sin A}$

$(\because \sin ^{2} A+\cos ^{2} A=1)$

$=\sec A cosec A+1$.

~~ 34. $\tan 5 \theta=\tan (2 \theta+3 \theta)=\frac{\tan 2 \theta+\tan 3 \theta}{1-\tan 2 \theta \tan 3 \theta}$

$ [\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}] $

$ \begin{aligned} & =\frac{\frac{2 \tan \theta}{1-\tan ^{2} \theta}+\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}}{1-(\frac{2 \tan \theta}{1-\tan ^{2} \theta}) \cdot(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta})} \\ & =\frac{2 \tan \theta(1-3 \tan ^{2} \theta)+(3 \tan \theta-\tan ^{3} \theta)(1-\tan ^{2} \theta)}{(1-\tan ^{2} \theta)(1-3 \tan ^{2} \theta)-(2 \tan \theta)(3 \tan \theta-\tan ^{3} \theta)} \end{aligned} $

$ \begin{aligned} & =\frac{2 \tan \theta-6 \tan ^{3} \theta+3 \tan \theta-\tan ^{3} \theta-3 \tan ^{3} \theta+\tan ^{5} \theta}{1-\tan ^{2} \theta-3 \tan ^{2} \theta+3 \tan ^{4} \theta-6 \tan ^{2} \theta+2 \tan ^{4} \theta} \\ & =\frac{\tan ^{5} \theta-\mathbf{1 0 \operatorname { t a n } ^ { 3 }} \theta+\mathbf{5} \tan \theta}{\mathbf{1}-\mathbf{1 0} \tan ^{2} \theta+\mathbf{5} \tan ^{4} \theta} . \end{aligned} $

~~ 35. Given, $\tan \alpha=2 \tan \beta \Rightarrow \frac{\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin \beta}{\cos \beta}}=2$

$\Rightarrow \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}=\frac{2}{1}$

Using componendo and dividendo, we get

$ \begin{aligned} & \frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \cos \beta-\cos \alpha \sin \beta}=\frac{2+1}{2-1} \\ \Rightarrow & \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{3}{1} \Rightarrow \sin (\alpha+\beta)=\mathbf{3} \sin (\alpha-\beta) \end{aligned} $

~~ 36. $\sin x+cosec x=2$

Squaring both the sides, we have

$\sin ^{2} x+cosec^{2} x+2=4$

$\Rightarrow \sin ^{2} x+cosec^{2} x=2$, i.e., for $n=2, \sin ^{n} x+\cos ^{n} x=2$

Cubing both the sides of $(i)$, we have

$\sin ^{3} x+cosec^{3} x+3 \sin x cosec x(\sin x+cosec x)=8$

$\Rightarrow \sin ^{3} x+cosec^{3}+3 \times 2=8$

$\Rightarrow \sin ^{3} x+cosec^{3}=8-6=2$,

i.e., for $n=3, \sin ^{n} x+\cos ^{n} x=2$

Squaring both the sides of (ii), we have

$\sin ^{4} x+cosec^{4} x+2=4$

$\Rightarrow \sin ^{4} x+cosec^{4} x=2$, i.e., for $n=4, \sin ^{n} x+\cos ^{n} x=2$

Proceeding in the same way, we see that $\sin ^{n} x+cosec^{n} x=2$ for all $n \in N$.

~~ 37. $\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$

$=\frac{1}{2 \sin \frac{2 \pi}{15}} \cdot(2 \sin \frac{2 \pi}{15} \cdot \cos \frac{2 \pi}{15}) \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$

(On multiplying and dividing the expression by $2 \sin \frac{2 \pi}{15}$ )

$=\frac{1}{2 \sin \frac{2 \pi}{15}} \cdot \sin \frac{4 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$

$(\because \sin 2 A=2 \sin A \cos A)$

$=\frac{1}{4 \sin \frac{2 \pi}{15}} \cdot(2 \sin \frac{4 \pi}{15} \cdot \cos \frac{4 \pi}{15}) \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$

$=\frac{1}{4 \sin \frac{2 \pi}{15}} \cdot \sin \frac{8 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$ $=\frac{1}{8 \sin \frac{2 \pi}{15}} \cdot(2 \sin \frac{8 \pi}{15} \cdot \cos \frac{8 \pi}{15}) \cdot \cos \frac{16 \pi}{15}$

$=\frac{1}{8 \sin \frac{2 \pi}{15}} \cdot \sin \frac{16 \pi}{15} \cdot \cos \frac{16 \pi}{15}$

$=\frac{1}{16 \sin \frac{2 \pi}{15}} \cdot \sin \frac{32 \pi}{15}$

$=\frac{1}{16 \sin \frac{2 \pi}{15}} \cdot(\sin 2 \pi+\frac{2 \pi}{15})($ Note the step)

$=\frac{1}{16 \sin \frac{2 \pi}{15}} \cdot \sin \frac{2 \pi}{15} \quad(\because \sin (360^{\circ}+\theta)=\sin \theta)$

$=\frac{1}{16}$.

~~ 38. $\sin ^{2} \alpha+\sin ^{2} \beta-\sin ^{2} \gamma$

$=\sin ^{2} \alpha+\sin (\beta+\gamma) \sin (\beta-\gamma)$

[Using $.\sin ^{2} A-\sin ^{2} B=\sin (A+B) \sin (A-B)]$

$=\sin ^{2} \alpha+\sin (\pi-\alpha) \sin (\beta-\gamma)(\because \alpha+\beta+\gamma=\pi)$

$=\sin ^{2} \alpha+\sin \alpha \sin (\beta-\gamma) \quad(\because \sin (180^{\circ}-\theta)=\sin \theta)$

$=\sin \alpha[\sin \alpha+\sin (\beta-\gamma)]$

$=\sin \alpha[\sin (\pi-(\beta+\gamma)+\sin (\beta-\gamma)]$

$=\sin \alpha[\sin (\beta+\gamma)+\sin (\beta-\gamma)]$

$=\sin \alpha[\sin \beta \cos \gamma+\cos \beta \sin \gamma+\sin \beta \cos \gamma-\cos \beta \sin \gamma]$ $(\because \sin (A+B)=\sin A \cos B+\cos A \sin B)$

$=2 \sin \alpha \sin \beta \cos \gamma$.

~~ 39. We know that $\cos 2 \theta=2 \cos ^{2} \theta-1$

Putting $\theta=15^{\circ}$ in the above expression,

$\cos 30^{\circ}=2 \cos ^{2} 15^{\circ}-1$

$\Rightarrow \frac{\sqrt{3}}{2}+1=2 \cos ^{2} 15^{\circ}$

$\Rightarrow \cos ^{2} 15^{\circ}=\frac{\sqrt{3}+2}{4} \Rightarrow \cos 15^{\circ}=\frac{1}{2} \sqrt{\sqrt{3}+2}$.

~~ 40. $\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 A}}}$

$=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4 A}}}$

$=\sqrt{2+\sqrt{2+\sqrt{4 \cos ^{2} 2 A}}}$

$=\sqrt{2+\sqrt{2+2 \cos 2 A}}$

$(\begin{matrix} \because \cos 2 A=2 \cos ^{2} A-1 \\ \text{ or } 1+\cos 2 A=2 \cos ^{2} A\end{matrix} )$

$=\sqrt{2+\sqrt{2(1+\cos 2 A)}}$

$=\sqrt{2+\sqrt{2(2 \cos ^{2} A)}}=\sqrt{2+2 \cos A}$

$=\sqrt{2(1+\cos A)}=\sqrt{2 \times 2 \cos ^{2} \frac{A}{2}}$

$=\sqrt{4 \cos ^{2} A / 2}=2 \cos A / 2$.

~~ 41. $\frac{\sin x}{1+\cos x}=\frac{2 \sin x / 2 \cos x / 2}{1+2 \cos ^{2} x / 2-1}$

$(\because \sin 2 \theta=2 \sin \theta \cos \theta.$ and $.\cos 2 \theta=2 \cos ^{2}-1)$ $=\frac{2 \sin x / 2 \cos x / 2}{2 \cos ^{2} x / 2}=\frac{\sin x / 2}{\cos x / 2}=\tan x / 2$.

~~ 42. $\frac{\sin 70^{\circ}+\cos 40^{\circ}}{\cos 70^{\circ}+\sin 40^{\circ}}=\frac{\sin 70^{\circ}+\cos (90^{\circ}-50^{\circ})}{\cos 70^{\circ}+\sin (90^{\circ}-50^{\circ})}$

$ \begin{aligned} & =\frac{\sin 70^{\circ}+\sin 50^{\circ}}{\cos 70^{\circ}+\cos 50^{\circ}} \quad \begin{bmatrix} \because \sin (90^{\circ}-\theta)=\cos \theta \\ \cos (90^{\circ}-\theta)=\sin \theta \end{bmatrix} \\ & =\frac{2 \sin (\frac{70^{\circ}+50^{\circ}}{2}) \cos (\frac{70^{\circ}-50^{\circ}}{2})}{2 \cos (\frac{70^{\circ}+50^{\circ}}{2}) \cos (\frac{70^{\circ}-50^{\circ}}{2})} \end{aligned} $

$ \begin{bmatrix} \because \sin A+\sin B=2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2}) \\ \cos A+\cos B=2 \cos (\frac{A+B}{2}) \cos (\frac{A-B}{2}) \end{bmatrix} $

$=\frac{2 \sin 60^{\circ} \cos 10^{\circ}}{2 \cos 60^{\circ} \cos 10^{\circ}}=\tan 60^{\circ}=\sqrt{3}$.

~~ 43. $\tan ^{2} \alpha=\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}=\frac{2 \sin ^{2} \alpha}{2 \cos ^{2} \alpha}=\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}$

$ \begin{aligned} & (\because 1-\cos 2 \theta=2 \sin ^{2} \theta, 1+\cos 2 \theta=2 \cos ^{2} \theta) \\ = & \frac{1-\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}}{1+\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}}=\frac{3-\cos 2 \beta-3 \cos 2 \beta+1}{3-\cos 2 \beta+3 \cos 2 \beta-1} \\ = & \frac{4-4 \cos 2 \beta}{2+2 \cos 2 \beta}=\frac{4(1-\cos 2 \beta)}{2(1+\cos 2 \beta)} \\ = & 2(\frac{.\sin ^{2} \beta)}{.2 \cos ^{2} \beta)})=2 \tan ^{2} \beta \end{aligned} $

$\therefore \tan ^{2} \alpha=2 \tan ^{2} \beta \Rightarrow \tan \alpha=\sqrt{2} \tan \beta$.

~~ 44. $\frac{\sin 5 \theta}{\sin \theta}=\frac{\sin (2 \theta+3 \theta)}{\sin \theta}$

$=\frac{1}{\sin \theta}\lbrace \sin 2 \theta \cos 3 \theta+\cos 2 \theta \sin 3 \theta\rbrace $

$=\frac{1}{\sin \theta}\lbrace 2 \sin \theta \cos \theta(4 \cos ^{3} \theta-3 \cos \theta).$

$ .+(2 \cos ^{2} \theta-1)(3 \sin \theta-4 \sin ^{3} \theta)\rbrace $

$(\because \cos 3 A=4 \cos ^{3} A-3 \cos A, \sin 3 A=3 \sin A-4 \sin ^{3} A)$ $=\lbrace 2 \cos \theta(4 \cos ^{3} \theta-3 \cos \theta)+(2 \cos ^{2} \theta-1)(3-4 \sin ^{2} \theta)\rbrace $.

(Note the step)

$=8 \cos ^{4} \theta-6 \cos ^{2} \theta+6 \cos ^{2} \theta-3-8 \cos ^{2} \theta \sin ^{2} \theta+4 \sin ^{2} \theta$

$=8 \cos ^{4} \theta-3-8 \cos ^{2} \theta(1-\cos ^{2} \theta)+4(1-\cos ^{2} \theta)$

$(\because \sin ^{2} \theta+\cos ^{2} \theta=1)$

$=8 \cos ^{4} \theta-3-8 \cos ^{2} \theta+8 \cos ^{4} \theta+4-4 \cos ^{2} \theta$

$=16 \cos ^{4} \theta-12 \cos ^{2} \theta+1$.

~~ 45. $\sin A+\sin B=a$

$\Rightarrow 2 \sin (\frac{A+B}{2}) \cos (\frac{A-B}{2})=a$

$\cos A-\cos B=b$

$\Rightarrow 2 \sin (\frac{A+B}{2}) \sin (\frac{B-A}{2})=b$

$\Rightarrow \quad 2 \sin (\frac{A+B}{2}) \sin [-(\frac{A-B}{2})]=b$

$\Rightarrow-2 \sin (\frac{A+B}{2}) \sin (\frac{A-B}{2})=b$

$(\because \sin (-\theta)=-\sin \theta)$

Dividing eqn (ii) by eqn ( ), we get

$\frac{-\sin (\frac{A-B}{2})}{\cos (\frac{A-B}{2})}=\frac{b}{a}$

$\Rightarrow-\tan (\frac{A-B}{2})=b / a \Rightarrow \tan (\frac{A-B}{2})=-b / a$

$\therefore \quad \cos (A-B)=\frac{1-\tan ^{2}(\frac{A-B}{2})}{1+\tan ^{2}(\frac{A-B}{2})}$

$(\because \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta})$

$=\frac{1-b^{2} / a^{2}}{1+b^{2} / a^{2}}=\frac{\boldsymbol{a}^{2}-\boldsymbol{b}^{2}}{\boldsymbol{a}^{2}+\boldsymbol{b}^{2}}$.

~~ 46. $\sin 16^{\circ}+\cos 16^{\circ}=\sin (15^{\circ}+1^{\circ})+\cos (15^{\circ}+1^{\circ})$

$=(\sin 15^{\circ} \cos 1^{\circ}+\cos 15^{\circ} \sin 1^{\circ})$

$+(\cos 15^{\circ} \cos 1^{\circ}-\sin 15^{\circ} \sin 1^{\circ})$

$=\sin 1^{\circ}(\cos 15^{\circ}-\sin 15^{\circ})+\cos 1^{\circ}(\sin 15^{\circ}+\cos 15^{\circ})$

Now, $\cos 15^{\circ}=\cos (45^{\circ}-30^{\circ})$

$=\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}$

$=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

and $\sin 15^{\circ}=\sin (45^{\circ}-30^{\circ})$

$=\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}$

$=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$

$\therefore \sin 16^{\circ}+\cos 16^{\circ}$

$=\sin 1^{\circ}(\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}})+\cos 1^{\circ}(\frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{\sqrt{3}+1}{2 \sqrt{2}})$

$=\sin 1^{\circ}(\frac{2}{2 \sqrt{2}})+\cos 1^{\circ}(\frac{2 \sqrt{3}}{2 \sqrt{2}})$

$=\frac{1}{\sqrt{2}} \sin 1^{\circ}+\frac{\sqrt{3}}{\sqrt{2}} \cos 1^{\circ}=\frac{1}{\sqrt{2}}(\sin 1^{\circ}+\sqrt{3} \cos 1^{\circ})$.

~~ 47. $(1+\cos \frac{\pi}{8})(1+\cos \frac{3 \pi}{8})(1+\cos \frac{5 \pi}{8})(1+\cos \frac{7 \pi}{8})$

$=(1+\cos \frac{\pi}{8})(1+\cos \frac{3 \pi}{8})(1+\cos (\pi-\frac{3 \pi}{8}))$

$(1+\cos (\pi-\frac{\pi}{8}))$

$=(1+\cos \frac{\pi}{8})(1+\cos \frac{3 \pi}{8})(1-\cos \frac{3 \pi}{8})(1-\cos \frac{\pi}{8})$

$(\because \cos (\pi-\theta)=-\cos \theta)$

$=(1-\cos ^{2} \frac{\pi}{8})(1-\cos ^{2} \frac{3 \pi}{8})=\sin ^{2} \frac{\pi}{8} \sin ^{2} \frac{3 \pi}{8}$

$=\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8} \quad(\because \sin \frac{3 \pi}{8}=\sin (\frac{\pi}{2}-\frac{\pi}{8})=\cos \frac{\pi}{8})$

$=\frac{1}{4}(4 \sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8})=\frac{1}{4}(2 \sin \frac{\pi}{8} \cos \frac{\pi}{8})^{2}$

$=\frac{1}{4}(\sin \frac{\pi}{4})^{2} \frac{1}{4}(\frac{1}{\sqrt{2}})^{2}=\frac{1}{8}$.

(Using, $2 \sin \theta \cos \theta=\sin 2 \theta$ )

~~ 48. $\cos 2(\theta+\phi)+4 \cos (\theta+\phi) \sin \theta \sin \phi+2 \sin ^{2} \phi$

$=\lbrace \cos 2 \theta \cos 2 \phi-\sin 2 \theta \sin 2 \phi\rbrace $

$+4\lbrace (\cos \theta \cos \phi-\sin \theta \sin \phi) \sin \theta \sin \phi\rbrace +2 \sin ^{2} \phi$

$=\lbrace (1-2 \sin ^{2} \theta)(1-2 \sin ^{2} \phi)-2 \sin \theta \cos \theta .2 \sin \phi \cos \phi\rbrace $

$+[4 \cos \theta \cos \phi \sin \theta \sin \phi-4 \sin ^{2} \theta \sin ^{2} \phi]+2 \sin ^{2} \phi$

$=1-2 \sin ^{2} \theta-2 \sin ^{2} \phi+4 \sin ^{2} \theta \sin ^{2} \phi$

$-4 \sin \theta \sin \phi \cos \theta \cos \phi+4 \cos \theta \cos \phi \sin \theta \sin \phi$

$-4 \sin ^{2} \theta \sin ^{2} \phi+2 \sin ^{2} \phi$

$=1-2 \sin ^{2} \theta=\cos 2 \theta$.

~~ 49. $\tan A-\tan B=x$

$\Rightarrow \frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}=x$

$\Rightarrow \frac{\sin A \cos B-\sin B \cos A}{\cos A \cos B}=x$

$\Rightarrow \frac{\sin (A-B)}{\cos A \cos B}=x$

$\Rightarrow \frac{1}{x}=\frac{\cos A \cos B}{\sin (A-B)}$

$\cot B-\cot A=y$

$\Rightarrow \frac{\cos B}{\sin B}-\frac{\cos A}{\sin A}=y$

$\Rightarrow \frac{\cos B \sin A-\cos A \sin B}{\sin B \sin A}=y$

$\Rightarrow \frac{\sin (A-B)}{\sin A \sin B}=y \Rightarrow \frac{1}{y}=\frac{\sin A \sin B}{\sin (A-B)}$

Adding ( $i$ ) and (ii), we get

$\frac{1}{x}+\frac{1}{y}=\frac{\cos A \cos B+\sin A \sin B}{\sin (A-B)}$

$\Rightarrow \frac{\cos (A-B)}{\sin (A-B)}=\cot (\boldsymbol{A}-\boldsymbol{B})$.

~~ 50. $\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha$

$=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+\frac{8}{\tan 8 \alpha}$

$=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+\frac{8}{\frac{2 \tan 4 \alpha}{1-\tan ^{2} 4 \alpha}}$

$ (\because \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}) $

$=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+\frac{4(1-\tan ^{2} 4 \alpha)}{\tan 4 \alpha}$

$=\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+\frac{4}{\tan 4 \alpha}-4 \tan 4 \alpha$

$ \begin{aligned} & =\tan \alpha+2 \tan 2 \alpha+\frac{4}{\tan 4 \alpha} \\ & =\tan \alpha+2 \tan 2 \alpha+\frac{4}{\frac{2 \tan 2 \alpha}{1-\tan ^{2} 2 \alpha}} \\ & =\tan \alpha+2 \tan 2 \alpha+\frac{2(1-\tan ^{2} 2 \alpha)}{\tan 2 \alpha} \\ & =\tan \alpha+\frac{2}{\tan 2 \alpha}=\tan \alpha+\frac{2}{\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}} \\ & =\tan \alpha+\frac{1-\tan ^{2} \alpha}{\tan \alpha}=\frac{1}{\tan \alpha}=\cot \alpha . \end{aligned} $

SELF ASSESSMENT SHEET

If $q cosec \theta=p$ and $\theta$ is acute, then what is the value of $\sqrt{p^{2}-q^{2}} \tan \theta$ ?

(a) $p$

(b) $q$

(d) $\sqrt{p^{2}+q^{2}}$

~~ 2. If $0<x<45^{\circ}$ and $45^{\circ}<y<90^{\circ}$, then which one of the following is correct?

(a) $\sin x=\sin y$

(b) $\sin x<\sin y$

(d) $\sin x \leq \sin y$

~~ 3. If $\tan ^{2} y cosec^{2} x-1=\tan ^{2} y$, then which one of the following is correct?

(a) $x-y=0$

(b) $x=2 y$

(d) $x-y=1^{\circ}$

~~ 4. If $\alpha$ and $\beta$ the complementary angles, then what is $\sqrt{cosec \alpha \cdot cosec \beta}(\frac{\sin \alpha}{\sin \beta}+\frac{\cos \alpha}{\cos \beta})^{-1 / 2}$ equal to

(a) 0

(b) 1

(d) None of there

~~ 5. What is $(\frac{\sec 18^{\circ}}{\sec 144^{\circ}}+\frac{cosec 18^{\circ}}{cosec 144^{\circ}})$ equal to?

(CDS 2011)

(a) $\sec 18^{\circ}$

(b) $cosec 18^{\circ}$

(d) $-cosec 18^{\circ}$

~~ 6. If an angle $\alpha$ is divided into two parts $A$ and $B$ such that $A-B=x$ and $\tan A: \tan B=2: 1$, then what is $\sin x$ equal to?

(a) $3 \sin \alpha$

(b) $\frac{2 \sin \alpha}{3}$

(d) $2 \sin \alpha$

(NDA/NA 2011)

~~ 7. If $\cos \theta=\frac{8}{17}$ and $\theta$ lies in the first quadrant, then the value of $\cos (30^{\circ}+\theta)+\cos (45^{\circ}-\theta)+\cos (120^{\circ}-\theta)$ is

(a) $\frac{23}{17}(\frac{\sqrt{3}-1}{2}-\frac{1}{\sqrt{2}})$

(b) $\frac{23}{17}(\frac{\sqrt{3}+1}{2}-\frac{1}{\sqrt{2}})$

(d) $\frac{23}{17}(\frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}})$

(DCE 2008)

~~ 8. Let $A, B, C$ be the angles of a plain triangle $\tan (A / 2)=\frac{1}{3}, \tan (\frac{B}{2})=\frac{2}{3}$. Then $\tan (C / 2)$ is equal to

(a) $\frac{2}{9}$

(b) $\frac{1}{3}$

(d) $\frac{2}{3}$

~~ 9. If $\frac{1+\cos A}{1-\cos A}=\frac{m^{2}}{n^{2}}$, then $\tan A$ is equal to

(a) $\pm \frac{2 m n}{m^{2}-n^{2}}$

(b) $\pm \frac{2 m n}{m^{2}+n^{2}}$

(d) $\frac{m^{2}-n^{2}}{m^{2}+n^{2}}$

(J&K CET 2007)

~~ 10. If $\alpha, \beta, \gamma$, are in A.P, then $\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}$. is equal to

(a) $\sin \alpha$

(b) $\cos \alpha$

(d) $\cot \beta$

ANSWERS

~~ 2. (b)

~~ 3. (a)

~~ 4. (b)

~~ 5. (a)

~~ 6. (c)

~~ 7. (c)

~~ 8. (c)

~~ 9. (a)

~~ 10. (d)

HINTS AND SOLUTIONS

$q cosec \theta=p$

$\therefore \sqrt{p^{2}-q^{2}} \tan \theta$

$=\sqrt{q^{2} cosec^{2} \theta-q^{2}} \cdot \tan \theta=\sqrt{q^{2}(cosec^{2} \theta-1)} \cdot \tan \theta$

$=\sqrt{q^{2} \cot ^{2} \theta} \cdot \tan \theta=q \cot \theta \cdot \tan \theta=\boldsymbol{q}$.

~~ 2. Since $\sin \theta$ in $cosec$ from $\theta=0^{\circ}$ to $\theta=90^{\circ}$, i.e., where $0^{\circ}<\theta<90^{\circ}$ then $0<\sin \theta<1$. $\therefore \boldsymbol{\operatorname { s i n }} \boldsymbol{x}<\sin \boldsymbol{y}$.

~~ 3. $\tan ^{2} y cosec^{2} x-1=\tan ^{2} y$

$\Rightarrow \tan ^{2} y cosec^{2} x-\tan ^{2} y=1$

$\Rightarrow \tan ^{2} y(cosec^{2} x-1)=1 \Rightarrow \tan ^{2} y \cdot \cot ^{2} x=1$

$\Rightarrow \cot ^{2} x=\cot ^{2} y \Rightarrow x=y \Rightarrow \boldsymbol{x}-\boldsymbol{y}=\mathbf{0}$.

~~ 4. Given $\alpha+\beta=90^{\circ}$

$\therefore \sqrt{cosec \alpha \cdot cosec \beta}(\frac{\sin \alpha}{\sin \beta}+\frac{\cos \alpha}{\cos \beta})^{-1 / 2}$

$=\frac{1}{(\sin \alpha \sin \beta)^{1 / 2}}(\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \beta \cos \beta})^{-1 / 2}$

$=\frac{1}{(\sin \alpha \sin \beta)^{1 / 2}}(\frac{\sin (\alpha+\beta)}{\sin \beta \cos \beta})^{-1 / 2}$

$=\frac{1}{(\sin \alpha \sin \beta)^{1 / 2}}(\frac{\sin 90^{\circ}}{\sin \beta \cos (90^{\circ}-\alpha)})^{-1 / 2}$

$=\frac{1}{(\sin \alpha \sin \beta)^{\frac{1}{2}}}(\frac{1}{\sin \beta \sin \alpha})^{-\frac{1}{2}}$

$=\frac{1}{1} \times(\sin \alpha \sin \beta)^{\frac{1}{2}}=\mathbf{1}$.

$(\sin \alpha \sin \beta)^{\overline{2}}$

~~ 5. $\frac{\sec 18^{\circ}}{\sec 144^{\circ}}+\frac{cosec 18^{\circ}}{cosec 144^{\circ}}$

$=\frac{\sec 18^{\circ}}{\sec (180^{\circ}-36^{\circ})}+\frac{cosec 18^{\circ}}{cosec(180^{\circ}-36^{\circ})}$

$=-\frac{\sec 18^{\circ}}{\sec 36^{\circ}}+\frac{cosec 18^{\circ}}{cosec 36^{\circ}} \quad(\begin{matrix} \because \sec (180^{\circ}-\theta)=-\sec \theta \\ cosec(180^{\circ}-\theta)=cosec \theta\end{matrix} )$

$=\frac{\sin 36^{\circ}}{\sin 18^{\circ}}-\frac{\cos 36^{\circ}}{\cos 18^{\circ}}=\frac{\sin 36^{\circ} \cos 18^{\circ}-\cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 18^{\circ}}$

$(\because \sin (A-B)=\sin A \cos B-\cos A \sin B)$

$=\frac{\sin (36^{\circ}-18^{\circ})}{\sin 18^{\circ} \cos 18^{\circ}}=\frac{\sin 18^{\circ}}{\sin 18^{\circ} \cos 18^{\circ}}=\frac{1}{\cos 18^{\circ}}=\sec 18^{\circ}$.

~~ 6. $\alpha=A+B$

$x=A-B$

$\Rightarrow A=\frac{\alpha+x}{2}$ and $B=\frac{\alpha-x}{2}$ Given, $\frac{\tan A}{\tan B}=\frac{2}{1} \Rightarrow \frac{\tan (\frac{\alpha+x}{2})}{\tan (\frac{\alpha-x}{2})}=\frac{2}{1}$

$\Rightarrow \frac{\sin (\frac{\alpha+x}{2}) \cdot \cos (\frac{\alpha-x}{2})}{\cos (\frac{\alpha+x}{2}) \cdot \sin (\frac{\alpha-x}{2})}=\frac{2}{1}$

$\Rightarrow \frac{2 \sin (\frac{\alpha+x}{2}) \cos (\frac{\alpha-x}{2})}{2 \cos (\frac{\alpha+x}{2}) \sin (\frac{\alpha-x}{2})}=\frac{2}{1}$

$\Rightarrow \frac{\sin \alpha+\sin x}{\sin \alpha-\sin x}=\frac{2}{1}$

$(\begin{matrix} \because 2 \sin (\frac{C+D}{2}) \cos (\frac{C-D}{2})=\sin C+\sin D \\ \text{ and } 2 \cos (\frac{C+D}{2}) \sin (\frac{C-D}{2})=\sin C-\sin D\end{matrix} )$

$\Rightarrow \sin \alpha+\sin x=2 \sin \alpha-2 \sin x$

$\Rightarrow 3 \sin x=\sin \alpha \Rightarrow \sin x=\frac{\sin \alpha}{3}$.

~~ 7. $\cos (30^{\circ}+\theta)+\cos (45^{\circ}-\theta)+\cos (120^{\circ}-\theta)$

$=\cos 30^{\circ} \cos \theta-\sin 30^{\circ} \sin \theta+\cos 45^{\circ} \cos \theta$

$+\sin 45^{\circ} \sin \theta+\cos 120^{\circ} \cos \theta+\sin 120^{\circ} \sin \theta$

$(\cos \theta=\frac{8}{17} \Rightarrow \sin \theta=\sqrt{1-(\frac{8}{17})^{2}}=\sqrt{\frac{289-64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17})$

$=\frac{\sqrt{3}}{2} \times \frac{8}{17}-\frac{1}{2} \times \frac{15}{17}+\frac{1}{\sqrt{2}} \times \frac{8}{17}+\frac{1}{\sqrt{2}} \times \frac{15}{17}$

$+(\frac{-1}{2}) \times \frac{8}{17}+\frac{\sqrt{3}}{2} \times \frac{15}{17}$

$ \begin{bmatrix} \because \cos 120^{\circ}=\cos (180^{\circ}-60^{\circ})=-\cos 60^{\circ}=-\frac{1}{2} \\ \sin 120^{\circ}=\sin (180^{\circ}-60^{\circ})=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\end{bmatrix} $

$=\frac{8}{17}(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2})+\frac{15}{17}(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2})$

$=\frac{8}{17}(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})+\frac{15}{17}(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})$

$=\frac{23}{17}(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})$.

~~ 8. $A+B+C=180^{\circ}$ (Angle sum property of a triangle)

$\Rightarrow A+B=180^{\circ}-C$

$\Rightarrow \frac{A+B}{2}=90^{\circ}-\frac{C}{2}$

$\Rightarrow \tan (\frac{A}{2}+\frac{B}{2})=\tan (90^{\circ}-\frac{C}{2})$

$\Rightarrow \frac{\tan A / 2+\tan B / 2}{1-\tan A / 2 \tan B / 2}=\cot C / 2$

$\Rightarrow \frac{\frac{1}{3}+\frac{2}{3}}{1-\frac{1}{3} \times \frac{2}{3}}=\cot C / 2 \Rightarrow \frac{1}{1-\frac{2}{9}}=\cot C / 2$

$\Rightarrow \frac{9}{7}=\cot C / 2 \Rightarrow \tan C / 2=7 / 9$.

~~ 9. $\frac{1+\cos A}{1-\cos A}=\frac{m^{2}}{n^{2}}$

$\Rightarrow \frac{2 \cos ^{2} A / 2}{2 \sin ^{2} A / 2}=\frac{m^{2}}{n^{2}}$

$(\because 1+\cos 2 A=2 \cos ^{2} A, 1-\cos ^{2} A=2 \sin ^{2} A)$

$ \begin{aligned} & \Rightarrow \tan ^{2} A / 2=\frac{n^{2}}{m^{2}} \Rightarrow \tan A / 2= \pm \frac{n}{m} \\ & \therefore \tan A=\frac{2 \tan A / 2}{1-\tan ^{2} A / 2}= \pm \frac{\frac{2 n}{m}}{1-\frac{n^{2}}{m^{2}}} \\ & \quad= \pm \frac{\frac{2 n}{m}}{\frac{m^{2}-n^{2}}{m^{2}}}= \pm \frac{\mathbf{2} \boldsymbol{n} \boldsymbol{m}}{\boldsymbol{m}^{2}-\boldsymbol{n}^{2}} \end{aligned} $

~~ 10. Since $\alpha, \beta$ and $\gamma$ are in A.P.

$ \begin{aligned} & \alpha+\gamma=2 \beta \\ & \Rightarrow \beta=\frac{\alpha+\gamma}{2} \\ & \therefore \frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}=\frac{2 \cos (\frac{\alpha+\gamma}{2}) \sin (\frac{\alpha-\gamma}{2})}{2 \sin (\frac{\alpha+\gamma}{2}) \sin (\frac{\alpha-\gamma}{2})} \\ & =\cot (\frac{\alpha+\gamma}{2})=\cot \beta . \end{aligned} $

11
Coordinate Geometry

KEY FACTS

Co-ordinate Geometry is that branch of geometry in which two numbers, called co-ordinates are used to indicate the position of a point in a plane and which make use of algebraic methods in the study of geometric figures.
Rectangular Axes: The position of a point in a plane is fixed by selecting the axes of reference which are formed by combining the two number scales at right angles to each other so that their zero points coincide. The horizontal number scale is called the $\boldsymbol{x}$-axis and the vertical number scale, the $\boldsymbol{y}$-axis. The point where the two scales cross each other is the origin. The two together are called rectangular axes.
Co-ordinates (Abscissa and Ordinate): The position of each point of the plane is determined with reference to the rectangular axes by means of a pair of numbers called co-ordinates which are distances of the point from the respective axes. The distance of the point from the $y$-axis is called the $x$-coordinate or abscissa and the distance of the point from the $x$-axis is called $y$-coordinate or ordinate. The position of point $P$ in the plane with respect to the co-ordinate axes is represented by the ordered pair $(x, y)$, where $x$ is abscissa and $y$ is the ordinate.

Co-ordinate of the origin are $(0,0)$.

The co-ordinates of any point on the $x$-axis are $(\boldsymbol{x}, \mathbf{0})$ and the co-ordinates of any point of the $y$-axis are $(0, \boldsymbol{y})$

Quadrants: The co-ordinate axes separate the plane into four regions called the quadrants. By custom the quadrants are numbered I, II, III, IV, in the counter clockwise direction as shown in the figure.

(i) For distances along the $x$-axis, positive values are measured to the right of the origin and negative values to its left.

(ii) For distances along the $y$-axis, positive values are measured upward above $x$-axis and negative values downward below $x$-axis.

Study the table given below for clear understanding.

Region	Quadrant Name	Value of $\boldsymbol{x}$ and $\boldsymbol{y}$	Sign of $(\boldsymbol{x}, \boldsymbol{y})$
$XOY$	1st quadrant	$x>0, y>0$	$(+,+)$
$YOX^{\prime}$	2nd quadrant	$x<0, y>0$	$(-,+)$
$X^{\prime} OY^{\prime}$	3rd quadrant	$x<0, y<0$	$(-,-)$
$XOY^{\prime}$	4th quadrant	$x>0, y<0$	$(+,-)$

Distance Formula:

(a) The distance $\boldsymbol{d}$ between any two points say $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by:

$ d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} $

$ \Rightarrow \quad d^{2}=(x_2-x_1)^{2}+(y_2-y_1)^{2} \Rightarrow d= \pm \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} $

(b) The distance of a point $P(x_1, y_1)$ form the origin

$ =\sqrt{(x_1-0)^{2}+(y_1-0)^{2}}=\sqrt{x_1^{2}+y_1^{2}} $

Section Formula:

(a) Internal division: If $P(x, y)$ divides the line segment formed by the joining of the points $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1: m_2$. Then

$ x=\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \quad \text{ and } \quad y=\frac{m_1 y_2+m_2 y_1}{m_1+m_2} $

(b) External division: If $P(x, y)$ divides the line segment formed by the joining of the points $A(x_1, y_1)$ and $B(x_2, y_2)$ externally in the ratio $m_1: m_2$, then

$ x=\frac{m_1 x_2-m_2 x_1}{m_1-m_2} \quad \text{ and } \quad y=\frac{m_1 y_2-m_2 y_1}{m_1-m_2} $

(c) Mid-point formula: Putting the ratio as $m_1: m_2=1: 1$, the co-ordinates of the mid-point are $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.

Some Important Points Relating to a Triangle:

(a) Centroid of a triangle (Point of intersection of medians of a $\Delta$ ). If $(x_1, y_1),(x_2, y_2)$ and $(x_3, y_3)$ are the vertices of a triangle, then the co-ordinates of its centroid are

$ (\frac{(x_1+x_2+x_3)}{3}, \frac{(y_1+y_2+y_3)}{3}) $

$(\begin{matrix} \text{ Centroid } G \text{ divides median from the vertex } \\ \text{ side in the ratio } 2: 1 \text{, i.e., } A G: G L=2: 1\end{matrix} )$

(b) Incentre (Point of concurrence of the internal bisectors of the angles of a $\triangle$ ).

If $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ are the vertices of a triangle $A B C$ such that $B C=a, C A=b$ and $A B=c$, then the co-ordinates of the incentre are $(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c})$

(c) Circumcentre. If ’ $O$ ’ is the circumcentre of $\triangle A B C$, then $O A=O B=O C$ and $O A$ is called the circumradius. So, to find the circumcentre of $\triangle A B C$, we use the relation $O A=O B=O C$. This gives two simultaneous liner equations and their solution provides the co-ordinates of the circumcentre.

(d) Orthocentre (Point of concurrence of the altitudes of a $\Delta$ ).

To determine the orthocentre, first we find equations of lines passing through the vertices and perpendicular to the opposite sides. Solving any two of these three equations we get the co-ordinates of the orthocentre.

(a) Area of a triangle: The area of a triangle $A B C$ those vertices are given as $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$

$ =\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| $

Condition of collinearity of three points: If three points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ be on a straight line, the area of the triangle

formed by them is zero, i.e.

$ \frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0 . \quad \stackrel{\bullet}{\stackrel{A}{A}} \underset{(x_1, y_1)}{.\stackrel{B}{B}, y_2)} \quad \underset{(x_3, y_3)}{\stackrel{C}{C}} $

(b) Area of a quadrilateral: The area of the quadrilateral where vertices taken in order $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ and $D(x_4, y_4)$ can be calculated by the formula:

Area of quad. $A B C D=\frac{1}{2}|(x_1 y_2-x_2 y_1)+(x_2 y_3-x_3 y_2)+(x_3 y_4-x_4 y_3)+(x_4 y_1-x_1 y_4)|$

THE STRAIGHT LINE

Slope or Gradient of a Straight Line: The slope (or gradient) of a line is the tangent of the angle which the part of the line above the $x$-axis makes with the positive direction of the $x$-axis. The slope of a line is indicated by the letter ’ $m$ ‘.

Thus, if $A P$ be the portion of the line above the $x$-axis and $\angle X A P=\theta$, then $\boldsymbol{m}=\boldsymbol{\operatorname { t a n }} \boldsymbol{\theta}$.

Notes:

(a) $\theta$ is measured positively, i.e., in the anti-clockwise direction from $0^{\circ}$ to $180^{\circ}$.

(b) The slope of a line is positive if it makes an acute angle in the anti-clockwise direction with the $x$-axis.

(c) The slope of the line is negative if the line makes an obtuse angle in the anti-clockwise direction with the $x$-axis.

(d) The slope of a line is zero if the line is parallel to the $x$-axis.

(e) The slope of a line parallel to the $y$-axis is not defined ( $\tan 90^{\circ}$ does not exist) or is taken loosely as $\infty$.

(i)

(ii)

(f) (i) If $\theta=45^{\circ}$, then $m=45^{\circ}=1$

(ii) If $\theta=135^{\circ}$, then $m=\tan 135^{\circ}=-1$

Hence, slope of a line making equal angles with the axes is given by $\boldsymbol{m}= \pm \mathbf{1}$.

Slope of a line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by

$ \begin{aligned} \boldsymbol{m} & =\tan \theta=\frac{y_2-y_1}{x_2-x_1} \\ & =\frac{(\text{ difference of ordinates })}{(\text{ difference of abscissas) }} \end{aligned} $

Slope of a line whose equation is $a x+b y+c=0$ is given by

$ m=\frac{-a}{b}=\frac{- \text{ coefficient of } x}{\text{ cofficient of } y} $

Angle between two lines whose slopes are given: The angle $\theta$ between two lines whose slopes $m_1$ and $m_2$ is given by

$ \tan \theta=|\frac{m_1-m_2}{1+m_1 m_2}|=|\frac{\text{ difference of slopes }}{1+\text{ product of slopes }}| $

Particular Cases:

(i) Two lines are parallel when $\boldsymbol{m}_1=\boldsymbol{m}_2$, i.e., their slopes are equal.

(ii) Two lines are perpendicular when $m_1 m_2=-1$, i.e., $m_2=-\frac{1}{m_1}$.

Equation of a Straight Line (Different Forms):

(a) Equation of $x$-axis is $\boldsymbol{y}=\mathbf{0}$ and that of $y$-axis is $\boldsymbol{x}=\mathbf{0}$.

(b) Equation of a line parallel to $x$-axis is $\boldsymbol{y}=\boldsymbol{b}$, where $b$ is the distance of the line from the $x$-axis.

(c) Equation of a line perpendicular to y-axis is $\boldsymbol{x}=\boldsymbol{a}$, where a is the distance of the line from the $y$-axis.

(d) Equation of a line whose slope is $\boldsymbol{m}$ and which cuts off an intercept ’ $c$ ’ from the $y$-axis is $\boldsymbol{y}=\boldsymbol{m} \boldsymbol{x}+\boldsymbol{c}$.

In particular, equation of a line whose slope is $\boldsymbol{m}$ and which passes through the origin is $\boldsymbol{y}=\boldsymbol{m} \boldsymbol{x}$.

Fig. for $(d)$

(e) The equation of a line whose slope is $\boldsymbol{m}$ and which passes through the $(x_1, y_1)$ is $\boldsymbol{y}-\boldsymbol{y}_1=\boldsymbol{m}(\boldsymbol{x}-\boldsymbol{x}_1)$.

(f) The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $(\boldsymbol{y}-\boldsymbol{y}_1) = \frac{y_2 - y_1}{x_2 - x_1}(x-x_1)$.

(g) Intercept form. The equation of a line cutting off intercepts ’ $\boldsymbol{a}$ ’ and ’ $\boldsymbol{b}$ ’ respectively from the $x$-axis and $y$-axis is $\frac{\boldsymbol{x}}{\boldsymbol{a}}+\frac{\boldsymbol{y}}{\boldsymbol{b}}=\mathbf{1}$.

The length $P Q$ of the line intercepted between the co-ordinates axis $=\sqrt{\boldsymbol{a}^{2}+\boldsymbol{b}^{2}}$.

(h) Perpendicular form: The equation of the line on which the length of the perpendicular from the origin is ’ $p$ ’ and the perpendicular makes an angle ’ $\alpha$ ’ with the $x$-axis is $\boldsymbol{x} \boldsymbol{\operatorname { c o s }} \alpha+\boldsymbol{y} \sin \alpha=p$.

Fig. for $(g)$

Fig. for $(h)$

General Equation of a Straight Line: The general equation of a straight line is $\boldsymbol{a x}+\boldsymbol{b} \boldsymbol{y} \boldsymbol{c}=\mathbf{0}$, where $a, b, c \in R$ and $a$ and $b$ are not simultaneously zero.

(a) To transform $a x+b y+c=0$ to slope-intercept form, i.e., $y=m x+c$.

$ a x+b y+c=0 \Rightarrow y=\frac{-a}{b} x-\frac{c}{b} \Rightarrow \text{ Slope }=\frac{-\boldsymbol{a}}{\boldsymbol{b}} \text{ and } y \text{-intercept }=\frac{-\boldsymbol{c}}{\boldsymbol{b}} $

(b) To transform $a x+b y+c=0$ to intercept form $(\frac{x}{a}+\frac{y}{b}=1)$

$ a x+b y+c=0 \Rightarrow a x+b y=-c \Rightarrow \frac{a}{-c} x+\frac{b}{-c} y=1 \Rightarrow \frac{\boldsymbol{x}}{(-\boldsymbol{c} / \boldsymbol{a})}+\frac{\boldsymbol{y}}{(-\boldsymbol{c} / \boldsymbol{b})}=\mathbf{1} $

Divide throughout by $\sqrt{a^{2}+b^{2}}$, i.e.,

$ a x+b y+c=0 \Rightarrow \frac{a}{\sqrt{a^{2}+b^{2}}} x+\frac{b}{\sqrt{a^{2}+b^{2}}} y+\frac{c}{\sqrt{a^{2}+b^{2}}}=0 . $

Transpose the constant team to the right hand side and make it positive by changing the sign if necessary.

i.e., $\frac{-a}{\sqrt{a^{2}+b^{2}}} x+\frac{-b}{\sqrt{a^{2}+b^{2}}} y=\frac{c}{\sqrt{a^{2}+b^{2}}}$

where $\cos \alpha=\frac{-a}{\sqrt{a^{2}+b^{2}}}, \sin \alpha=\frac{-b}{\sqrt{a^{2}+b^{2}}}, p=\frac{c}{\sqrt{a^{2}+b^{2}}}$

Identical Lines:

If two equation $a x+b y+c=0$ and $a^{\prime} x+b^{\prime} y+c^{\prime}=0$ represent the same straight line, then $\frac{\boldsymbol{a}}{\boldsymbol{a}^{\prime}}=\frac{\boldsymbol{b}}{\boldsymbol{b}^{\prime}}=\frac{\boldsymbol{c}}{\boldsymbol{c}^{\prime}}$, i.e., ratio of coefficients of $x=$ ratio of coefficients of $y=$ ratio of constant terms

Position of Two Points Relative to a Line: Two points lie on the same side or opposite sides of the given line $a x+b y+c=0$ according as $a x+b y+c$ is of the same sign or opposite signs when the co-ordinates of the given points are substituted successively for $x$ and $y$ in $a x+b y+c$.

Example: To check whether the points $(2,3)$ and $(1,3)$ are on the same side or on opposite sides of the line $x-2 y=-3$.

The equation of the line can be written as $x-2 y+3=0$

Substituting co-ordinates $(2,3)$ on the left side of $(i)$, we get

$ 2-2(3)+3=2-6+3=-1 $

Substituting co-ordinates $(1,3)$ on the left side of $(i)$, we get

$ 1-2(3)+3=1-6+4=-2 $

Since both the results are of the same sign, the points lie on the same side of the line.

Intersection of Straight Lines: The co-ordinates of the point or points of intersection of two lines or two curves satisfy the equations of both of them and so their co-ordinates are obtained by solving their equations simultaneously.
Perpendicular distance of a point $(x_1, y_1)$ from the line $a x+b y+c=0$ is given by

$ d=\frac{|a x_1+b y_1+c|}{\sqrt{a^{2}+b^{2}}} $

Distance between two parallel lines: Let $a x+b y+c=0$ and $a x+b y+c^{\prime}=0$ be the given parallel lines.

To find the distance between them, choose a point on any one of the lines and find the perpendicular distance from this point on the other line.

A convenient point can be obtained by letting $x=0$ in $a x+b y+c=0$ and getting $y=\frac{-c}{b}$. Therefore, the point

would be $(0,-c / b)$ $\therefore \quad$ Hence required distance between the lines $=\frac{|a \cdot 0+b \cdot(\frac{-c}{b})+c^{\prime}|}{\sqrt{a^{2}+b^{2}}}=\frac{|\boldsymbol{c}^{\prime}-\boldsymbol{c}|}{\sqrt{\boldsymbol{a}^{2}+\boldsymbol{b}^{2}}}$.

If $u=0$ and $v=0$ be two given lines, then the line through their intersection is $\boldsymbol{u}+\boldsymbol{k} \boldsymbol{v}=\mathbf{0}$.
Equation of a line parallel to a given line $\boldsymbol{a x}+\boldsymbol{b} \boldsymbol{y}+\boldsymbol{c}=\mathbf{0}$ is $\boldsymbol{a} \boldsymbol{x}+\boldsymbol{b y}+\boldsymbol{k}=\mathbf{0}$, where $k$ is a constant.
Equation of a line perpendicular to a given line $a x+b y+c=0$ is $\boldsymbol{b x}-\boldsymbol{a y}+\boldsymbol{k}=\mathbf{0}$.

SOLVED EXAMPLES

Ex. 1 . If the points $(2,1)$ and $(1,-2)$ are equidistant from the point $(x, y)$, show that $x+3 y=0$.

Sol. $(x, y)$ is equidistant from the points $(2,1)$ and $(1,-2)$

$\Rightarrow$ Distance between $(x, y)$ and $(2,1)=$ Distance between $(x, y)$ and $(1,-2)$

$ \begin{aligned} & \Rightarrow \sqrt{(x-2)^{2}+(y-1)^{2}}=\sqrt{(x-1)^{2}+(y+2)^{2}} \Rightarrow x^{2}-4 x+4+y^{2}-2 y+1=x^{2}-2 x+1+y^{2}+4 y+4 \\ & \Rightarrow-4 x+2 x-2 y-4 y=0 \Rightarrow-2 x-6 y=0 \Rightarrow \boldsymbol{x}+\mathbf{3} y=\mathbf{0} \end{aligned} $

Ex. 2 . Find the co-ordinates of the points on the $x$-axis which are at a distance of 10 units from the point $(-4,8)$ ?

Sol. Let the co-ordinates of any point on the $x$-axis be $(x, 0)$. Then distance between $(x, 0)$ and $(-4,8)$ is 10 units.

$ \begin{aligned} & \Rightarrow \sqrt{(x+4)^{2}+(0-8)^{2}}=10 \Rightarrow x^{2}+8 x+16+64=100 \Rightarrow x^{2}+8 x-20=0 \\ & \Rightarrow(x+10)(x-2)=0 \Rightarrow x=-10 \text{ or } 2 \end{aligned} $

$\therefore$ The required points are $(-\mathbf{1 0}, \mathbf{0})$ and $(\mathbf{2}, \mathbf{0})$.

Ex. 3 . Prove that the points $(1,-1),(\frac{-1}{2}, \frac{1}{2})$ and $(1,2)$ are the vertices of an isosceles triangle.

Sol. Let $P(1,-1), Q(\frac{-1}{2}, \frac{1}{2})$ and $R(1,2)$ be the vertices of the $\triangle P Q R$.

Then, $P Q=\sqrt{(\frac{-1}{2}-1)^{2}+(\frac{1}{2}+1)^{2}}=\sqrt{\frac{9}{4}+\frac{9}{4}}=\sqrt{\frac{18}{4}}=\frac{3 \sqrt{2}}{2}$

$Q R=\sqrt{(1+\frac{1}{2})^{2}+(2-\frac{1}{2})^{2}}=\sqrt{\frac{9}{4}+\frac{9}{4}}=\sqrt{\frac{18}{4}}=\frac{3 \sqrt{2}}{2}$

$P R=\sqrt{(1-1)^{2}+(2+1)^{2}}=\sqrt{9}=3$

$\because P Q=Q R$, the triangle $P Q R$ is isosceles.

Ex. 4 . Prove that the points $(2,-2),(-2,1)$ and $(5,2)$ are the vertices of a right angled triangle. Also find the length of the hypotenuse and the area of the triangle.

Sol. Let $A \equiv(2,-2), B \equiv(-2,1), C \equiv(5,2)$. Then,

$ \begin{aligned} & A B=\sqrt{(-2-2)^{2}+(1+2)^{2}}=\sqrt{16+9}=\sqrt{25}=5 \\ & B C=\sqrt{(5+2)^{2}+(2-1)^{2}}=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2} \end{aligned} $

$A C=\sqrt{(5-2)^{2}+(2+2)^{2}}=\sqrt{9+16}=\sqrt{25}=5$

$\Rightarrow A B^{2}+A C^{2}=B C^{2}$ as $\mathbf{2 5}+\mathbf{2 5}=\mathbf{5 0}$

$\therefore \quad \triangle A B C$ is right angled at $A$ and hyp. $B C=5 \sqrt{2}$ units

Area of $\triangle A B C=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times A B \times A C=\frac{1}{2} \times 5 \times 5=\mathbf{1 2 . 5}$ sq. units.

Ex. 5 . Let $P(-3,2), Q(-5,-5), R(2,-3)$ and $S(4,4)$ be four points in a plane. Then show that $P Q R S$ is a rhombus.

Is it a square? Also find the area of the rhombus.

Sol. $P \equiv(-3,2), Q \equiv(-5,-5), R \equiv(2,-3), S \equiv(4,4)$

$\begin{aligned} \therefore & P Q=\sqrt{(-5+3)^{2}+(-5-2)^{2}}=\sqrt{4+49}=\sqrt{53} \\ & Q R=\sqrt{(2+5)^{2}+(-3+5)^{2}}=\sqrt{49+4}=\sqrt{53} \\ & R S=\sqrt{(4-2)^{2}+(4+3)^{2}}=\sqrt{4+49}=\sqrt{53} \\ & P S=\sqrt{(4+3)^{2}+(4-2)^{2}}=\sqrt{49+4}=\sqrt{53}\end{aligned}$

$\because P Q=Q R=R S=P S$, therefore $P Q R S$ is a rhombus. For $P Q R S$ to be a square, diagonals $P R$ and $Q S$ should be equal.

$P R=\sqrt{(2+3)^{2}+(-3-2)^{2}}=\sqrt{25+25}=\sqrt{50}=5 \sqrt{2}$

$Q S=\sqrt{(4+5)^{2}+(4+5)^{2}}=\sqrt{81+81}=\sqrt{162}=9 \sqrt{2}$

As $P R \neq Q S$, so $P Q R S$ is not a square.

Area of rhombus $=\frac{1}{2} \times$ (Product of length of diagonals) $=\frac{1}{2} \times 5 \sqrt{2} \times 9 \sqrt{2}=\mathbf{4 5}$ sq. units.

Ex. 6 . Find the co-ordinates of the circumcentre of the triangle whose vertices are $(3,0),(-1,-6)$ and $(4,-1)$. Also find its circum-radius.

Sol. The circumcentre of a triangle is equidistant from the vertices of a triangle.

Let $A(3,0), B(-1,-6)$, and $C(4,-1)$ be the vertices of $\triangle A B C$ and $P(x, y)$ be the circumcentre of this triangle.

Then, $P A=P B=P C \Rightarrow P A^{2}=P B^{2}=P C^{2}$

Now, $\quad P A^{2}=P B^{2} \Rightarrow(x-3)^{2}+(y-0)^{2}=(x+1)^{2}+(y+6)^{2}$

$ \begin{aligned} & \Rightarrow x^{2}-6 x+9+y^{2}=x^{2}+2 x+1+y^{2}+12 y+36 \\ & \Rightarrow-6 x+9-2 x-1-12 y-36=0 \\ & \Rightarrow 8 x+12 y+28=0 \Rightarrow 2 x+3 y=-7 \end{aligned} $

$P B^{2}=P C^{2} \Rightarrow(x+1)^{2}+(y+6)^{2}=(x-4)^{2}+(y+1)^{2}$

$ \begin{aligned} & \Rightarrow x^{2}+2 x+1+y^{2}+12 y+36=x^{2}-8 x+16+y^{2}+2 y+1 \\ & \Rightarrow 10 x+10 y+20=0 \Rightarrow x+y=-2 \end{aligned} $

Solving eqn (i) and eqn (ii) simultaneously, we have

(i) $-2 \times(i i) \Rightarrow(2 x+3 y)-(2 x+2 y)=-7+4 \Rightarrow y=-3 \Rightarrow x-3=-2 \Rightarrow x=1$

$\therefore$ Co-ordinates of circumcentre $P \equiv(\mathbf{1},-\mathbf{3})$

$\therefore$ Circumradius $=P A=P B=P C=\sqrt{(1-3)^{2}+(-3-0)^{2}}=\sqrt{4+9}=\sqrt{\mathbf{1 3}}$ units.

Ex. 7 . Let the opposite angular points of a square be $(3,4)$ and $(1,-1)$, Find the co-ordinates of the remaining angular points.

Sol. Let $A B C D$ be the given square and let $A \equiv(3,4)$ and $C \equiv(1,-1)$. Also let $B \equiv(x, y)$.

$A B C D$ being a square,

$A B=B C \quad \Rightarrow A B^{2}=B C^{2}, \quad \angle A B C=90^{\circ}$

$\Rightarrow(\sqrt{(x-3)^{2}+(y-4)^{2}})^{2}=(\sqrt{(x-1)^{2}+(y+1)^{2}})^{2}$

$\Rightarrow x^{2}-6 x+9+y^{2}-8 y+16=x^{2}-2 x+1+y^{2}+2 y+1$

$\Rightarrow 4 x+10 y-23=0 \Rightarrow x=\frac{23-10 y}{4}$

Also, in $\triangle A B C, \angle B=90^{\circ} \Rightarrow A B^{2}+B C^{2}=A C^{2}$

$\Rightarrow(x-3)^{2}+(y-4)^{2}+(x-1)^{2}+(y+1)^{2}=(3-1)^{2}+(4+1)^{2}$

$\Rightarrow x^{2}-6 x+9+y^{2}-8 y+16+x^{2}-2 x+1+y^{2}+2 y+1=4+25$

$\Rightarrow 2 x^{2}+2 y^{2}-8 x-6 y+27=29 \Rightarrow 2 x^{2}+2 y^{2}-8 x-6 y-2=0 \Rightarrow x^{2}+y^{2}-4 x-3 y-1=0$

Now substitute the value of $x$ from $(i)$, we have

$ \begin{aligned} & \quad(\frac{23-10 y}{4})^{2}+y^{2}-(23-10 y)-3 y-1=0 \Rightarrow(23-10 y)^{2}+16 y^{2}-16(23-10 y)-48 y-16=0 \\ & \Rightarrow 529-460 y+100 y^{2}+16 y^{2}-368+160 y-48 y-16=0 \Rightarrow 116 y^{2}-348 y+145=0 \\ & \Rightarrow 4 y^{2}-12 y+5=0 \Rightarrow(2 y-1)(2 y-5)=0 \Rightarrow y=\frac{1}{2} \text{ or } \frac{5}{2} \\ & \text{ Putting } y=\frac{1}{2} \text{ and } \frac{5}{2} \text{ respectively in }(i) \text{, we get } x=\frac{9}{2} \text{ and } x=-\frac{1}{2} . \end{aligned} $

$\therefore$ The required vertices of the square $are(\frac{\mathbf{9}}{\mathbf{2}}, \frac{\mathbf{1}}{\mathbf{2}})$ and $(-\frac{1}{\mathbf{2}}, \frac{\mathbf{5}}{\mathbf{2}})$

Ex. 8 . Find the coordinates of the point which divides the join of the points $(8,9)$ and $(-7,4)$ internally in the ratio $2: 3$.

Sol. Let the co-ordinates of the point of internal division $A$ be $(x, y)$. Then,

$x=\frac{2 \times(-7)+3 \times 8}{2+3}=\frac{-14+24}{5}=\frac{10}{5}=2$

$y=\frac{2 \times 4+3 \times 9}{2+3}=\frac{8+27}{5}=\frac{35}{5}=7$

$\therefore$ Co-ordinates of the point for internal division are $(2,7)$.

Ex. 9 . In what ratio is the line joining the points $A(4,4)$ and $B(7,7)$ divided by $P(-1,-1)$ ?

Sol. Let $P$ divide $A B$ in the ratio $k: 1$.

Then, co-ordinates of $P$ are $(\frac{7 k+4}{k+1}, \frac{7 k+4}{k+1})$

But $P \equiv(-1,-1)$

$\therefore \frac{7 k+4}{k+1}=-1 \Rightarrow 7 k+4=-k-1 \Rightarrow 8 k=-5 \Rightarrow k=-\frac{5}{8}$

$\because k$ is negative, it means that the division is external.

$\therefore A B$ is divided by $P$ externally in the ratio $\frac{5}{8}: 1$, i.e. $5: 8$.

Ex. 10 . Find the ratio in which the $x$-axes divides the line joining the points $(-2,5)$ and $(1,-9)$ ?

Sol. Let the $x$-axis divide the line joining the points $(-2,5)$ and $(1,-9)$ in the ratio $k: 1$. Let the point of division on $x$-axis is $P$

Then,

$x=\frac{k-2}{k+1}, y=\frac{-9 k+5}{k+1}$

As this is a point on the $x$-axis, $y=0$

$\therefore \frac{-9 k+5}{k+1}=0 \quad \Rightarrow 9 k=5 \quad \Rightarrow k=\frac{5}{9}$

$k$ being positive, the division is internal.

$\therefore x$-axis divides the given line internally in the ratio $5: 9$.

Ex. 11 . Find the point of trisection of the line segment joining the points $(1,2)$ and $(11,9)$ ?

Sol. Let $A(1,2)$ and $B(11,9)$ be the given points. Let the points of trisection be $P$ and $Q$. Then,

$ \begin{aligned} & A P=P Q=Q B=k(\text{ say }) \\ \Rightarrow & A Q=A P+P Q=2 k \text{ and } P B=P Q+Q B=2 k \\ \therefore & A P: P B=k: 2 k=1: 2 \text{ and } A Q: Q B=2 k: k=2: 1 \end{aligned} $

$\Rightarrow P$ divides $A B$ internally in the ratio $1: 2$ and $Q$ divides $A B$ internally in the ratio $2: 1$.

$\therefore$ Coordinates of $P$ are $[\frac{1 \times 11+2 \times 1}{1+2}, \frac{1 \times 9+2 \times 2}{1+2}]$,i.e $(\frac{13}{3}, \frac{13}{3})$

Coordinates of $Q$ are $[\frac{2 \times 11+1 \times 1}{2+1}, \frac{2 \times 9+1 \times 2}{2+1}]$, i.e $(\frac{23}{3}, \frac{20}{3})$

Hence, the two points of trisection are $(\frac{\mathbf{1 3}}{\mathbf{3}}, \frac{\mathbf{1 3}}{\mathbf{3}})$ and $(\frac{\mathbf{2 3}}{\mathbf{3}}, \frac{\mathbf{2 0}}{\mathbf{3}})$.

Ex. 12 . If the points $A(a,-11), B(5, b), C(2,15)$ and $D(1,1)$ are the vertices of a parallelogram $A B C D$, find the values of $a$ and $b$.

Sol. $A B C D$ is a parallelogram, if the mid-points of diagonals $A C$ and $B D$ have the same co-ordinates

( $\because$ Diagonals of a parallelogram bisect each other)

Co-ordinates of mid-point of $A C$ are $(\frac{a+2}{2}, \frac{-11+15}{2})=(\frac{a+2}{2}, 2)$

Co-ordinates of mid-point of $B D$ are $(\frac{5+1}{2}, \frac{b+1}{2})=(3, \frac{b+1}{2})$

Here, $\quad \frac{a+2}{2}=3$ and $2=\frac{b+1}{2} \quad \Rightarrow \boldsymbol{a}=\mathbf{4}, \boldsymbol{b}=\mathbf{3}$.

C $(2,15)$

Ex. 13 . Determine the ratio in which $2 x+3 y-30=0$ divides the join of $A(3,4)$ and $B(7,8)$ and at what point?

Sol. Let the line $2 x+3 y-30=0$ divide the join of $A(3,4)$ and $B(7,8)$ at point $C(p, q)$ in the ratio $k: 1$. Then, $p=\frac{7 k+3}{k+1}, q=\frac{8 k+4}{k+1}$

As the point $C$ lies on the line $2 x+3 y-30=0$, it satisfies the given equation, i.e,

$2 \times(\frac{7 k+3}{k+1})+3(\frac{8 k+4}{k+1})-30=0$

$\Rightarrow 14 k+6+24 k+12-30 k-30=0$

$\Rightarrow 8 k-12=0 \Rightarrow k=\frac{12}{8}=\frac{3}{2}$

$\therefore$ The line $2 x+3 y-30=0$ divides the line joining $A(3,4)$ and $B(7,8)$ in the ratio $\frac{3}{2}: 1$, i.e. $3: 2$ at $\boldsymbol{C}$.

Now the co-ordinates of $C$ are $(\frac{7 \times \frac{3}{2}+3}{3 / 2+1}, \frac{8 \times \frac{3}{2}+4}{3 / 2+1})=(\frac{\mathbf{2 7}}{\mathbf{5}}, \frac{\mathbf{3 2}}{\mathbf{5}})$.

Ex. 14 . The co-ordinates of mid-points of sides of a triangle are $(1,2),(0,-1)$ and $(2,-1)$. Find its centroid.

Sol. Let $A(1,2), B(0,-1)$ and $C(2,-1)$ be the mid-points of the sides $P Q, Q R$ and $R P$ of the triangle $P Q R$. Let the co-ordinates of $P, Q$ and $R$ be $(x_1, y_1),(x_2, y_2)$ and $(x_3, y_3)$ respectively. Then, by the mid-point formula.

$ \begin{matrix} \frac{x_1+x_2}{2}=1 \text{ and } \frac{y_1+y_2}{2}=2 & \Rightarrow x_1+x_2=2 \text{ and } y_1+y_2=4 \\ \frac{x_1+x_3}{2}=2 \text{ and } \frac{y_1+y_3}{2}=-1 & \Rightarrow x_1+x_3=4 \text{ and } y_1+y_3=-2 \\ \frac{x_2+x_3}{2}=0 \text{ and } \frac{y_2+y_3}{2}=-1 & \Rightarrow x_2+x_3=0 \text{ and } y_2+y_3=-2 \end{matrix} $

From (i), (ii) and (iii)

$ \begin{aligned} & \quad x_1+x_2+x_1+x_3+x_2+x_3=2+4+0 \text{ and } y_1+y_2+y_1+y_3+y_2+y_3=4+(-2)+(-2) \\ & \Rightarrow x_1+x_2+x_3=3 \text{ and } y_1+y_2+y_3=0 \\ & \text{ Now, }(x_1+x_2+x_3)-(x_1+x_2)=3-2 \Rightarrow x_3=1 \text{ and }(y_1+y_2+y_3)-(y_1+y_2)=0-4 \Rightarrow y_3=-4 \\ & \qquad \begin{matrix} (x_1+x_2+x_3)-(x_1+x_3)=3-4 \Rightarrow x_2=-1 \text{ and }(y_1+y_2+y_3)-(y_1+y_3)=0-(-2) \Rightarrow y_2=2 \\ (x_1+x_2+x_3)-(x_2+x_3)=3-0 \Rightarrow x_1=3 \text{ and }(y_1+y_2+y_3)-(y_2+y_3)=0-(-2) \Rightarrow y_1=2 \end{matrix} \end{aligned} $

$\therefore \quad$ The vertices of the $\triangle P Q R$ are $P(3,2), Q(-1,2)$ and $R(1,-4)$

Hence, co-ordinates of centroid of $\triangle P Q R=(\frac{3+(-1)+1}{3}, \frac{2+2+(-4)}{3})=(\mathbf{1 , 0})$.

Ex. 15 . In the adjoining figure, $P$ and $Q$ have co-ordinates $(4,6)$ and $(0,3)$ respectively. Find $(i)$ the co-ordinates of $R$ (ii) Area of quadrilateral $O A P Q$.

Sol. Let $O R=x$ units

(i) $\triangle Q O R \sim \triangle P A R$

$\Rightarrow \frac{P A}{A R}=\frac{Q O}{O R} \Rightarrow \frac{6}{x+4}=\frac{3}{x} \Rightarrow \frac{x}{x+4}=\frac{3}{6} \Rightarrow \frac{x}{x+4}=\frac{1}{2} \Rightarrow 2 x=x+4 \Rightarrow x=4$

$\therefore$ Co-ordinates of $R$ are $(-4,0)$

(ii) Quadrilateral $O A P Q$ is a trapezium.

Area of trapezium $O A P Q=\frac{1}{2}(O Q+A P) \times O A=\frac{1}{2}(3+6) \times 4=\frac{1}{2} \times 9 \times 4=\mathbf{1 8}$ sq. units.

Ex. 16 . Find the co-ordinates of the in-centre of the triangle whose vertices are $(-36,7),(20,7)$ and $(0,-8)$.

Sol. Let $A(-36,7), B(20,7)$ and $C(0,-8)$ be the vertices of the given triangle.

Then, $\quad a=B C=\sqrt{(0-20)^{2}+(-8-7)^{2}}=\sqrt{400+225}=\sqrt{625}=25$

$ \begin{aligned} & \text{ छ } A C=\sqrt{(0+36)^{2}+(-8-7)^{2}}=\sqrt{1296+225}=\sqrt{1521}=39 \\ & \text{ € } A B=\sqrt{(20+36)^{2}+(7-7)^{2}}=\sqrt{56^{2}}=56 . \end{aligned} $

The co-ordinates of the incentre of the $\triangle A B C$ are

$ \begin{aligned} & {[\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}]=(\frac{25(-36)+39(20)+56 \times 0}{25+39+56}, \frac{25(7)+39(7)+56(-8)}{25+39+56})} \\ & =(\frac{-900+780}{120}, \frac{175+273-448}{120})=(\frac{-120}{120}, \frac{0}{120})=(\mathbf{1 , 0 )} \end{aligned} $

Ex. 17 . Find the area of a triangle whose vertices are $(1,3),(2,4)$ and $(5,6)$.

Sol. Let $A(x_1, y_1) \equiv(1,3), B(x_2, y_2) \equiv(2,4), C(x_3, y_3) \equiv(5,6)$ be the vertices of $\triangle A B C$

$ \text{ Area of } \begin{aligned} \triangle A B C & =\frac{1}{2}|{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}| \\ & =\frac{1}{2}|{1(4-6)+2(6-3)+5(3-4)}|=\frac{1}{2}|{-2+6-5}|=\frac{\mathbf{1}}{\mathbf{2}} \text{ sq. units. } \end{aligned} $

Ex. 18 . Find the area of the quadrilateral whose vertices are $(3,4),(0,5),(2,-1)$ and $(3,-2)$.

Sol. Let $A(x_1, y_1)=(3,4), B(x_2, y_2) \equiv(0,5), C(x_3, y_3) \equiv(2,-1)$ and $D(x_4, y_4) \equiv(3,-2)$ be the vertices of quadrilateral $A B C D$.

Area of quad. $A B C D=\frac{1}{2}|{(x_1 y_2-x_2 y_1)+(x_2 y_3-x_3 y_2)+(x_3 y_4-x_4 y_3)+(x_4 y_1-x_1 y_4)}|$

$ \begin{aligned} & =\frac{1}{2}|{(3 \times 5-0 \times 4)+(0 \times(-1)-2 \times 5)+(2 \times(-2)-3 \times(-1))+(3 \times 4-3 \times(-2))}| \\ & =\frac{1}{2}|{(15-0)+(0-10)+(-4+3)+(12+6)}| \\ & =\frac{1}{2}|{15-11+0+18}|=\frac{1}{2} \times 22=11 \text{ sq. units. } \end{aligned} $

Ex. 19 . Show that the points $(a, b+c),(b, c+a),(c, a+b)$ are collinear.

Sol. For three points to be collinear, the area of the triangle formed by the three points should be zero.

$\therefore$ Area of $\Delta$ formed by the given three points

$ \begin{aligned} & =\frac{1}{2}[a((c+a)-(a+b))+b((a+b)-(b+c))+c((b+c)-(c+a))] \\ & =\frac{1}{2}[a(c-b)+b(a-c)+c(b-a)]=\frac{1}{2}[a c-a b+b a-b c+c b-c a]=\mathbf{0} . \end{aligned} $

Hence $(a, b+c),(b, c+a)$ and $(c, a+b)$ are collinear.

Ex. 20 . The area of a triangle is 5 . Two of its vertices are $(2,1)$ and $(3,-2)$. The third vertex is $(x, y)$ where $y=x+3$. Find the co-ordinates of the third vertex.

Sol. Let $A \equiv(x, y), B \equiv(2,1), C \equiv(3,-2)$

$ \text{ Area of } \begin{aligned} \triangle A B C & =\frac{1}{2}|{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}| \\ & =\frac{1}{2}|x(1+2)+2(-2-y)+3(y-1)|=\frac{1}{2}|3 x-4-2 y+3 y-3|=\frac{1}{2}|3 x+y-7| \end{aligned} $

Given $\frac{1}{2}|3 x+y-7|=5$

$\Rightarrow|3 x+y-7|=10 \Rightarrow 3 x+y-7=10$ or $-(3 x+y-7)=10 \Rightarrow 3 x+y=17$ or $3 x+y=-3$

Case I. $3 x+y=17$. Also given $y=x+3$

$\therefore 3 x+x+3=17 \Rightarrow 4 x=14 \Rightarrow x=\frac{7}{2} \Rightarrow y=\frac{7}{2}+3=\frac{13}{2}$

Case II. $3 x+y=-3, y=x+3$

$\therefore 3 x+x+3=-3 \Rightarrow 4 x=-6 \Rightarrow x=\frac{-3}{2} \Rightarrow y=\frac{-3}{2}+3=\frac{3}{2}$

$\therefore$ Co-ordinates of $A$ are $(\frac{7}{2}, \frac{13}{2})$ or $(\frac{-3}{2}, \frac{3}{2})$

Ex. 21 . Find the centre of a circle passing through the points $(6,-6),(3,-7)$ and $(3,3)$.

Sol. Let $C(x, y)$ be the centre of the circle passing through the points $P(6,-6), Q(3,-7)$ and $R(3,3)$ Then, $P C=Q C=R C$ (Being radius of the same circle)

$ \begin{aligned} & P C^{2}=Q C^{2} \\ \Rightarrow & (x-6)^{2}+(y+6)^{2}=(x-3)^{2}+(y+7)^{2} \\ \Rightarrow & x^{2}-12 x+36+y^{2}+12 y+36=x^{2}-6 x+9+y^{2}+14 y+49 \\ \Rightarrow & -12 x+12 y+6 x-14 y+72-58=0 \\ \Rightarrow & -6 x-2 y+14=0 \Rightarrow 3 x+y-7=0 \end{aligned} $

Also, $Q C^{2}=R C^{2}$

$\Rightarrow(x-3)^{2}+(y+7)^{2}=(x-3)^{2}+(y-3)^{2}$

$\Rightarrow x^{2}-6 x+9+y^{2}+14 y+49=x^{2}-6 x+9+y^{2}-6 y+9$

$\Rightarrow 14 y+6 y=9-49 \Rightarrow 20 y=-40 \Rightarrow y=-2$

Putting $y=-2$ in $(i)$, we get

$ 3 x+(-2)-7=0 \Rightarrow 3 x=9 \Rightarrow x=3 $

$\therefore$ The centre is $(\mathbf{3},-\mathbf{2})$.

Ex. 22 . Find the slope and inclination of the line which passes through the points $(1,2)$ and $(5,6)$ ?

Sol. Slope $(m)=\frac{(y_2-y_1)}{(x_2-x_1)}=\frac{6-2}{5-1}=\frac{4}{4}=1$

Also slope $(m)=\tan \theta$, where $\theta$ is the inclination of the line to the positive direction of the $x$-axis in the anticlockwise direction. $\tan \theta=1 \quad \Rightarrow \theta=\tan ^{-1} 1=\mathbf{4 5}^{\circ}$.

Ex. 23 . Find the value of $a$, if the line passing through $(-5,-8)$ and $(3,0)$ is parallel to the line passing through $(6,3)$ and $(4, a)$.

Sol. Two lines are parallel if their slopes are equal

$\therefore \frac{0-(-8)}{3-(-5)}=\frac{a-3}{4-6} \Rightarrow \frac{8}{8}=\frac{a-3}{-2} \Rightarrow a-3=-2 \Rightarrow \boldsymbol{a}=\mathbf{1}$.

Ex. 24 . Without using Pythagoras’ theorem, show that the points $A(0,4), B(1,2)$ and $C(3,3)$ are the vertices of a right angle triangle.

Sol. Slope of $A B=\frac{2-4}{1-0}=-2$, Slope of $B C=\frac{3-2}{3-1}=\frac{1}{2}$

Slope of $A C=\frac{3-4}{3-0}=-\frac{1}{3}$

Slope of $A B \times$ Slope of $B C=-2 \times \frac{1}{2}=-1$

$\therefore A B \perp B C$, i.e, $\angle B=90^{\circ} \Rightarrow \triangle A B C$ is a right angled.

Ex. 25 . If the points $(x, 1),(1,2)$ and $(0, y+1)$ are collinear show that $\frac{1}{x}+\frac{1}{y}=1$.

Sol. There are two ways to prove it.

1st way: Area of triangle formed by the given points $=0$ if they are collinear.

$\therefore \quad \frac{1}{2}[x(2-(y+1))+1((y+1)-1)+0(1-2)]=0$

$\Rightarrow \frac{1}{2}[2 x-x y-x+y+0]=0 \Rightarrow x+y-x y=0 \Rightarrow x+y=x y \quad \Rightarrow \frac{x}{x y}+\frac{y}{x y}=1 \Rightarrow \frac{\mathbf{1}}{\boldsymbol{y}}+\frac{\mathbf{1}}{\boldsymbol{x}}=\mathbf{1}$.

2nd way: Slope of the lines formed by joining these points are equal.

Slope of line joining $(x, 1),(1,2)=\frac{(2-1)}{1-x}$

Slope of line joining $(1,2),(0, y+1)=\frac{y+1-2}{0-1}$

$\Rightarrow \frac{1}{1-x}=\frac{y-1}{-1} \Rightarrow-1=(1-x)(y-1) \Rightarrow-1=y-x y-1+x \Rightarrow x+y=x y \Rightarrow \frac{\mathbf{1}}{\boldsymbol{x}}+\frac{\mathbf{1}}{\boldsymbol{y}}=\mathbf{1}$.

Ex. 26 . What is the equation of the line having the $y$-intercept -1 and parallel to the line $y=5 x-7$ ?

Sol. Comparing $y=5 x-7$ with $y=m x+c$, the slope of given line $=m=5$

$\therefore$ Equation of a line parallel to $y=5 x-7$ having $y$-intercept $=-1$ is $\boldsymbol{y}=\mathbf{5} \boldsymbol{x}-\mathbf{1}$.

Ex. 27 . Find the equation of the straight line passing through the point $(4,5)$ and perpendicular to $3 x-2 y+5=0$.

Sol. $3 x-2 y+5=0 \Rightarrow-2 y=-3 x-5 \Rightarrow y=\frac{3}{2} x+\frac{5}{2}$

On comparing with $y=m x+c$, we see that slope of given line $=\frac{3}{2}$

As the required line is perpendicular to the given line,

Slope of required line $=\frac{-2}{3}$

$\therefore$ Equation of required line: $(y-5)=\frac{-2}{3}(x-4)$

$\Rightarrow 3(y-5)=-2 x+8 \Rightarrow 3 y-15=-2 x+8 \Rightarrow \mathbf{3} \boldsymbol{y}+\mathbf{2} \boldsymbol{x}-\mathbf{2 3}=\mathbf{0}$

Ex. 28 . The line segment joining $P(5,-2)$ and $Q(9,6)$ is divided in the ratio $3: 1$ by a point $A$ on it. Find the equation of a line through the point $A$ parallel to the line $x-3 y+4=0$.

Sol. Co-ordinates of $A$ are $(\frac{3 \times 9+1 \times 5}{3+1}, \frac{3 \times 6+1 \times-2}{3+1})=(\frac{32}{4}, \frac{16}{4})$, i.e. $(8,4)$

Now, $x-3 y+4=0 \Rightarrow-3 y=-x-4 \Rightarrow y=\frac{x}{3}+\frac{4}{3}$

$\therefore$ Slope of given line $=\frac{1}{3}$

$\Rightarrow$ Slope of required line $=\frac{1}{3} \quad$ (Since lines are parallel)

$\therefore$ Equation of line through $(8,4)$ with slope $\frac{1}{3}$ is

$ \begin{matrix} (y-4)=\frac{1}{3}(x-8) \quad[\text{ Using, } y-y_1=m(x-x_1)] \\ \Rightarrow 3 y-12=x-8 \Rightarrow 3 y-x=4 . \end{matrix} $

Ex. 29 . $(-2,-1)$ and $(4,-5)$ are the co-ordinates of vertices $B$ and $D$ respectively of rhombus $A B C D$. Find the equation of the diagonal $A C$.

Sol. Diagonals of a rhombus bisect each other at right angles $\Rightarrow$ Co-ordinates of mid-points of $A C$ and $B D$ are equal $\therefore 0 \equiv(\frac{4+(-2)}{2}, \frac{-5+(-1)}{2})=(1,-3)$

Slope of $B D=\frac{-5+1}{4+2}=\frac{-4}{6}=\frac{-2}{3}$

$\because A C \perp B D$, Slope of $A C=\frac{3}{2}$

$\therefore$ Equation of $A C$, passing through $O(1,-3)$ having slope $\frac{3}{2}$ is

$ y+3=\frac{3}{2}(x-1) $

$\Rightarrow 2 y+6=3 x-3 \Rightarrow \mathbf{2} \boldsymbol{y}=\mathbf{3} \boldsymbol{x}-\mathbf{9}$.

Ex. 30 . The line $x-4 y=6$ is the perpendicular bisector of the segment $A B$ and the co-ordinates of $B$ are $(1,3)$.

Find the co-ordinates of $A$.

Sol. Let the co-ordinates of $A$ be $(h, k)$.

Now, slope of given line, i.e., $x-4 y=6$, i.e., $y=\frac{x}{4}+\frac{-6}{4}$ is $\frac{1}{4}$

As $A B \perp$ given line, product of their slopes $=-1$

Slope of $A B=\frac{3-k}{1-h}$

$\Rightarrow \frac{3-k}{1-h} \times \frac{1}{4}=-1 \Rightarrow 3-k=-4(1-h) \Rightarrow 3-k=-4+4 h \Rightarrow 4 h+k=7$

Mid-point of $A B=(\frac{h+1}{2}, \frac{k+3}{2})$

This mid-point also lies on the line $x-4 y=6$

$ \begin{aligned} & \therefore(\frac{h+1}{2})-4(\frac{k+3}{2})=6 \\ & \Rightarrow h+1-4(k+3)=12 \Rightarrow h-4 k-11=12 \Rightarrow h-4 k=23 \end{aligned} $

Now solving ( $i$ ) and (ii) simultaneously, we have

From $(i), k=7-4 h$, putting the value of $k$ in (ii), we have

$ h-4(7-4 h)=23 \Rightarrow h-28+16 h=23 \Rightarrow 17 h=51 \Rightarrow h=3 $

$\therefore k=7-12=-5$

$\therefore$ Coordinates of $A$ are $(h, k)$ i.e., $(\mathbf{3},-\mathbf{5})$.

Ex. 31 . What is the equation of the straight line passing through the point $(4,3)$ and making intercepts on the co-ordinates axes whose sum is -1 ?

Sol. Let the $x$-intercept $=a$. Then $y$-intercept $=-1-a$

The equation of the required line is $\frac{x}{a}+\frac{y}{-1-a}=1$

Given, it passes through $(4,3)$, so,

$\frac{4}{a}+\frac{3}{-1-a}=1 \Rightarrow-4-4 a+3 a=-a-a^{2} \Rightarrow a^{2}-4=0 \Rightarrow(a+2)(a-2)=0 \quad \Rightarrow a=-2$, or 2 .

When $a=2$, required line is $\frac{x}{2}-\frac{y}{3}=1$

When $a=-2$, required line is $\frac{x}{-2}+\frac{y}{3}=1$

Ex. 32 . Find the equation of the line through the point $(3,2)$ which makes an angle of $45^{\circ}$ with the line $x-2 y=3$ ?

(Kerala PET 2007)

Sol. Given line: $x-2 y=3 \Rightarrow y=\frac{x}{2}-\frac{3}{2}$

$\therefore$ Its slope $=m_1=\frac{1}{2}$

Let $m_2$ be the slope of line through $(3,2)$. Since this line is inclined at $45^{\circ}$ to line $(i)$,

$ \begin{aligned} & \tan 45^{\circ}=|\frac{m_1-m_2}{1+m_1 m_2}|=|\frac{\frac{1}{2}-m_2}{1+\frac{1}{2} m_2}| \\ & \Rightarrow \frac{\frac{1}{2}-m_2}{1+\frac{1}{2} m_2}= \pm 1 \Rightarrow \frac{1-2 m_2}{2+m_2}= \pm 1 \Rightarrow 1-2 m_2=2+m_2 \text{ or } 1-2 m_2=-2-m_2 \\ & \Rightarrow 3 m_2=-1 \text{ or }-m_2=-3 \Rightarrow m_2=-\frac{1}{3} \text{ or } m_2=3 \end{aligned} $

Since the line passes through $(3,2)$, the equation of the required line is

$ \begin{aligned} & (y-2)=-\frac{1}{3}(x-3) \quad \text{ or }(y-2)=3(x-3) \\ & \Rightarrow 3 y-6=-x+3 \text{ or } y-2=3 x-9 \Rightarrow \mathbf{3} \boldsymbol{y}+\boldsymbol{x}-\mathbf{9}=\mathbf{0} \quad \text{ or } \boldsymbol{y}-\mathbf{3} \boldsymbol{x}+\mathbf{7}=\mathbf{0} \text{. } \end{aligned} $

Ex. 33 . If $p$ is the length of the perpendicular drawn from the origin to the line $\frac{x}{a}+\frac{y}{b}=1$, then show that $\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

(NDA/NA 2011)

Sol. $\because$ Length of perpendicular from point $(x_1, y_1)$ to line $a x+b y+c=0=\frac{|a x_1+b y_1+c|}{\sqrt{a^{2}+b^{2}}}$

$\therefore$ Length of perpendicular from $(0,0)$ to $\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{|\frac{1}{a} \times 0+\frac{1}{b} \times 0-1|}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=p$

$\Rightarrow \frac{|-1|}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=p \Rightarrow \frac{1}{p}=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}} \Rightarrow \frac{\mathbf{1}}{p^{2}}=\frac{\mathbf{1}}{\boldsymbol{a}^{2}}+\frac{\mathbf{1}}{\boldsymbol{b}^{2}}$.

Ex. 34 . Find the perpendicular distance between the lines $9 x+40 y-20=0$ and $9 x+40 y+21=0$.

Sol. Putting $x=0$ in equation of one of the lines say $9 x+40 y-20=0$, we get $y=\frac{1}{2}$.

$\therefore A$ point on $9 x+40 y-20=0$ is $(0, \frac{1}{2})$

$\therefore$ Distance of $(0, \frac{1}{2})$ from $9 x+40 y+21=0$ is $\frac{|9 \times 0+40 \times \frac{1}{2}+21|}{\sqrt{9^{2}+40^{2}}}=\frac{|41|}{\sqrt{1681}}=\frac{41}{41}=\mathbf{1}$.

Ex. 35 . Show that the equation of the parallel line midway between the parallel lines $a x+b y+c_1=0$ and $a x+b y+c_2=0$ is $a x+b y+\frac{c_1+c_2}{2}=0$.

Sol. The two given lines are $a x+b y+c_1=0$ and $a x+b y+c_2=0$.

Any line parallel to these two lines and midway between them is

$ a x+b y+c=0 $

Putting $x=0, y=-\frac{c}{b}$ is a point on line $(i)$

It is equidistant from the given lines but in opposite directions, so

$\frac{|a \times 0+b \times-\frac{c}{b}+c_1|}{\sqrt{a^{2}+b^{2}}}=-\frac{|a \times 0+b \times-\frac{c}{b}+c_2|}{\sqrt{a^{2}+b^{2}}} \Rightarrow-c+c_1=c-c_2 \Rightarrow c=\frac{c_1+c_2}{2}$.

$\therefore$ Required equation is $\boldsymbol{a x}+\boldsymbol{b} \boldsymbol{y}+\frac{\boldsymbol{c}_1+\boldsymbol{c}_2}{2}=\mathbf{0}$.

PRACTICE SHEET

The point whose abscissa is equal to its ordinate and which is equidistant from $A(-1,0)$ and $B(0,5)$ is

(a) $(1,1)$

(b) $(2,2)$

(d) $(3,3)$

(NDA/NA 2013)

~~ 2. What is the perimeter of the triangle with the vertices $A(-4,2)$, $B(0,-1)$ and $C(3,3)$ ?

(a) $7+3 \sqrt{2}$

(b) $10+5 \sqrt{2}$

(d) $5+\sqrt{2}$

(NDA/NA 2012)

~~ 3. The point $P$ is equidistant from $A(1,3), B(-3,5)$ and $C(5,-1)$. Then $P B$ is equal to :

(a) $5 \sqrt{2}$

(b) 5

(d) $5 \sqrt{10}$

~~ 4. If $M(x, y)$ is equidistant from $A(a+b, b-a)$ and $B(a-b$, $a+b)$, then

(a) $b x=a y$

(b) $a x=b y$

(d) $x=y$

~~ 5. The points $(a, a),(-a,-a)$ and $(-\sqrt{3 a},+\sqrt{3 a})$ are the vertices of……triangle whose area is…….

(a) Isosceles, $2 \sqrt{2} a^{2}$ sq. units

(b) Equilateral, $2 \sqrt{3} a^{2}$ sq. units

(d) None of these

~~ 6. The coordinates of the circumcentre of the triangle whose vertices are $(8,6),(8,-2)$ and $(2,-2)$.

(a) $(5,2)$

(b) $(-2,-5)$

(d) $(5,0)$

~~ 7. Name the quadrilateral $A B C D$, the coordinates of whose vertices are $A(3,5), B(1,1), C(5,3)$ and $D(7,7)$.

(a) Square

(b) Rhombus

(d) Trapezium

~~ 8. The vertices $A(4,5), B(7,6), C(4,3)$ and $D(1,2)$ from the quadrilateral $A B C D$ whose special name is

(a) Rectangle

(b) Square

(d) Rhombus

~~ 9. $A B C D$ is a rectangle formed by the points $A(-1,-1)$, $B(-1,4), C(5,4)$ and $D(5,-1) . P, Q, R$ and $S$ are mid-points of $A B, B C, C D$ and $D A$ respectively. The quadrilateral $P Q R S$ is a

(a) Square

(b) Rectangle

(d) None of these

~~ 10. The points $P(12,8), Q(-2, a)$ and $R(6,0)$ are the vertices of a right angled triangle where $\angle R=90^{\circ}$. The value of $a$ is

(a) 5

(b) -6

(d) 6

~~ 11. Point $A(5,1)$ is the centre of the circle with radius 13 units. $A B \perp$ chord $P Q . B$ is $(2,-3)$. The length of chord $P Q$ is

(a) 10 units

(b) 20 units

(d) 24 units

~~ 12. If $A(3,5), B(-5,-4), C(7,10)$ are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are:

(a) $(10,12)$

(b) $(12,17)$

(d) $(18,21)$

(IIT 1998)

~~ 13. The points $A(2,3), B(3,5), C(7,7)$ and $D(5,6)$ are such that:

(a) $A, B, C$ and $D$ are collinear

(b) $A B C D$ is a parallelogram

(d) $D$ lies on the boundary of $\triangle A B C$

(AMU 2007)

~~ 14. $(0,-1)$ and $(0,3)$ are the two opposite vertices of a square. The other two vertices are:

(a) $(0,1),(0,-3)$

(b) $(2,1)(-2,1)$

(d) $(2,2)(1,1)$ (KCET 2005)

~~ 15. Find the coordinates of the point which divides externally the join of the points $(3,4)$ and $(-6,2)$ in the ratio $3: 2$.

(a) $(-24,-2)$

(b) $(-24,-4)$

(d) $(24,2)$

~~ 16. In what ratio is the line joining the points $(2,-3)$ and $(5,6)$ divided by the $x$-axis.

(a) $1: 3$

(b) $1: 2$

(d) $3: 1$

~~ 17. The ratio in which the line $3 x+4 y=7$ divides the line joining the points $(-2,1)$ and $(1,2)$ is

(a) $2: 3$

(b) $4: 5$

(d) $9: 4$

(Gujarat CET 2008)

~~ 18. Determine the ratio in which the point $P(m, 6)$ divides the join of $A(-4,3)$ and $B(2,8)$. Also find the value of $m$.

(a) $2: 5 ; m=3$

(b) $3: 2 ; m=-\frac{2}{5}$

(d) $2: 1 ; m=\frac{2}{3}$

~~ 19. For an equilateral triangle $\triangle A B C$ with vertices, $A(1,2)$, $B(2,3)$, its incentre is $(\frac{9+\sqrt{3}}{6}, \frac{15-\sqrt{3}}{6})$. The coordinates of vertex $C$ are:

(a) $(\frac{3-\sqrt{3}}{2}, \frac{5+\sqrt{3}}{2})$

(b) $(\frac{3-\sqrt{3}}{6}, \frac{5+\sqrt{3}}{6})$

(d) $(\frac{3+\sqrt{3}}{6}, \frac{5-\sqrt{3}}{6})$

(Gujarat CET 2007)

~~ 20. If the co-ordinates of the mid-points of the sides of a triangle are $(1,1),(2,-3),(3,4)$, find its incentre.

(a) $(0,0)$

(b) $(\frac{2 \sqrt{13}+20 \sqrt{2}}{\sqrt{13}+\sqrt{17}+5 \sqrt{2}}, \frac{8 \sqrt{13}-6 \sqrt{17}}{\sqrt{13}+\sqrt{17}+5 \sqrt{2}})$

(c) $(\frac{2 \sqrt{2}+20 \sqrt{5}}{\sqrt{2}+\sqrt{5}+\sqrt{13}}, \frac{8 \sqrt{2}-10 \sqrt{13}}{\sqrt{2}+\sqrt{5}+\sqrt{13}})$

(d) $(-2,2)$

~~ 21. If $(0,0)$ and $(2,0)$ are the two vertices of a triangle whose centroid is $(1,1)$, then the area of the triangle is:

(a) 1

(b) $\frac{\sqrt{3}}{2}$

(d) 3

(Kerala PET 2002)

~~ 22. If the three points $(k, 2 k),(2 k, 3 k)$ and $(3,1)$ are collinear then $k$ is equal to

(a) -2

(b) $-\frac{1}{2}$

(d) $\frac{3}{2}$

~~ 23. If the points $A(1,2), B(0,0)$ and $C(a, b)$ are collinear, then

(a) $a=b$

(b) $a=2 b$

(d) $2 a=b$

~~ 24. The vertices of a $\triangle A B C$ are $A(4,6), B(1,5)$ and $C(7,2)$. A line is drawn to intersect sides $A B$ and $A C$ at $D$ and $E$ respectively such that $\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}$. Then ratio, $\frac{\text{ Area of } \triangle A D E}{\text{ Area of } \triangle A B C}$ is

(a) $1: 2$

(b) $1: 4$

(d) $1: 16$

~~ 25. If the area of the quadrilateral whose angular points $A, B$, $C, D$ taken in order are $(1,2),(-5,6),(7,-4)$ and $(-2, k)$ be zero, then the value of $k$ is

(a) 1

(b) 3

(d) -1

~~ 26. $A, B, C$ are the points $(-1,5),(3,1)$ and $(5,7)$ respectively. $D, E, F$ are the mid-points of $B C, C A$ and $A B$ respectively. Then, Area of $\triangle A B C:$ Area of $\triangle D E F$ is equal to:

(a) $2: 1$

(b) $3: 1$

(d) $8: 1$

~~ 27. The line $3 x+4 y-24=0$ cuts the $x$-axis at $A$ and $y$-axis at $B$. Then the in centre of the $\triangle A O B$, where $O$ is the origin is

(a) $(1,2)$

(b) $(2,2)$

(d) $(12,12)$

(Rajasthan PET 2006)

~~ 28. The slope of a line perpendicular to the line which passes through the points $(-k, h)$ and $(b,-f)$ is

(a) -1

(b) $\frac{f-h}{b+k}$

(d) $\frac{-b+k}{f-h}$

~~ 29. The medians $A D$ and $B E$ of the triangle with vertices $A(0, b), B(0,0)$ and $C(a, 0)$ are mutually perpendicular if

(a) $b=-\sqrt{2} a$

(b) $a= \pm \sqrt{2} b$

(d) $a=b$

~~ 30. Let $P S$ be the median of the triangle with vertices $P(2,2), Q(6,-1)$ and $R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to $P S$ is

(a) $2 x-9 y-11=0$

(b) $2 x-9 y-7=0$

(d) $2 x+9 y+7=0$

(IIT 2000)

~~ 31. What are the co-ordinates of the foot of the perpendicular from the point $(2,3)$ on the line $x+y-11=0$ ?

(a) $(2,9)$

(b) $(5,6)$

(d) $(6,5)$

(NDA/NA 2011)

~~ 32. What is the equation of the line joining the origin with the point of intersection of the lines $4 x+3 y=12$ and $3 x+4 y$ $=12$ ?

(a) $x+y=1$

(b) $x-y=1$

(d) $x=y$

(NDA/NA 2011)

~~ 33. If $(-5,4)$ divides the line segment between the co-ordinate axes in the ratio $1: 2$, then what is its equation?

(a) $8 x+5 y+20=0$

(b) $5 x+8 y-7=0$

(d) $5 x-8 y+57=0$.

(NDA/NA 2010)

~~ 34. If $x \cos \theta+y \sin \theta=2$ is perpendicular to the line $x-y=3$, then what is the value of $\theta$ ?

(a) $\pi / 6$

(b) $\pi / 4$

(d) $\pi / 3$

(NDA/NA 2009)

~~ 35. The co-ordinates of $P$ and $Q$ are $(-3,4)$ and $(2,1)$ respectively. If $P Q$ is extended to $R$ such that $P R=2 Q R$, then what are the co-ordinates of $R$ ?

(a) $(3,7)$

(b) $(2,4)$

(d) $(7,-2)$

~~ 36. If the points with the co-ordinates ${a, m a},{b,(m+1) b}$, ${c,(m+2) c}$ are collinear, then which of the following is correct?

(a) $a, b, c$ are in A.P. only for all $m$

(b) $a, b, c$ are in G.P. only for all $m$

(d) $a, b, c$ are in A.P. only for all $m=1$

(NDA/NA 2007)

~~ 37. The line through the points $(4,3)$ and $(2,5)$ cuts off intercepts of lengths $\lambda$ and $\mu$ on the axes. Which one of the following is correct?

(a) $\lambda>\mu$

(b) $\lambda<\mu$

(d) $\lambda=\mu$

(NDA/NA 2008)

~~ 38. The equation of the line bisecting the join of $(3,-4)$ and $(5,2)$ and having its intercepts on the $x$-axis and $y$-axis in the ratio $2: 1$ is

(a) $2 x-y=9$ (b) $x-2 y=4$

(d) $x+2 y=2$

(KCET 2005)

~~ 39. A line passes through the point of intersection of the lines $100 x+50 y-1=0$ and $75 x+25 y+3=0$ and makes equal intercepts on the axes. Its equation is

(a) $25 x-25 y+6=0$

(b) $5 x-5 y+3=0$

(d) $5 x+5 y-7=0$

(Kerala PET 2008)

~~ 40. The image of the origin with reference to the line $4 x+3 y$ $-25=0$ is

(a) $(-4,-3)$

(b) $(-3,4)$

(d) $(6,-8)$

~~ 41. What is the angle between the lines whose equations are: $3 x+y-7=0$ and $x+2 y+9=0$.

(a) $45^{\circ}$

(b) $\tan ^{-1}(\frac{5}{2})$

(d) $\tan ^{-1}(\frac{3}{4})$

~~ 42. Find the equation of the line which passes through the point of intersection of the lines $2 x-y+5=0$ and $5 x+3 y-4=0$ and is perpendicular to the line $x-3 y+21=0$.

(a) $2 x+y+10=0$

(b) $3 x+y+21=0$

(d) $3 y-x+21=0$

~~ 43. What is the acute angle between the lines $A x+B y=A+B$ and $A(x-y)+B(x+y)=2 B$ ?

(a) $45^{\circ}$

(b) $\tan ^{-1}(\frac{A}{\sqrt{A^{2}+B^{2}}})$

(d) $60^{\circ}$

(NDA/NA 2007)

~~ 44. The line $L$ is given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is

(a) $\sqrt{17}$

(b) $\frac{17}{\sqrt{15}}$

(d) $\frac{23}{\sqrt{17}}$

(AIEEE 2010)

~~ 45. A straight line passes through the points $(a, 0)$ and $(0, b)$. The length of the line segment contained between the axes is 13 and the product of the intercepts is 60 . Find the equation of the straight line

(a) $5 x+12 y=60$

(b) $7 x-12 y=50$

(d) Both (a) and (c)

~~ 46. Find the equation of the straight line with a positive gradient which passes through the point $(-5,0)$ and is at a perpendicular distance of 3 units from the origin.

(a) $3 x+4 y-15=0$

(b) $4 x-3 y+15=0$

(d) $3 y-x+10=0$

~~ 47. A straight line is parallel to the lines $3 x-y-3=0$ and $3 x-y+5=0$ and lies between them. Find its equation if its distances from these lines are in the ratio $3: 5$.

(a) $3 x-y+10=0$

(b) $3 x-y=0$

(d) $3 y-x-10=0$

~~ 48. The point $A(0,0), B(1,7)$ and $C(5,1)$ are the vertices of a triangle. Find the length of the perpendicular from $A$ to $B C$ and hence the area of the $\triangle A B C$.

(a) $\frac{17}{\sqrt{84}} ; 17$ sq. units

(b) $\frac{17}{\sqrt{84}} ; 13$ sq. units

(d) $\frac{17}{\sqrt{13}} ; 17$ sq. units

~~ 49. If $p$ be the length of the perpendicular from the origin on the straight line $a x+b y=p$ and $b=\frac{\sqrt{3}}{2}$, then what is the angle between the perpendicular and the positive direction of $x$ ?

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $90^{\circ}$

(NDA/NA 2007)

~~ 50. The orthocentre of a triangle whose vertices are $(0,0)$, $(3,0)$ and $(0,4)$ is

(a) $(2,1)$

(b) $(-1,0)$

(d) $(0,0)$

(Rajasthan PET)

ANSWERS

1. (b)	2. (b)	3. $(d)$	4. $(a)$	5. $(b)$	6. (a)	7. $(b)$	8. (c)	9. (c)	10. $(d)$
11. $(d)$	12. (c)	13. $(d)$	14. $(b)$	15. (a)	16. $(b)$	17. $(d)$	18. $(b)$	19. (c)	20. (b)
21. (d)	22. (a)	23. $(d)$	24. $(d)$	25. $(b)$	26. $(c)$	27. (b)	28. (c)	29. (b)	30. $(d)$
31. (b)	32. $(d)$	33. (c)	34. $(b)$	35. $(d)$	36. (c)	37. $(d)$	38. $(d)$	39. (c)	40. $(c)$
41. (a)	42. $(c)$	43. (a)	44. $(d)$	45. $(d)$	46. $(c)$	47. (b)	48. $(d)$	49. $(c)$	50. $(d)$

HINTS AND SOLUTIONS

Let the point be $P$ whose abscissa $=$ ordinate $=a$.

$\therefore P \equiv(a, a)$

Given, $P A=P B$

$\Rightarrow(a+1)^{2}+a^{2}=a^{2}+(a-5)^{2}$

$\Rightarrow 2 a^{2}+2 a+1=2 a^{2}-10 a+25$

$\Rightarrow 12 a=24 \Rightarrow a=2$.

$\therefore \quad$ The point is $(\mathbf{2}, \mathbf{2})$.

~~ 2. Perimeter of $\triangle A B C=A B+B C+C A$

$ \begin{aligned} & =\sqrt{(0+4)^{2}+(-1-2)^{2}}+\sqrt{(3-0)^{2}+(3+1)^{2}} \\ & \quad+\sqrt{(3+4)^{2}+(3-2)^{2}} \\ & =\sqrt{16+9}+\sqrt{9+16}+\sqrt{49+1} \\ & =\sqrt{25}+\sqrt{25}+\sqrt{50}=5+5+5 \sqrt{2}=\mathbf{1 0}+\mathbf{5} \sqrt{\mathbf{2}} . \end{aligned} $

~~ 3. Let $P \equiv(x, y)$. Then, $P A=P B$ and $P B=P C$.

$\therefore \quad P A^{2}=P B^{2} \Rightarrow(x-1)^{2}+(y-3)^{2}=(x+3)^{2}+(y-5)^{2}$

$\Rightarrow x^{2}-2 x+1+y^{2}-6 y+9=x^{2}+6 x+9+y^{2}-10 y+25$

$\Rightarrow-8 x+4 y-24=0 \Rightarrow 2 x-y+6=0$

$P B^{2}=P C^{2} \Rightarrow(x+3)^{2}+(y-5)^{2}=(x-5)^{2}+(y+1)^{2}$

$\Rightarrow x^{2}+6 x+9+y^{2}-10 y+25$

$ =x^{2}-10 x+25+y^{2}+2 y+1 $

$\Rightarrow 16 x-12 y+8=0 \Rightarrow 4 x-3 y+2=0$

From $(i), y=2 x+6$. Putting in (ii), we have

$ \begin{aligned} & 4 x-3(2 x+6)+2=0 \\ \Rightarrow \quad & 4 x-6 x-18+2=0 \Rightarrow-2 x-16=0 \Rightarrow x=-8 \\ \therefore \quad & y=2 \times(-8)+6=-10 . \\ \therefore \quad & P B=\sqrt{(-8+3)^{2}+(-10-5)^{2}} \\ & =\sqrt{(-5)^{2}+(-15)^{2}}=\sqrt{25+225}=\sqrt{250}=\mathbf{5} \sqrt{\mathbf{1 0} .} \end{aligned} $

~~ 4. Given, $A M=B M$

$\Rightarrow A M^{2}=B M^{2}$

$\Rightarrow[x-(a+b)]^{2}+[y-(b-a)]^{2}$

$ =[x-(a-b)]^{2}+(y-(a+b))^{2} $

$\Rightarrow x^{2}-2(a+b) x+(a+b)^{2}+y^{2}-2(b-a) y+(b-a)^{2}$

$=x^{2}-2(a-b) x+(a-b)^{2}+y^{2}-2(a+b) y+(a+b)^{2}$

$\Rightarrow-2 a x-2 b x-2 b y+2 a y=-2 a x+2 b x-2 a y-2 b y$

$\Rightarrow-4 b x=-4 a y \Rightarrow b \boldsymbol{x}=\boldsymbol{a y}$.

~~ 5. Let $A(a, a), B(-a,-a)$ and $C(-\sqrt{3} a, \sqrt{3} a)$ be the vertices of $\triangle A B C$. Then,

$ \begin{aligned} A B & =\sqrt{(a+a)^{2}+(a+a)^{2}}=\sqrt{4 a^{2}+4 a^{2}} \\ & =\sqrt{8 a^{2}}=2 \sqrt{2} a \end{aligned} $

$ \begin{aligned} & B C=\sqrt{(-a+\sqrt{3} a)^{2}+(-a-\sqrt{3} a)^{2}} \\ & =\sqrt{a^{2}-2 \sqrt{3} a+3 a^{2}+a^{2}+2 \sqrt{3} a+3 a^{2}}=\sqrt{8 a^{2}}=2 \sqrt{2} a \\ & A C=\sqrt{(a+\sqrt{3} a)^{2}+(a-\sqrt{3} a)^{2}} \\ & =\sqrt{a^{2}+2 \sqrt{3} a+3 a^{2}+a^{2}-2 \sqrt{3} a+3 a^{2}}=\sqrt{8 a^{2}}=2 \sqrt{2} a \\ & \because A B=B C=A C, \triangle A B C \text{ is equilateral. } \\ & \quad \text{ Area }=\frac{\sqrt{3}}{4}(\text{ side })^{2}=\frac{\sqrt{3}}{4} \times(2 \sqrt{2} a)^{2} \\ & \quad=\frac{\sqrt{3}}{4} \times 8 a^{2}=2 \sqrt{3} a^{2} . \end{aligned} $

~~ 6. Let $A(8,6), B(8,-2)$ and $C(2,-2)$ be the vertices of the given triangle and $P(x, y)$ be the circum-centre of this triangle. Then, $P A^{2}=P B^{2}=P C^{2}$

Now, $P A^{2}=P B^{2} \Rightarrow(x-8)^{2}+(y-6)^{2}=(x-8)+(y+2)^{2}$

$\Rightarrow x^{2}-16 x+64+y^{2}-12 y+36$

$ =x^{2}-16 x+64+y^{2}+4 y+4 $

$\Rightarrow 16 y=32 \Rightarrow y=2$.

Now, $P B^{2}=P C^{2} \Rightarrow(x-8)^{2}+(y+2)^{2}=(x-2)^{2}+(y+2)^{2}$

$\Rightarrow x^{2}-16 x+64+y^{2}+4 y+4=x^{2}-4 x+4+y^{2}+4 y+4$

$\Rightarrow 12 x=60 \Rightarrow x=5$.

$\therefore$ Co-ordinates of the circumcentre are $(5,2)$.

~~ 7. $A B=\sqrt{(3-1)^{2}+(5-1)^{2}}=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}$

$B C=\sqrt{(1-5)^{2}+(1-3)^{2}}=\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}$

$C D=\sqrt{(5-7)^{2}+(3-7)^{2}}=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}$

$A D=\sqrt{(3-7)^{2}+(5-7)^{2}}=\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}$

$A C=\sqrt{(3-5)^{2}+(5-3)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

$B D=\sqrt{(1-7)^{2}+(1-7)^{2}}=\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}$

Now, $A B=B C=C D=A D \Rightarrow$ All sides are equal

Also, $A C \neq B D \quad \Rightarrow$ Diagonals are not equal.

$\Rightarrow A B C D$ is a rhombus.

~~ 8. $A B=\sqrt{(4-7)^{2}+(5-6)^{2}}=\sqrt{9+1}=\sqrt{10}$

$B C=\sqrt{(7-4)^{2}+(6-3)^{2}}=\sqrt{9+9}=3 \sqrt{2}$

$C D=\sqrt{(4-1)^{2}+(3-2)^{2}}=\sqrt{9+1}=\sqrt{10}$

$A D=\sqrt{(4-1)^{2}+(5-2)^{3}}=\sqrt{9+9}=3 \sqrt{2}$

$A C=\sqrt{(4-4)^{2}+(5-3)^{2}}=\sqrt{4}=2$ $B D=\sqrt{(7-1)^{2}+(6-2)^{2}}=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13}$

$A B=C D, B C=A D$ and $A C \neq B D \Rightarrow$ opposite sides are equal and diagonals are not equal.

$\Rightarrow A B C D$ is a parallelogram.

~~ 9. Co-ordinates of $P$ are $(\frac{-1-1}{2}, \frac{-1+4}{2})$ i.e. $(-1, \frac{3}{2})$

Co-ordinates of $Q$ are $(\frac{-1+5}{2}, \frac{4+4}{2})$, i.e, $(2,4)$

Co-ordinates of $R$ are $(\frac{5+5}{2}, \frac{4-1}{2})$,i.e, $(5, \frac{3}{2})$

Co-ordinates of $S$ are $(\frac{-1+5}{2}, \frac{-1-1}{2})$,i.e, $(2,-1)$

Now, $P Q=\sqrt{(2+1)^{2}+(4-\frac{3}{2})^{2}}=\sqrt{3^{2}+(\frac{5}{2})^{2}}$

$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}=\sqrt{\frac{61}{4}}$

$Q R=\sqrt{(5-2)^{2}+(\frac{3}{2}-4)^{2}}=\sqrt{3^{2}+(\frac{-5}{2})^{2}}$

$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}=\sqrt{\frac{61}{4}}$

$R S=\sqrt{(2-5)^{2}+(-1-\frac{3}{2})^{2}}=\sqrt{(-3)^{2}+(\frac{-5}{2})^{2}}$

$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}=\sqrt{\frac{61}{4}}$

$S P=\sqrt{(-1-2)^{2}+(\frac{3}{2}+1)^{2}}=\sqrt{(-3)^{2}+(\frac{5}{2})^{2}}$

$=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}=\sqrt{\frac{61}{4}}$

$\Rightarrow P Q=Q R=R S=S P \Rightarrow$ All sides are equal

Also, $P R=\sqrt{(5+1)^{2}+(\frac{3}{2}-\frac{3}{2})^{2}}=\sqrt{36}=6$

$ S Q=\sqrt{(2-2)^{2}+(4+1)^{2}}=\sqrt{25}=5 $

$\Rightarrow P R \neq S Q \Rightarrow$ Diagonals are not equal

$\Rightarrow P Q R S$ is a rhombus.

~~ 10. $\triangle P Q R$ is right angle at $R$

$\Rightarrow P R^{2}+R Q^{2}=P Q^{2}$

$ \begin{aligned} & \text{ (Pythagoras’ Theorem) } \\ & \Rightarrow(6-12)^{2}+(0-8)^{2}+(-2-6)^{2} \\ & +(a-0)^{2}=(-2-12)^{2}+(a-8)^{2} \\ & \Rightarrow 36+64+64+a^{2}=196+a^{2}-16 a+64 \\ & \Rightarrow 16 a=196-(64+36)=196-100=96 \\ & \Rightarrow a=6 \end{aligned} $

~~ 11. $A B \perp$ chord $P Q \Rightarrow A B$ bisects chord $P Q \Rightarrow P Q=2 P B$.

$ \begin{aligned} A B & =\sqrt{(2-5)^{2}+(-3-1)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}} \\ & =\sqrt{9+16}=\sqrt{25}=5 \\ A P & =\text{ radius of circle }=13 \end{aligned} $

$\therefore \quad$ By Pythagoras’ Theorem, $P B=\sqrt{A P^{2}-A B^{2}}$

$ =\sqrt{169-25}=\sqrt{144}=12 \text{ units } $

$\therefore P Q=2 \times P B=\mathbf{2 4}$ units.

~~ 12. Let $\triangle(a, b)$ be the fourth vertex of the parallelogram $A B C D$. The diagonals of a parallelogram bisect each other at point $O$ (say), so the diagonals $A C$ and $B D$ have the same midpoint.

~~ 13. $\because$ Mid-point of $B C=(\frac{7+3}{2}, \frac{7+5}{2})$,i.e., $(5,6)$, we can easily show that $D$ lies on the boundary of $\triangle A B C$.

~~ 14. Let $P Q R S$ be the required square $P(0,-1)$ and $P(0,-1)$ and $R(0,3)$ be its two opposite vertices.

Length of diagonal $P R$

$=\sqrt{(0-0)^{2}+(3+1)^{2}}=\sqrt{16}=4$

$\therefore$ Length of each side $=\frac{\text{ diagonal }}{\sqrt{2}}$

$=\frac{4}{\sqrt{2}}=2 \sqrt{2}$.

$(0,3)$

Let the co-ordinates of another vertex of the square say $Q$ be $(a, b)$

$\therefore \quad$ Its distance from vertex $P$ should be equal to its distance from vertex $R$.

$\therefore \sqrt{(a-0)^{2}+(b+1)^{2}}=\sqrt{(a-0)^{2}+(b-3)^{2}}$

$\Rightarrow a^{2}+b^{2}+2 b+1=a^{2}+b^{2}-6 b+9$

$\Rightarrow 8 b=8 \Rightarrow b=1$

Also, this distance $Q P=2 \sqrt{2}$

$\therefore \sqrt{(a-0)^{2}+(b+1)^{2}}=2 \sqrt{2}$

$\Rightarrow a^{2}+b^{2}+2 b+1=8 \Rightarrow a^{2}+4=8 \Rightarrow a^{2}=4 \Rightarrow a= \pm 2$.

$\therefore$ The other two vertices of the square are $(+2,1)$ and $(-2,1)$.

~~ 15. Co-ordinates of the point of external division are

$(\frac{m_1 x_2-m_2 x_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2})$,i.e.,

$\therefore \quad$ Required point $=(\frac{3 \times-6-2 \times 4}{3-2}, \frac{3 \times 2-2 \times 4}{3-2})$

$=(\frac{-24}{1}, \frac{-2}{1})$, i.e., $(\mathbf{- 2 4},-\mathbf{2})$.

~~ 16. Any point on the $x$-axis is $(a, 0)$.

Let the point $(a, 0)$ divide the join of $A(2,-3)$ and $B(5,6)$ in the ratio $k: 1$. Then the co-ordinates of the point of division are $(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1})$

Now, $\frac{5 k+2}{k+1}=a$ and $\frac{6 k-3}{k+1}=0$

$\therefore \quad 6 k-3=0 \Rightarrow k=\frac{1}{2}$

$\therefore$ Required ratio is $k: 1 \Rightarrow \frac{1}{2}: 1=\mathbf{1}: \mathbf{2}$.

~~ 17. Let the line $3 x+4 y=7$ divide the join of $(-2,1)$ and $(1,2)$ in the ratio $k: 1$. Then the co-ordinates of the point of division are : $(\frac{k-2}{k+1}, \frac{2 k+1}{k+1})$

This point lies on the line $3 x+4 y=7$, so satisfies the equation of the given line, i.e., $3(\frac{k-2}{k+1})+4(\frac{2 k+1}{k+1})=7$

$\Rightarrow 3 k-6+8 k+4=7 k+7$

$\Rightarrow 4 k=9 \Rightarrow k=\frac{9}{4}$.

$\therefore$ Required ratio $=\frac{9}{4}: 1=9: 4$.

~~ 18. Let $P(m, 6)$ divides $A B$ in the ratio $k: 1$.

Then co-ordinates of $P$ are $(\frac{2 k-4}{k+1}, \frac{8 k+3}{k+1})$

Given, co-ordinates of $P$ are $(m, 6) \Rightarrow \frac{2 k-4}{k+1}=m$

and $\frac{8 k+3}{k+1}=6 \Rightarrow 8 k+3=6 k+6 \Rightarrow 2 k=3 \Rightarrow k=\frac{3}{2}$

$\therefore$ Required ratio is $\frac{3}{2}: 1=\mathbf{3}: \mathbf{2}$

Now, $\frac{2 k-4}{k+1}=m \Rightarrow \frac{2 \times \frac{3}{2}-4}{\frac{3}{2}+1}=\frac{3-4}{5 / 2}=\frac{-2}{5}$

$\therefore \quad m=\frac{-\mathbf{2}}{\mathbf{5}}$.

~~ 19. Let the co-ordinates of vertex $C$ are $(x, y)$.

Since the incentre and centroid of an equilateral triangle coincide, co-ordinates of centroid of

$\triangle A B C \equiv(\frac{9+\sqrt{3}}{6}, \frac{15-\sqrt{3}}{6})$

$\therefore \quad \frac{x+1+2}{3}=\frac{9+\sqrt{3}}{6}$ and $\frac{2+3+y}{3}=\frac{15-\sqrt{3}}{6}$

$\Rightarrow 2(x+3)=9+\sqrt{3}$ and $2(5+y)=15-\sqrt{3}$

$\Rightarrow 2 x=3+\sqrt{3}$ and $10+2 y=15-\sqrt{3}$

$\Rightarrow x=\frac{3+\sqrt{3}}{2}$ and $2 y=5-\sqrt{3} \Rightarrow y=\frac{5-\sqrt{3}}{2}$

$\therefore$ Required co-ordinates are $(\frac{3+\sqrt{3}}{2}, \frac{5-\sqrt{3}}{2})$.

~~ 20. Let $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ be the vertices of $\triangle A B C$ the coordinates who mid-points are given as $P(1,1)$, $Q(3,4)$ and $R(2,-3)$

Now, $\frac{x_1+x_2}{2}=1 \quad$ and $\quad \frac{y_1+y_2}{2}=1$

$\frac{x_2+x_3}{2}=3$ and $\quad \frac{y_2+y_3}{2}=4$

$\frac{x_1+x_3}{2}=2$ and $\frac{y_1+y_3}{2}=-3$

$\therefore \quad \frac{x_1+x_2+x_2+x_3+x_1+x_3}{2}=6$

and $\frac{y_1+y_2+y_2+y_3+y_1+y_3}{2}=2$

$\Rightarrow x_1+x_2+x_3=6$ and $y_1+y_2+y_3=2$

Now, $(x_1+x_2+x_3)-(x_1+x_2)=6-2 \Rightarrow x_3=4$

and $(y_1+y_2+y_3)-(y_1+y_2)=2-2 \Rightarrow y_3=0$

$(x_1+x_2+x_3)-(x_2+x_3)=6-6 \Rightarrow x_1=0$

and $(y_1+y_2+y_3)-(y_2+y_3)=2-8 \Rightarrow y_1=-6$

$(x_1+x_2+x_3)-(x_1+x_3)=6-4 \Rightarrow x_2=2$

and $(y_1+y_2+y_3)-(y_1+y_3)=2+6 \Rightarrow y_2=8$

$\therefore \quad A(0,-6), B(2,8), C(4,0)$ are the vertices of $\triangle A B C$

Now, $\quad a=B C=\sqrt{(4-2)^{2}+(0-8)^{2}}=\sqrt{4+64}$

$ =\sqrt{68}=2 \sqrt{17} $

$b=A C=\sqrt{(4-0)^{2}+(0+6)^{2}}=\sqrt{16+36}=\sqrt{52}=2 \sqrt{13}$

$c=A B=\sqrt{(2-0)^{2}+(8+6)^{2}}=\sqrt{4+196}=\sqrt{200}=10 \sqrt{2}$ $\therefore$ Co-ordinates of incentre of $\triangle A B C$ are

$(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c})$

$=(\frac{2 \sqrt{17} \times 0+2 \sqrt{13} \times 2+10 \sqrt{2} \times 4}{2 \sqrt{17}+2 \sqrt{13}+10 \sqrt{2}}.$,

$.\frac{2 \sqrt{17} \times-6+2 \sqrt{13} \times 8+10 \sqrt{2} \times 0}{2 \sqrt{17}+2 \sqrt{13}+10 \sqrt{2}})$

$=(\frac{4 \sqrt{13}+40 \sqrt{2}}{2 \sqrt{17}+2 \sqrt{13}+10 \sqrt{2}}, \frac{-12 \sqrt{17}+16 \sqrt{13}}{2 \sqrt{17}+2 \sqrt{13}+10 \sqrt{2}})$

$=(\frac{2 \sqrt{13}+20 \sqrt{2}}{\sqrt{17}+\sqrt{13}+5 \sqrt{2}}, \frac{8 \sqrt{13}-6 \sqrt{17}}{\sqrt{17}+\sqrt{13}+5 \sqrt{2}})$.

~~ 21. Let $(x, y)$ be the co-ordinates of the third vertex of the triangle.

Then $\frac{0+2+x}{3}=1$ and $\frac{0+0+y}{3}=1$

$\Rightarrow 2+x=3$ and $y=3$

$\Rightarrow x=1, y=3$.

$\therefore$ Co-ordinates of vertices of the triangle are $(0,0),(2,0)$ and $(1,3)$.

$\therefore \quad$ Area of triangle $=\frac{1}{2}[0(0-3)+2(3-0)+1(0-0)]$

$[\because.$ Area of $\Delta=\frac{1}{2}(x_1(y_2-y_3)+x_2(y_2-y_3)+x_3(y_1-y_2)]$

$=\frac{1}{2}[0+6+0]=\frac{6}{2}=3$.

~~ 22. For three points to be collinear, area of the triangle formed by the three points should be equal to zero, i.e.

$ \begin{aligned} & \frac{1}{2}[k(3 k-1)+2 k(1-2 k)+3(2 k-3 k)]=0 \\ \Rightarrow & \frac{1}{2}[3 k^{2}-k+2 k-4 k^{2}-3 k]=0 \\ \Rightarrow & k^{2}+2 k=0 \Rightarrow k=0 \text{ or }-2 \end{aligned} $

Neglecting $k=0$, as then $(k, 2 k)$ and $(2 k, 3 k)$ will be the same point, we take $k=-2$.

~~ 23. Points $A(1,2), B(0,0)$ and $C(a, b)$ are collinear if area of $\triangle A B C=0$

$\Rightarrow \frac{1}{2}[1(0-b)+0(b-2)+a(2-0)]=0$

$\Rightarrow-b+2 a=0 \Rightarrow \mathbf{2 a}=\boldsymbol{b}$.

~~ 24. Given $\frac{A D}{A B}=\frac{1}{4} \Rightarrow \frac{A D}{D B}=\frac{1}{3}$, i.e., $D$ divides $A B$ internally in the ratio $1: 3$.

$\therefore$ Co-odinates of $D$ are

$(\frac{1+12}{1+3}, \frac{5+18}{1+3})$ i.e., $(\frac{13}{4}, \frac{23}{4})$

Also, $\frac{A E}{A C}=\frac{1}{4} \Rightarrow \frac{A C}{E C}=\frac{1}{3}$, i.e., $E$ divides $A C$ internally in the ratio $1: 3$. Co-ordinates of $E$ are $(\frac{7+12}{1+3}, \frac{2+18}{1+3})$,i.e., $(\frac{19}{4}, 5)$

Now, area of $\triangle A B C=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]$

$ =\frac{1}{2}[12-4+7]=\frac{15}{2} \text{ sq. units } $

Area of $\triangle A D E$, where $A(4,6), D(\frac{13}{4}, \frac{23}{4}), E(\frac{19}{4}, 5)$ is

$\frac{1}{2}[4(\frac{23}{4}-5)+\frac{13}{4}(5-6)+\frac{19}{4}(6-\frac{23}{4})]$

$=\frac{1}{2}[4 \times \frac{3}{4}+\frac{13}{4} \times-1+\frac{19}{4} \times \frac{1}{4}]=\frac{1}{2}[3-\frac{13}{4}+\frac{19}{16}]$

$=\frac{1}{2}[\frac{48-52+19}{16}]=\frac{15}{32}$ sq. units

$\therefore \quad \frac{\text{ Area of } \triangle A D E}{\text{ Area of } \triangle A B C}=\frac{15 / 32}{15 / 2}=\mathbf{1 : 1 6}$.

~~ 25. Area of quadrilateral whose vertices are $(x_1, y_1),(x_2, y_2)$, $(x_3, y_3)$ and $(x_4, y_4)$ taken in order is

$ \begin{aligned} & \begin{matrix} \frac{1}{2}[(x_1 y_2-x_2 y_1)+(x_2 y_3-x_3 y_2)+(x_3 y_4-x_4 y_3). \\ .+(x_4 y_1-x_1 y_4)] \end{matrix} \\ & =\frac{1}{2}[(1 \times 6)-(-5) \times 2]+[(-5 \times-4)-(7 \times 6)]+[(7 \times k) \\ & -(-2) \times(-4)]+[(-2 \times 2)-(1 \times k)] \\ & =\frac{1}{2}[(6+10)+(20-42)+(7 k-8)+(-4-k)] \\ & =\frac{1}{2}[16-22+7 k-8-4-k] \\ & =\frac{1}{2}[-18+6 k]=0 \Rightarrow 6 k=18 \Rightarrow \boldsymbol{k}=\mathbf{3} . \end{aligned} $

~~ 26. Type Q. 24 .

~~ 27. The line $3 x+4 y-24=0$ cuts the axis at $A$. To obtain the co-ordinates of $A$ put $y=0$, as on $x$-axis, $y=0$.

$\therefore A \equiv(8,0)$

Also, on $y$-axis, $x=0$, therefore $B \equiv(0,6)$

$\therefore$ The three vertices of

$\triangle A O B$ are $A(8,0), O(0,0), B(0,6)$

$\therefore a=O B=6, b=O A=8$,

$c=A B=\sqrt{(0-8)^{2}+(6-0)^{2}}=\sqrt{64+36}=\sqrt{100}=10$.

$\therefore$ Co-ordinates of the incentre

$=(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c})$

$=(\frac{6 \times 8+8 \times 0+10 \times 0}{6+8+10}, \frac{6 \times 0+8 \times 6+10 \times 0}{6+8+10})$

$=(\frac{48}{24}, \frac{48}{24})=(2,2)$.

~~ 28. Let the slope of the lin passing through the points $(-k, h)$ and $(b,-f)$ be $m_1$. Then $m_1=\frac{-f-h}{b+k}=-(\frac{f+h}{b+k})$

$ [\text{ Slope }=\frac{y_2-y_1}{x_2-x_1}] $

$\therefore \quad$ Slope of line perpendicular to the given line $=-\frac{1}{m_1}$

$ =\frac{-1}{-(\frac{f+h}{b+k})}=(\frac{\boldsymbol{b}+\boldsymbol{k}}{\boldsymbol{f}+\boldsymbol{h}}) $

~~ 29. Let $D$ be the mid-point of $B C$.

Then $D \equiv(\frac{a+0}{2}, \frac{0}{2})$ i.e. $(\frac{a}{2}, 0)$

Let $E$ be the mid-point of $A C$, then

$E \equiv(\frac{a+0}{2}, \frac{0+b}{2})=(\frac{a}{2}, \frac{b}{2})$

$A D \perp B E$ if slope of

$A D \times$ Slope of $B E=-1 \ldots(i)$

Slope of $A D=\frac{0-b}{a / 2-0}=\frac{-b}{a / 2}=\frac{-2 b}{a}$

Slope of $B E=(\frac{b / 2-0}{a / 2-0})=\frac{b / 2}{a / 2}=\frac{b}{a}$.

$\therefore \quad$ From $(i), \frac{-2 b}{a} \times \frac{b}{a}=-1$

$\Rightarrow 2 b^{2}=a^{2} \Rightarrow a= \pm \sqrt{2} b$.

~~ 30. $P S$ being the median of $\triangle P Q R, S$ is the mid-point of $Q R$, i.e.,

Coordinates of $S \equiv$

$(\frac{6+7}{2}, \frac{-1+3}{2})=(\frac{13}{2}, 1)$

Slope of line parallel to $P S=$ Slope of $P S$

$ =\frac{1-2}{\frac{13}{2}-2}=\frac{-1}{9 / 2}=\frac{-2}{9} . $

$\therefore \quad$ Equation of a line parallel to $P S$ passing through the point $(1,-1)$ is

$(y+1)=\frac{-2}{9}(x-1)$, i.e., $9 y+9=-2 x+2$

$\Rightarrow 2 x+9 y+7=0$.

~~ 31. Let the foot of the perpendicular be $M(x_1, y_1)$

Slope of line $A B$, i.e., $y=-x+11=-1$

Slope of line $P M=\frac{y_1-3}{x_1-2}$

Now, $P M \perp A B$

$\Rightarrow(\frac{y_1-3}{x_1-2}) \times-1=-1$

$\Rightarrow y_1-3=x_1-2 \Rightarrow x_1-y_1=-1$

Also, $(x_1, y_1)$ lies on $A B$

$\Rightarrow x_1+y_1-11=0$

$\Rightarrow x_1+y_1=11$

Adding (i) and (ii), we get $2 x_1=10 \Rightarrow x_1=5$

Putting $x_1$ in (ii), we get $y_1=6$.

$\therefore$ Required foot of the perpendicular $M$ is $(\mathbf{5 , 6})$.

~~ 32. The equations of the given lines are:

$ \begin{aligned} & 4 x+3 y=12 \\ & 3 x+4 y=12 \end{aligned} $

Solving the simultaneous equations (i) and (ii), we get

$ x=\frac{12}{7}, y=\frac{12}{7} $

$\therefore$ Point of the intersection of the given lines is $(\frac{12}{7}, \frac{12}{7})$

Now equation of the line passing through $(0,0)$ and $(\frac{12}{7}, \frac{12}{7})$ is

$ y-0=(\frac{\frac{12}{7}-0}{\frac{12}{7}-0})(x-0) \text{, i.e., } \quad \boldsymbol{y}=\boldsymbol{x} . $

~~ 33. Let $A B$ be the given line between the co-ordinates axes $x=0, y=0$.

Let $A(a, 0)$ and $B(0, b)$ be two points of the given line on the co-ordinates axes. $(-5,4)$ divides $A B$ in the ratio $1: 2$.

$\therefore$ Co-ordinates of point of

division are $(\frac{1 \times 0+2 \times a}{1+2}, \frac{1 \times b+2 \times 0}{1+2})$,i.e., $(\frac{2 a}{3}, \frac{b}{3})$

Given, $\frac{2 a}{3}=-5$, and $\frac{b}{3}=4$

$\Rightarrow a=\frac{-15}{2}, b=12$

$\therefore$ The points $A$ and $B$ are $(\frac{-15}{2}, 0)$ and $(0,12)$ respectively.

Equation of line $A B:(y-0)=\frac{12-0}{0-(\frac{-15}{2})}(x+\frac{15}{2})$

$\Rightarrow y=\frac{12 \times 2}{15}(x+\frac{15}{2})$

$ =\frac{24}{15 \times 2}(2 x+15)=\frac{4}{5}(2 x+15) $

$\Rightarrow 5 y=8 x+60 \Rightarrow \mathbf{8 x}-\mathbf{5 y}+\mathbf{6 0}=\mathbf{0}$ is the equation of $A B$.

~~ 34. $x \cos \theta+y \sin \theta=2$

$\Rightarrow y \sin \theta=2-x \cos \theta \Rightarrow y=-x \cot \theta+2$

$\Rightarrow$ Slope of this line $=-\cot \theta$

Also, given $x-y=3 \Rightarrow y=x+3 \Rightarrow$ Slope of this line $=1$

$\therefore$ These lines are perpendicular, $-\cot \theta \times 1=-1$ $\Rightarrow \cot \theta=1 \Rightarrow \cot \theta=\cot \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4}$

~~ 35. Let the co-ordinates of $R$ be $(x, y)$.

As can be easily seen, it is a point of external division

Also, $P R=2 Q R$

$\Rightarrow R$ divides the join of $P$ and $Q$ externally in the ratio 2:1.

$\therefore \quad x=\frac{2 \times 2-1 \times-3}{2-1}, y=\frac{2 \times 1-1 \times 4}{2-1}$

$\Rightarrow x=4+3=7$ and $y=2-4=-2$.

$\therefore$ Co-ordinates of $R$ are $\mathbf{( 7 , - 2 )}$.

~~ 36. As the points $A(a, m a), B[b,(m+1) b]$ and $C[c,(m+2) c]$ are collinear.

Area of $\triangle A B C$ should be equal to zero.

$\Rightarrow \frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$

$\Rightarrow \frac{1}{2}[a{(m+1) b-(m+2) c}+b{(m+2) c-m a}$

$+c{m a-(m+1) b}]=0$

$\Rightarrow m a b+a b-m a c-2 a c+m b c+2 b c-m a b$

$+m a c-m b c-b c=0$

$\Rightarrow a b-2 a c+2 b c-b c=0 \Rightarrow a b+b c=2 a c$

$\Rightarrow b=\frac{2 a c}{a+c}$

$\therefore a, b, c$ are harmonic progression (H.P.) for all $m$.

~~ 37. Let the equation of the line in the intercept from be

$ \frac{x}{\lambda}+\frac{y}{\mu}=1 $

Since it passes through $(4,3)$ and $(2,5)$

$ \frac{4}{\lambda}+\frac{3}{\mu}=1 $

and $\frac{2}{\lambda}+\frac{5}{\mu}=1$

Multiplying (ii) by 2 and subtracting from (i), we get

$ \begin{aligned} & (\frac{4}{\lambda}+\frac{3}{\mu})-(\frac{4}{\lambda}+\frac{10}{\mu})=1-2 \\ \Rightarrow & \frac{-7}{\mu}=-1 \Rightarrow \mu=7 \end{aligned} $

Putting $\mu=7$ in (i), we get $\frac{4}{\lambda}+\frac{3}{7}=1$

$\Rightarrow \frac{4}{\lambda}=1-\frac{3}{7}=\frac{4}{7} \Rightarrow \lambda=7$

$\therefore \lambda=\mu=7$.

~~ 38. Let the required equation make intercept on $x$-axis $=2 a$

$\Rightarrow$ intercept made on $y$-axis $=a$

$\therefore$ Eqn of the given line in the intercept from: $\frac{x}{2 a}+\frac{y}{a}=1$

Since the line given by $(i)$ bisects the join of $(3,-4)$ and $(5,2)$, the mid-point of the line joining $(3,-4)$ and $(5,2)$ lies on $(i)$.

Mid- point of join of $(3,-4)$ and $(5,2)=(\frac{3+5}{2}, \frac{-4+2}{2})$

$ =(4,-1) $

Putting $(4,-1)$ in $(i)$, we get

$\frac{4}{2 a}+\frac{-1}{a}=1 \Rightarrow \frac{4-2}{2 a}=1 \Rightarrow \frac{2}{2 a}=1 \Rightarrow a=1$.

$\therefore$ Required equation of line : $\frac{x}{2 \times 1}+\frac{y}{1}=1 \Rightarrow \boldsymbol{x}+\mathbf{2} \boldsymbol{y}=\mathbf{2}$.

~~ 39. The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously.

$ \begin{aligned} & 100 x+50 y=1 \\ & 75 x+25 y=-3 \end{aligned} $

Eqn $(i)-2 \times$ Eqn (ii)

$\Rightarrow(100 x+50 y)-(150 x+50 y)=1-(-6)$

$\Rightarrow-50 x=7 \Rightarrow x=\frac{-7}{50}$

Putting the value of $x$ in (i), we get

$100 \times \frac{-7}{50}+50 y=1 \Rightarrow 50 y=1+14=15 \Rightarrow y=\frac{15}{50}=\frac{3}{10}$

Let the required line make $x$-intercept $=y$-intercept $=a$.

Then eqn of required line is $\frac{x}{a}+\frac{y}{a}=1 \Rightarrow x+y=a$

Since it passes through $(\frac{-7}{50}, \frac{3}{10})$, therefore

$\Rightarrow \frac{-7}{50}+\frac{3}{10}=a \Rightarrow \frac{-7+15}{50}=a \Rightarrow a=\frac{8}{50}=\frac{4}{25}$

$\therefore \quad$ Eqn of required line: $x+y=\frac{4}{25} \Rightarrow \mathbf{2 5}+\mathbf{2 5 y}-\mathbf{4}=\mathbf{0}$.

~~ 40. Let $A B$ be the given line $4 x+$ $3 y=25$

Let $O^{\prime}(a, b)$ be the image of $O$ in the given line $A B$.

Let $O O^{\prime}$ cut $A B$ in point $P$. Also $O P \perp A B$ and $P$ is the mid-point of $O O^{\prime}$.

$\therefore$ Co-ordinates of $P$ are $(\frac{a}{2}, \frac{b}{2})$

$\because P$ lies on $A B, 4 \times \frac{a}{2}+3 \times \frac{b}{2}-25=0$

$\Rightarrow 4 a+3 b=50$

Also, $O O^{\prime} \perp A B \Rightarrow$ slope of $O O^{\prime} \times$ slope of $A B=-1$

$\Rightarrow(\frac{b-0}{a-0}) \times \frac{-4}{3}=-1 \quad(\because.$ Line $.A B: y=\frac{-4 x}{3}+\frac{25}{3})$ $\Rightarrow \frac{b}{a}=\frac{3}{4} \Rightarrow 3 a=4 b$

From (ii), $a=\frac{4 b}{3}$. Putting this value of $a$ in (i), we get

$ 4 \times \frac{4 b}{3}+3 b=50 $

$\Rightarrow 16 b+9 b=150 \Rightarrow 25 b=150 \Rightarrow b=6$

$\Rightarrow a=\frac{4 \times 6}{3}=8$

$\therefore \quad$ The image of the point $O(0,0)$ in the line

$ 4 x+3 y-25=0 \text{ is }(\mathbf{8}, \mathbf{6}) \text{. } $

~~ 41. $3 x+y-7=0 \Rightarrow y=-3 x+7 \Rightarrow$ Slope $(m_1)=-3$

$x+2 y+9=0 \Rightarrow y=\frac{-x}{2}-\frac{9}{2} \Rightarrow$ Slope $(m_2)=-\frac{1}{2}$

If $\theta$ is the angle between the given lines, then

$\tan \theta=|\frac{m_1-m_2}{1+m_1 m_2}|=|\frac{-3-(\frac{1}{2})}{1+(-3)(-\frac{1}{2})}|=|\frac{-\frac{5}{2}}{1+\frac{3}{2}}|$

$=|\frac{-\frac{5}{2}}{\frac{5}{2}}|=1$

$\therefore \quad \theta=\mathbf{4 5}^{\circ}$.

~~ 42. The equations of the two lines whose point of intersection is needed are:

$ \begin{aligned} & 2 x-y=-5 \\ & 5 x+3 y=4 \end{aligned} $

$3 \times$ Eqn $(i)+$ Eqn $(i i) \Rightarrow(6 x-3 y)+(5 x+3 y)=-15+4$

$\Rightarrow 11 x=-11 \Rightarrow x=-1$

Putting $x=-1$ in $(i)$, we get $-2-y=-5 \Rightarrow y=3$.

$\therefore$ Point of intersection is $(-1,3)$.

Slope of line $x-3 y+21=0$, i.e., $y=\frac{x}{3}+7$ is $\frac{1}{3}$.

$\Rightarrow$ Slope of line perpendicular to line $x-3 y+21=0$ is -3

$[\because m_1 \times m_2=-1]$

$\therefore \quad$ Equation of a line through $(-1,3)$ with slope -3 is $y-3=-3(x+1) \Rightarrow y-3=-3 x-3 \Rightarrow \mathbf{3} \boldsymbol{x}+\boldsymbol{y}=\mathbf{0}$.

~~ 43. The equations of the given lines are:

$A x+B y=A+B \Rightarrow B y=-A x+(A+B)$

$\Rightarrow y=-\frac{A}{B} x+\frac{(A+B)}{B}$

and $A(x-y)+B(x+y)=2 B \Rightarrow(A+B) x+(B-A) y=2 B$

$\Rightarrow y=\frac{-(A+B)}{(B-A)} x+\frac{2 B}{(B-A)}$

$\therefore \quad$ Slope of line $(i)=m_1=-\frac{A}{B}$

Slope of line $($ ii $)=m_2=-\frac{(A+B)}{B-A}$ Let $\theta$ be the angle between both the lines, then

$ \begin{aligned} \tan \theta & =|\frac{m_1-m_2}{1+m_1 m_2}|=|\frac{-\frac{A}{B}+(\frac{A+B}{B-A})}{1+\frac{A}{B}(\frac{A+B}{B-A})}| \\ & =|\frac{-A(B-A)+B(A+B)}{B(B-A)+A(A+B)}| \\ & =|\frac{-A B+A^{2}+B A+B^{2}}{B^{2}-B A+A^{2}+A B}|=|\frac{A^{2}+B^{2}}{A^{2}+B^{2}}|=1 \end{aligned} $

$\therefore \tan \theta=1 \Rightarrow \theta=\tan ^{-1}(1)=\mathbf{4 5}^{\circ}$.

~~ 44. The given lines are:

$L: \frac{x}{5}+\frac{y}{b}=1$

$K: \frac{x}{c}+\frac{y}{3}=1$

Since line $L$ passes through $(13,32)$,

$\frac{13}{5}+\frac{32}{b}=1 \Rightarrow \frac{32}{b}=1-\frac{13}{5}=\frac{-8}{5} \Rightarrow b=\frac{32 \times 5}{-8}=-20$.

$\therefore \quad$ Line $L$ is $\frac{x}{5}-\frac{y}{20}=1$, i.e., $y=4 x-20$

$\Rightarrow$ Slope of line $L=4$

As $L | K$, slope of line $K=$ Slope of line $L$.

Eqn of line $K$ can be written as $y=\frac{-3 x}{c}+3$

$\therefore$ Slope of $K=-\frac{3}{c}$.

Given, $\frac{-3}{c}=4 \Rightarrow c=\frac{-3}{4}$

$\therefore$ Equation of line $K:(\frac{-4}{3}) x+\frac{y}{3}=1 \Rightarrow 4 x-y+3=0$.

$\therefore \quad$ Distance between the lines $L \equiv 4 x-y-20=0$ and $K \equiv 4 x-y+3=0$ is $d=|\frac{3-(-20)}{\sqrt{4^{2}+(-1)^{2}}}|=\frac{\mathbf{2 3}}{\sqrt{\mathbf{1 7}}}$. $(\begin{matrix} \text{ Distance between parallel lines } \\ a x+b y+c_1=0 \text{ and } a x+b y+c_2=0 \text{ is } d=\frac{|c_2-c_1|}{\sqrt{a^{2}+b^{2}}}\end{matrix} )$

~~ 45. Since the line passes through $A(a, 0)$ and $B(0, b)$, it makes intercepts $a$ and $b$ on $x$-axis and $y$-axis respectively. Let the equation of this line in the intercept from be $\frac{x}{a}+\frac{y}{b}=1$

By the given condition, $A B=\sqrt{a^{2}+b^{2}}=13$ and $a b=60$

$\therefore a^{2}+b^{2}=169 \Rightarrow a^{2}+b^{2}+2 a b=169+120$

$\Rightarrow(a+b)^{2}=289 \Rightarrow a+b= \pm 17$

Also, $a^{2}+b^{2}-2 a b=169-120 \Rightarrow(a-b)^{2}=49$

$\Rightarrow a-b= \pm 7$.

$\therefore$ From (i) and (ii) $a=12, b=5$ and $a=-12, b=-5$

$\therefore$ The required equations of the straight line are:

$ \begin{aligned} & \frac{x}{12}+\frac{y}{5}=1 \text{ and } \frac{x}{-12}+\frac{y}{-5}=1 \\ \Rightarrow 5 x+12 y=60 & \text{ and } \mathbf{5 x}+\mathbf{1 2 y}+\mathbf{6 0}=\mathbf{0 .} \end{aligned} $

~~ 46. Let $(m>0)$ be the gradient (slope) of the required line. Then, Equation of any line through $(-5,0)$ having slope $=m$ is $y-0=m(x-(-5))$ or $m x-y+5 m=0$

Its perpendicular distance from origin is 3

$\Rightarrow \frac{ \pm|m \cdot 0-0+5 m|}{\sqrt{m^{2}+(-1)^{2}}}=3 \Rightarrow|5 m|=3 \sqrt{m^{2}+1}$

$(\because.$ Distance of $(x_1, y_1)$ from line $.a x+b y+c=\frac{|a x_1+b y_1+c|}{\sqrt{a^{2}+b^{2}}})$

$\Rightarrow 25 m^{2}=9(m^{2}+1) \Rightarrow 16 m^{2}=9 \Rightarrow m=\frac{3}{4}(\because m$ is $+ve)$

$\therefore$ Required equation: $y=\frac{3}{4}(x+5)$

$\Rightarrow 3 x-4 y+15=0$.

~~ 47. Given lines are $3 x-y-3=0$ and $3 x-y+5=0$.

Line parallel to the given lines can be written as

$ 3 x-y+c=0 $

Let us taken a point, say, $(0, c)$ on $(i)$

(Putting $x=0$ in $(i)$, we get $y=c$ )

$\therefore \quad \frac{\text{ Distance of }(0, c) \text{ from } 3 x-y-3}{\text{ Distance of }(0, c) \text{ from } 3 x-y+5}=\frac{3}{5}$

$\Rightarrow \frac{\frac{|3 \times 0-c-3|}{\sqrt{3^{2}+1^{2}}}}{\frac{|3 \times 0-c+5|}{\sqrt{3^{2}+1^{2}}}}=\frac{3}{5} \Rightarrow \frac{c+3}{-c+5}=\frac{3}{5}$

$\Rightarrow 5 c+15=-3 c+15 \Rightarrow 8 c=0 \Rightarrow c=0$.

Substituting $c=0$ in $(i)$, the required equation is $3 \boldsymbol{x}-\boldsymbol{y}=\mathbf{0}$.

~~ 48. Equation of line $B C$ :

$y-7=\frac{1-7}{5-1}(x-1)$

$\Rightarrow y-7=\frac{-6}{4}(x-1)$

$\Rightarrow 2(y-7)=-3(x-1)$

$\Rightarrow 2 y-14=-3 x+3 \Rightarrow 3 x+2 y-17=0$.

$\therefore$ Distance of perpendicular from $A(0,0)$ on $B C \equiv 3 x+2 y-11=0$ is

$ \frac{|3 \times 0+2 \times 0-17|}{\sqrt{3^{2}+2^{2}}}=\frac{17}{\sqrt{13}} $

Area of $\triangle A B C=\frac{1}{2} \times B C \times$ perpendicular distance

$ B C=\sqrt{(5-1)^{2}+(1-7)^{2}}=\sqrt{4^{2}+6^{2}}=\sqrt{16+36} $

$ =\sqrt{52}=2 \sqrt{13} $

$\therefore \quad$ Required area $=\frac{1}{2} \times \frac{17}{\sqrt{13}} \times 2 \sqrt{13}=17$ sq. units.

~~ 49. Since $p$ is the length of perpendicular from origin on the straight line $a x+b y-p=0$.

$p=\frac{|a \cdot 0+b \cdot 0-p|}{\sqrt{a^{2}+b^{2}}}$

$\Rightarrow 1=\sqrt{a^{2}+b^{2}} \Rightarrow 1=a^{2}+\frac{3}{4}$

$(\because b=\frac{\sqrt{3}}{2})$

$\Rightarrow a^{2}=\frac{1}{4} \Rightarrow a=\frac{1}{2}$.

$\therefore \quad$ Equation of the straight line is $\frac{1}{2} x+\frac{\sqrt{3}}{2} y=p$

$\Rightarrow x \cos 60^{\circ}+y \sin 60^{\circ}=p$

Hence required angle is $60^{\circ}$, which is the angle between the perpendicular and the positive direction of $x$-axis.

~~ 50. Let the vertices of $\triangle A B C$ be given as:

$A(0,0), B(3,0)$ and $C(0,4)$

The orthocentre $O$ is the point of intersection of the altitudes drawn from the vertices of $\triangle A B C$ on the opposite sides.

Slope of $B C=\frac{4-0}{0-3}=\frac{-4}{3}$

$\therefore$ Slope of $A D \perp B C=\frac{3}{4}$ (Slope of $B C \times$ Slope of $A D=-1$ )

Equation of line through $A(0,0)$ having slope $\frac{3}{4}$ is

$ y-0=\frac{3}{4}(x-0) \Rightarrow 4 y=3 x $

Similarly, slope of $A C=\frac{4-0}{0-0}=\infty$

$\therefore \quad$ Slope of line $B E \perp A C=\frac{1}{\infty}=0$

$\therefore$ Equation of $B E:(y-0)=0(x-3) \Rightarrow y=0$

From $(i), x=0$

Hence, the orthocentre is $(\mathbf{0 , ~} \mathbf{0})$.

SELF ASSESSMENT SHEET

If the sum of the squares of the distances of the point $(x, y)$ from the points $(a, 0)$ and $(-a, 0)$ be $2 b^{2}$, then which of the following is correct?

(a) $x^{2}+a^{2}=b^{2}+y^{2}$

(b) $x^{2}+a^{2}=2 b^{2}-y^{2}$

(d) $x^{2}+a^{2}=b^{2}-y^{2}$

(NDA/NA 2011)

~~ 2. If the points $A(1,2), B(2,4)$ and $C(3, a)$ are collinear, what is the length of $B C$ ?

(a) $\sqrt{2}$ units

(b) $\sqrt{3}$ units

(d) 5 units

~~ 3. The mid-point of the line joining the points $(-10,8)$ and $(-6,12)$ divides the line joining the points $(4,-2)$ and $(-2,4)$ in the ratio

(a) $1: 2$ internally

(b) $1: 2$ externally

(d) $2: 1$ externally

(Kerala PET 2007)

~~ 4. The two vertices of a triangle are $(2,-1),(3,2)$ and the third vertex lies on the line $x+y=5$. The area of the triangle is 4 units. The third vertex is

(a) $(0,5)$ or $(1,4)$

(b) $(5,0)$ or $(4,1)$

(d) $(0,5)$ or $(4,1)$

~~ 5. What is the equation of the straight line which passes through $(3,4)$ and the sum of whose $x$-intercept and $y$-intercept is 14 ?

(a) $4 x+3 y=24$

(b) $x+y=14$

(d) $3 x+4 y=25$

(NDA/NA 2013)

~~ 6. What is the equation of the line passing through $(2,-3)$ and parallel to the $y$-axis?

(a) $y=-3$

(b) $y=2$

(d) $x=-3$

(NDA/NA 2011)

~~ 7. The acute angle which the perpendicular from the origin on the line $7 x-3 y=4$ makes with the $x$-axis is:

(a) zero

(b) positive but not $\pi / 4$

(d) $\pi / 4$ (NDA/NA2012)

~~ 8. A straight line passes through the points $(5,0)$ and $(0,3)$. The length of the perpendicular from the point $(4,4)$ on the line is:

(a) $\frac{\sqrt{17}}{2}$

(b) $\frac{\sqrt{17}}{2}$

(d) $\frac{17}{2}$

(NDA/NA 2013)

~~ 9. What is the acute angle between the two straight lines $y=(2-\sqrt{3}) x+5$ and $y=(2+\sqrt{3}) x-7$ ?

(a) $60^{\circ}$

(b) $45^{\circ}$

(d) $15^{\circ}$

~~ 10. What is the product of the perpendiculars from the two points $( \pm \sqrt{b^{2}-a^{2}}, 0)$ to the line $a x \cos \phi+b y \sin \phi=a b$ ?

(a) $a^{2}$

(b) $b^{2}$

(d) $a / b$

~~ 11. If $(-2,6)$ is the image of the point $(4,2)$ with respect to the line $L=0$, then $L$ is equal to

(a) $2 x+3 y-5$

(b) $3 x-2 y+5$

(d) $6 x-4 y+5$

~~ 12. The straight line $a x+b y+c=0$ and the co-ordinate axes form an isosceles triangle under which of the following conditions?

(a) $|a|=|b|$

(b) $|a|=|c|$

(d) None of these

ANSWERS

$($ d)

~~ 2. (c)

~~ 3. $(d)$

~~ 4. $(c)$

~~ 5. (a)

~~ 6. (c)

~~ 7. (c)

~~ 8. (b)

~~ 9. (a)

~~ 10. (a)

~~ 11. (b)

~~ 12. (a)

HINTS AND SOLUTIONS

Let $P(x, y), A(a, 0)$ and $B(-a, 0)$ be the given points. Then, $P A^{2}+P B^{2}=2 b^{2}$

$\Rightarrow(x-a)^{2}+(y-0)^{2}+(x+a)^{2}+(y-0)^{2}=2 b^{2}$

$\Rightarrow x^{2}-2 a x+a^{2}+y^{2}+x^{2}+2 a x+a^{2}+y^{2}=2 b^{2}$

$\Rightarrow x^{2}+a^{2}+y^{2}=b^{2}$

$\Rightarrow x^{2}+a^{2}=b^{2}-y^{2}$.

~~ 2. Area of $\triangle A B C=0$ for collinearity of $A, B, C$.

$\Rightarrow \frac{1}{2}[1(4-a)+2(a-2)+3(2-4)]=0$

$\Rightarrow 4-a+2 a-4+6-12=0$

$\Rightarrow a-6=0 \Rightarrow a=6$.

$\therefore$ Point $C \equiv(3,6)$

$\Rightarrow B C=\sqrt{(3-2)^{2}+(6-4)^{2}}$

$=\sqrt{1+4}=\sqrt{5}$ units.

~~ 3. The mid-point of the line joining the points $(-10,8)$ and $(-6,12)$ is $(\frac{-10+(-6)}{2}, \frac{8+12}{2})$, i.e., $(-8,10)$. Let $(-8,10)$ divide the join of $(4,-2)$ and $(-2,4)$ in the ratio $k: 1$.

Then, $-8=\frac{-2 k+4}{k+1}$ and $10=\frac{4 k-2}{k+1}$

$\Rightarrow-8 k-8=-2 k+4 \Rightarrow-6 k=12 \Rightarrow k=-2$

Since the value of $k$ is negative, it is a case of external division and the ratio is $\mathbf{2} \mathbf{:} \mathbf{1}$.

~~ 4. Let the third vertex of the triangle be $P(a, b)$.

Since it lies on the line $x+y=5, a+b=5$

Also, given area of triangle formed by the points $(2,-1)$, $(3,2)$ and $(a, b)=4$ units

$\Rightarrow \frac{1}{2}[2(2-b)+3(b+1)+a(-1-2)]= \pm 4$

$\Rightarrow[4-2 b+3 b+3-3 a]= \pm 8$

$\Rightarrow-3 a+b=1 \quad \ldots$ (ii) or $\quad-3 a+b=-15$

Eqn $(i)-Eqn(i i) \Rightarrow(a+b)-(-3 a+b)=5-1$

$\Rightarrow 4 a=4 \Rightarrow a=1 \Rightarrow b=4$

Eqn $(i)-Eqn(i i i) \Rightarrow(a+b)-(-3 a+b)=5+15$ $\Rightarrow 4 a=20 \Rightarrow a=5 \Rightarrow b=0$.

$\therefore$ The points are $(\mathbf{1 , 4})$ and $(\mathbf{5}, \mathbf{0})$.

~~ 5. Let the $x$-intercept $=a$. Then, $y$-intercept $=14-a$

$\therefore$ Eqn of the straight line is $\frac{x}{a}+\frac{y}{14-a}=1$

Since it passes through $(3,4)$, so

$ \begin{aligned} & \frac{3}{a}+\frac{4}{14-a}=1 \\ \Rightarrow & 3(14-a)+4 a=a(14-a) \\ \Rightarrow & 42-3 a+4 a=14 a-a^{2} \\ \Rightarrow & a^{2}-13 a+42=0 \\ \Rightarrow & (a-7)(a-6)=0 \Rightarrow a=7 \text{ or } 6 . \\ \therefore & \text{ Eqn is } \frac{x}{7}+\frac{y}{7}=1 \Rightarrow x+y=7 \\ & \text{ or } \frac{x}{6}+\frac{y}{8}=1 \Rightarrow 8 x+6 y=48 \Rightarrow \mathbf{4 x + 3 y = 2 4 .} \end{aligned} $

~~ 6. Slope of $y$-axis $=\tan 90^{\circ}$

( $\because y$-axis $\perp x$-axis)

$\therefore$ Equation of line passing through $(2,-3)$ parallel to $y$-axis is $(y+3)=\tan 90^{\circ}(x-2)$

$\Rightarrow(y+3)=\infty(x-2) \Rightarrow(x-2)=\frac{1}{\infty}(y+3)=0 \Rightarrow \boldsymbol{x}=\mathbf{2}$.

~~ 7. As the line from the origin is perpendicular to the line

$7 x-3 y=4$, so its slope $=\frac{-1}{\text{ slope of } 7 x-3 y=4}$

Slope of $7 x-3 y-4=\frac{7}{3}$

$\therefore$ Slope of line from origin $=\frac{-1}{+7 / 3}=\frac{-3}{7}$

$\therefore \tan \theta=\frac{-3}{7}$, where $\theta$ is the angle the line makes with the + ve direction of $x$-axis

$\Rightarrow \theta=\tan ^{-1}(\frac{-3}{7})=-\tan ^{-1}(\frac{3}{7})$

~~ 8. Equation of the line through the points $(5,0)$ and $(0,3)$

$y-0=\frac{3-0}{0-5}(x-5)$

$\Rightarrow y=\frac{-3}{5}(x-5)$

$\Rightarrow 5 y+3 x-15=0$

$\therefore$ Distance of perpendicular from point $(4,4)$ on the line

$ \begin{aligned} & 5 y+3 x-15=0 \text{ is }|\frac{5 \times 4+3 \times 4-15}{\sqrt{5^{2}+3^{2}}}| \\ & =\frac{|20+12-15|}{\sqrt{25+9}}=\frac{17}{\sqrt{34}} \text{ units. }=\sqrt{\frac{\mathbf{1 7}}{\mathbf{2}}} \text{ units. } \end{aligned} $

~~ 9. The two lines are:

$ \begin{aligned} & y=(2-\sqrt{3}) x+5 \\ & y=(2+\sqrt{3}) x-7 \end{aligned} $

Slope of line $(i), m_1=2-\sqrt{3}$

Slope of line (ii), $m_2=2+\sqrt{3}$ If $\theta$ is the angle between the two lines, then

$ \begin{aligned} \tan \theta & =|\frac{m_1-m_2}{1+m_1 m_2}|=\lvert, \frac{(2-\sqrt{3})-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}. \\ & =|\frac{-2 \sqrt{3}}{1+1}|=\sqrt{3} \\ \therefore \quad \theta & =\tan ^{-1}(\sqrt{3})=\mathbf{6 0}^{\circ} . \end{aligned} $

~~ 10. Given, $a x \cos \phi+b y \sin \phi-a b=0$.

$\therefore$ Perpendicular distance of the given line from

$ \begin{aligned} & (+\sqrt{b^{2}-a^{2}}, 0) \\ & d_1=|\frac{a \sqrt{b^{2}-a^{2}} \cos \phi-a b}{\sqrt{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}}| \end{aligned} $

$\therefore$ Perpendicular distance of the given line from $(-\sqrt{b^{2}-a^{2}}, 0)$

$ \begin{aligned} d_2= & |\frac{-a \sqrt{b^{2}-a^{2}} \cos \phi-a b}{\sqrt{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}}| \\ \therefore d_1 d_2 & =|-\frac{[a^{2}(b^{2}-a^{2}) \cos ^{2} \phi-a^{2} b^{2}]}{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}| \\ & =|-\frac{a^{2}(-b^{2} \cos ^{2} \phi-a^{2} \cos ^{2} \phi)}{a^{2} \cos ^{2} \phi+b^{2} \sin ^{2} \phi}|=\boldsymbol{a}^{2} . \end{aligned} $

~~ 11. Let the equation of the $L$ be $y=m x+c$

Since $O^{\prime}(-2,6)$ is the image of the point $O(4,2)$ in line $L=0$, the mid-point of $O O^{\prime}$, i.e., $(\frac{-2+4}{2}, \frac{6+2}{2})$, i.e., $(1,4)$ will lie on the given line.

Also, $L \perp O O^{\prime}$, so

Slope of $L \times$ Slope of $O O^{\prime}=-1$

$\Rightarrow m \times(\frac{6-2}{-2-4})=-1 \Rightarrow m=\frac{-1}{\frac{-2}{3}}=\frac{3}{2}$

$\therefore$ Equation of $L$ is $y=\frac{3}{2} x+c$

$\because$ It passes through $(1,4)$

$ 4=\frac{3}{2} \times 1+c \Rightarrow c=4-\frac{3}{2}=\frac{5}{2} . $

$\therefore$ Required equation is $y=\frac{3}{2} x+\frac{5}{2}$

$\Rightarrow 3 x-2 y+5=0$

$\therefore L=3 x-2 y+5$

~~ 12. The equation of line $A B$, i.e., $a x+b y+c=0$ in intercept form is $a x+b y=-c$

$\Rightarrow \frac{x}{(-c / a)}+\frac{y}{(-c / b)}=1$

$\triangle A O B$ is isosceles $\triangle$ if $O A=O B$, i.e., $x$-intercept

$=y-$ intercept

$\Rightarrow \frac{-c}{a}=\frac{-c}{b} \Rightarrow \frac{1}{a}=\frac{1}{b} \Rightarrow|\boldsymbol{a}|=|\boldsymbol{b}|$.

Area and Perimeter

KEY FACTS

SQUARE

Perimeter of a square of side $a=4 a=4 \times$ side
Diagonal of a square $=a \sqrt{2}=side \sqrt{2}$
Area of a square $=a^{2}=side^{2}$
Also, Area of a square $=a^{2}=\frac{(a \sqrt{2})^{2}}{2}=\frac{(\text{ Diagonal })^{2}}{2}$

RECTANGLE

Area of a rectangle $=$ length $\times$ breadth $=l \times b$
Perimeter of a rectangle $=2($ length + breadth $)=2(l+b)$
Diagonal of a rectangle $=\sqrt{\text{ length }^{2}+\text{ breadth }^{2}}$

$ =\sqrt{l^{2}+b^{2}} $

TRIANGLE

Given base and height

Area of a triangle $=\frac{1}{2} \times$ base $(b) \times$ height $(h)$, where the height of a triangle is the perpendicular distance from a vertex to the base of the triangle.

Given length of the three sides

(Heron’s formula) :

Area $=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{1}{2}(a+b+c)$,

where $a, b$ and $c$ are the lengths of the sides and $s$ is the semi-perimeter of the triangle.

Given side, angle, side

If we are given the lengths of two sides of a triangle and the size of the angle between them, then

Area of triangle $=\frac{1}{2} a b \sin C$ or $\frac{1}{2} b c \sin A$ or $\frac{1}{2} a c \sin B$

Equilateral triangle

Area $=\frac{\sqrt{3}}{4}(\text{ side })^{2}$

Altitude $(h)=\frac{\sqrt{3}}{2}$ side; $\quad$ Perimeter $=3 \times$ side

Isosceles triangle

Area $=\frac{1}{4} \times b \times \sqrt{4 a^{2}-b^{2}}$

where each of equal sides $=a$, other side $=b$.

Perimeter $=2 a+b$

Circumscribed circle of a triangle

Isosceles Triangle

If $a, b, c$ are the sides of a triangle and $R$ the radius of the circumscribed circle, then

Area of triangle $=\frac{a \times b \times c}{4 R}$

or Radius of circumscribed circle $(\boldsymbol{R})=\frac{\boldsymbol{a} \times \boldsymbol{b} \times \boldsymbol{c}}{4(\text{ Area of triangle })}$

For an equilateral triangle, circum-radius $(R)=\frac{a \times a \times a}{4 \cdot \frac{\sqrt{3}}{4} a^{2}}=\frac{a}{\sqrt{3}}=\frac{\text{ side }}{\sqrt{3}}$

Circumscribed Circle

Circle inscribed in a triangle

If $A=$ area of the given triangle, $r=$ radius of inscribed circle, $a, b, c$ are sides of the triangle, then

$ s=\frac{a+b+c}{2} \text{ and } A=r . s \text{ or } r=\frac{A}{s}=\frac{\text{ Area }}{\text{ Semi-perimeter }} $

For an equilateral triangle, in-radius $=\frac{\frac{\sqrt{3}}{4} \cdot a^{2}}{\frac{3 a}{2}}=\frac{a}{2 \sqrt{3}}=\frac{\text{ side }}{2 \sqrt{3}}$

Inscribed Circle

For similar triangles, $A B C$ and $D E F$
$\quad$ Ratio of areas $=area(\triangle A B C): area(\triangle D E F)$

$ =a^{2}: a^{\prime 2} \text{ or } b^{2}: b^{\prime 2} \text{ or } c^{2}: c^{\prime 2} $

$=$ Ratio of square of corresponding sides

$\quad$ Ratio of perimeters $=$ Ratio of corresponding sides.
Area of a triangle, given its medians

Area $=\frac{4}{3} \sqrt{s_m(s_m-m_1)(s_m-m_2)(s_m-m_3)}$,

where $m_1, m_2, m_3$ are the medians of the triangle and $s_m=\frac{m_1+m_2+m_3}{2}$

[See Q. 3 in Practice Sheet]

QUADRILATERAL

Area $=\frac{1}{2} \times$ diagonal $\times$ sum of perpendiculars from the opposite vertices on it

$ =\frac{\mathbf{1}}{\mathbf{2}} \times \boldsymbol{d} \times(\boldsymbol{h}_1+\boldsymbol{h}_2), \text{ where } A C=d $

Area $=\frac{1}{2} \times$ product of diagonals $\times$ sine of angle between them

$ =\frac{1}{2} d_1 d_2 \sin \theta_1=\frac{1}{2} d_1 d_2 \sin \theta_2 $

where $A C=d_1$ and $B D=d_2$

PARALLELOGRAM

Area $=$ base $(b) \times$ height $(h)$
Area $=$ product of any two adjacent sides $\times$ sine of included angle $=1$

$ =A B \cdot A D \cdot \sin \theta $

Perimeter $=2 \times$ (sum of any two adjacent sides)
1. RHOMBUS
Area $=\frac{1}{2} \times($ Product of diagonals $)=\frac{1}{2} d_1 d_2$
Area $=$ Product of any two adjacent sides $\times$ sine of included angle $=a^{2} \sin \theta$
Perimeter $=4 a=4 \times$ side
1. TRAPEZIUM
$A$ rea $=\frac{1}{2} \times($ sum of parallel sides $) \times$ height $=\frac{1}{2}(a+b) h$
Perimeter $=$ Sum of all four sides

CIRCLE (radius $=R$ )

Area $=\pi r^{2}$
Circumference $=2 \pi R$
Length of an $arc(l)=A B=2 \pi R(\frac{\theta}{360^{\circ}})$

Area of a sector $O A B=\pi R^{2}(\frac{\theta}{360^{\circ}})=\frac{1}{2}(arc$ length $\times R)$
Perimeter of a sector $=l+2 R$
Area of a sector $O A B=\pi R^{2}(\frac{\theta}{360^{\circ}})=\frac{1}{2}(arc$ length $\times R)$
Perimeter of a sector $=l+2 R$

Area of segment $A P B=$ Area of sector $O A P B$ - Area of $\triangle O A B$

$ =\pi R^{2}(\frac{\theta}{360^{\circ}})-\frac{R^{2}}{2} \sin \theta $

Perimeter of segment $=Arc A P B+Chord A B$

$ =2 R[\frac{\pi \theta}{360^{\circ}}+\sin (\frac{\theta}{2})] $

$ A M=R \sin \frac{\theta}{2}, A B=2 A M=2 R \sin \frac{\theta}{2} $

SEMI-CIRCLE

Area $=\frac{\pi R^{2}}{2}(R=$ radius $)$
Perimeter $=Arc A P B+$ Diameter $A B=\pi R+2 R$

QUADRANT

Area $=\frac{\pi R^{2}}{4}(R=$ radius $)$
Perimeter $=\frac{1}{4}($ circumference $)+2 R=\frac{1}{4} \times 2 \pi R+2 R=\frac{1}{2} \pi R+2 R$

Area of a ring or annulus

$=\pi(R^{2}-r^{2})$, where $R=$ outer radius, $r=$ inner radius.

Area of a regular hexagon

$ =\frac{3 \sqrt{3}}{2} a^{2}=\frac{3 \sqrt{3}}{2} \text{ side }^{2} $

Perimeter $=6 a=6 \times$ side

Area of a regular octagon $=2 a^{2}(1+\sqrt{2})$, where $a=$ side.

SOLVED EXAMPLES

Ex. 1 . If the sides of a triangle are $3 cm, 4 cm$ and $5 cm$, then what is the radius of the circum-circle?

Sol. Semi-perimeter of triangle $(s)=\frac{3+4+5}{2} cm=6 cm$

$\therefore$ Area of triangle $A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{6 \times 3 \times 2 \times 1} cm^{2}=6 cm^{2}$

$\therefore$ Radius of circum-circle $=\frac{a b c}{4(\text{ Area of } \Delta)}=\frac{3 \times 4 \times 5}{4 \times 60} cm=\mathbf{2 . 5} \mathbf{~ c m}$

Ex. 2 . If the length of hypotenuse of a right angled triangle is $5 cm$ and its area is $6 sq cm$, then what are the lengths of the remaining sides?

(CDS 2001)

Sol. Let one of the remaining sides be $x cm$.

Then, other side $=\sqrt{5^{2}-x^{2}} cm$

$\therefore$ Area $=\frac{1}{2} \times x \times \sqrt{25-x^{2}}=6$

$\Rightarrow x \sqrt{25-x^{2}}=12 \Rightarrow x^{2}(25-x^{2})=144$

$\Rightarrow 25 x^{2}-x^{4}=144 \Rightarrow x^{4}-25 x^{2}+144=0$

$\Rightarrow(x^{2}-16)(x^{2}-9)=0$

$\Rightarrow x^{2}=16$ or $x^{2}=9 \Rightarrow x=4$ or 3

$\therefore$ The two sides are $4 cm$ and $3 cm$.

Ex. 3 . $A C B$ is an arc of a circle of radius $3 cm$. It subtends an angle of $120^{\circ}$ at the centre as shown in the figure. What is the area of the shaded portion $A C B$ ?

Sol. Area of shaded portion $A C B=$ Area of sector $O A C B-$ Area of $\triangle O A B$

$ \begin{aligned} & ={\frac{120}{360} \times \pi(3)^{2}}-{\frac{1}{2} \times 3 \times 3 \sin 120^{\circ}} \\ & =3 \pi-\frac{9}{2} \times \frac{\sqrt{3}}{2}=(3 \pi-\frac{9 \sqrt{3}}{4}) \mathbf{c m}^{2} . \end{aligned} $

Ex. 4 . There are two concentric hexagons. Each of the side of both the hexagons are parallel. Each side of internal regular hexagon is $8 cm$. What is the area of the shaded region, if the distance between the corresponding parallel sides is $2 \sqrt{3} cm$ ?

Sol. In the second figure, $\triangle O P Q$ is an equilateral triangle as a regular hexagon is divided into six equilateral triangles.

$\therefore \angle O P C=60^{\circ}$

$\Rightarrow \frac{O C}{O P}=\sin 60^{\circ} \Rightarrow O C=O P \sin 60^{\circ} \Rightarrow O C=8 \times \frac{\sqrt{3}}{2} cm=4 \sqrt{3} cm$.

Now in similar $\triangle s O P C$ and $O A D, \frac{O C}{O D}=\frac{O P}{O A} \Rightarrow \frac{4 \sqrt{3}}{6 \sqrt{3}}=\frac{8}{O A} \Rightarrow O A=12 cm \Rightarrow A B=O A=12 cm$

$(\because O D=O C+2 \sqrt{3}=4 \sqrt{3}+2 \sqrt{3}=6 \sqrt{3} cm)$

$\therefore$ Required area $=$ Area of outer hexagon - Area of inner hexagon

$ =\frac{3 \sqrt{3}}{2}(12^{2}-8^{2}) cm^{2}=\mathbf{1 2 0} \sqrt{\mathbf{3}} \mathbf{c m}^{2} . \quad \text{ [Area of a regular hexagon }=\frac{3 \sqrt{3}}{2} \text{ side }^{2} \text{ ] } $

Ex. 5 . $\triangle A B C$ is an isosceles right angled triangle. If $r$ is its in radius and $R$ its circum radius, then what is $\frac{R}{r}$ equal to?

(CDS 2007)

Sol. Let $A B C$ be the right $\angle d$ isosceles triangle, where $A B=B C=x$ and $\angle A B C=90^{\circ}$.

$ \begin{aligned} & \therefore \quad A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{x^{2}+x^{2}}=x \sqrt{2} \\ & \quad S=\text{ Semi-perimeter of } \triangle A B C=\frac{x+x+x \sqrt{2}}{2}=\frac{2 x+x \sqrt{2}}{2}=\frac{x \sqrt{2}(\sqrt{2}+1)}{2} \\ & \therefore \text{ In-radius of } \triangle A B C(r)=\frac{\text{ Area }}{\text{ Semi-perimeter }}=\frac{x^{2} / 2}{x \sqrt{2}(1+\sqrt{2}) / 2}=\frac{x}{\sqrt{2}(1+\sqrt{2})} \\ & \text{ Circum radius of } \triangle A B C(R)=\frac{a b c}{4 \times A r e a}=\frac{x \times x \times x \sqrt{2}}{4 \times x^{2} / 2}=\frac{x \sqrt{2}}{2}=\frac{x}{\sqrt{2}} \\ & \therefore \text{ Required ratio }=\frac{R}{r}=\frac{x / \sqrt{2}}{x / \sqrt{2}(1+\sqrt{2})}=1+\sqrt{2} . \end{aligned} $

Ex. 6 . In a right-angled triangle $A B C, D$ is the foot of the perpendicular from $B$ on the hypotenuse $A C$. If $A B=3 cm$ and $B C=4 cm$, what is the area of $\triangle A B D$ ?

Sol. Area of $\triangle A B C=\frac{1}{2} \times 3 \times 4 cm^{2}=6 cm^{2}$. Also, $A C=\sqrt{3^{2}+4^{2}}=5 cm$.

$\therefore$ Area of $\triangle A B C=\frac{1}{2} \times B D \times A C \Rightarrow 6=\frac{1}{2} \times B D \times 5 \Rightarrow B D=\frac{12}{5} cm$.

Now in $\triangle A B D, A D=\sqrt{A B^{2}-B D^{2}}=\sqrt{3^{2}-(\frac{12}{5})^{2}}=\sqrt{9-\frac{144}{25}}$

$=\sqrt{\frac{225-144}{25}}=\sqrt{\frac{81}{25}}=\frac{9}{5} cm$

$\therefore$ Area of $\triangle A B D=\frac{1}{2} \times A D \times B D=\frac{1}{2} \times \frac{9}{5} \times \frac{12}{5}=\frac{\mathbf{5 4}}{\mathbf{2 5}} \mathbf{c m}^{2}$.

Ex. 7 . One side of an equilateral triangle is $24 cm$. The mid-points of its sides are joined to form another triangle whose mid-points are in turn joined to form still another triangle. This process continues indefinitely. Find the sum of the perimeters of all the triangles.

(SNAP 2011)

Sol. Perimeter of the largest (outermost) equilateral triangle $=3 \times 24=72 cm$.

Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle.

$\therefore$ Required sum $=72+36+18+$ upto infinite terms

This is an infinite $G P$, where first term $\boldsymbol{a}=72$ and common ratio $\boldsymbol{r}=\frac{1}{2}$.

$\therefore$ Required sum $=\frac{a}{1-r}=\frac{72}{1-\frac{1}{2}}=\frac{72}{\frac{1}{2}}=72 \times 2=144 \mathbf{c m}$.

Ex. 8 . In the adjoining figure, points $A, B, C$ and $D$ lie on a circle. $A D=24$ and $B C=12$. What is the ratio of the area of the triangle $C B E$ to that of triangle $A D E$ ?

(CAT)

Sol. $A D=24, B C=12$.

In $\triangle C B E$ and $\triangle A D E$,

$\angle C B A=\angle C D A, \angle B C E=\angle D A E$ (Angles in the same segment are equal)

$\angle B E C=\angle D E A$ (vertical opposite angles are equal)

$\Rightarrow \triangle B C E$ and $\triangle D E A$ are similar $\triangle s$ with sides in the ratio $1: 2$.

$\therefore$ Ratio of areas $=$ Ratio of square of sides $=1^{2}: 2^{2}=\mathbf{1}: \mathbf{4}$

Ex. 9 . If a circular sheet of perimeter $2 \pi r$ touching each side of a given quadrilateral sheet of perimeter $2 p$ is cut off from the quadrilateral, then what is the area of the remaining sheet?

Sol. Let $A B C D$ be the quadrilateral from which a circular sheet is cut off touching each side of the quadrilateral.

Also, given $A B+B C+C D+D A=2 p$

Circumference of the circle $=2 \pi r \Rightarrow$ Radius of circle $=r$

$\therefore$ Area of quadrilateral $=$ Area of $(\triangle O A B+\triangle O B C+\triangle O C D+\triangle O D A)$

$ \begin{aligned} & =\frac{1}{2} r(A B)+\frac{1}{2} r(B C)+\frac{1}{2} r(C D)+\frac{1}{2} r(D A) \\ & =\frac{1}{2} r(A B+B C+C D+D A)=\frac{1}{2} r \cdot 2 p=p r . \end{aligned} $

$\therefore$ Required remaining area $=$ Area of quadrilateral - Area of circle $=p r-\pi r^{2}=\boldsymbol{r}(\boldsymbol{p}-\boldsymbol{\pi} \boldsymbol{r})$.

PRACTICE SHEET

The diagonal of a square $A$ is $(a+b)$. The diagonal of a square whose area is twice the area of square $A$ is.

(a) $2(a+b)$

(b) $2(a+b)^{2}$

(d) $\sqrt{2}(a-b)$

(SSC 2002)

~~ 2. The lengths of the perpendiculars drawn from any point in the interior of an equilateral triangle to the respective sides are $p_1, p_2$ and $p_3$. The length of each side of the triangle is

(a) $\frac{2}{\sqrt{3}}(p_1+p_2+p_3)$

(b) $\frac{1}{3}(p_1+p_2+p_3)$

(d) $\frac{4}{\sqrt{3}}(p_1+p_2+p_3)$

(SSC 2004)

~~ 3. The lengths of three medians of a triangle are $9 cm, 12 cm$ and $15 cm$. The area (in sq. $cm$ ) of this triangle is

(a) 24

(b) 72

(d) 144

(SSC 2012)

~~ 4. The perimeter of a triangular field is $240 m$. If two of its sides are $78 m$ and $50 m$, then what is the length of the perpendicular on the side of length $50 m$ from the opposite vertex?

(a) $43 m$

(b) $52.2 m$

(d) $70 m$

(CDS 2011)

~~ 5. What is the radius of a circle inscribed in a triangle having side lengths $35 cm, 44 cm$ and $75 cm$ ?

(a) $3 cm$

(b) $4 cm$

(d) $6 cm$

(CDS 2010)

~~ 6. The percentage increase in the area of a triangle if each of its side is doubled is

(a) $200 %$

(b) $50 %$

(d) $100 %$

~~ 7. If the area of a circle, inscribed in an equilateral triangle is $4 \pi cm^{2}$, then what is the area of the triangle?

(a) $12 \sqrt{3} cm^{2}$

(b) $9 \sqrt{3} cm^{2}$

(d) $18 cm^{2}$

(CDS 2008)

~~ 8. A circle is inscribed in an equilateral triangle of side $a$. What is the area of any square inscribed in this circle?

(a) $\frac{a^{2}}{3}$

(b) $\frac{a^{2}}{4}$

(d) $\frac{a^{2}}{8}$

(CDS 2007)

~~ 9. If the distance from the vertex to the centroid of an equilateral triangle is $6 cm$, then what is the area of the triangle?

(a) $24 cm^{2}$

(b) $27 \sqrt{3} cm^{2}$

(d) $12 \sqrt{3} cm^{2}$

(CDS 2007)

~~ 10. In the given figure, the area of the shaded portion is $(\pi-1) R^{2}$, where $R$ is the radius of the circle with centre at $O$. What is one of the acute angles of $\triangle A B C$ ?

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) Cannotbe determined.

(CDS 2007)

~~ 11. The area of an isosceles triangle is $a$, when the angle included between the two equal sides is $60^{\circ}$. What will be the area if the angle included between the two equal sides becomes $120^{\circ}$ ? (Keeping the length of equal sides same as before)

(a) $a / 2$

(b) $a$

(d) $2 a$

(CDS 2006)

~~ 12. If $O$ is the centre of the circle and chord $A B=O A$ and the area of triangle $A O B=4 \sqrt{3} cm^{2}$, then the area of the circle will be

(a) $4 \pi cm^{2}$

(b) $16 \pi cm^{2}$

(d) $24 \pi \sqrt{3} cm^{2}$

(CDS)

~~ 13. A square and a rhombus have the same base. If the rhombus is inclined at $60^{\circ}$, find the ratio of the area of the square to the area of the rhombus.

(a) $2 \sqrt{3}: 3$

(b) $1: \sqrt{3}$

(d) None of these

~~ 14. Find the area of regular octagon with each side ’ $a$ ’ $cm$.

(a) $2 a^{2}(1+\sqrt{2})$

(b) $\sqrt{2} a(1+\pi)$

(d) None of these

~~ 15. What is the ratio of the area of a circum circle of equilateral triangle to the area of the square with same side length as equilateral triangle?

(a) $\pi: \sqrt{3}$

(b) $\sqrt{3}: 2$

(d) $\pi: 3$

~~ 16. $A B C D$ is a square of side $a cm$. $A B$, $B C, C D$ and $A D$ are all chords of circles with equal radii each. If the chords subtend an angle of $120^{\circ}$ at their respective centres, find the total area of the given figure, where arcs are a part of the circle.

(a) $[a^{2}+4[\frac{\pi a^{2}}{9}-\frac{a^{2}}{3 \sqrt{2}}]]$

(b) $[a^{2}+4(\frac{\pi a^{2}}{9}-\frac{a^{2}}{4 \sqrt{3}})]$

(d) None of these

~~ 17. $A B C D$ is a square. Another square $E F G H$ with the same area is placed on the square $A B C D$ such that the point of intersection of diagonals of square $A B C D$ and square $E F G H$ coincide and the sides of the square $E F G H$ are parallel to the diagonals of square $A B C D$. Thus a new figure is formed as shown in the figure. What is the area enclosed by the given figure, if each side of the square is $4 cm$ ?

(a) $32(2-\sqrt{2})$

(b) $16(\frac{3+\sqrt{2}}{2+\sqrt{2}})$

(d) None of these

~~ 18. Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points of the adjacent sides of a square of side ’ $a$ ‘.

(a) $\frac{3 a^{2}}{16}$

(b) $\frac{3 \sqrt{3} a^{2}}{16}$

(d) $\frac{3 \sqrt{3} a^{2}}{32}$

~~ 19. A goat is tethered to one end of a rope of length $20 m$, while the other end is fixed at the centre of a large circular field. There is a square elevated platform with sides of $10 m$ on the field such that one corner of the elevated square platform councides with centre of the circular field. If the goat is unable to mount the square elevated platform, what is the total area the goat will be able to graze?

(a) $3 \times 100 \pi$

(b) $3.5 \times 100 \pi$

(d) $100(3 \pi+1)$

(JMET 2009)

~~ 20. A semi-circle of diameter $14 cm$ has three chords of equal length connecting the two end points of the diameter so as to form a trapezoid inscribed within a semicircle. What is the value of the area enclosed by the trapezoid?

(a) $(\frac{157 \times \sqrt{3}}{4}) cm^{2}$

(b) $(49 \times \sqrt{3}) cm^{2}$

(d) $(\frac{100}{\sqrt{3}}) cm^{2}$

(JMET 2009)

~~ 21. In the given figure, $A B C$ is a triangle in which $C D E F G$ is a pentagon. Triangles $A D E$ and $B F G$ are equilateral triangles each with side $2 cm$ and $E F=2 cm$. Find the area of the pentagon.

(a) $8 \sqrt{3} cm^{2}$

(b) $7 \sqrt{3} cm^{2}$

(d) $11.28 cm^{2}$

~~ 22. The hypotenuse of an isosceles right-angled triangle is $q$. If we describe equilateral triangles (outwards) on all its three sides, then the total area of the re-entrant hexagon thus obtained is.

(a) $q^{2}(\sqrt{3}+2)$

(b) $q^{2} \frac{(2 \sqrt{3}+1)}{4}$

(d) $\frac{q^{2}(5 \sqrt{3}-2)}{4}$

~~ 23. In the adjoining figure, $A B C$ is an equilateral triangle inscribing a square of maximum possible area. Again in this squares there is an equilateral triangle whose side is same as that of the square. Further the smaller equilateral triangle inscribes a square of maximum possible area. What is the area of the innermost square if each side of the outermost triangle be $0.01 m$ ?

(a) $(873-504 \sqrt{3}) cm^{2}$

(b) $(738-504 \sqrt{3}) cm^{2}$

(d) None of these

~~ 24. A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was $35 %$ reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle?

(a) 16.665

(b) 16.565

(d) 14.365

(CAT 2004)

ANSWERS

1. (c)	2. (a)	3. (b)	4. (c)	5. $(d)$	6. (c)	7. (a)	8. (c)	9. $(b)$	10. (b)
11. (b)	12. (b)	13. (a)	14. (a)	15. $(d)$	16. $(b)$	17. (a)	18. $(d)$	19. (b)	20. (c)
21. (b)	22. (b)	23. (a)	24. $(d)$

HINTS AND SOLUTIONS

Area of new square $=2($ Area of the square $A)$

$ =2 \times \frac{(\text{ diagonal })^{2}}{2}=2 \times \frac{(a+b)^{2}}{2}=(a+b)^{2} $

$\therefore$ Each side of new square $=(a+b)$

$\Rightarrow$ Diagonal of new square $=$ side $\sqrt{2}=(\boldsymbol{a}+\boldsymbol{b}) \sqrt{\mathbf{2}}$.

~~ 2. Let each side of equilateral $\triangle P Q R=\boldsymbol{a}$ units. $O$ is any point in the interior of $\triangle P Q R$ $\Rightarrow O D=p_1, O E=p_2$ and $O F=p_3$ are perpendiculars on sides $P Q$, $P R$ and $Q R$ respectively.

$\therefore$ Area of $\triangle P Q R$

$ \begin{aligned} & =\text{ Area of } \triangle O P Q+\text{ Area of } \triangle O P R+\text{ Area of } \triangle O Q R \\ & =\frac{1}{2} \times a \times p_1+\frac{1}{2} \times a \times p_2+\frac{1}{2} \times a \times p_3 \\ & =\frac{a}{2}(p_1+p_2+p_3) \\ & \Rightarrow \frac{\sqrt{3}}{4} a^{2}=\frac{a}{2}(p_1+p_2+p_3) \Rightarrow a=\frac{2}{\sqrt{3}}(p_1+p_2+p_3) \end{aligned} $

~~ 3. Here $s_m=\frac{9+12+15}{2}=18 cm$, where lengths of medians are $m_1=9 cm, m_2=12 cm, m_3=15 cm$.

$\therefore$ Area of triangle $=\frac{4}{3} \sqrt{18(18-9)(18-12)(18-15)} cm^{2}$

[Refer to key facts]

$ =\frac{4}{3} \sqrt{18 \times 9 \times 6 \times 3} cm^{2}=\frac{4}{3} \times 9 \times 6 cm^{2}=72 cm^{2} . $

~~ 4. Given $2 s=a+b+c \Rightarrow 240=78+50+$ Third side

$\Rightarrow$ Third side $=240 m-128 m=112 m$.

$\therefore$ Area of $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{120 \times(120-78) \times(120-50) \times(120-112)} \\ & =\sqrt{120 \times 42 \times 70 \times 8}=1680 \end{aligned} $

Also, Area of $\Delta=\frac{1}{2} \times b \times h$

$\therefore \frac{1}{2} \times 50 \times h=1680 \Rightarrow h=\frac{1680}{25}=\mathbf{6 7 . 2} \mathbf{~ m .}$

~~ 5. Let $a=35 cm, b=44 cm, c=75 cm$. Then

$ s=\frac{a+b+c}{2}=\frac{35+44+75}{2} cm=77 cm $

$\therefore$ Area of triangle $=\sqrt{77(77-35)(77-44)(77-75)} cm^{2}$

$ \begin{aligned} & =\sqrt{77 \times 42 \times 33 \times 2} cm^{2} \\ & =(7 \times 11 \times 6) cm^{2}=462 cm^{2} \end{aligned} $

$\therefore$ Radius of incircle $=\frac{\text{ Area }}{\text{ semi-perimeter }}=\frac{462}{77} cm=6 \mathbf{c m}$.

~~ 6. Let $a, b, c$ be the sides of the original triangle and $s$ be its semi-perimeter.

Then, $s=\frac{1}{2}(a+b+c)$

Let $s_1$ be the semi-perimeter of the new triangle. Then,

$ s_1=\frac{1}{2}(2 a+2 b+2 c)=(a+b+c)=2 s $

$\therefore$ Area $A$ of original $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ and Area $A_1$ of new $\Delta=\sqrt{s_1(s_1-2 a)(s_1-2 b)(s_1-2 c)}$

$ \begin{aligned} & =\sqrt{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)} \\ & =4 \sqrt{s(s-a)(s-b)(s-c)}=4 A . \end{aligned} $

$\therefore$ Percentage increase in area $=(\frac{A_1-A}{A} \times 100) %$

$ =\frac{(4 A-A)}{A} \times 100 %=\mathbf{3 0 0 %} . $

~~ 7. Since area of circle $=4 \pi$

$\Rightarrow \pi r^{2}=4 \pi \Rightarrow r=2 cm$

$\therefore$ In $\triangle O A D$,

$\tan 30^{\circ}=\frac{O D}{A D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{A D}$

$\Rightarrow A D=2 \sqrt{3} cm$

$\therefore A B=2 A D=4 \sqrt{3} cm$

$\therefore$ Area of equilateral $\triangle A B C=\frac{\sqrt{3}}{4}(A B)^{2}$

$=\frac{\sqrt{3}}{4}(4 \sqrt{3})^{2}=\mathbf{1 2} \sqrt{\mathbf{3}} \mathbf{c m}^{2}$.

~~ 8. If ’ $a$ ’ is length of the side of $\triangle A B C$, then

Area of $\triangle A B C=\frac{\sqrt{3}}{4} a^{2}$

semi-perimeter of $\triangle A B C=\frac{3 a}{2}$

$\therefore$ Radius of in-circle $=\frac{\text{ Area }}{\text{ semi-perimeter }}=\frac{\sqrt{3}}{4} a^{2} \times \frac{2}{3 a}=\frac{a}{2 \sqrt{3}}$

$\therefore$ Diagonal of square $P Q R S=$ Diameter of incircle.

$ =2 \times \frac{a}{2 \sqrt{3}}=\frac{a}{\sqrt{3}} $

$\therefore$ Area of square $=\frac{(\text{ diagonal })^{2}}{2}=\frac{(a / \sqrt{3})^{2}}{2}=\frac{\boldsymbol{a}^{2}}{\mathbf{6}}$.

~~ 9. Let $G$ be the centroid of $\triangle P Q R$.

Then, $P G=6 cm$.

Now, $\frac{P G}{G S}=\frac{2}{1} \Rightarrow G S=3 cm$

$\therefore P S=P G+G S=9 cm$

$\therefore$ If $a$ is the length of a side of

$\triangle P Q R$, then $\triangle P Q R$ being equilateral, $P S \perp Q R$

$\therefore$ Altitude $P S=\frac{\sqrt{3}}{2} a=9($ From $(i))$

$\Rightarrow a=\frac{9 \times 2}{\sqrt{3}}=6 \sqrt{3} cm$

$\therefore$ Area of equilateral $\triangle P Q R=\frac{\sqrt{3}}{4}(a)^{2}$

$ =\frac{\sqrt{3}}{4} \times(6 \sqrt{3})^{2} cm^{2}=27 \sqrt{3} cm^{2} . $

~~ 10. Area of triangle $A B C$

$=$ Area of circle - Area of shaded region

$=\pi R^{2}-(\pi-1) R^{2}=R^{2}$

But Area of $\triangle A B C=\frac{1}{2} \times A B \times A C$

( $\angle A B C=90^{\circ}$, angle in a semicircle is a right angle)

$\Rightarrow \frac{1}{2} \times A B \times B C=R^{2} \Rightarrow A B \times B C=2 R^{2}$

In $\triangle A B C, A C^{2}=A B^{2}+B C^{2}$

$\Rightarrow \quad(2 R)^{2}=A B^{2}+B C^{2}$

$\Rightarrow 2 \times A B \times B C=A B^{2}+B C^{2}$

$\Rightarrow A B^{2}+B C^{2}-2 A B B C=0$

$\Rightarrow(A B-B C)^{2}=0$

$\Rightarrow A B=B C=\sqrt{2} R$

$\therefore \tan A=\frac{B C}{A B}=\frac{\sqrt{2} R}{\sqrt{2} R}=1 \Rightarrow \angle A=45^{\circ}$.

~~ 11. In isosceles $\triangle A B C, \frac{B C}{x}=\sin 30^{\circ}$

$\Rightarrow B C=x \sin 30^{\circ}$

$\Rightarrow B D=2 x \sin 30^{\circ}$

and $\frac{A C}{A B}=\cos 30^{\circ}$

$\Rightarrow A C=x \cos 30^{\circ}$

$\therefore$ Area of $\triangle A B C=a=\frac{1}{2} \times B D \times A C$

$=\frac{1}{2} \times 2 x \sin 30^{\circ} \times x \cos 30^{\circ}$

$=x^{2} \times \frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} x^{2}=a$ (given)

$\Rightarrow \quad x^{2}=\frac{4}{\sqrt{3}} a$

Now angle between equal sides $=120^{\circ}$

$\therefore$ Area of triangle $=\frac{1}{2} x^{2} \sin 120^{\circ}$

[Using Area of $\Delta=\frac{1}{2} A B \cdot A D \sin 120^{\circ}$ ]

$ =\frac{1}{2} \times(\frac{4 a}{\sqrt{3}})^{2} \times \frac{\sqrt{3}}{2}=a . $

~~ 12. $A B=O A \Rightarrow A B=O A=O B$ (radii of circle are equal)

$\Rightarrow \triangle A O B$ is equilateral.

$\therefore$ If ’ $r$ ’ is the radius of the circle, then area of $\triangle A O B=\frac{\sqrt{3}}{4}$ side $^{2}$

$\Rightarrow \frac{\sqrt{3}}{4}(r)^{2}=4 \sqrt{3}$ (given)

~~ 13. In the given rhombus, $\frac{h}{x}=\sin 60^{\circ} \Rightarrow \frac{h}{x}=\frac{\sqrt{3}}{2} \Rightarrow h=\frac{x \sqrt{3}}{2}$

$\therefore$ Required ratio $=\frac{\text{ Area of square }}{\text{ Area of rhombus }}$

$=\frac{x^{2}}{x \times \frac{x \sqrt{3}}{2}} \begin{bmatrix} \text{ Using, Area of rhombus } \\ =\text{ base } \times \text{ height }\end{bmatrix} $

$=\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}=2 \sqrt{3}: 3$

~~ 14. Let $A B C D E F G H$ be the regular octagon of side $a cm$.

Now if we produce the sides of the octagon on both the sides, we get a square $P Q R S$.

Given, $B C=D E=F G=H A=a cm$.

Also,

$B Q=Q C=D R=R E=F S=S G=H P=P A=\frac{a}{\sqrt{2}}$

$ (\because B Q C, D R E, F S G, H P A \text{ are rt. } \angle d \Delta s) $

$\therefore$ Side of square $=\frac{a}{\sqrt{2}}+a+\frac{a}{\sqrt{2}}=a(1+\sqrt{2}) cm$.

$\therefore$ Area of square $=a^{2}(1+\sqrt{2})^{2} cm^{2}=a^{2}(3+2 \sqrt{2}) cm^{2}$

Each of the shaded $\triangle s, A P H, B C Q, D E R, F G S$ is an isosceles right angled $\triangle$, whose area $=\frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}} cm^{2}=\frac{a^{2}}{4} cm^{2}$

$\therefore$ Total area of shaded region $=4 \times \frac{a^{2}}{4}=a^{2} cm^{2}$

$\therefore$ Area of octagon $=$ Area of square $P Q R S-$ Total area of shaded isosceles $\Delta s=a^{2}(3+2 \sqrt{2})-a^{2}$

$ =2 a^{2}(1+\sqrt{2}) cm^{2} \text{. } $

~~ 15. Let each side of the equilateral $\Delta$ be $\boldsymbol{a}$ units.

Then, circumradius of the circle $=\frac{\text{ side }}{\sqrt{3}}=\frac{a}{\sqrt{3}}$ units

$\therefore$ Area of circumcircle $=\pi(\frac{a}{\sqrt{3}})^{2}=\frac{\pi a^{2}}{3}$ sq units

Area of square of side $a$ units $=a^{2}$ sq units

$\therefore \quad$ Required ratio $=\frac{\pi a^{2}}{3} / a^{2}=\frac{\pi}{3}=\pi: 3$.

~~ 16. As shown in the given figures, if ’ $a$ ’ is each side of the square, then $\angle D O C=120^{\circ} \Rightarrow \angle O D C=\angle O C D=30^{\circ}$

(i) (ii)

(iii)

Now in fig. (iii), $\frac{P C}{O C}=\sin 60^{\circ} \Rightarrow \frac{a / 2}{O C}=\frac{\sqrt{3}}{2}$

$\Rightarrow O C=\frac{a}{\sqrt{3}}$, i.e., radius of $arc CD$

Also, $O P=\frac{a}{2} \cot 60^{\circ}=\frac{a}{2 \sqrt{3}}$

Now area of $\triangle O D C=\frac{1}{2} \times C D \times O P=\frac{1}{2} \times a \times \frac{a}{2 \sqrt{3}}=\frac{a^{2}}{4 \sqrt{3}}$

Area of sector $D O C=\pi \times(\frac{a}{\sqrt{3}})^{2} \times \frac{120}{360}=\frac{\pi a^{2}}{9}$

$ [\text{ Using Area of sector }=\pi R^{2}(\frac{\theta}{360^{\circ}})] $

$\therefore \quad$ Area of segment $D L C$

$=$ Area of sector $D O C-$ Area of $\triangle D O C$

$=\frac{\pi a^{2}}{9}-\frac{a^{2}}{4 \sqrt{3}}$

$\therefore$ Total area of the figure

$=$ Area of square + Total area of 4 segments

$ =a^{2}+4 \times(\frac{\pi a^{2}}{9}-\frac{a^{2}}{4 \sqrt{3}}) . $

~~ 17. As is seen in the given figure, the sides of one square are parallel to the diagonals of another square. Also, square $A B C D$ and $E F G H$ have same area.

$\Rightarrow$ Sides of square $A B C D$ and square $E F G H$ are $4 cm$ each.

Let $D P=a$ units

As $D P=P G=G Q=Q C=a$ units and $\angle G=90^{\circ}$

$\therefore P Q=a \sqrt{2}$ units

$\therefore \quad D C=D P+P Q+Q C=(a+a \sqrt{2}+a)$ units

$=a(2+\sqrt{2})$ units

Area of $\triangle P G Q=\frac{1}{2} \times a \times a=\frac{a^{2}}{2}$ sq units.

Area of all triangles outside square $A B C D=4 \times \frac{a^{2}}{2}$

$=2 a^{2}$ sq units

Also, side of square $=4 \Rightarrow a(2+\sqrt{2})=4 \Rightarrow a=\frac{4}{2+\sqrt{2}}$

$\therefore \quad$ Total area of the four $\triangle s$ outside square $A B C D$

$ \begin{aligned} & =2 \times(\frac{4}{2+\sqrt{2}})^{2}=\frac{2 \times 16}{(4+4 \sqrt{2}+2)} \\ & =\frac{16}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}=16(3-2 \sqrt{2}) \end{aligned} $

$\therefore$ Total area of the figure $=$ Area of square $A B C D+$ Area

$=16+16(3-2 \sqrt{2})$ of the four $\triangle s$ outside $A B C D$

$=16(4-2 \sqrt{2})=32(2-\sqrt{2}) cm^{2}$.

~~ 18. Let $A B=a$ be the side of the outermost square.

Then $A G=A H=\frac{a}{2}$

$\Rightarrow \quad G H=\sqrt{\frac{a^{2}}{4}+\frac{a^{2}}{4}}=\frac{a}{\sqrt{2}}$

$\therefore$ Diameter of circle $=\frac{a}{\sqrt{2}}$

$\Rightarrow$ Radius of circle $=\frac{a}{2 \sqrt{2}}$

If $O$ is the centre of the circle, then $\angle P O Q=120^{\circ}$.

$\therefore$ Area of $\triangle P O Q=\frac{1}{2} \times P O \times O Q \times \sin 120^{\circ}$

$ =\frac{1}{2} \times \frac{a}{2 \sqrt{2}} \times \frac{a}{2 \sqrt{2}} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3} a^{2}}{32} $

$\therefore$ Area of $\triangle P Q R=3($ Area of $\triangle P O Q)=\frac{3 \sqrt{3} a^{2}}{32}$.

~~ 19. The goat can graze the area $C A B E O D C+$ quarter circle $C D F+$ quarter circle $F E B$ $=\frac{3}{4} \times \pi \times(20)^{2}+\frac{1}{4} \times \pi$ $\times(10)^{2}+\frac{1}{4} \times \pi \times(10)^{2}$

$=300 \pi+25 \pi+25 \pi$

$=3.5 \times 100 \pi$.

~~ 20. Take the trapezoid $A B C D$ in the semi-circle with centre $O$ such that $A D=D C=C B$.

Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, $A D C B E F$ is a regular hexagon.

$\Rightarrow \angle D A O=60^{\circ}$

( $\triangle D A O$ is an equilateral triangle)

$\Rightarrow D A=A O=7 cm$.

$ (\because A B=14 cm) $

$\therefore \quad$ Area of regular hexagon $=\frac{3 \sqrt{3}}{2} \times(\text{ side })^{2}$

$ =\frac{3 \sqrt{3}}{2} \times 49 cm^{2} $

$\Rightarrow$ Area of trapezoid $=\frac{1}{2} \times$ Area of hexagon

$ =\frac{1}{2} \times \frac{3 \sqrt{3}}{2} \times 49=\frac{147 \sqrt{3}}{4} cm^{2} . $

~~ 21. $A B=6 cm, \angle C=60^{\circ}$

$\therefore \triangle A B C$ is an equilateral triangle

Area of $\triangle A B C=\frac{\sqrt{3}}{4} \times(6)^{2}=9 \sqrt{3}$

Area of $(\triangle A D E+\triangle B F G)$

$ =2 \times(\frac{\sqrt{3}}{4} \times(2)^{2})=2 \sqrt{3} $

$\therefore \quad$ Area of pentagon $=9 \sqrt{3}-2 \sqrt{3}=7 \sqrt{3} cm^{2}$.

~~ 22. $A C=q, \angle A B C=90^{\circ}$

$\Rightarrow q=\sqrt{A B^{2}+B C^{2}}$

$\Rightarrow q=\sqrt{2 x^{2}}$

$\Rightarrow q^{2}=2 x^{2} \Rightarrow x=\frac{q}{\sqrt{2}}$

$\therefore$ Area of the re-entrant hexagon

$=$ Sum of areas of $(\triangle A B C+\triangle A D C+\triangle B F C+\triangle A E B)$

$=\frac{1}{2} \times \frac{q}{\sqrt{2}} \times \frac{q}{\sqrt{2}}+\frac{\sqrt{3}}{4} q^{2}+\frac{\sqrt{3}}{4}(\frac{q}{\sqrt{2}})^{2}+\frac{\sqrt{3}}{4}(\frac{q}{\sqrt{2}})^{2}$

$=\frac{q^{2}}{4}+\frac{\sqrt{3}}{4} q^{2}+\frac{\sqrt{3}}{8} q^{2}+\frac{\sqrt{3} q^{2}}{8}=\frac{\boldsymbol{q}^{2}}{\mathbf{4}}(\mathbf{2} \sqrt{\mathbf{3}}+\mathbf{1})$.

~~ 23. Since $\angle C P O=\angle C O P=60^{\circ}$, therefore, $P C O$ is also an equilateral triangle.

Let each side of the square $M N O P$ be $x cm$.

Then $P C=C O=P O=x cm$

Then in $\triangle P A M$,

$\frac{P M}{P A}=\sin 60^{\circ}$

$\Rightarrow \frac{x}{P A}=\frac{\sqrt{3}}{2} \Rightarrow P A=\frac{2 x}{\sqrt{3}}$

$\therefore \quad A C=A P+P C=\frac{2 x}{\sqrt{3}}+x$

Given $\frac{2 x}{\sqrt{3}}+x=0.01 m=1 cm$

$\Rightarrow \quad x=\frac{\sqrt{3}}{(2+\sqrt{3})} cm=\sqrt{3}(2-\sqrt{3}) cm$

(On rationalising the denominator)

Now, let each side of the square $R S X Y$ be $y$. Then $R T=y$

$(\because R T S$ is an equilateral triangle) $\therefore \quad$ In $\triangle R Y M, \frac{R Y}{R M}=\sin 60^{\circ}$

$\Rightarrow \frac{y}{R M}=\frac{\sqrt{3}}{2} \Rightarrow R M=\frac{2 y}{\sqrt{3}}$

$\therefore \quad M T=M R+R T=\frac{2 y}{\sqrt{3}}+y=\frac{(2+\sqrt{3})}{\sqrt{3}} y$

Given $M T=x$, then

$x=(\frac{2+\sqrt{3}}{\sqrt{3}}) y \Rightarrow y=\frac{\sqrt{3} x}{2+\sqrt{3}}$

But from (i), $x=\sqrt{3}(2-\sqrt{3})$

$\therefore \quad y=\frac{\sqrt{3} \sqrt{3}(2-\sqrt{3})}{2+\sqrt{3}}=\frac{3 \cdot(2-\sqrt{3})}{(2+\sqrt{3})} \cdot \frac{(2-\sqrt{3})}{(2-\sqrt{3})}$

$\Rightarrow y=3(2-\sqrt{3})^{2}=3(7-4 \sqrt{3})$

$\therefore \quad$ Area of the inner most square $=y^{2}$

$ \begin{aligned} & =(3(7-4 \sqrt{3}))^{2} \\ & =9(49+48-56 \sqrt{3}) \\ & =(\mathbf{8 7 3}-\mathbf{5 0 4} \sqrt{\mathbf{3}}) \mathbf{c m}^{2} . \end{aligned} $

~~ 24. Given, $S T | R Q$

$\therefore \frac{\text{ Area of } \triangle S P T}{\text{ Area of } \triangle R P Q}=\frac{S T^{2}}{R Q^{2}}$

Also, given $S T=(1-\frac{35}{100}) R Q$

$=(0.65) R Q$

$\therefore \frac{S T}{R Q}=0.65 \Rightarrow(\frac{S T}{R Q})^{2}=0.4225$

$\Rightarrow \frac{\text{ Area of } \triangle S P T}{\text{ Area of } \triangle R P Q}=0.4225 \Rightarrow \frac{\text{ Area of } \triangle S P T}{34}=0.4225$

$\Rightarrow$ Area of $\triangle S P T=0.4225 \times 34$

SELF ASSESSMENT SHEET

An equilateral triangle is cut from its three vertices to form a regular hexagon. What is the percentage of area wasted?

(a) $20 %$

(b) $50 %$

(d) $66.66 %$

~~ 2. The area of a square and circle is same and the perimeter of square and equilateral triangle is same, then the ratio between the area of circle and area of equilateral triangle is:

(a) $\pi: 3$

(b) $9: 4 \sqrt{3}$

(d) None of these

~~ 3. The figure given besides shows five squares inside one another by joining the midpoints of the outer square. The side of square I is $a cm$. The total area of the five squares is: (a) $\frac{(4 \sqrt{2}-1) a^{2}}{(4 \sqrt{2}+1)}$ (b) $\frac{(4 \sqrt{2}+1) a^{2}}{4(\sqrt{2}-1)}$

(d) $(7+3 \sqrt{2}) a^{2}$

~~ 4. Three circles of radius $a, b, c$ touch each other externally. The area of the triangle formed by joining their centres is:

(a) $\sqrt{(a+b+c) a b c}$

(b) $(a+b+c) \sqrt{a b+b c+c a}$

(d) None of these

(SSC 2013)

~~ 5. If the lengths of a sides of a triangle are in the ratio $4: 5: 6$ and the in-radius of the triangle is $3 cm$, then the altitude of the triangle corresponding to the largest side as the base is:

(a) $8 cm$

(b) $7.5 cm$

(d) $10 cm$

(SSC 2013)

~~ 6. Find the ratio of the diameter of the circles inscribed in and circumscribing an equilateral triangle to its height?

(a) $1: 2: 1$

(b) $2: 4: 3$

(d) $3: 2: 1$

~~ 7. In an isosceles triangle, the measure of each of equal sides is $10 cm$ and the angle between them is $45^{\circ}$. The area of the triangle is:

(a) $25 cm^{2}$

(b) $\frac{25}{2} \sqrt{2} cm^{2}$

(d) $25 \sqrt{3} cm^{2}$

(SSC 2006)

~~ 8. Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the sides of the park are exactly double of the corresponding sides of the flower bed. The ratio of the area of the path to the flower bed is:

(a) $1: 1$

(b) $1: 2$

(d) $3: 1$

(SNAP 2007)

~~ 9. A cow is tethered at corner A by a rope. Neither the cow nor the rope is allowed to enter $\triangle A B C . \angle A=30^{\circ}, A B=A C=$ $10 m$ and $B C=6 cm$. What is the area that can be grazed by the cow if the length of the rope is $8 m$.

(a) $133 \frac{1}{6} \pi$ sq. $m$

(b) $121 \pi$ sq. $m$

(d) $176 / 3 \pi$ sq. $m$

~~ 10. In the figure given alongside, $A B C D E F$ is a regular hexagon and $\angle A O F=90^{\circ}$. $F O$ is parallel to $E D$. What is the ratio of the area of triangle $A O F$ to that of the hexagon $A B C D E F$ ?

(a) $\frac{1}{12}$

(b) $\frac{1}{6}$

(d) $\frac{1}{18}$

(CAT)

ANSWERS

~~ 2. (b)

~~ 3. (c)

~~ 4. (a)

~~ 5. (b)

~~ 6. (b)

~~ 7. (c)

~~ 8. (d)

~~ 9. (d)

~~ 10. (a)

HINTS AND SOLUTIONS

When an equilateral triangle is cut from its three vertices to form a regular hexagon then out of the 9 equilateral triangles that form $\triangle A B C$, three triangle, $\triangle A D E, \triangle F C G, \triangle I H B$ are cut off and 6 remain in the hexagon.

This means that $\frac{1}{3} rd$ of the area has been removed or wasted to get the hexagon.

$\therefore$ Area wasted $=(\frac{1}{3} \times 100) %=\mathbf{3 3 . 3 3} %$.

~~ 2. Let each side of the square $=a cm$. Then,

Area of square $=a^{2} cm^{2}$

Also, let $r$ be the radius of the circle. Then, $\pi r^{2}=a^{2}$

Let each side of the equilateral triangle $=b cm$. Then

$3 b=4 a \Rightarrow b=\frac{4 a}{3}$.

$\therefore$ Area of equilateral triangle

$ =\frac{\sqrt{3}}{4} b^{2}=\frac{\sqrt{3}}{4} \times(\frac{4 a}{3})^{2}=\frac{\sqrt{3}}{4} \times \frac{16 a^{2}}{9}=\frac{4 \sqrt{3} a^{2}}{9} $

$\therefore$ Required ratio between area of circle and area of equilateral $\Delta$ is $a^{2}: \frac{4 \sqrt{3} a^{2}}{9}=9: 4 \sqrt{3}$.

~~ 3. Side of the square $I=a$

$\therefore$ Area of square I $=a^{2}$

Side of square II $=\sqrt{(\frac{a}{2})^{2}+(\frac{a}{2})^{2}}$

$ =\sqrt{\frac{a^{2}}{4}+\frac{a^{2}}{4}}=\sqrt{\frac{2 a^{2}}{4}}=\frac{a}{\sqrt{2}} $

$\therefore$ Area of square II $=(\frac{a}{\sqrt{2}})^{2}=\frac{a^{2}}{2}$

Side of square III $=\sqrt{(\frac{a}{2 \sqrt{2}})^{2}+(\frac{a^{2}}{2 \sqrt{2}})^{2}}$

$ =\sqrt{\frac{a^{2}}{8}+\frac{a^{2}}{8}}=\sqrt{\frac{a^{2}}{4}}=\frac{a}{2} $

$\therefore$ Area of square III $=\frac{a^{2}}{4}$

Side of square IV $=\sqrt{(\frac{a}{4})^{2}+(\frac{a}{4})^{2}}$

$ =\sqrt{\frac{a^{2}}{16}+\frac{a^{2}}{16}}=\sqrt{\frac{2 a^{2}}{16}}=\frac{a}{2 \sqrt{2}} $

$\therefore$ Area of square IV $=(\frac{a}{2 \sqrt{2}})^{2}=\frac{a^{2}}{8}$

Side of square $V=\sqrt{(\frac{a}{4 \sqrt{2}})^{2}+(\frac{a}{4 \sqrt{2}})^{2}}$

$ =\sqrt{\frac{a^{2}}{32}+\frac{a^{2}}{32}}=\sqrt{\frac{2 a^{2}}{32}}=\frac{a}{4} $

$\therefore$ Area of square $V=(\frac{a}{4})^{2}=\frac{a^{2}}{16}$

$\therefore$ Sum of the areas of the five squares

$ \begin{aligned} & =a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}+\frac{a^{2}}{8}+\frac{a^{2}}{16} \\ & =a^{2}[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}]=\frac{\mathbf{3 1}}{\mathbf{1 6}} \boldsymbol{a}^{2} . \end{aligned} $

~~ 4. As shown in the figure, $A B=a+b, B C=b+c, C A=a+c$

$\therefore$ Area of $\triangle A B C=\sqrt{s(s-A B)(s-B C)(s-C A)}$

where, $s=\frac{1}{2}(A B+B C+C A)$

$ \begin{aligned} & =\frac{a+b+b+c+c+a}{2} \\ & =a+b+c \end{aligned} $

$\therefore$ Area of $\triangle A B C$

$\sqrt{(a+b+c)[(a+b+c)-(a+b)][(a+b+c)-(b+c)]}$ $[(a+b+c)-(c+a)]$

$=\sqrt{(a+b+c) \cdot a \cdot b \cdot c}$

~~ 5. Area of a triangle $=\frac{1}{2} \times$ base $\times$ height

$=$ In-radius $\times$ semi-perimeter of the $\Delta[.$ Using $.r=\frac{\Delta}{s}]$

Let the sides of triangle be $4 x, 5 x$ and $6 x$ respectively.

Given: In-radius $=3 cm$

Therefore, $3 \times \frac{(4 x+5 x+6 x)}{2}=\frac{1}{2} \times 6 x \times h$, where $h$ is the height.

$\Rightarrow 3 \times \frac{15 x}{2}=\frac{6 x}{2} \times h \Rightarrow h=\frac{45}{6}=7.5 cm$.

~~ 6. For an equilateral triangle of side $\boldsymbol{a}$ units,

In-radius $=\frac{a}{2 \sqrt{3}}$ units

$\Rightarrow$ Diameter of inscribed circle $=\frac{a}{\sqrt{3}}$ units

Circumradius $=\frac{a}{\sqrt{3}}$ units

$\Rightarrow$ Diameter of circumscrible circle $=\frac{2 a}{\sqrt{3}}$ units

Height $=\frac{\sqrt{3}}{2} a$ units. $\therefore$ Required ratio $=\frac{a}{\sqrt{3}}: \frac{2 a}{\sqrt{3}}: \frac{\sqrt{3}}{2} a=2 a: 4 a: 3 a=2: 4: 3$.

~~ 7. $\triangle A B C$ is an isosceles triangle with $A B=A C=10 cm$.

$\angle A=45^{\circ}$

$\therefore$ Area of $\triangle A B C$

$=\frac{1}{2} \times 10 \times 10 \times \sin 45^{\circ}$

[Using $\Delta=\frac{1}{2} b c \sin A$ ]

$ =\frac{50}{\sqrt{2}}=\frac{50}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=25 \sqrt{2} cm^{2} . $

~~ 8. Let $A B C$ be the triangular flower bed of side lengths $a, b$ and $c$ respectively. Then Area of $\triangle A B C$

$=\sqrt{s(s-a)(s-b)(s-c)}$

where $s=\frac{a+b+c}{2}$

Now according to the given condition,

$\triangle P Q R$ forms the park with side lengths $2 a, 2 b, 2 c$.

$\therefore$ Area of $\triangle P Q R=\sqrt{s^{\prime}(s^{\prime}-2 a)(s^{\prime}-2 b)(s^{\prime}-2 c)}$

where $s^{\prime}=\frac{2 a+2 b+2 c}{2}=a+b+c=2 s$

$\therefore \quad$ Area of $\triangle P Q R=\sqrt{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}$

$ \begin{aligned} & =4 \sqrt{s(s-a)(s-b)(s-c)} \\ & =4 \text{. Area of } \triangle A B C . \end{aligned} $

$\therefore \quad$ Area of path $=$ Area of $\triangle P Q R-$ Area of $\triangle A B C$

$ \begin{aligned} & =4 \times \text{ Area of } \triangle A B C-\text{ Area of } \triangle A B C \\ & =3(\text{ Area of } \triangle A B C) \end{aligned} $

$\therefore \quad$ Reqd. Ratio $=$ Area of Path : Area of $\triangle A B C=\mathbf{3}: \mathbf{1}$.

~~ 9. Given : $A B=A C=10 m, B C=6 m$ and $\angle A=30^{\circ}$. The area that the cow can graze is the area of the circle with centre $A$ and radius $8 m$. But the cow is not allowed to enter the sector $A D E$ formed inside $\triangle A B C$ with

$\angle A=30^{\circ}$ and $A D=A E=8 cm$.

$\therefore$ Area grazed $=$ Area of circle - Area of sector $A D E$

$ \begin{aligned} & =\pi r^{2}-\pi r^{2} \cdot \frac{30^{\circ}}{360^{\circ}}=\pi r^{2}(1-\frac{30^{\circ}}{360^{\circ}}) \\ & =\pi \times 64 \times \frac{11}{12}=\frac{\mathbf{1 7 6}}{\mathbf{3}} \pi \mathbf{s q} \cdot \mathbf{c m} \end{aligned} $

~~ 10. Area of $\triangle A O F=\frac{1}{2} \times O F \times A O$

$\therefore$ Area of hexagon $A B C D E F=4 \times($ Area of $\triangle A O F+$ Area of rectangle $A B D E$ )

$ \begin{aligned} & \quad=2 O F \times A O+E D \times A E \\ & =2 O F \times A O+2 O F \times 2 A O \\ & \quad=6(O F \times A O) \\ & \therefore \text{ Required ratio }=\frac{1 / 2}{6}=\frac{\mathbf{1}}{\mathbf{1 2}} . \end{aligned} $

Volume and Surface Area of Solids

KEY FACTS

Parallelopiped: A solid bounded by three pairs of parallel plane surfaces is called a parallelopiped. The plane surfaces are known as the faces of the parallelopiped.

(i) Cuboid : A parallelopiped whose faces are rectangles and adjacent faces are perpendicular is called a cuboid.

Volume of a cuboid $=(\boldsymbol{l} \times \boldsymbol{b} \times \boldsymbol{h})$ cu. . units, where $l=$ length, $b=$ breadth, $h=$ height
Whole surface of cuboid $=2(l b+b h+l h)$ sq. units
Diagonal of a cuboid $=\sqrt{l^{2}+b^{2}+h^{2}}$ units
Area of 4 walls $=2(l+b) h$ sq. units

(ii) Cube : It is a special type of cuboid whose length, breadth and height are all equal.

Volume of cube $=(\text{ edge })^{3}=l^{3}$ cu. units
Whole surface area of cube $=6(\text{ edge })^{2}=6 l^{2}$ sq. units
Diagonal $=\sqrt{3} l$ units

Prism : A prism is a polyhedron with two parallel faces called bases. The other faces are always parallelograms. The prism is named by the shape of its base.

(i)

Triangular Prism (ii)

(iii)

(iv)

Hexagonal Prism

Surface Area $=2 \times$ Area of base shape + Perimeter of base shape $\times$ height
Volume $=$ Area of base shape $\times$ height of prism

In case of a triangular prism, the area of the base triangles can be found out by :

(i) $A=\frac{1}{2} \times b \times h$, if base (b) and altitude ( $h$ ) of the triangle are known.

(ii) $A=\sqrt{s(s-a)(s-b)(s-c)}$, if all three sides $a, b, c$ are known where $s=\frac{a+b+c}{2}$

(iii) $A=\frac{\sqrt{3}}{4} a^{2}$, if the bases are equilateral triangles of side ’ $a$ ‘.

A prism is a regular prism if its bases are regular figures, i.e., with sides equal, i.e., equilateral triangle, square, regular hexagon, etc.

Note : Area of a regular hexagon $=\frac{\sqrt{3}}{2}(\text{ edge })^{2}$.

Cylinder

(a) Right circular cylinder: It is a solid generated by the revolution of a rectangle about one of its sides that remains fixed. So, a right circular cylinder is a closed solid that has two parallel circular bases connected by a curved surface.

Here $r=$ Radius of base of cylinder, $h=$ Perpendicular distance between bases.

$\quad$ Curved surface area of the cylinder $=$ Perimeter of base $\times$ height

$=2 \pi r h$ sq. units

$\quad$ Total surface area $=$ Curved surface area + Area of bases

$ =2 \pi r h+2 \pi r^{2}=2 \pi r(h+r) \text{ sq. units } $

$\quad$ Volume $=$ Area of base $\times$ Height $=\pi r^{2} h$ cu. units

(b) Volume and surface area of a hollow cylinder: A solid bounded by two coaxial cylinders of same height and different radii is called a hollow cylinder.

For a hollow cylinder, whose inner radius $=r$, outer radius $=R$, perpendicular height $=h$.

Curved (Lateral) surface area $=$ External curved surface area + Internal curved surface area

$ =2 \pi R h+2 \pi r h=2 \pi \boldsymbol{h}(\boldsymbol{R}+\boldsymbol{r}) \text{ sq. units } $

Area of bases $=2 \pi(R^{2}-r^{2})$ sq. units
Total surface area $=2 \pi h(R+r)+2 \pi(R^{2}-r^{2})$

$ =2 \pi(R+r)(h+R-r) \text{ sq. units } $

Volume of material = External volume - Internal volume used in making the cylinder

$ =\pi R^{2} h-\pi r^{2} h=\pi h(R^{2}-r^{2}) \text{ cu. units } $

Right circular cone is a solid that has a circular base which is connected to its vertex by a curved surface. It is generated by the revolution of a right angled triangle about one of its sides containing the right angle as the axis.

If $r=$ radius of base of the cone,

$h=$ perpendicular distance between the vertex and the base, then

Volume $=\frac{1}{3} \pi r^{2} h$ cu. units
Curved surface area $=\pi r l=\pi r \sqrt{\boldsymbol{h}^{2}+\boldsymbol{r}^{2}} \quad(l=$ slant height distance of any point on the circumference of the circle and the vertex)
Total surface area $=\pi r l+\pi r^{2}=\pi r(l+r)=\pi r(\sqrt{h^{2}+r^{2}}+r)$

Sphere: It is a solid geometric figure generated by the revolution of a semicircle about its diameter, i.e., it is a round solid figure, with every point on its surface equidistant from the centre.

For a sphere with $r$ as the radius,

Volume $=\frac{4}{3} \pi r^{3}$ cu. units
Surface area $=4 \pi r^{2}$ sq. units

Volume of a hollow sphere $=\frac{4}{3} \pi(R^{3}-r^{3})$ cu. units, where $R$ is the external radius and $r$, the internal radius.

Hemisphere : Any plane that contains the centre of the sphere divides it into two equal parts called hemispheres. For a hemisphere with $r$ as radius,

Volume $=\frac{2}{3} \pi r^{3}$ cu. units

Curved surface area $=2 \pi r^{2}$ sq. units
Total surface area $=2 \pi r^{2}+\pi r^{2}=3 \pi r^{2}$ sq. units
Volume of a hollow hemisphere $=\frac{2}{3} \pi(R^{3}-r^{3})$, where $R=$ external radius and $r=$ internal radius.
Curved surface area of a hollow hemisphere $=2 \pi(R^{2}+r^{2})$

Total surface area of a hollow hemisphere $=2 \pi(R^{2}+r^{2})+\pi(R^{2}-r^{2})$

Notes :

Curved surface area of a cone, when sector of a circle is converted into a cone $=\frac{\theta}{360^{\circ}} \times \pi r^{2}$, where $\theta$ is the sector angle, and $r$, the bounding radii.
Volume of water that flows out through a pipe $=($ Cross section area $\times$ Speed $\times$ Time $)$
When a solid is melted and converted into another solid, then the volume of both the solids remain the same, assuming there is no wastage in conversion.
Number of new solids obtained by recasting $=\frac{\text{ Volume of the solid that is melted }}{\text{ Volume of the solid that is made }}$.

SOLVED EXAMPLES

Ex. 1 . How many small cubes each of $96 cm^{2}$ surface area can be formed from the material obtained by melting a larger cube of $384 cm^{2}$ surface area?

(MAT 2007)

Sol. Let the edge of the bigger cube be $a cm$ and the edge of the smaller cube be $b cm$. Then,

$ \begin{aligned} & 6 a^{2}=384 \quad \text{ and } \quad 6 b^{2}=96 \\ & \Rightarrow \quad a^{2}=64 \quad \text{ and } \quad b^{2}=16 \\ & \Rightarrow \quad a=8 cm \text{ and } \quad b=4 cm \end{aligned} $

$\Rightarrow$ Volume of bigger cube $=a^{3}=(8)^{3} cm^{3}=512 cm^{3}$ and

Volume of smaller cube $=b^{3}=(4)^{3} cm^{3}=64 cm^{3}$.

$\therefore$ Total number of smaller cubes $=\frac{512}{64}=\mathbf{8}$.

Ex. 2 . The diagonals of the three faces of a cuboid are $x, y, z$ respectively. What is the volume of the cuboid?

(a) $\frac{x y z}{2 \sqrt{2}}$

(b) $\frac{\sqrt{(y^{2}+z^{2})(z^{2}+x^{2})(x^{2}+y^{2})}}{2 \sqrt{2}}$

(d) None of the above

(CDS 2010)

Sol. Let the length, breadth and height of the cuboid be $l, b$ and $h$ respectively. Then,

$ \begin{aligned} l^{2}+b^{2} & =x^{2} \\ b^{2}+h^{2} & =y^{2} \\ h^{2}+l^{2} & =z^{2} \end{aligned} $

Adding eqn. (i), (ii) and (iii), we get

$ \begin{aligned} 2(l^{2}+b^{2}+h^{2}) & =x^{2}+y^{2}+z^{2} \\ \Rightarrow \quad l^{2}+b^{2}+h^{2} & =\frac{1}{2}(x^{2}+y^{2}+z^{2}) \end{aligned} $

$\therefore$ From eqns. (i), (ii), (iii) and (iv), we get

$ \begin{aligned} h^{2} & =\frac{1}{2}(x^{2}+y^{2}+z^{2})-x^{2} \\ & =\frac{1}{2}(y^{2}+z^{2}-x^{2}) \Rightarrow h=\sqrt{\frac{y^{2}+z^{2}-x^{2}}{2}} \end{aligned} $

[Eqn. (iv) - Eqn. (i)]

Similarly, $\quad l=\sqrt{\frac{z^{2}+x^{2}-y^{2}}{2}}$ and $b=\sqrt{\frac{x^{2}+y^{2}-z^{2}}{2}}$

$\therefore$ Volume of the cuboid $=l b h=\sqrt{\frac{(z^{2}+x^{2}-y^{2})(x^{2}+y^{2}-z^{2})(y^{2}+z^{2}-x^{2})}{2 \times 2 \times 2}}$

$ =\frac{1}{2 \sqrt{2}} \sqrt{(y^{2}+z^{2}-x^{2})(z^{2}+x^{2}-y^{2})(x^{2}+y^{2}-z^{2})} $

Ex. 3 . The area of the curved surface and the area of the base of a right circular cylinder are $a$ square $cm$ and $b$ square $cm$ respectively. The height of the cylinder is

(a) $\frac{2 a}{\sqrt{\pi b}} cm$

(b) $\frac{a \sqrt{b}}{2 \sqrt{\pi}} cm$

(d) $\frac{a \sqrt{\pi}}{2 \sqrt{b}} cm$

(SSC 2012)

Sol. Let $r$ and $b$ be the radius of the base and height of the cylinder respectively. Then,

$C S A=2 \pi r h=a$ and Area of base $=\pi r^{2}=b$

$2 \pi r h=a \Rightarrow 4 \pi^{2} r^{2} h^{2}=a^{2} \Rightarrow 4 \pi b h^{2}=a^{2} \Rightarrow h^{2}=\frac{a^{2}}{4 \pi b} \Rightarrow h=\frac{\boldsymbol{a}}{2 \sqrt{\pi b}} \mathbf{c m}$.

Ex. 4 . A semicircular thin sheet of a metal of diameter $28 cm$ is bent and an open conical cup is made. What is the capacity of the cup?

(CDS 2010)

Sol. The radius $(r)$ of the semicircle $=14 cm$

Also, for the conical cup (slant height) $l=r=14 cm$

Let $R$ be the radius of the base of the conical cup. Then,

Circumference of the base of the cone

$=$ Circumference of the semicircle

$\Rightarrow \quad 2 \pi R=\pi r \Rightarrow 2 R=r \Rightarrow R=\frac{r}{2}=\frac{14}{2}=7 cm$

Let $h$ be the vertical height of the cone. Then,

$ \begin{aligned} & l^{2}=R^{2}+h^{2} \Rightarrow h^{2}=l^{2}-R^{2}=(14)^{2}-7^{2}=196-49=147 \\ \Rightarrow \quad & h=\sqrt{147}=7 \sqrt{3} cm \end{aligned} $

$\therefore$ Capacity of the cup $=\frac{1}{3} \pi R^{2} h=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \sqrt{3} cu cm=\frac{\mathbf{1 0 7 8}}{\mathbf{3}} \sqrt{\mathbf{3}} \mathbf{~ c u ~ c m .}$

Ex. 5 . A cylinder is filled to $\frac{4}{5}$ th of its volume. It is, then tilted so that the level of water coincides with one edge of its bottom and top edge of the opposite side. In the process $30 cc$ of the water is spilled. What is the volume of the cylinder?

(MAT 2008)

Sol. Volume of the cylinder $=\pi r^{2} h$. After tilting, by the given condition, cylinder is half full of water.

$ \begin{aligned} & \therefore \frac{4}{5} \pi r^{2} h-30=\frac{1}{2} \pi r^{2} h \Rightarrow(\frac{4}{5}-\frac{1}{2}) \pi r^{2} h=30 \\ & \Rightarrow \frac{3}{10} \pi r^{2} h=30 \Rightarrow \pi r^{2} h=\frac{30 \times 10}{3} cc=100 cc . \end{aligned} $

Cylinder after tilting

Ex. 6 . A sphere and a cube have the same surface area. What is the ratio of the square of volume of the sphere to the square of volume of the cube?

(CDS 2010)

Sol. By the given condition, surface area of sphere $=$ surface area of cube

$\Rightarrow \quad 4 \pi r^{2}=6 a^{2}$ (where $r=$ radius of sphere, $a=$ edge of cube)

$\Rightarrow \quad(\frac{r}{a})^{2}=\frac{3}{2 \pi}$

$\therefore \frac{(\text{ Volume of sphere })^{2}}{(\text{ Volume of cube })^{2}}=\frac{(\frac{4}{3} \pi r^{3})^{2}}{(a^{3})^{2}}=\frac{16}{9} \pi^{2}[(\frac{r}{a})^{3}]^{2}$

$ =\frac{16}{9} \pi^{2}[(\frac{r}{a})^{2}]^{3}=\frac{16}{9} \pi^{2}(\frac{3}{2 \pi})^{3}=\frac{16}{9} \pi^{2} \times \frac{27}{8 \pi^{3}}=\frac{6}{\pi} . $

Ex. 7 . A rectangular paper $11 cm$ by $8 cm$ can be exactly wrapped to cover the curved surface of a cylinder of height $8 cm$. What is the volume of the cylinder ?

Sol. Area of the curved surface $=$ Area of the rectangle $=(11 \times 8) cm^{2}=88 cm^{2}$.

$\Rightarrow 2 \pi r h=88 cm \Rightarrow 2 \times \frac{22}{7} \times r \times 8=88 \Rightarrow r=\frac{88 \times 7}{44 \times 8}=\frac{7}{4} cm$.

$\therefore \quad$ Volume of the cylinder $=\pi r^{2} h=\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 8=77 cm^{3}$.

Ex. 8 . A cylinder is within the cube touching all the vertical faces. A cone is inside the cylinder. If their heights are same with the same base, find the ratio of their volumes.

Sol. Let the length of each edge of the cube be $x$ units. Then,

$ V_1=\text{ Volume of the cube }=x^{3} \text{ cubic units. } $

Since a cylinder is within the cube and it touches all the faces of the cube,

$ \begin{aligned} & \text{ radius of the cylinder }=\frac{x}{2} \text{ units } \\ & \text{ height of the cylinder }=x \text{ units } \\ \therefore \quad V_2= & \text{ Volume of the cylinder }=\pi r^{2} h=\frac{22}{7} \times \frac{x^{2}}{4} \times x \text{ cubic units } \\ = & \frac{11}{14} x^{3} \text{ cubic units. } \end{aligned} $

Also, the cone stands on the same base as the cylinder and has the same height.

So, $\quad V_3=$ Volume of the cone $=\frac{1}{3} \pi r^{2} h$

$ =\frac{1}{3} \times \frac{11}{14} x^{3} \text{ cubic units }=\frac{11}{42} x^{3} \text{ cubic units } $

$\therefore$ Required ratio $=V_1: V_2: V_3=a^{3}: \frac{11}{14} a^{3}: \frac{11}{42} a^{3}=\mathbf{4 2}: \mathbf{3 3}: \mathbf{1 1}$.

Ex. 9 . Water is following at the rate of $5 km / hr$ through a pipe of diameter $14 cm$ into a rectangular tank which is $50 m$ long and $44 m$ wide. Determine the time in which the level of water in the tank will rise by $7 cm .(.$ Take $.\pi=\frac{22}{7})$

(SSC CPO 2009)

Sol. Rate of flow $=5 km / hr=5000 m /$ hour.

$\Rightarrow$ Length of cylinder for water flowing in one hour $=5000 m$.

$ \text{ Radius }=7 cm=\frac{7}{100} m . $

$\therefore$ Volume of water flowing through the pipe per hour $=\pi r^{2} h$

$ =\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 5000 m^{3}=77 m^{3} . $

Volume of water to be filled in the rectangular tank $=(50 \times 44 \times 7 / 100) m^{3}=154 m^{3}$

$\therefore \quad$ Reqd. time $=\frac{154}{77}=\mathbf{2}$ hours.

Ex. 10 . The outer and inner diameters of a circular pipe are $6 cm$ and $4 cm$ respectively. If its length is $10 cm$, then what is the total surface area in sq $cm$ ?

Sol. Outer radius $(R)=3 cm$, Inner radius $(r)=2 cm$, Height $(h)=10 cm$.

$\therefore$ Total surface area of the pipe $=$ Total area of both the bases + Outer $C S A+$ Inner CSA

$(C S A=$ Curves surface Area)

$ \begin{aligned} & =2 \pi(R^{2}-r^{2})+2 \pi R h+2 \pi r h \\ & =2 \pi(R+r)(R-r)+2 \pi h(R+r) \\ & =2 \pi(R+r)(R-r+h)=2 \pi(3+2)(3-2+10) cm^{2} \\ & =2 \pi(5)(11) cm^{2}=110 \pi cm^{2} \end{aligned} $

Ex. 11 . A right $\triangle A B C$ with sides $5 cm, 12 cm$ and $13 cm$ is revolved about the side $12 cm$. What is the volume of the solid so obtained?

(CDS 2009)

Sol. Here, radius $(r)$ of cone $=5 cm$, vertical height $(h)$ of the cone $=12 cm$.

$\therefore$ Volume of the cone $=\frac{1}{3} \pi r^{2} h$

$ \begin{aligned} & =\frac{1}{3} \pi \times 5^{2} \times 12 cu cm \\ & =\mathbf{1 0 0} \boldsymbol{\pi} \mathbf{c u ~ c m .} \end{aligned} $

Ex. 12 . A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is $r$ unit and the rate of painting is ₹ 8 per sq. unit. What is the total cost of painting the two halves of the sphere in rupees?

(CDS 2008)

Sol. Total surface area of the two halves $=3 \pi r^{2}+3 \pi r^{2}=6 \pi r^{2}$ sq units

$\therefore$ Cost of painting the two halves $=6 \pi r^{2} \times ₹ 8=₹ \mathbf{4 8} \pi r^{2}$.

Ex. 13 . A hemispherical bowl is made of steel $0.5 cm$ thick. The inside radius of the bowl is $4 cm$. What is the volume of steel used in making the bowl ?

(MAT 2003)

Sol. Inner radius of the bowl $(r)=4 cm$

Outer radius of the bowl $(R)=(4+0.5) cm=4.5 cm$

$\therefore$ Volume of steel used in making the bowl $=\frac{2}{3} \pi(R^{3}-r^{3})=\frac{2}{3} \times \frac{22}{7}((4.5)^{3}-4^{3})$

$ =\frac{2}{3} \times \frac{22}{7} \times(91.125-64)=\frac{2}{3} \times \frac{22}{7} \times 27.125=\mathbf{5 6 . 8 3} \mathbf{c m}^{3} . $

Ex. 14 . The volumes of two spheres are in the ratio $64: 27$. Find the difference of their surface areas, if the sum of their radii is $7 cm$.

Sol. Let the radii of the two spheres be $r_1 cm$ and $r_2 cm$ respectively and their respective volumes be $V_1$ and $V_2$. Then,

$ \frac{V_1}{V_2}=\frac{\frac{4}{3} \pi r_1^{3}}{\frac{4}{3} \pi r_2^{3}}=\frac{64}{27} \Rightarrow \frac{r_1^{3}}{r_2^{3}}=\frac{64}{27} \Rightarrow(\frac{r_1}{r_2})^{3}=(\frac{4}{3})^{3} \Rightarrow \frac{r_1}{r_2}=\frac{4}{3} \Rightarrow r_1=\frac{4}{3} r_2 $

Given, $\quad r_1+r_2=7 \Rightarrow \frac{4}{3} r_2+r_2=7 \Rightarrow \frac{7 r_2}{3}=7 \Rightarrow r_2=\mathbf{3} \mathbf{c m} . \therefore r_1=\frac{4}{3} \times 3=\mathbf{4} \mathbf{~ c m}$

$\therefore$ Difference in the surface areas of the two spheres $=4 \pi r_2^{2}-4 \pi r_1^{2}$

$ =4 \pi(r_2^{2}-r_1^{2})=4 \times \frac{22}{7} \times(16-9)=4 \times \frac{22}{7} \times 7 cm^{2}=\mathbf{8 8} \mathbf{c m}^{2} . $

Ex. 15 . The base of a right prism is an equilateral triangle with a side $6 cm$ and its height is $18 cm$. Find its volume, lateral surface area and total surface area?

Sol. Volume of a right prism $=$ Area of base $\times$ height.

Since the base is an equilateral triangle of side $6 cm$,

Area of base $=\frac{\sqrt{3}}{4} \times(\text{ side })^{2}=(\frac{\sqrt{3}}{4} \times 6^{2}) cm^{2}=\frac{\sqrt{3}}{4} \times 36 cm^{2}=9 \sqrt{3} cm^{2}$

$\therefore$ Volume $=(9 \sqrt{3} \times 18) cm^{3}=\mathbf{1 6 2} \sqrt{\mathbf{3}} \mathbf{~ c m}^{3}$

Lateral surface area $=$ Perimeter of the base $\times$ Height

$ =(6+6+6) cm \times 18 cm=18 cm \times 18 cm=\mathbf{3 2 4} \mathbf{c m}^{2} $

Total surface area $=$ Lateral surface area + Area of ends (bases)

$ =(324+2 \times 9 \sqrt{3}) cm^{2}=(324+18 \sqrt{3}) cm^{2}=(324+31.176) cm^{2}=\mathbf{3 5 5 . 1 7 6} \mathbf{c m}^{2} . $

Ex. 16 . The base of a right prism is a trapezium. The lengths of the parallel sides are $8 cm$ and $14 cm$ and the distance between the parallel sides is $8 cm$. If the volume of the prism is $1056 cm^{3}$, then what is the height of the prism?

(SSC 2011)

Sol. Volume of the prism $=$ Area of the base $\times$ height

$ \Rightarrow \quad 1056=\frac{1}{2}(8+14) \times 8 \times h \Rightarrow h=\frac{1056 \times 2}{22 \times 8}=\mathbf{1 2} \mathbf{~ c m} $

Ex. 17 . An equilateral triangle with side a is revolved about one of its sides as axis. What is the volume of the solid of revolution thus obtained?

(CDS 2006)

Sol. When an equilateral triangle is revolved about one of its sides say $B C$ then a double cone is generated, whose vertical radius $=a / 2 cm$ and slant height $=a cm$

$\therefore$ Height of the cone $=\sqrt{a^{2}-(a / 2)^{2}}=\sqrt{a^{2}-\frac{a^{2}}{4}}=\sqrt{\frac{3 a^{2}}{4}}=\frac{a}{2} \sqrt{3} cm$

Hence, volume of the solid $=2 \times \frac{1}{3} \pi r^{2} h=2 \times \frac{1}{3} \times \pi \times \frac{a}{2} \times(\frac{\sqrt{3} a}{2})^{2}=\frac{2}{3} \times \pi \times \frac{a}{2} \times \frac{3 a^{2}}{4}=\frac{\pi a^{3}}{4} \mathbf{c m}^{3}$.

Ex. 18 . There are two identical cubes. Out of one cube, a sphere of maximum volume $(V_S)$ is cut off. Out of the second cube, a cone of maximum volume $(V_C)$ is cut such that its base lies on one of the faces of the cube. Which one of the following is correct ?

(a) $V_S=V_C$

(b) $V_S=2 V_C$

(d) $3 V_S=4 V_C$

(CDS 2006)

Sol. Let the side of the cube be $a$ units. Since sphere is cut off from the cube, radius of the sphere $=a / 2$ units.

$\therefore$ Volume of the sphere $=\frac{4}{3} \pi(\frac{a}{2})^{3}$

$\Rightarrow \quad V_S=\frac{4}{3} \cdot \frac{\pi a^{3}}{8}=\frac{\pi a^{3}}{6}$ cu. units

Since the cone is cut off from an identical cube, radius of base of cone $=a / 2$ units height of cone $=a$ units.

$\therefore$ Volume of cone $V_C=\frac{1}{3} \pi(\frac{a}{2})^{2} \cdot a=\frac{1}{3} \pi \cdot \frac{a^{2}}{4} \cdot a=\frac{\pi a^{3}}{12}$

From eqn. (i) and (ii), we get $V_S \times \frac{1}{2}=V_C \Rightarrow \boldsymbol{V}_S=\mathbf{2} \boldsymbol{V} _{\boldsymbol{C}}$

Ex. 19 . In a sphere of radius $2 cm$ a cone of height $3 cm$ is inscribed. What is the ratio of volumes of the cone and sphere?

Sol. Given, radius of sphere $(R)=2 cm$. Height of cone $(h)=3 cm$.

Point $O$ is the centre of the sphere.

$ \begin{matrix} \therefore & O A=O B=R, A C=h \\ \Rightarrow & O C=A C-O A=h-R \end{matrix} $

Hence, in $r$. $\triangle O R C, r^{2}=R^{2}-(h-R)^{2}$, where $r$ is the radius of the cone

$ =2^{2}-(3-2)^{2}=4-1=3 \\ $

$\therefore r =\sqrt{3} . \\ $

$ \therefore \quad \frac{\text{ Volume of cone }}{\text{ Volume of sphere }} =\frac{\frac{1}{3} \pi r^{2} h}{\frac{4}{3} \pi R^{3}}=\frac{(\sqrt{3})^{2} \times 3}{4 \times 2^{3}}=\mathbf{9 : 3 2 .} $

Ex. 20 . A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area $S$. If $A$ be the surface area of the sphere, which one of the following is correct ?

(a) $A=3 S$

(b) $A=2 S$

(d) $A<S$

(CDS 2007)

Sol. Let side of the square $=a$ units. Then, radius of the sphere $=a$ units Surface area of sphere $(A)=4 \pi a^{2}$

Since the square revolves round one of its sides, a right circular cylinder is generated whose height and radius are $a$ units and $a$ units respectively. So,

Total surface area of the cylinder $(S)=2 \pi r(r+h)=2 \pi a(a+a)=4 \pi a^{2}$

$\therefore A=S$.

Ex. 21 . A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. What is the number of such spherical balls?

(SSC 2011)

Sol. Let the radius of the base of the cylinder be $r$ units. Height $=8 r$ units

Its volume $=\pi r^{2} \times 8 r=8 \pi r^{3}$ cu. units

Radius of sphere $=r / 2$ units $\quad \therefore$ Its volume $=\frac{4}{3} \pi(\frac{r}{2})^{3}=\frac{\pi r^{3}}{6}$ cu units.

$\therefore$ Number of spherical balls $=\frac{\text{ Volume of cylinder }}{\text{ Volume of sphere }}=\frac{8 \pi r^{3}}{\pi r^{3}} \times 6=\mathbf{4 8}$.

Ex. 22 . A hemispherical bowl has its external diameter equal to $10 cm$ and its thickness is $1 cm$. What is the whole surface area of the bowl?

(CDS 2005)

Sol. External radius of hemispherical bowl $=5 cm$

Internal radius of the bowl $=(5-1) cm=4 cm$

Surface area of external portion $=2 \pi(5)^{2}=50 \pi$ sq. $cm$

Surface area of internal portion $=2 \pi(4)^{2}=32 \pi$ sq. $cm$

Area of the top circular portion $=\pi(5^{2}-4^{2})=9 \pi$ sq. $cm$

$\therefore$ Total surface area of the bowl $=50 \pi+32 \pi+9 \pi=91 \pi$ sq. $cm=(91 \times \frac{22}{7})$ sq. $cm$

$ =13 \times 22 \text{ sq. } cm=\mathbf{2 8 6} \mathbf{c m}^{2} . $

Ex. 23 . A metallic sheet is of rectangular shape with dimensions $28 m \times 36 m$. From each of its corners, a square is cut off so as to make an open box. The volume of the box is $x \mathbf{~ m}^{3}$, when the length of the square is $8 m$. What is the value of $x$ ?

(MAT 2003)

Sol. After cutting the squares from all the four corners, the dimensions of the box are $20 m, 12 m, 8 m$.

$\therefore$ Volume of the box $=(20 \times 12 \times 8) m^{3}$

$ =1920 m^{3} . $

PRACTICE SHEET

Two cans have the same height equal to $21 cm$. One can is cylindrical, the diameter of whose base is $10 cm$. The other can has a square base of side $10 cm$. What is the difference in their capacities?

(a) $350 cm^{3}$

(b) $250 cm^{3}$

(d) $300 cm^{3}$

(MAT 2010)

~~ 2. Water flows in a tank $150 m \times 100 m$ at the base, through a pipe whose cross-section is $2 dm$ by $1.5 dm$, at a speed of $15 km$ per hour. In what time will the water be 3 metre deep?

(a) 100 hour

(b) 120 hour

(d) 150 hour (SSC FCI 2012)

~~ 3. A rectangular box has dimensions $x, y$ and $z$ units, where $x<y<z$. If one dimension is only increased by one unit, then the increase in volume is

(a) Greatest when $x$ is increased

(b) Greatest when $y$ is increased

(d) The same regardless of which dimension is increased.

~~ 4. If three cubes of copper, each with an edge $6 cm, 8 cm$ and $10 cm$ respectively are melted to form a single cube, then what is the diagonal of the new cube?

(a) $18.8 cm$

(b) $22.8 cm$

(d) $24.8 cm$

(MAT 2012)

~~ 5. The length and width of a swimming pool are 50 metres and 15 metres respectively. If the depth of the swimming pool at one end is 10 metres and at the other end 20 metres, then find the volume of water in the swimming pool?

(a) $10000 m^{3}$

(b) $11250 m^{3}$

(d) $8000 m^{3}$

~~ 6. A cone of height $7 cm$ and base radius $1 cm$ is carved from a cuboidal block of wood $10 cm \times 5 cm \times 2 cm$. Assuming $\pi=\frac{22}{7}$. What is the percentage of wood wasted in the process?

(a) $92 \frac{2}{3} %$

(b) $46 \frac{1}{3} %$

(d) $7 \frac{1}{3} %$

(SSC 2002)

~~ 7. A cylindrical container of height $14 m$ and base $12 m$ contains oil. The oil is to be transferred to one cylindrical can, one conical can and a spherical can. The base radius of all the containers is same. The height of the conical can is $6 m$. While pouring some oil is dropped and hence only $\frac{3}{4}$ th of cylindrical can could be filled. How much oil is dropped?

(a) $54 \pi m^{3}$

(b) $36 \pi m^{3}$

(d) $50 \pi m^{3}$

(MAT 2010)

~~ 8. A hemispherical bowl is $176 cm$ round the brim. Supposing it to be half full, how many persons may be served from it in hemispherical glasses $4 cm$ in diameter at the top ?

(a) 1372

(b) 1272

(d) 1472

(MAT 2009)

~~ 9. The length, breadth and height of a rectangular parallelopiped are in the ratios $6: 3: 1$. If the surface area of a cube is equal to the surface area of this parallelopiped, then what is the ratio of the volume of the cube to the volume of the parallelopiped?

(a) $1: 1$

(b) $5: 4$

(d) $3: 2$

(CDS 2010)

~~ 10. A copper wire $4 mm$ in diameter is evenly wound about a cylinder whose length is $24 cm$ and diameter $20 cm$ so as to cover the surface. Find the length and weight of the wire assuming the specific gravity to be $8.88 gm / cm^{3}$.

(a) $1100 \pi cm, 545 \pi^{2} gm$

(b) $1200 \pi cm, 441 \pi^{2} gm$

(d) $1400 \pi cm, 426.24 \pi^{2} gm$

~~ 11. A sector of a circle of radius $15 cm$ has the angle $120^{\circ}$. It is rolled up so that the two bounding radii are joined together to form a cone. The volume of the cone is

(a) $(250 \sqrt{2} \pi) cm^{3}$

(b) $[(500 \sqrt{2}) \pi / 3] cm^{3}$

(d) $[(1000 \sqrt{2}) \pi / 3] cm^{3}$

(MAT 2010)

~~ 12. The radius and height of a right solid circular cone $(A B C)$ are respectively $6 cm$ and $2 \sqrt{7} cm$. A coaxial cone $(D E F)$ of radius $3 cm$ and height $\sqrt{7} cm$ is cut out of the cone as shown in the given figure. What is the whole surface area of the

solid thus formed?

(a) $96 \pi cm^{2}$

(b) $87 \pi cm^{2}$

(d) $36 \pi cm^{2}$

(CDS 2006)

~~ 13. A right circular cylinder and a right circular cone have equal bases and equal volumes. But the lateral surface area of the right circular cone is $15 / 8$ times the lateral surface area of the right circular cylinder. What is the ratio of radius to height of the cylinder?

(a) $3: 4$

(b) $9: 4$

(d) $8: 15$

(CDS 2007)

~~ 14. From a wooden cylindrical block, whose diameter is equal to its height, a sphere of maximum possible volume is carved out. What is the ratio of the utilised wood to that of the wasted wood?

(a) $2: 1$

(b) $1: 2$

(d) $3: 2$

(CDS 2008)

~~ 15. The volume of a cube is numerically equal to sum of its edges. What is the total surface area in square units?

(a) 12

(b) 36

(d) 144

(CDS 2012)

~~ 16. The base in a right prism is an equilateral triangle of side $8 cm$ and the height of the prism is $10 cm$. The volume of the prism is

(a) $150 \sqrt{3}$ cubic $cm$

(b) $300 \sqrt{3}$ cubic $cm$

(d) $160 \sqrt{3}$ cubic $cm$

(SSC 2012)

~~ 17. The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the volume of the original prism to the new prism.

(a) $1: 1$

(b) $1: 2$

(d) $2: 3$

~~ 18. What is the volume of a right prism standing on a triangular base of sides $5 cm, 5 cm$ and $8 cm$ whose lateral surface area is $828 cm^{2}$ ?

(a) $680 cm^{3}$

(b) $552 cm^{3}$

(d) $728 cm^{3}$

~~ 19. A sphere, a cylinder and a cone respectively are of the same radius and same height. Find the ratio of their curved surfaces.

(a) $3: 2: 1$

(b) $4: \sqrt{3}: 4$

(d) $3: 3: \sqrt{5}$

~~ 20. The sum of the radii of two spheres is $10 cm$ and the sum of their volumes is $880 cm^{3}$. What will be the product of their radii?

(a) 21

(b) $26 \frac{1}{3}$

(d) 70

(SSC 2005)

~~ 21. If $S$ denotes the area of the curved surface of a right circular cone of height $h$ end semi-vertical angle $\alpha$, then $S$ equals

(a) $\pi h^{2} \tan ^{2} \alpha$

(b) $\frac{1}{3} \pi h^{2} \tan ^{2} \alpha$

(d) $\frac{1}{3} \pi h^{2} \sec \alpha \tan \alpha$

~~ 22. The base of a right prism is a pentagon whose sides are in the ratio $1: \sqrt{2}: \sqrt{2}: 1: 2$ and its height is $10 cm$. If the longest side of the base be $6 cm$, the volume of the prism is

(a) $270 cm^{3}$

(b) $360 cm^{3}$

(d) None of these

~~ 23. If three cylinders of radius $r$ and height $h$ are placed vertically such that the curved surface of each cylinder touches the curved surfaces of the other two cylinders tangentially, then the volume of the air space left between the three cylinders is

(a) $h r^{2}(3+\pi)$

(b) $h r^{2}(\sqrt{3}+\frac{\pi}{2})$

(d) $h r^{2}(\sqrt{3}-\pi)$

~~ 24. If two rectangular sheets each of dimensions $2 x$ and $2 y$ form the curved surfaces of two different cylinders, then the ratio of the quotient of the volumes $(\frac{V_1}{V_2})$ to that of the areas of the curved surfaces $(\frac{S_1}{S_2})$ of the two cylinders is

(a) $\frac{x^{2}}{y}: 1$ or $\frac{y^{2}}{x}: 1$

(b) $\frac{2 x}{y}: 1$ or $\frac{2 y}{x}: 1$

(d) $\frac{x}{y}: 1$ or $\frac{y}{x}: 1$

~~ 25. A sphere and a right circular cone of same radius have equal volumes. By what percentage does the height of the cone exceed its diameter?

(a) $50 %$

(b) $66 \frac{1}{3} %$

(d) $43.75 %$

~~ 26. If the base of right rectangular prism remains constant and the measures of the lateral edges are halved, then its volume will be reduced by :

(a) $50 %$

(b) $33.33 %$

(d) None of these

~~ 27. A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. What is the number of spherical balls?

(a) 2

(b) 3

(d) 5

(CDS 2011)

~~ 28. The diameter of a solid mettalic right circular cylinder is equal to its height. After culting out the largest possible solid sphere $S$ from this cylinder, the remaining material is recast to form a solid sphere $S_1$. What is the ratio of the radius of the sphere $S$ to that of the sphere $S_1$.

(a) $1: 2^{\frac{1}{3}}$

(b) $2^{\frac{1}{3}}: 1$

(d) $3^{\frac{1}{2}}: 2^{\frac{1}{2}}$

(CDS 2007)

~~ 29. A right circular solid cone of maximum possible volume is cut off from a solid metallic right circular cylinder of volume $V$. The remaining metal is melt and recast into four identical solid spheres. What is the volume of each sphere?

(a) $\frac{V}{12}$

(b) $\frac{V}{9}$

(d) $\frac{V}{6}$

(CDS 2006)

~~ 30. A solid right circular cylinder of radius $8 cm$ and height $2 cm$ is melted and cast into a right circular cone of height 3 times that of the cylinder. Find the curved surface of the cone?

(a) $54 \pi cm^{2}$

(b) $80 \pi cm^{2}$

(d) $77 \pi cm^{2}$.

~~ 31. A spherical metal of radius $10 cm$ is melted and made into 1000 smaller spheres of equal sizes. In this process the surface area of the metal is increased by

(a) 1000 times

(b) 100 times

(d) None of these

(XAT 2012)

~~ 32. A tank internally measuring $150 cm \times 120 cm \times 100 cm$ has $1281600 cm^{3}$ of water in it. Porous bricks are placed in the water until the tank is full upto to its brim. Each brick absorbs one-tenth of its volume of water. How many bricks of dimensions $20 cm \times 6 cm \times 4 cm$ can be put in the tank without spilling over the water.

(a) 1100

(b) 1200

(d) 1250 (e) None of these

(XAT 2010)

~~ 33. A child consumed an ice-cream of inverted right-circular conical shape from the top and left only $12.5 %$ of the cone for her mother. If the height of the ice-cream cone was $8 cm$, what was the height of the remaining ice-cream cone?

(a) $2.5 cm$

(b) $3.0 cm$

(d) $4.0 cm$

(JMET 2009)

~~ 34. 27 drops of water form a big drop of water. If the radius of each smaller drop is $0.2 cm$, then what is the radius of the bigger drop?

(a) $0.4 cm$

(b) $0.6 cm$

(d) $1.0 cm$

(CDS 2007)

~~ 35. The curved surface of a cylinder is developed into a square whose diagonal is $2 \sqrt{2} cm$. The area of the base of the cylinder (in $cm^{2}$ ) is

(a) $3 \pi$

(b) $\frac{1}{\pi}$

(d) $6 \pi$

ANSWERS

1. (c)	2. $(a)$	3. (a)	4. $(c)$	5. $(b)$	6. $(d)$	7. (b)	8. (a)	9. (d)	10. $(c)$
11. (c)	12. $(b)$	13. $(b)$	14. (a)	15. $(c)$	16. $(d)$	17. $(b)$	18. $(b)$	19. (c)	20. (b)
21. (c)	22. (a)	23. (c)	24. $(d)$	25. $(c)$	26. (a)	27. $(b)$	28. $(b)$	29. $(d)$	30. (b)
31. $(d)$	32. $(b)$	33. $(d)$	34. (b)	35. $(b)$

HINTS AND SOLUTIONS

Required difference in capacities $=\frac{22}{7} \times(5)^{2} \times 21 \sim(10)^{2} \times 21$

$ \begin{aligned} & =(1650 \sim 2100) cm^{3} \\ & =450 cm^{3} . \end{aligned} $

~~ 2. Time required (in hours) $=\frac{\text{ Volume of water in the tank }}{\text{ Volume of water flowing }}$ through the pipe per hour

$ =\frac{150 \times 100 \times 3}{0.2 \times 0.15 \times 15000}=\mathbf{1 0 0} \text{ hours } $

~~ 3. We can check by an example. Let $x=2, y=3, z=4$. Then Volume of box $=2 \times 3 \times 4=24$ cu. units.

If we increase $x$ by 1 unit, keeping the other same i.e, $x=3$, then new volume $=3 \times 3 \times 4=36$ cu. units.

If we increase $y$ by 1 unit keeping the others same, i.e., $y=4$, then new volume $=2 \times 4 \times 4=32$ cu. units

If we increase $z$ by 1 unit keeping the other same, i.e., $z=5$, then new volume $=2 \times 3 \times 5=30 cu$. units.

Hence the increase in volume is greatest, when $x$ increases. You can check with other values also.

~~ 4. Let the edge of the single cube be ’ $a$ ’ $cm$. Then, total volume melted $=$ Volume of cube formed

$\Rightarrow(6)^{3}+(8)^{3}+(10)^{3}=a^{3}$

$\Rightarrow a^{3}=216+512+1000=1728 \Rightarrow a=12 cm$.

$\therefore$ Diagonal of the new cube $=\sqrt{3} a=(\sqrt{3} \times 12) cm$

$=20.8 \mathbf{~ c m}$ (approx.)

~~ 5. If we take the vertical crossection of the face of the swimming pool, then it is a trapezium $A B C D$, with parallel sides $A D$ and $B C$ respectively of lengths

$10 m$ and $20 m$ and distance between parallel sides as $50 m$. The width of the swimming pool is $15 m$.

$\therefore \quad$ Volume of water in the swimming pool

$=$ Area of vertical cross section $(A B C D) \times$ width of swimming pool

$=[\frac{1}{2}(10+20) \times 50] \times 15 m^{3}$

$=(15 \times 50 \times 15) m^{3}=\mathbf{1 1 2 5 0} \mathbf{~ m}^{3}$.

~~ 6. Volume of cone $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \frac{22}{7} \times 1 \times 7=\frac{22}{3} cu . cm$

Volume of cubical block $=(10 \times 5 \times 2) cm^{3}=100 cm^{3}$

$\therefore \quad %$ Wastage of wood $=\frac{(100-\frac{22}{3})}{100} \times 100$

$ =\frac{278}{3} %=92 \frac{2}{3} % $

~~ 7. Volume of oil $=\pi \times(6)^{2} \times 14=504 \pi m^{3}$

Volume of conical can $=\frac{1}{3} \times \pi \times(6)^{2} \times 6=72 \pi m^{3}$

Volume of spherical can $=\frac{4}{3} \times \pi \times(6)^{3}=288 \pi m^{3}$

Remaining oil $=$ Vol. of oil - Vol. of oil in (conical can + spherical can)

$ =504 \pi-(72 \pi+288 \pi)=144 \pi m^{3} $

Volume of the cylindrical can $=\pi \times(6)^{2} \times h=144 \pi$

$\Rightarrow h=4 m$.

As only $\frac{3}{4}$ th of the cylindrical can could be filled,

Vol. of oil dropped $=\frac{1}{4} \times \pi \times(6)^{2} \times 4=\mathbf{3 6} \pi \mathbf{m}^{3}$.

~~ 8. Given, perimeter of the circular base of the cup $=176 cm$. If $r$ is the radius of the hemispherical bowl, then $2 \pi r=176$

$\Rightarrow r=\frac{176 \times 7}{22 \times 2}=28 cm$.

Total number of persons who can be served

$ \begin{aligned} & =\frac{\frac{1}{2} \times \text{ Volume of hemispherical bowl }}{\text{ Volume of one hemispherical glass }} \\ & =\frac{\frac{1}{2} \times \frac{2}{3} \times \pi \times(28)^{3}}{\frac{2}{3} \times \pi \times(2)^{3}}=\frac{2744}{2}=\mathbf{1 3 7 2 .} \end{aligned} $

~~ 9. Let the length, breadth and height of the rectangular parallelopiped be $6 x, 3 x$ and $x$.

Let the side of the cube be $a$.

$\therefore$ By the given condition,

Surface area of a cube

$=$ Surface area of rectangular parallelopiped

$ \begin{aligned} & 6(a)^{2}=2(6 x \times 3 x+3 x \times x+x \times 6 x) \\ & \Rightarrow \quad 6 a^{2}=2(18 x^{2}+3 x^{2}+6 x^{2}) \Rightarrow 6 a^{2}=54 x^{2} \Rightarrow a^{2}=9 x^{2} \\ & \Rightarrow \quad a=3 x \end{aligned} $

Now, Volume of cube : Volume of rectangular parallelopiped $=a^{3}:(6 x \times 3 x \times x)=(3 x)^{3}: 18 x^{3}=27: 18=\mathbf{3}: \mathbf{2}$.

~~ 10. One round of wire covers $4 mm=\frac{4}{10} cm$ in thickness of the surface of the cylinder

Length of the cylinder $=24 cm$

$\therefore$ Number of rounds to cover $24 cm=\frac{24}{4 / 10}=\frac{24 \times 10}{4}=60$

Diameter of the cylinder $=20 cm$

$\Rightarrow$ Radius of cylinder $=10 cm$

Length of the wire in completing one round

$ =2 \pi r=2 \pi \times 10 cm=20 \pi cm . $

$\therefore \quad$ Length of the wire in covering the whole surface $=$ Length of the wire in completing 60 rounds $=(20 \pi \times 60) cm=1200 \pi cm$

Radius of copper wire $=2 mm=\frac{2}{10} cm$

$\therefore$ Volume of wire $=(\pi \times \frac{2}{10} \times \frac{2}{10} \times 1200 \pi) cm^{3}=48 \pi^{2} cm^{3}$

So, weight of wire $=(48 \pi^{2} \times 8.88) gm=\mathbf{4 2 6 . 2 4} \boldsymbol{\pi}^{2} \mathbf{~ g m}$.

~~ 11. Curved surface area of cone $=$ Area of the sector of circle

Volume of the cone $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \pi \times 25 \times 10 \sqrt{2}$

$ =\frac{250 \sqrt{2} \pi}{3} cm^{3} $

~~ 12. Radius and height of cone $A B C$ are $6 cm$ and $2 \sqrt{7} cm$ respectively.

$\therefore$ Slant height of cone $A B C=\sqrt{6^{2}+(2 \sqrt{7})^{2}}$

$ =\sqrt{36+28}=\sqrt{64}=8 cm $

$\therefore$ Curved surface area of cone $A B C=\pi \times 6 \times 8=48 \pi cm^{2}$

Also radius and height of cone $D E F$ are $3 cm$ and $\sqrt{7}$ respectively

$\therefore \quad$ Slant height of cone $D E F=\sqrt{3^{2}+(\sqrt{7})^{2}}=\sqrt{9+7}$

$ =\sqrt{16}=4 cm . $

$\therefore$ Curved surface area of cone $D E F=\pi \times 3 \times 4=12 \pi cm^{2}$

$\therefore \quad$ Whole surface area of remaining solid

$=$ Curved surface area of cone $A B C+$ Area of base

Curved surface area of cone $D E F$

$=48 \pi+\pi(6^{2}-3^{2})+12 \pi$

$=48 \pi+\pi(36-9)+12 \pi=87 \boldsymbol{\pi} \mathbf{~ c m}^{2}$.

~~ 13. Let $r$ and $h$ be the radius and height of the cone and $r$ and $H$ be the radius and height of the cylinder. $\therefore \quad$ Volume of cone $=\frac{1}{3} \pi r^{2} h$

Volume of cylinder $=\pi r^{2} H$

By the given condition, $\frac{1}{3} \pi r^{2} h=\pi r^{2} H$

$\Rightarrow h=3 H$

Lateral surface area of cone $=\pi r l$

Lateral surface area of cylinder $=2 \pi r H$

By the given condition, $\pi r l=\frac{15}{8} \times 2 \pi r H$

$\Rightarrow l=\frac{15}{4} H \Rightarrow l^{2}=\frac{225}{16} H^{2}$

$\Rightarrow r^{2}+h^{2}=\frac{225}{16} H^{2}$

$\Rightarrow r^{2}+9 H^{2}=\frac{225}{16} H^{2}$

[Using $h=3 H$ ]

$\Rightarrow r^{2}=\frac{225}{16} H^{2}-9 H^{2}=\frac{81}{16} H^{2}$

$\Rightarrow \frac{r^{2}}{H^{2}}=\frac{81}{16} \Rightarrow \frac{r}{H}=\frac{9}{4} \Rightarrow r: H=\mathbf{9}: 4$.

~~ 14. Let $r$ be the radius of the cylindrical block, then height of the block $=2 r$.

Volume of the block $=\pi(r^{2})(2 r)=2 \pi r^{3}$

A sphere of maximum possible volume is carved out of the cylindrical block, so the radius of the sphere $=r$.

$\therefore \quad$ Volume of sphere $=\frac{4}{3} \pi r^{3}=$ Volume of utilised wood

$\therefore$ Volume of wasted wood $=2 \pi r^{3}-\frac{4}{3} \pi r^{3}=\frac{2 \pi r^{3}}{3}$

$\therefore$ Required ratio $=\frac{4}{3} \pi r^{3}: \frac{2}{3} \pi r^{3}=\mathbf{2}: \mathbf{1}$.

~~ 15. Let $a$ be the length of an edge of a cube. Then,

$ a^{3}=12 a \quad(\because \text{ A cube has } 12 \text{ edges }) $

$\Rightarrow a^{2}=12 \quad \Rightarrow \quad a=2 \sqrt{3}$

$\therefore$ Total surface area of cube $=6 a^{2}=6 \times(2 \sqrt{3})^{2}=\mathbf{7 2}$.

~~ 16. Volume of a prism $=$ Base Area $\times$ Height

$ \begin{aligned} & =(\frac{\sqrt{3}}{4} \times(8)^{2} \times 10) cm^{3} \\ & =(\frac{\sqrt{3}}{4} \times 64 \times 10) cm^{3}=160 \sqrt{\mathbf{3}} cm^{3} . \end{aligned} $

~~ 17. Let each side of the base of the original prism be $a$ units and the height of the prism be $h$ units. Then

$ \begin{aligned} \text{ Required ratio } & =\frac{\text{ Vol. of original prism }}{\text{ Vol. of new prism }} \\ & =\frac{\frac{\sqrt{3}}{4} \times(a)^{2} \times h}{\frac{\sqrt{3}}{4} \times(2 a)^{2} \times h / 2}=\frac{2 a^{2} h}{4 a^{2} h}=\mathbf{1}: \mathbf{2} . \end{aligned} $

~~ 18. Lateral surface area of a prism $=$ Perimeter of base $\times$ Height

$ \begin{matrix} \Rightarrow & 840=(5+5+8) \times \text{ Height } \\ \Rightarrow & \text{ Height }=\frac{840}{18}=46 cm . \end{matrix} $

Semi perimeter of the triangular base $=\frac{18}{2}=9 cm$

$\therefore \quad$ Area of triangle $=\sqrt{9(9-5)(9-5)(9-8)}$

(Herons Formula)

$ =\sqrt{9 \times 4 \times 4 \times 1}=12 cm^{2} $

$\therefore$ Required volume of prism $=$ Area of base $\times$ Height

$ =(12 \times 46) cm^{3}=\mathbf{5 5 2} \mathbf{c m}^{3} \text{. } $

~~ 19. Let $r$ be the common radius of the sphere, cone and cylinder. Then, Height of the cone $=$ Height of the cylinder $=$ Height of the sphere $=2 r$

Let $l$ be the slant height of the cone. Then,

$ l=\sqrt{h^{2}+r^{2}}=\sqrt{4 r^{2}+r^{2}}=\sqrt{5 r^{2}}=r \sqrt{5} $

Now, $C S A$ of sphere $=4 \pi r^{2}$

CSA of cylinder $=2 \pi r .2 r=4 \pi r^{2}$

CSA of cone $=\pi r l=\pi r . \sqrt{5} r=\sqrt{5} \pi r^{2}$

$\therefore$ Required ratio $=4 \pi r^{2}: 4 \pi r^{2}: \sqrt{5} \pi r^{2}=4: 4: \sqrt{5}$.

~~ 20. Given, $r_1+r_2=10$

and $\frac{4}{3} \pi r_1^{3}+\frac{4}{3} \pi r_2^{3}=880$

$\Rightarrow \frac{4}{3} \pi(r_1^{3}+r_2^{3})=880 \Rightarrow r_1^{3}+r_2^{3}=\frac{880 \times 3 \times 7}{4 \times 22}=210$

Taking the cube of both the sides of eqn. (i), we have $(r_1+r_2)^{3}=1000 \Rightarrow r_1^{3}+r_2^{3}+3 r_1 r_2(r_1+r_2)=1000$

$\Rightarrow 210+3 r_1 r_2(10)=1000$

$\Rightarrow 30 r_1 r_2=1000-210=790 \Rightarrow r_1 r_2=\frac{790}{30}=\mathbf{2 6} \frac{\mathbf{1}}{\mathbf{3}}$.

~~ 21. $S=\pi r l$, where $r=$ radius, $l=$ slant height, $h=$ height

Now $\frac{r}{h}=\tan \alpha \Rightarrow r=h \tan \alpha$

$ \begin{aligned} l & =\sqrt{h^{2}+r^{2}} \\ & =\sqrt{h^{2}+h^{2} \tan ^{2} \alpha} \\ & =\sqrt{h^{2}(1+\tan ^{2} \alpha)} \\ & =\sqrt{h^{2} \sec ^{2} \alpha}=h \sec \alpha \end{aligned} $

$\therefore \quad S=\pi . h \tan \alpha . h \sec \alpha=\pi \boldsymbol{h}^{2} \sec \boldsymbol{\alpha} \tan \boldsymbol{\alpha}$.

~~ 22. The base of the prism is a pentagon as shown.

If the longest side is $6 cm$, then the other sides are $3 cm, 3 \sqrt{2} cm$, $3 \sqrt{2} cm, 3 cm$

Since $A B C$ is an isosceles triangle, $A F \perp B C$ and $\angle A F B=90^{\circ}$.

$\therefore B F=F C=3 cm$

In $\triangle A B F, A F=\sqrt{(3 \sqrt{2})^{2}-3^{2}}=\sqrt{18-9}=\sqrt{9}=3$.

$\therefore$ Total area of the base

$=$ Area of $\triangle A B C+$ Area of rect. $B C D E$

$=(\frac{1}{2} \times 6 \times 3+6 \times 3) cm^{2}=(9+18) cm^{2}=27 cm^{2}$.

$\therefore$ Volume of prism $=$ Area of base $\times$ height

$ =(27 \times 10) cm^{3}=\mathbf{2 7 0} \mathbf{c m}^{3} . $

~~ 23. The bases of the three cylinders when placed as given are as shown in the figure :

Let the radius of the base of each cylinder $=r cm$.

We are required to find the volume of air.

Space left between the

cylinders $=$ Area of shaded portion $\times$ height of cylinder Now, area of shaded portion

$=$ Area of $\triangle A B C-$ Sum of areas of sectors of the three bases $\triangle A B C$, as can be seen is an equilateral triangle of side $2 r$.

$\therefore \quad$ Area of $\triangle A B C=\frac{\sqrt{3}}{4} \times(2 r)^{2}=\sqrt{3} r^{2}$

Area of (sector $A E F+$ sector $B E D+$ sector $C F D$ )

$ \begin{gathered} (\because \text{ sector angles } \angle A=\angle B=\angle C=60^{\circ}) \\ \quad=3 \times \frac{60^{\circ}}{360^{\circ}} \times \pi r^{2}=\frac{\pi r^{2}}{2} \end{gathered} $

$\therefore$ Required volume $=(\sqrt{3} r^{2}-\frac{\pi}{2} r^{2}) h=(\sqrt{3}-\frac{\pi}{2}) \boldsymbol{r}^{2} \boldsymbol{h}$.

~~ 24. Let the radii of the two cylinders be $r_1$ and $r_2$ respectively. Then, $2 \pi r_1=2 x$ and $2 \pi r_2=2 y$

$\Rightarrow r_1=\frac{x}{\pi} \quad$ and $\quad r_2=\frac{y}{\pi}$

Also, the height $h_1=2 y$ and $h_2=2 x$

$\therefore$ If $V_1, V_2$ and $S_1, S_2$ be the respective volumes and curved surface areas of these cylinders, then,

$ \begin{aligned} & V_1=\pi r_1^{2} h_1=\pi(\frac{x}{\pi})^{2} 2 y=\frac{2 x^{2} y}{\pi} \\ & V_2=\pi r_2^{2} h_2=\pi(\frac{y}{\pi}) \cdot 2 x=\frac{2 y^{2} x}{\pi} \\ & S_1=2 \pi r_1 h_1=2 \pi(\frac{x}{\pi}) \cdot 2 y=4 x y \\ & S_2=2 \pi r_2 h_2=2 \pi(\frac{y}{\pi}) \cdot 2 x=4 x y \\ & \therefore \quad \frac{V_1}{V_2}: \frac{S_1}{S_2}=\frac{\boldsymbol{x}}{\boldsymbol{y}}: \mathbf{1} . \end{aligned} $

~~ 25. Let $r$ be the radius of the cone and sphere.

Let $h$ be the height of the cone.

Then, according to the question,

$ \frac{4}{3} \pi r^{3}=\frac{1}{3} \pi r^{2} h \Rightarrow h=4 r $

$\therefore \quad$ Required $%$ difference $=(\frac{4 r-2 r}{2 r} \times 100) %=\mathbf{1 0 0 %}$.

~~ 26. Volume of a prism $=$ Area of base $\times$ height.

Since, the base area is constant and the height is halved, so the volume will also be halved, i.e. volume will be reduced by $50 %$.

~~ 27. Let the height of the cylindrical iron rod be $h cm$. Then, Radius of iron rod $=$ Radius of spherical ball

$ =\frac{h}{4} cm $

$\therefore \quad$ Number of spherical balls

$=\frac{\text{ Volume of cylindrical rod }}{\text{ Volume of spherical ball }}$

$ =\frac{\pi \times(\frac{h}{4})^{2} \times h}{\frac{4}{3} \times \pi \times(\frac{h}{4})^{3}}=3 . $

~~ 28. Let the height of the cylinder $=h cm$.

Then, radius of cylinder $=$ radius of sphere $=\frac{h}{2} cm$.

$\therefore \quad$ Volume of remaining material

$=$ Volume of cylinder - Volume of sphere

$ =\pi(\frac{h}{2})^{2} \cdot h-\frac{4}{3} \pi(\frac{h}{2})^{3}=\frac{\pi h^{3}}{4}-\frac{\pi h^{3}}{6}=\frac{\pi h^{3}}{12} $

Let the recasted solid sphere have radius $R cm$. Then, volume of recasted sphere $=$ Volume of remaining material.

$\Rightarrow \frac{4}{3} \pi R^{3}=\frac{\pi h^{3}}{12} \Rightarrow R^{3}=\frac{h^{3}}{16} \Rightarrow R=\frac{h}{2 \cdot(2^{1 / 3})}$

$\therefore \quad$ Required ratio $=r: R=h / 2: h / 2(2^{1 / 3})=\mathbf{2}^{1 / 3}: \mathbf{1}$.

~~ 29. Volume of remaining metal

$=$ Volume of cylinder - Volume of cone

$=\pi r^{2} h-\frac{1}{3} \pi r^{2} h=\frac{2}{3} \pi r^{2} h=\frac{2}{3} V$

$\therefore \quad$ Volume of 4 spheres $=\frac{2}{3} V$

$\Rightarrow$ Volume of one sphere $=\frac{\frac{2}{3} V}{4}=\frac{V}{\mathbf{6}}$.

~~ 30. For cylinder, radius $(r_1)=8 cm$, height $(h_1)=2 cm$

For cone, radius $(r_2)=$ ?, height $(h_2)=6 cm$

Now, Volume of cone $=$ Volume of cylinder

$\Rightarrow \frac{1}{3} \pi r_2^{2} h_2=\pi r_1^{2} h_1 \Rightarrow \frac{1}{3} \pi r_2^{2} \times 6=\pi \times 8 \times 8 \times 2$

$\Rightarrow r_2^{2}=\frac{\pi \times 8 \times 8 \times 2 \times 3}{\pi \times 6}=64 \Rightarrow r_2=8 cm$

Curved surface of cone $=\pi r_2 l=\pi r_2 \sqrt{h_2^{2}+r_2^{2}}$

$=\pi \times 8 \times \sqrt{36+64}=(\pi \times 8 \times 10) cm^{2}=\mathbf{8 0} \boldsymbol{\pi} \mathbf{c m}^{2}$.

~~ 31. Volume of bigger sphere $=\frac{4}{3} \times \pi \times(10)^{3} cm^{3}=\frac{4000}{3} \pi cm^{3}$

Volume of each smaller sphere $=\frac{\frac{4000}{3}}{1000} cm^{3}=\frac{4}{3} \pi cm^{3}$

$\therefore$ If $r$ is the radius of each smaller sphere, then

$\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \Rightarrow r=1 cm$.

Surface area of bigger sphere $=4 \times \pi \times(10)^{2} cm^{2}=400 \pi cm^{2}$

Surface area of 1000 smaller spheres $=1000 \times(4 \times \pi \times 1) cm^{2}$ $=4000 \pi cm^{2}$

So, the total surface area increases by $\frac{4000 \pi-400 \pi}{400 \pi}$ times

$ =9 \text{ times. } $

~~ 32. Volume of a brick $=20 cm \times 6 cm \times 4 cm=480 cm^{3}$.

Water absorbed by one brick $=(\frac{1}{10} \times 480) cm^{3}=48 cm^{3}$.

Let $x$ bricks be placed in the water.

Then, $x$ bricks absorb $48 x cm^{3}$ of water.

$\therefore$ Vol. of water + Volume of $x$ bricks - Volume of absorbed water $=$ Volume of water tank

$\Rightarrow 1281600+480 x-48 x=150 \times 120 \times 100$

$ \begin{matrix} \Rightarrow & 432 x & =1800000-1281600=518400 \\ \Rightarrow \quad x & =\mathbf{1 2 0 0 .} \end{matrix} $

~~ 33. Let $A D E$ be the remaining portion of the cone. Then, from similar $\triangle s A F E$ and $A G C$

$\frac{A F}{A G}=\frac{F E}{G C} \Rightarrow F E=\frac{A F . G C}{A G}$

$\Rightarrow r_1=\frac{h r}{8}$

where $r_1=$ radius of cone $A D E$,

$h=$ height of cone $A D E$ and $r=$ radius of filled cone $A B C$ Now, volume of smaller cone

$A D E=\frac{1}{3} \pi r_1^{2} h=\frac{1}{3} \pi(\frac{h r}{8})^{2} . h$

$=\frac{1}{3} \times \pi \times \frac{h^{3} r^{2}}{64}$

Volume of bigger cone $A B C=\frac{1}{3} \pi r^{2} \times 8$

Given, $\frac{1}{3} \times \pi \times \frac{h^{3} r^{2}}{64}=12.5 %$ of $(\frac{8}{3} \pi r^{2})$

$\Rightarrow \frac{\pi}{3} \times \frac{h^{3} r^{2}}{64}=\frac{125}{1000} \times \frac{8}{3} \pi r^{2} \Rightarrow h^{3}=64 \Rightarrow h=4$.

~~ 34. By the given condition,

$27 \times$ Volume of smaller drops $=$ Volume of bigger drop

$\Rightarrow 27 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$

$\Rightarrow 27 \times \frac{4}{3} \pi(0.2)^{3}=\frac{4}{3} \pi R^{3} \Rightarrow R^{3}=(3 \times 0.2)^{3} \Rightarrow \boldsymbol{R}=\mathbf{0 . 6} \mathbf{c m}$.

~~ 35. Each edge of the square $=\frac{1}{\sqrt{2}} \times 2 \sqrt{2}=2 cm$.

If ’ $r$ ’ is the radius of the cylinder, then circumference of base of cylinder $=$ edge of square

$\Rightarrow 2 \pi r=2 \Rightarrow r=\frac{1}{\pi}$

$\therefore \quad$ Area of base $=\pi r^{2}=\pi \times(\frac{1}{\pi})^{2}=\frac{1}{\pi} \mathbf{c m}^{2}$.

SELF ASSESSMENT SHEET

If $S$ is the total surface area of a cube and $V$ is its volume, then which of the following is correct?

(a) $V^{3}=216 S^{2}$

(b) $S^{3}=216 V^{2}$

(d) $S^{2}=36 V^{3}$

(CDS 2011)

~~ 2. The volume of a certain rectangular solid is $8 cm^{3}$. Its total surface area is $32 cm^{2}$ and its three dimensions are in geometric progression. The sum of the lengths in $cm$ of all the edges of this solid is

(a) 28

(b) 32

(d) 44 (e) 40

~~ 3. The volume of the metal of a cylindrical pipe is $748 cm^{3}$. The length of the pipe is $14 cm$ and its external radius is $9 cm$. What is its thickness? (Take $\pi=22 / 7$ )

(a) $1 cm$

(b) $5.2 cm$

(d) $3.7 cm$

(SSC 2008)

~~ 4. A sphere and a cone have equal bases. If their heights are also equal, the ratio of their curved surface will be :

(a) $1: \sqrt{3}$

(b) $4: \sqrt{5}$

(d) $\sqrt{5}: 1$

(SSC 2012)

~~ 5. The magnitude of the volume of a closed right circular cylinder of unit height divided by the magnitude of the total surface area of the cylinder ( $r$ being the radius of the cylinder) is equal to

(a) $\frac{1}{2}(1+\frac{1}{r})$

(b) $\frac{1}{2}(1+\frac{1}{r+1})$

(d) $\frac{1}{2}(1-\frac{1}{r+1})$

~~ 6. Three identical balls fit snugly into a cylindrical can. The radius of the spheres is equal to the radius of the can and the balls just touch the bottom and the top of the can. What fraction of the volume of the can is taken up by the balls?

(a) $\frac{1}{3}$

(b) $\frac{3}{7}$

(d) $\frac{1}{4}$

~~ 7. The total surface area of a right triangular prism of height $4 cm$ is $72 \sqrt{3} cm^{2}$. If the base of the prism is an equilateral triangle, find its volume.

(a) $36 \sqrt{3} cm^{3}$

(b) $42 \sqrt{3} cm^{3}$

(d) $54 \sqrt{3} cm^{3}$

~~ 8. A right triangular prism of height $18 cm$ and of base sides $5 cm, 12 cm$ and $13 cm$ is transformed into another right triangular prism on a base of sides $9 cm, 12 cm$ and $15 cm$. Find the height of the new prism and the change in the whole surface area.

(a) $10 cm, 120 cm^{2}$

(b) $8 cm, 132 cm^{2}$

(d) $8 cm, 120 cm^{2}$

~~ 9. There are two prisms, one has equilateral triangle as a base and the other has a regular hexagon as a base. If both the prisms have equal height and volume, the ratio of length of each side of the equilateral triangle to the length of each side of the regular hexagon is

(a) $\sqrt{3}: 1$

(b) $\sqrt{6}: 1$

(d) $2: \sqrt{3}$

~~ 10. A spherical iron shell with external diameter $21 cm$ weighs $22775 \frac{5}{21}$ grams. Find the thickness of the shell if the metal weighs $10 gms$ per cu cm.

(a) $3 cm$

(b) $1 cm$

(d) $2.5 cm$.

ANSWERS

~~ 2. (b)

~~ 3. (a)

~~ 4. (b)

~~ 5. (d)

~~ 6. (c)

~~ 7. $(c)$

~~ 8. (c)

~~ 9. (b)

~~ 10. (c)

HINTS AND SOLUTIONS

Let the side of the cube be $x$ units. Then, $S=6 x^{2}$ and $V=x^{3}$

$\therefore \quad S^{3}=6^{3} \cdot(x^{2})^{3}=6^{3} \cdot(x^{3})^{2}=\mathbf{2 1 6} \mathbf{V}^{2}$.

~~ 2. Let the edges of the solid be $a, a r, a r^{2}$. Then,

Volume $=a \times a r \times a r^{2}=a^{3} r^{3}=(a r)^{3}$.

Given $(a r)^{3}=8 \Rightarrow a r=2$

Also, surface area $=2(a \times a r+a r \times a r^{2}+a \times a r^{2})$

$ \begin{aligned} & =2(a^{2} r+a^{2} r^{3}+a^{2} r^{2}) \\ & =2 a r(a+a r+a r^{2}) \end{aligned} $

Given,

$ 2 a r(a+a r+a r^{2})=32 $

$\Rightarrow 4(a+a r+a r^{2})=32$

$\Rightarrow$ Sum of lengths of all edges $=\mathbf{3 2}$.

~~ 3. Let the thickness of the pipe be $a cm$.

External Radius $=9 cm$.

$\therefore \quad$ Internal radius $=(9-a) cm$.

Given, Volume of metal $=748 cm^{3}$

$\Rightarrow(\pi \times 9^{2} \times 14)-(\pi \times(9-a)^{2} \times 14)=748$ $\Rightarrow \pi \times 14 \times[81-(81+a^{2}-18 a)]=748$

$\Rightarrow-a^{2}+18 a=\frac{748}{\pi \times 14}=\frac{748 \times 7}{22 \times 14}=17$

$\Rightarrow a^{2}-18 a+17=0$

$\Rightarrow(a-17)(a-1)=0$

$\Rightarrow \boldsymbol{a}=\mathbf{1}$ or $\mathbf{1 7}$, but $a=17$ is inadmissible.

~~ 4. Let $r=$ radius of cone and hemisphere.

Then, $h$ (height of cone) $=2 r$

$\therefore \quad$ Slant height $(l)$ of cone $=\sqrt{h^{2}+r^{2}}=\sqrt{5 r^{2}}=r \sqrt{5}$

$\therefore$ Required ratio $=4 \pi r^{2}: \pi r l=4 \pi r^{2}: \pi \times r \times r \sqrt{5}=\mathbf{4}: \sqrt{\mathbf{5}}$.

~~ 5. Volume of cylinder $=\pi r^{2} h=\pi r^{2}$

$ (\because h=1) $

Total surface area of cylinder $=2 \pi r(r+h)=2 \pi r(r+1)$

$\therefore \quad$ Required magnitude $=\frac{\pi r^{2}}{2 \pi r(r+1)}=\frac{r}{2(r+1)}$

$ =\frac{1}{2}[1-\frac{1}{r+1}] \text{. } $

~~ 6. If $r$ is the radius of the cylinder, then its height $(h)=3 \times 2 r=6 r$.

$\therefore$ Volume of cylinder $=\pi r^{2} \times 6 r=6 \pi r^{3}$

Volume of the three balls

$ =3 \times \frac{4}{3} \pi r^{3}=4 \pi r^{3} $

$\therefore$ Required fraction of volume

$ =\frac{4 \pi r^{3}}{6 \pi r^{3}}=\frac{2}{3} . $

~~ 7. Let each side of the base of the prism be $a cm$.

Total surface area $=72 \sqrt{3} cm^{2}$

$\Rightarrow$ (Perimeter of the base $\times$ Height) +2 (Area of base)

$\Rightarrow 3 a \times 4+2(\frac{\sqrt{3}}{4} a^{2})=72 \sqrt{3}$

$=72 \sqrt{3}$

$\Rightarrow \sqrt{3} a^{2}+24 a-144 \sqrt{3}=0 \Rightarrow a^{2}+8 \sqrt{3} a-144=0$

$\Rightarrow \quad(a+12 \sqrt{3})(a-4 \sqrt{3})=0$

$\Rightarrow a=-12 \sqrt{3}$ or $4 \sqrt{3} \Rightarrow a=4 \sqrt{3}$ as $a>0$

$\therefore$ Volume of the prism $=$ Area of the base $\times$ Height

$ =\frac{\sqrt{3}}{4} \times(4 \sqrt{3})^{2} \times 4 cm^{2}=48 \sqrt{3} cm^{2} . $

~~ 8. For the first prism, $a=5 cm, b=12 cm, c=13 cm$

$\Rightarrow s=\frac{a+b+c}{2}=\frac{5+12+13}{2}=15 cm$

Area of the base $=\sqrt{s(s-a)(s-b)(s-c)}$

$ =\sqrt{15 \times 10 \times 3 \times 2} cm^{2}=30 cm^{2} $

Let $V_1$ be the volume of the prism. Then,

$ \begin{aligned} V_1 & =\text{ Area of the base } \times \text{ height } \\ \Rightarrow \quad V_1 & =(30 \times 18) cm^{3}=540 cm^{3} \end{aligned} $

Let $S_1$ be the total surface area of the prism. Then

$S_1=$ Lateral surface area +2 (Area of base)

$=($ Perimeter of the base $\times$ height $)+2($ Area of base $)$

$=(30 \times 18+2 \times 30) cm^{2}=600 cm^{2}$ Let $h$ be the height of new prism. Then, for the new prism $a=9 cm, b=12 cm, c=15 cm \Rightarrow s=\frac{9+12+15}{2}=18 cm$

Area of the base $=\sqrt{s(s-a)(s-b)(s-c)}$

$ =\sqrt{18 \times 9 \times 6 \times 3} cm^{2}=54 cm^{2} $

$\therefore \quad V_2=$ Volume of new prism $=(54 \times h) cm^{3}$

$ V_1=V_2 \Rightarrow 54 \times h=540 \Rightarrow h=10 cm $

Let $S_2$ be the total surface area of the new prism. Then,

$ S_2=(36 \times 10+2 \times 54) cm^{2}=468 cm^{2} $

Change in the whole surface area $=S_1-S_2$

$ \begin{aligned} & =(600-468) cm^{2} \\ & =\mathbf{1 3 2} \mathbf{c m}^{2} \end{aligned} $

~~ 9. Let the length of each side of the equilateral triangle be $a cm$.

Also, let the length of each side of the regular hexagon be $b cm$.

Area of equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$

Area of regular hexagon $=6 \times \frac{\sqrt{3}}{4} b^{2}=\frac{3 \sqrt{3}}{2} b^{2}$

Since, volume of prism with equilateral triangle base

$=$ Volume of prism with hexagonal base

$ \begin{aligned} & \Rightarrow \frac{\sqrt{3}}{4} a^{2} \times h=\frac{3 \sqrt{3}}{2} b^{2} \times h \\ & \Rightarrow \frac{a^{2}}{b^{2}}=\frac{3 \sqrt{2}}{2} \div \frac{\sqrt{3}}{4}=\frac{6}{1} \Rightarrow \boldsymbol{a}: \boldsymbol{b}=\sqrt{\mathbf{6}}: \mathbf{1} . \end{aligned} $

~~ 10. Let the internal radius of the shell be $r cm$.

$\therefore \quad$ Internal volume of the shell $=\frac{4}{3} \pi r^{3}$ cu. cm

External radius of the shell $=\frac{21}{2} cm$.

External volume of the shell $=\frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} cu . cm$

$ =4851 cu . cm . $

Weight of $1 cu$. $cm$ of metal $=10 g$.

$\therefore \quad$ Volume of the metal in the shell $=22775 \frac{5}{21} \times \frac{1}{10}$ cu. cm

$ =\frac{478280}{21} \times \frac{1}{10} \text{ cu. } cm=\frac{47828}{21} \text{ cu. } cm $

$\therefore \quad$ Internal vol. of the shell $=4851-\frac{47828}{21}=\frac{54043}{21} cu . cm$

$\Rightarrow \frac{4}{3} \pi r^{3}=\frac{54043}{21} \Rightarrow r^{3}=\frac{54043 \times 3 \times 7}{21 \times 4 \times 22}=614.125$

$\Rightarrow \quad r=8.5 cm$

$\therefore \quad$ Thickness of shell $=10.5 cm-8.5 cm=\mathbf{2} \mathbf{~ c m}$.

पिछला

A Compact & Comprehensive Book of

S.Chand’s IIT Foundation Series

IIT Foundation Mathematics

S.Chand’s IIT Foundation Series

EURASIA PUBLISHING HOUSE

C) 2014. S.K. Gupta, Anubhuti Gangal

PREFACE AND A NOTE FOR THE STUDENTS

ARE YOU ASPIRING TO BECOME AN ENGINEER AND AN IIT SCHOLAR?

SALIENT FEATURES

CONTENTS

Logarithms

KEY FACTS

SOLVED EXAMPLES

PRACTICE SHEET

LEVEL-1

LEVEL-2

LEVEL-3

ANSWERS

HINTS AND SOLUTIONS

SELF ASSESSMENT SHEET

SELF ASSESSMENT SHEET

HINTS AND SOLUTIONS

Polynomials

KEY FACTS

SOLVED EXAMPLES

PRACTICE SHEET

LEVEL-1

LEVEL-2

LEVEL-3

ANSWERS

HINTS AND SOLUTIONS

SELF ASSESSMENT SHEET

ANSWERS

HINTS AND SOLUTIONS

Quadratic Equations

KEY FACTS

SOLVED EXAMPLES

Alternatively,

PRACTICE SHEET

(Rajasthan PET 2002)

ANSWERS

HINTS AND SOLUTIONS

SELF ASSESSMENT SHEET

ANSWERS

HINTS AND SOLUTIONS

Inequalities

KEY FACTS

SOLVED EXAMPLES

PRACTICE SHEET

ANSWERS

HINTS AND SOLUTIONS

SELF ASSESSMENT SHEET

ANSWERS

HINTS AND SOLUTIONS

Relations

KEY FACTS

SOLVED EXAMPLES

PRACTICE SHEET

ANSWERS

HINTS AND SOLUTIONS

$\therefore \quad R$ is not transitive

$R$ is not transitive

SELF ASSESSMENT SHEET

ANSWERS

HINTS AND SOLUTIONS

Plane Geometry -Triangles

KEY FACTS

XI. Properties of Similar Triangles:

XII. Some Useful Results for a Triangle:

SOLVED EXAMPLES

PRACTICE SHEET

ANSWERS

HINTS AND SOLUTIONS

7

Quadrilaterals

KEY FACTS

RECTANGLE:

RHOMBUS:

SQUARE:

TRAPEZIUM:

IIT Foundation
Mathematics