S. Chand’s IIT Foundation Series

A Compact and Comprehensive Book of IIT Foundation Series

$ \text{ Class }-X $

S.Chand’s IIT Foundation Series

A Compact and Comprehensive Book of

IIT Foundation
Mathematics

CLASS - X

S.Chand’s IIT Foundation Series

A Compact and Comprehensive Book of IIT Foundation Mathematics

CLASS - X

S.K. GUPTA
ANUBHUTI GANGAL

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C 2014. S.K. Gupta, Anubhuti Gangal

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First Published in 2014

Reprints 2016

ISBN : 978-93-837-4653-8

Code : 1014A 681

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 Facts’. 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 $\boldsymbol{b} \boldsymbol{a} \boldsymbol{c} \boldsymbol{k}$-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.

AUTHORS¹Chapter 1. Sequence and Series ….. $1-1$ to $1-52$ Chapter 2. Functions ….. $2-1$ to $2-26$ Chapter 3. Matrices and Determinants ….. 3-1 to $3-44$ Chapter 4. Binomial Theorem ….. $4-1$ to $4-20$ Chapter 5. Mathematical Induction ….. 5-1 to $5-11$ Chapter 6. Plane Geometry: Circle ….. 6-1 to $6-32$ Chapter 7. Probability (Further Continued from Class IX) ….. 7-1 to $7-13$ Chapter 8. Trigonometric Equations ….. $8-1$ to $8-13$ Chapter 9. Heights and Distances ….. 9-1 to $9-22$ Chapter 10. Circles: Area and Perimeter ….. 10-1 to $10-11$ Chapter 11. Volume and Surface Area of Solids (Further Continued from Class IX) ….. $11-1$ to $11-21$

CHAPTER 1 Sequence and Series

ARITHMATIC PROGRESSION (A.P.)

KEY FACTS

A sequence is a set of numbers specified in a definite order by some assigned rule or law. Ex. 2 , 7, 12, $17 \ldots$.

(Each succeeding term is obtained by adding 5 to the preceding term)

$1,2,4,8,16 \ldots$

(Each succeeding term is obtained by multiplying the preceding term by 2 )

A finite sequence is that which ends or has a last term.

Ex. 5 , 9, 13, 17, 21.

An infinite sequence is one which has no last term.

Ex. 3 , 6, 12, 24, 48…

In general, $a_n$ or $T_n$ denotes the nth term of a sequence.

An expression consisting of the term of a sequence, alternating with the symbol ’ + ’ is called a series.

Ex. The sequence $\frac{2}{3}, \frac{4}{5}, \frac{6}{7}, \ldots$ expressed as a series is $\frac{2}{3}+\frac{4}{5}+\frac{6}{7}+\ldots$.

Arithmetic Progression (A.P.): A sequence is called an arithmetic progression if its terms continually increase or decrease by the same number. The fixed number by which they increase or decrease is called the common difference. Three quantities $a, b, c$ will be in A.P. if $b-a=c-b$, i.e., $\boldsymbol{2} \boldsymbol{b}=\boldsymbol{a}+\boldsymbol{c}$.

(a) nth term of an A.P.: The $n$th term of an A.P. $a, a+d, a+2 d, a+3 d, \ldots$ is

$ T_n=a+(n-1) d $

where $T_n$ denotes $n$th term, n the first term, $symbol{n}$ the number of terms and $\boldsymbol{d}$ the common difference.

Also, common difference $d=T_n-T _{n-1}$.

Ex. The 9 th term of the A.P.: $2,5,8 \ldots$ is

$ T_9=2+(9-1) \times 3=2+24=\mathbf{2 6} . $

Note: Here $a=2, d=3$.

(b) Sum of n terms of an A.P.

Let the A.P. be $a, a+d, a+2 d, \ldots$ Let $l$ be the last term and $S$ the required sum. Then,

$ \begin{aligned} \boldsymbol{S} & =\frac{\boldsymbol{n}}{\mathbf{2}}(\boldsymbol{a}+\boldsymbol{l})=\frac{\text{ Number of terms }}{2}(\text{ First term }+ \text{ Last term }) \\ & =\frac{n}{2}(a+a+(n-1) d)=\frac{n}{2}(2 a+(n-1) d) \end{aligned} $

where $n$ is the number of terms, $a$ is first term and $d$ is common difference.

Also, $n$th term $=$ Sum of $n$ terms $-Sum$ of $(n-1)$ terms

i.e., $ T_n=S_n-S _{n-1} $

Ex. The sum of 20 terms of the the A.P. $1,3,5,7,9 \ldots$ is

$ \begin{aligned} S _{20} & =\frac{20}{2}(2 \times 1+(20-1) \times 2) \quad(\because a=1, d=2, n=20) \\ & =10 \times 40=400 . \end{aligned} $

(c) Arithmetic Mean: The Arithmetic Mean between two numbers is the number which when placed between them forms an arithmetic progression with them. Thus if $\boldsymbol{x}$ is the arithmetic mean of two given numbers $\boldsymbol{a}$ and $\boldsymbol{b}$, then $a, x, b$ form an A.P.

$\therefore x-a=b-x \Rightarrow \boldsymbol{x}=\frac{\boldsymbol{a}+\boldsymbol{b}}{\mathbf{2}}$

Ex. (a) The arithmetic mean between -4 and 6 is $\frac{6+(-4)}{2}=\mathbf{1}$.

(b) Find 4 arithmetic means between 3 and 23.

Let $A_1, A_2, A_3, A_4$, be the four arithmetic means between 3 and 23 .

Then, $3, A_1, A_2, A_3, A_4$, 23 form an A.P.

Here, First term $=a=3$

Sixth term $=T_6=23$

Number of terms $=n=6$

Common difference $=d=$ ?

$ T_6=a+(n-1) \times d $

$\Rightarrow 23=3+(6-1) \times d$

$\Rightarrow 23=3+5 d \Rightarrow 5 d=20 \Rightarrow d=4$.

$\therefore A_1=3+4=7, A_2=7+4=11, A_3=11+4=15, A_4=15+4=19$.

(d) Some useful facts about an A.P.

I. If each term of a given A.P. is increased or decreased or multiplied or divided by the same number, the resulting progression is also an $A . P$.

II. If a, b, c are in A.P., then $\frac{a-b}{b-c}=1$, i.e., $\frac{\text{ First term - Second term }}{\text{ Second term - Third term }}=1$.

(e) If we have to find an odd number of terms in A.P. whose sum is given, it is convenient to take $a$ as the middle term and $d$ as the common difference. Thus, three terms may by taken as $\boldsymbol{a}-\boldsymbol{d}, \boldsymbol{a}, \boldsymbol{a}+\boldsymbol{d}$ and five terms as $\boldsymbol{a}-\mathbf{2} \boldsymbol{d}, \boldsymbol{a}-\boldsymbol{d}, \boldsymbol{a}, \boldsymbol{a}+\boldsymbol{d}, \boldsymbol{a}+\mathbf{2} d$. [Solved Ex. 13 , 14]. If we have to find even number of terms, we take $a-d, a+d$ as the middle terms and $2 d$ as the common difference. Then, four terms are taken as $\boldsymbol{a}-\mathbf{3} \boldsymbol{d}, \boldsymbol{a}-\boldsymbol{d}, \boldsymbol{a}+\boldsymbol{d}, \boldsymbol{a}+\mathbf{3} \boldsymbol{d}$.

SOLVED EXAMPLES

Ex. 1. The 8th term of a series in A.P. is 23 and the 102th term is 305. Find the series.

Sol. Let $a$ be the first term and $d$ be the common difference.

$ \text{ Then, } \begin{aligned} T_8 & =a+(8-1) d & \Rightarrow & 23=a+7 d \\ T _{102} & =a+(102-1) d & \Rightarrow & 305=a+101 d \end{aligned} $

$\therefore \quad$ Eqn $(i i)-Eqn(i)$

$\Rightarrow \quad 94 d=282 \quad \Rightarrow \quad d=3$

Now substituting $d=3$ in $(i)$, we get $23=a+21 \Rightarrow a=2$.

$\therefore a=2, d=3 \Rightarrow$ Series is $2+5+8+11+\ldots$.

Ex. 2. If $a, b$ and $c$ be respectively the $p$ th, $q$ th and $r$ th terms of an A.P., prove that $a(q-r)+b(r-p)+c(p-q)=0$.

Sol. Let $A$ be the first term and $D$ the common difference of the given A.P.

Then,

$ \begin{aligned} & T_p=A+(p-1) D=a \\ & T_q=A+(q-1) D=b \\ & T_r=A+(r-1) D=c \end{aligned} $

$\therefore(i) \times(q-r)+(i i) \times(r-p)+(i i i) \times(p-q)$

$\Rightarrow A[(q-r)+(r-p)+(p-q)]+D[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]$

$ =a(q-r)+b(r-p)+c(p-q) $

$\Rightarrow A[0]+D[p q-q-p r+r+q r-r-p q+p+p r-p-q r+q]=a(q-r)+b(r-p)+c(p-q)$

$\Rightarrow a(q-r)+b(r-p)+c(p-q)=A \times 0+D \times 0=\mathbf{0}$.

Ex. 3. Which term of the progression $19,18 \frac{1}{5}, 17 \frac{2}{5}, \ldots$. is the first negative term?

Sol. Here $a=19, d=18 \frac{1}{5}-19=-\frac{4}{5}$

Let the $n$th term be the first negative term. Then,

$ \begin{aligned} T_n<0 & \Rightarrow a+(n-1) d<0 \\ & \Rightarrow 19+(n-1)(-4 / 5)<0 \\ & \Rightarrow 19-\frac{4}{5} n+\frac{4}{5}<0 \\ & \Rightarrow \frac{99}{5}-\frac{4}{5} n<0 \Rightarrow n>\frac{99}{5} \times \frac{5}{4} \Rightarrow n>\frac{99}{4}=24 \frac{3}{4} \\ & \Rightarrow n=\mathbf{2 5 .} \end{aligned} $

Ex. 4. Find the sum of the series $101+99+97+\ldots .+47$.

Sol. In this case, we have to first find the number of terms.

Here $a=101, l=T_n=47, d=99-101=-2$

$\therefore \quad 47=101+(n-1) \times(-2)$, where $n=$ number of terms

$\Rightarrow 47=101-2 n+2$

$\Rightarrow 2 n=103-47=56 \Rightarrow n=28$

$\therefore S_n=\frac{n}{2}(a+l)$

$\Rightarrow S _{28}=\frac{28}{2}(101+47)=14 \times 148=2072$.

Ex. 5. The sums of $n$ terms of two arithmetic series are in the ratio $2 n+1: 2 n-1$. Find the ratio of their 10 th terms.

Sol. Let the two arithmetic series be $a, a+d, a+2 d \ldots .$. and $A, A+D, A+2 D, \ldots$.

Given that, $\frac{n / 2[2 a+(n-1) d]}{n / 2[2 A+(n-1) D]}=\frac{2 n+1}{2 n-1}$

$\Rightarrow \frac{2 a+(n-1) d}{2 A+(n-1) D}=\frac{2 n+1}{2 n-1}$

Ratio of the 10th terms of these series $=\frac{t _{10}}{T _{10}}=\frac{a+9 d}{A+9 D}=\frac{2 a+18 d}{2 A+18 D}$

$\therefore \quad$ Putting $n=19$ in $(i)$, we have $\frac{t _{10}}{T _{10}}=\frac{2 a+18 d}{2 A+18 D}=\frac{2 \times 19+1}{2 \times 19-1}=\frac{\mathbf{3 9}}{\mathbf{3 7}}$.

Ex. 6. The sum of the first $n$ terms of the arithmetical progression $3,5 \frac{1}{2}, 8, \ldots$ is equal to the $2 n$th term of the A.P. $16 \frac{1}{2}, 28 \frac{1}{2}, 40 \frac{1}{2}, \ldots$. Calculate the value of $n$.

Sol. For the A.P. : $3,5 \frac{1}{2}, 8, \ldots . \quad a=3, d=2 \frac{1}{2}$, number of terms $=n$

$ \Rightarrow \quad S_n=\frac{n}{2}[2 a+(n-1) d]=\frac{n}{2}(6+(n-1) \times \frac{5}{2}) $

For the A.P.: $16 \frac{1}{2}, 28 \frac{1}{2}, 40 \frac{1}{2} \ldots \quad a=16 \frac{1}{2}, d=12$.

$\therefore \quad T _{2 n}=a+(2 n-1) d=16 \frac{1}{2}+(2 n-1) \times 12$

Given, $S_n=T _{2 n} \Rightarrow \frac{n}{2}(6+(n-1) \frac{5}{2})=\frac{33}{2}+(2 n-1) \times 12$

(From (i) and (ii))

$\Rightarrow \frac{6 n}{2}+\frac{5 n^{2}}{4}-\frac{5 n}{4}=\frac{33}{2}+24 n-12$

$\Rightarrow \frac{7 n}{4}+\frac{5 n^{2}}{4}=\frac{9}{2}+24 n$

$\Rightarrow \frac{5 n^{2}}{4}-\frac{89 n}{4}-\frac{9}{2}=0$

$\Rightarrow 5 n^{2}-89 n-18=0$

$\Rightarrow 5 n^{2}-90 n+n-18=0$

$\Rightarrow 5 n(n-18)+1(n-18)=0$

$\Rightarrow(n-18)(5 n+1)=0$

$\Rightarrow n=18$ or $-\frac{1}{5}$

$\Rightarrow$ Neglecting - ve value, we have $n=\mathbf{1 8}$.

Ex. 7. Let $a_1, a_2, a_3, \ldots .$. be the terms of an A.P. If $\frac{a_1+a_2+a_3+\ldots .+a_p}{a_1+a_2+a_3+\ldots .+a_q}=\frac{p^{2}}{q^{2}}(p \neq q)$, then find $\frac{a_6}{a _{21}}$.

(AIEEE 2006)

Sol. Let $d$ be the common difference for the A.P.; $a_1, a_2, a_3, \ldots$

Then, $\frac{S_p}{S_q}=\frac{p / 2[2 a_1+(p-1) d]}{q / 2[2 a_1+(q-1) d]}=\frac{p^{2}}{q^{2}}$

$\Rightarrow \frac{2 a_1+(p-1) d}{2 a_1+(q-1) d}=\frac{p}{q}$

$\Rightarrow q[2 a_1+(p-1) d]=p[2 a_1+(q-1) d]$

$\Rightarrow 2 a_1 q+p q d-q d=2 a_1 p+p q d-p d$

$\Rightarrow p d-q d=2 a_1 p-2 a_1 q$

$\Rightarrow d(p-q)=2 a_1(p-q)$

$\Rightarrow d=2 a_1$

Now $\frac{\text{ Term } 6}{\text{ Term } 21}=\frac{a_6}{a _{21}}=\frac{a_1+(6-1) d}{a_1+(21-1) d}=\frac{a_1+5 \times 2 a_1}{a_1+20 \times 2 a_1}=\frac{11 a_1}{41 a_1}=\frac{\mathbf{1 1}}{\mathbf{4 1}}$.

Ex. 8. If the number of terms of an A.P. is $(2 n+1)$, then what is the ratio of the sum of the odd terms to the sum of even terms?

(NDA/NA 2008)

Sol. Let the A.P. be $a, a+d, a+2 d, \ldots, a+(2 n-1) d, a+2 n d$

Then, the progression of odd terms is $a, a+2 d, a+4 d, \ldots, a+2 n d$.

This progression has $(n+1)$ terms.

Its sum $=\frac{n+1}{2}[a+a+2 n d]=\frac{n+1}{2}[2 a+2 n d]=(n+1)(a+n d)$

The progression of even terms is $a+d, a+3 d, \ldots \ldots . a+(2 n-1) d$.

This progression has $n$ terms.

Its sum $=\frac{n}{2}[(a+d)+(a+(2 n-1) d]=\frac{n}{2}[2 a+2 n d]=n(a+n d).$

$\therefore \quad \frac{\text{ Sum of odd terms }}{\text{ Sum of even terms }}=\frac{(n+1)(a+n d)}{n(a+n d)}=\frac{\boldsymbol{n}+\mathbf{1}}{\boldsymbol{n}}$.

Ex. 9. If the sum of the roots of the quadratic equation $a x^{2}+b x+c=0$ is equal to the sum of the squares of their reciprocals, then show that $a b^{2}, c a^{2}, b c^{2}$ are in A.P.

(DCE)

Sol. 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, Sum of roots $=$ Sum of squares of reciprocals of roots

$ \begin{matrix} \Rightarrow \alpha+\beta=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}} & \Rightarrow \alpha+\beta=\frac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}} & \Rightarrow \alpha+\beta=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}} \\ \Rightarrow-b / a=\frac{(-b / a)^{2}-\frac{2 c}{a}}{(c / a)^{2}} & \Rightarrow-\frac{b}{a}=\frac{b^{2}-2 c a}{c^{2}} & \Rightarrow-b c^{2}=a b^{2}-2 c a^{2} \\ \Rightarrow 2 c a^{2}=a b^{2}+b c^{2} & \Rightarrow a b^{2}, c a^{2}, b c^{2} \text{ are in A.P. } & (\because a, b, c \text{ in A.P. } \Rightarrow 2 b=a+c) \end{matrix} $

Ex. 10. Find the value of $n$, if $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ is the arithmetic mean between $a$ and $b$.

(WBJEE 2009)

Sol. The arithmetic mean between $a$ and $b$ is $\frac{a+b}{2}$.

Given, $\frac{a+b}{2}=\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$

$\Rightarrow(a^{n}+b^{n})(a+b)=2(a^{n+1}+b^{n+1}) \quad \Rightarrow a^{n+1}+a^{n} b+b^{n} a+b^{n+1}=2 a^{n+1}+2 b^{n+1}$

$\Rightarrow a^{n+1}+b^{n+1}=a^{n} b+b^{n} a$

$\Rightarrow(a^{n+1}-b a^{n})-(b^{n} a-b^{n+1})=0$

$\Rightarrow a^{n}(a-b)-b^{n}(a-b)=0$

$\Rightarrow(a^{n}-b^{n})(a-b)=0$

$\Rightarrow(a^{n}-b^{n})=0$ or $(a-b)=0$

But $a \neq b \Rightarrow a-b \neq 0$

$ \therefore a^{n}-b^{n}=0 \Rightarrow a^{n}=b^{n} \Rightarrow(\frac{a}{b})^{n}=1 \Rightarrow(\frac{a}{b})^{n}=(\frac{a}{b})^{0} \Rightarrow \boldsymbol{n}=\mathbf{0} $

Ex. 11. If $\log _{10} 2, \log _{10}(2^{x}-1)$ and $\log _{10}(2^{x}+3)$ be three consecutive terms of an A.P., then find the value of $x$.

(AMU 2012)

Sol. Given, $\log _{10} 2, \log _{10}(2^{x}-1), \log _{10}(2^{x}+3)$ are in A.P. Then,

$\log _{10}(2^{x}-1)-\log _{10} 2=\log _{10}(2^{x}+3)-\log _{10}(2^{x}-1)$

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

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

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

$\Rightarrow 2^{2 x}-2.2^{x}+1=2.2^{x}+6$

$\Rightarrow 2^{2 x}-4.2^{x}-5=0$

$\Rightarrow 2^{2 x}-5.2^{x}+2^{x}-5=0$

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

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

$\Rightarrow 2^{x}=5$

$(\because 2^{x} \neq-1)$

$\Rightarrow x=\log _2 \mathbf{5}$.

Ex. 12. If $a_1, a_2, a_3, \ldots a_n$ be an A.P. of non-zero terms, then find the sum: $\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots+\frac{1}{a _{n-1} a_n}$ (AMU 2009)

Sol. Let $a_2-a_1=a_3-a_2=\ldots \ldots . .=a_n-a _{n-1}=d$ (common difference)

Then,

$ \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + …… +\frac{1}{a _{n-1} a_n} = \frac{1}{d}[\frac{d}{a_1 a_2} + \frac{d}{a_2 a_3}+…..+\frac{d}{a _{n-1} a_n}] $

$ =\frac{1}{d}[\frac{a_2-a_1}{a_1 a_2}+\frac{a_3-a_2}{a_2 a_3}+ ….. +\frac{a_n -a _{n-1}}{a _{n-1} a_n}]=\frac{1}{d}[\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+\ldots \ldots . .+\frac{1}{a _{n-1}}-\frac{1}{a_n}]$

$ =\frac{1}{d}[\frac{1}{a_1}-\frac{1}{a_n}]=\frac{1}{d}[\frac{a_n - a_1}{a_1 a_n}]=\frac{1}{d}[\frac{(a_1+(n-1) d)-a_1}{a_1 a_n}]=\frac{n-1}{a_1 a_n}$

Ex. 13. If the sides of a right angled triangle form an A.P., then find the sines of the acute angles. (VITEE 2008)

Sol. Let the $\triangle A B C$ be right angled at $C$.

Then, $A B=c, B C=a, A C=b$

Given, the sides of the right angled $\Delta$ are in A.P. $\Rightarrow a, b, c$ are in A.P.

Now, let $a=x-d, b=x, c=x+d$

( $\boldsymbol{d}$ being a + ve quantity as $c$ being the hypotenuse is the greatest side)

$\therefore c^{2}=a^{2}+b^{2}$ (Pythagoras’ Theorem)

$\Rightarrow(x+d)^{2}=(x-d)^{2}+x^{2}$

$\Rightarrow x^{2}+2 x d+d^{2}=x^{2}-2 x d+d^{2}+x^{2}$

$\Rightarrow 4 x d=x^{2} \Rightarrow d=\frac{x}{4}$.

$\therefore \quad a=x-d=x-\frac{x}{4}=\frac{3 x}{4}, b=x, c=x+d=x+x / 4=\frac{5 x}{4}$

$C$ being the right angle, $A$ and $B$ are the acute angles.

$\therefore \sin A=\frac{a}{c}=\frac{3 x / 4}{5 x / 4}=\frac{3}{5}$ and $\sin B=\frac{b}{c}=\frac{x / 5}{x / 4}=\frac{x}{\frac{5 x}{4}}=\frac{4}{5}$

$\therefore$ The sines of the acute angles are $\frac{\mathbf{3}}{\mathbf{5}}, \frac{\mathbf{4}}{\mathbf{5}}$.

Ex. 14. $a_1, a_2, a_3, a_4, a_5$ are the first five terms of an A.P. such that $a_1+a_3+a_5=-12$ and $a_1 \cdot a_2 \cdot a_3=8$. Find the first term and common difference.

Sol. Let $a_1=a_3-2 d, a_2=a_3-d, a_3=a_3, a_4=a_3+d, a_5=a_3+2 d$

Then $a_1+a_3+a_5=-12$, (given) $\Rightarrow a_3-2 d+a_3+a_3+2 d=-12 \Rightarrow 3 a_3=-12 \Rightarrow a_3=-4$

Also, $a_1 \cdot a_2 \cdot a_3=8$ (given)

$\Rightarrow a_1 \cdot a_2=-2(\because a_3=-4) \Rightarrow(a_3-2 d)(a_3-d)=-2$

$\Rightarrow(-4-2 d)(-4-d)=-2 \Rightarrow(2+d)(4+d)=1$

$\Rightarrow d^{2}+6 d+9=0 \quad \Rightarrow(d+3)^{2}=0 \Rightarrow d=-3$

$\therefore a_1=a_3-2 d=-4-2(-3)=\mathbf{2}$.

Ex. 15. If $a, b, c$ are in A.P. show that $\frac{a(b+c)}{b c}, \frac{b(c+a)}{c a}, \frac{c(a+b)}{a b}$ are in A.P.

Sol. $a, b, c$ are in A.P.

$\Rightarrow \frac{a}{a b c}, \frac{b}{a b c}, \frac{c}{a b c}$ are in A.P.

(Dividing each term by $a b c$ ) $\Rightarrow \frac{1}{b c}, \frac{1}{a c}, \frac{1}{a b}$ are in A.P.

$\Rightarrow \frac{a b+b c+c a}{b c}, \frac{a b+b c+c a}{a c}, \frac{a b+b c+c a}{a b}$ are in A.P.

(Multiplying each term by $a b+b c+c a$ )

$\Rightarrow \frac{a b+c a}{b c}+1, \frac{a b+b c}{a c}+1, \frac{b c+c a}{a b}+1$ are in A.P.

$\Rightarrow \frac{a b+c a}{b c}, \frac{a b+b c}{a c}, \frac{b c+c a}{a b}$ are in A.P.

(Subtracting 1 from each term)

$\Rightarrow \frac{a(b+c)}{b c}, \frac{b(a+c)}{a c}, \frac{c(b+a)}{a b}$ are in A.P.

Ex. 16. If $\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in A.P., prove that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are also in A.P.

Sol. $\frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in A.P.

$\Rightarrow \frac{b+c-a}{a}+2, \frac{c+a-b}{b}+2, \frac{a+b-c}{c}+2$ are in A.P.

(Adding 2 to each term of A.P.)

$\Rightarrow \frac{b+c+a}{a}, \frac{c+a+b}{b}, \frac{a+b+c}{c}$ are in A.P.

$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

(Dividing each term by $a+b+c$ )

Ex. 17. If $a, b, c$ are in A.P., show that $\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P.

(NDA/NA 2010)

Sol. Given $a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$

Now $\frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P., if

$\frac{1}{\sqrt{c}+\sqrt{a}}-\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{1}{\sqrt{a}+\sqrt{b}}-\frac{1}{\sqrt{c}+\sqrt{a}}$

$\Rightarrow \frac{2}{\sqrt{c}+\sqrt{a}}=\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{a}+\sqrt{b}} \Rightarrow \frac{2}{\sqrt{c}+\sqrt{a}}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{b}+\sqrt{c}}{(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})}$

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

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

$\Rightarrow 2 \sqrt{a b}+2 b+2 \sqrt{a c}+2 \sqrt{b c}=2 \sqrt{a c}+2 \sqrt{b c}+c+a+2 \sqrt{b a}$

$\Rightarrow 2 b=a+c$, which is true as $a, b, c$ are in A.P.

$\Rightarrow \frac{1}{\sqrt{b}+\sqrt{c}}, \frac{1}{\sqrt{c}+\sqrt{a}}, \frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P.

Ex. 18. What is the sum of all two-digit numbers which leave remainder 5 when they are divided by 7 ?

Sol. The two digit natural numbers which leave a remainder 5 , when divided by 7 are 12, 19, $26 \ldots ., 89,96$. $\therefore 12,19,26, \ldots ., 89,96$ is an A.P. whose first term $a=12$ and common difference $d=7$.

Let the last or $n$th term be $T_n$.

Then, $T_n=a+(n-1) d$, where $n$ is the number of terms in A.P.

$\Rightarrow 96=12+(n-1) 7$

$ \begin{aligned} & \Rightarrow 84=(n-1) 7 \Rightarrow n-1=12 \Rightarrow n=13 \\ & \therefore \text{ Required Sum }=\frac{n}{2}(a+l)=\frac{13}{2}(12+96)=\frac{13}{2} \times 108=13 \times 54=\mathbf{7 0 2} . \end{aligned} $

Ex. 19. What is the sum of the series $\mathbf{1}^{2}-2^{2}+3^{2}-4^{2}+\ldots \ldots .+99^{2}-100^{2}$.

(Orissa JEE 2006)

Sol. $1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+\ldots \ldots .+99^{2}-100^{2}$

$ \begin{aligned} & =(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+\ldots \ldots+(99-100)(99+100) \\ & =-(1+2)-(3+4)-(5+6) \ldots . .-(99+100) \\ & =-(1+2+3+4+5+6+\ldots .+99+100)=-(\frac{100}{2}(1+100)) \quad(\because S_n=\frac{n}{2}(\text{ first term }+ \text{ last term })) \\ & =-50 \times 101=-\mathbf{5 0 5 0} . \end{aligned} $

Ex. 20. If the first, second and last terms of an arithmetic series are $a, b$ and $c$ respectively, then what is the number of terms?

(MPPET 2009)

Sol. First term = a $\Rightarrow$ common difference $(d)=b-a$

Second term = b $\Rightarrow$ common difference $(d)=b-a$

Last term = c

Let the number of terms be $n$. Then,

$c=a+(n-1) \times d \Rightarrow c=a+(n-1) \times(b-a) \Rightarrow c-a=(n-1)(b-a)$

$\Rightarrow(n-1)=\frac{c-a}{b-a} \Rightarrow n=\frac{c-a}{b-a}+1=\frac{c-a+b-a}{b-a}=\frac{\boldsymbol{c}+\boldsymbol{b}-\mathbf{2} \boldsymbol{a}}{\boldsymbol{b}-\boldsymbol{a}}$.

Ex. 21. After inserting $x$ A.M’s. between 2 and 38, the sum of the resulting progression is 200 . What is the value of $x$ ?

(AMU 2001)

Sol. After inserting $x$ A.M’s between 2 and 38, we get an A.P. of $(x+2)$ terms with first term as 2 and last term as 38 .

Now, sum of $n$ terms of an A.P. $=\frac{n}{2}(a+l)$, where $a=$ first term, $l=$ last term

$ \begin{matrix} \therefore \text{ Here, } 200=\frac{(x+2)}{2}(2+38) & \Rightarrow 200=20(x+2) \\ \Rightarrow \quad 20 x+40=200 & \Rightarrow 20 x=160 \Rightarrow x=\mathbf{8} . \end{matrix} $

PRACTICE SHEET

Find the 26th term of the A.P: $10,6,2,-2,-6,-10, \ldots$. ?

(a) -86

(b) 96

(d) -106

The 2nd, 31 st and last term of an A.P. are $7 \frac{3}{4}, \frac{1}{2}$ and $-6 \frac{1}{2}$ respectively. The number of terms of the A.P. is

(a) 48

(b) 60

(d) 59

If $p$ times the $p$ th term of an A.P. is $q$ times the $q$ th term, then what is $(p+q)$ th term equal to?

(a) $p+q$

(b) $p q$

(d) 0

(NDA/NA2010)

The 59th term of an A.P. is 449 and the 449th term is 59 . Which term is equal to 0 ?

(a) $501 st$ term

(b) 502nd term

(d) 509th term

(NDA/NA2010)

Let $T_r$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$. If for some positive integers $m$ and $n, T_m=\frac{1}{n}$ and $T_n=\frac{1}{m}$, then $(a-d)$ equals

(a) $1 / m n$

(b) 1

(d) $\frac{1}{m}+\frac{1}{n}$

(AIEEE 2004, UPSEE 2007)

If $a, b, c$ be in Arithmetic Progression, then the value of $(a+2 b-c)(2 b+c-a)(a+2 b+c)$ is

(a) $3 a b c$

(b) $4 a b c$

(d) $16 a b c$

(WBJEE 2008)

Find the sum of 24 terms of the series $2 \frac{1}{2}, 3 \frac{1}{3}, 4 \frac{1}{6}, 5, \ldots \ldots$ ?

(a) 200

(b) 185

(d) 250

If the sum of the 12 th and 22 nd terms of an A.P. is 100 , then the sum of the first 33 terms of the A.P. is

(a) 1650

(b) 2340

(d) 3400

(Kerala PET 2008)

If $S_1=a_2+a_4+a_6+\ldots+$ upto 100 terms and $S_2=a_1+a_3$ $+a_5+\ldots+$ upto 100 terms of a certain A.P., then its common difference is (a) $S_1-S_2$ (b) $S_2-S_1$ (c) $\frac{S_1-S_2}{2}$ (d) None of these

(AMU 2010)

The sum of $n$ terms of an A.P. is $2 n+3 n^{2}$. Which term of this A.P. is equal to 299 ?

(a) 11 th

(b) 50 th

(d) 29th

If $S_n$ denotes the sum of first $n$ terms of an A.P. $a_1+a_2+a_3+\ldots$, such that $\frac{S_m}{S_n}=\frac{m^{2}}{n^{2}}$, then $\frac{a_m}{a_n}=$ ?

(a) $m-1: n-1$

(b) $m-n: m+n$

(d) $m+1: n+1$

(J&K CET 2013)

If the sum of the first ten terms of an A.P. is 4 times the sum of the first five terms, then the ratio of the first term to the common difference is:

(a) $1: 2$

(b) $2: 1$

(d) $4: 1$

(NDA/NA 2003)

What is the sum of numbers lying between 107 and 253 , which are divisible by 5 ?

(a) 5250

(b) 5210

(d) 5000

The ratio between the sum of $n$ terms of two A.P.’s is $(7 n+1):(4 n+27)$. The ratio of their 11 th terms is

(a) $125: 106$

(b) $148: 111$

(d) $127: 108$

Let $S_n$ denote the sum of first $n$ terms of an A.P. If $S _{2 n}=3 S_n$, then the ratio $\frac{S _{3 n}}{S_n}$ is equal to

(a) 4

(b) 6

(d) 10

(MAT 2002, Rajasthan PET 2006)

If the sum of $2 n$ terms of the A.P. $2,5,8,11, \ldots$ is equal to the sum of $n$ terms of A.P. $57,59,61,63, \ldots$, then $n$ is equal to

(a) 10

(b) 11

(d) 13

(IIT 2001)

If $S_1, S_2, S_3$ denote respectively the sum of first $n_1, n_2$ and $n_3$ terms of an A.P., then

$\frac{S_1}{n_1}(n_2-n_3)+\frac{S_2}{n_2}(n_3-n_1)+\frac{S_3}{n_3}(n_1-n_2)$ is equal to

(a) 0

(b) $n_1+n_2+n_3$

(d) $S_1 S_2 S_3$

(DCE 2007)

If $S_n=n P+\frac{n}{2}(n-1) Q$, where $S_n$ denotes the sum of first $n$ terms of an A.P., then the common difference of the A.P. is

(a) $Q$

(b) $P+Q$

(d) $2 P+3 Q$

(DCE 2001)

$n$ arithmetic means are inserted between 3 and 17. If the ratio of the last and the first arithmetic mean is $3: 1$, then $n$ is equal to

(a) 5

(b) 6

(d) 9

(DEC 2008)

If $a_1, a_2, a_3, \ldots, a_n$ are in A.P. and $a_1=0$, then the value of $(\frac{a_3}{a_2}+\frac{a_4}{a_3}+\ldots+\frac{a_n}{a _{n-1}})-a_2(\frac{1}{a_2}+\frac{1}{a_3}+\ldots+\frac{1}{a _{n-2}})$ is equal to

(a) $n+\frac{1}{n}$

(b) $n+\frac{1}{n-1}$

(d) $(n-2)+\frac{1}{(n-2)}$

(Kerala PET 2011)

Let $a_1, a_2, a_3, a_4$ be in A.P. If $a_1+a_4=10$ and $a_2 a_3=24$, then the least term of them is

(a) 1

(b) 2

(d) 4

(Kerala PET 2013)

An A.P. has a property that the sum of first ten terms is half the sum of next ten terms. If the second term is 13 , then the common difference is

(a) 3

(b) 2

(d) 4

(Kerala PET 2013)

The sides of a triangle are in A.P. and its area is $\frac{3}{5}$ th the area of an equilateral triangle of same perimeter. The sides of the triangle are in the ratio.

(a) $1: 2: \sqrt{7}$

(b) $2: 3: 5$

(d) $3: 5: 7$

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

(a) 0

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

(d) 3

(IIT 1990)

If $a_1, a_2, a_3, \ldots, a_n$ are in A.P with common difference $d \neq 0$, then the value of $\sin d(cosec a_1 cosec a_2+cosec a_2.$ $cosec a_3+\ldots .+cosec a _{n-1} cosec a_n$ ) will be

(a) $\sec a_1-\sec a_n$

(b) $\tan a_1-\tan a_n$

(d) $\cot a_1-\cot a_n$

(Rajasthan PET 2000)

ANSWERS


1. (c)	2. (d)	3. (d)	4. (c)	5. (c)
6. (d)	7. (c)	8. (a)	9. (d)	10. (b)
11. (c)	12. (a)	13. (c)	14. (b)	15. (b)
16. (b)	17. (a)	18. (a)	19. (b)	20. (d)
21. (b)	22. (b)	23. (d)	24. (d)	25. (d)

HINTS AND SOLUTIONS

1. $T_n=a+(n-1) d \Rightarrow T _{26}=a+25 d$

Here $a=10, d=-4$

$ \therefore T _{26}=10+25 \times(-4)=10-100=-\mathbf{9 0} \text{. } $

2. Let the first term, common difference and number of terms of the A.P. be $a, d$ and $n$ respectively. Then,

$ \begin{aligned} a+d & =7 \frac{3}{4} \\ a+30 d & =\frac{1}{2} \end{aligned} $

and $a+(n-1) d=-6 \frac{1}{2}$

Eqn (ii) - Eqn (i)

$ \begin{aligned} \Rightarrow & 29 d & =\frac{1}{2}-\frac{31}{4}=-\frac{29}{4} \\ \Rightarrow & & d=-\frac{1}{4} \end{aligned} $

Putting $d=-\frac{1}{4}$ in $(i)$, we get

$ \begin{aligned} a-\frac{1}{4}=7 \frac{3}{4} & \Rightarrow a=7 \frac{3}{4}+\frac{1}{4} \\ & \Rightarrow a=8 \end{aligned} $

$\therefore$ Putting the values of $a$ and $d$ in (iii), we have

$ \begin{matrix} 8+(n-1)(-\frac{1}{4}) & =-\frac{13}{2} \\ & \Rightarrow & 8+\frac{1}{4}-\frac{1}{4} n & =-\frac{13}{2} \\ \Rightarrow & -\frac{1}{4} n & =-\frac{13}{2}-\frac{33}{4} \\ \Rightarrow & -\frac{1}{4} n & =-\frac{59}{4} \\ \Rightarrow & n & =\mathbf{5 9 .} \end{matrix} $

3. Given, $p(a+(p-1) d)=q(a+(q-1) d)$, where $a$ and $d$ are the first term and common difference of the A.P.

$ \begin{aligned} & \Rightarrow \quad(p-q) a=(q^{2}-q-p^{2}+p) d \\ & \Rightarrow \quad-(q-p) a=(q-p)((q+p)-1) d \\ & \Rightarrow \quad-a=((q+p)-1) d \\ & \Rightarrow a+((p+q)-1) d=0 \\ & \Rightarrow \quad \boldsymbol{t} _{p+q}=\mathbf{0} \end{aligned} $

4. Let $a$ and $d$ be the first term and common difference of the given A.P.

Then,

$ \begin{aligned} a+58 d & =449 \\ a+448 d & =59 \end{aligned} $

Solving eqns ( $i$ ) and (ii) simultaneously, we get

$ a=507, d=-1 $

Now assume that the $n$th term is zero.

$ \therefore 0 =a+(n-1) d $

$\Rightarrow 0 =507+(n-1)(-1) $

$\Rightarrow 507 =n-1 $

$\Rightarrow \boldsymbol{n} =\mathbf{5 0 8 .} $

5. Given,

$ \begin{aligned} & T_m=\frac{1}{n} \Rightarrow a+(m-1) d=\frac{1}{n} \\ & T_n=\frac{1}{m} \Rightarrow a+(n-1) d=\frac{1}{m} \end{aligned} $

Eq. (ii) - Eq. (i) $\quad \Rightarrow(n-m) d=\frac{1}{m}-\frac{1}{n}$

$ \Rightarrow(n-m) d=\frac{n-m}{m n} \Rightarrow d=\frac{1}{m n} $

Putting $d=\frac{1}{m n}$ in $(i)$, we get $a=\frac{1}{m n}$

$\therefore \quad a-d=\frac{1}{m n}-\frac{1}{m n}=\mathbf{0}$.

6. $a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$

$ \begin{aligned} \therefore \quad & (a+2 b-c)(2 b+c-a)(a+2 b+c) \\ = & (a+a+c-c)(a+c+c-a)(2 b+2 b) \\ = & 2 a \cdot 2 c \cdot 4 b=16 a b c . \end{aligned} $

7. Here $a=\frac{5}{2}, d=\frac{10}{3}-\frac{5}{2}=\frac{20-15}{6}=\frac{5}{6}$

$ \begin{aligned} \therefore S _{24} & =\frac{24}{2}[2 \times \frac{5}{2}+(24-1) \frac{5}{6}] \\ & {[\because S_n=\frac{n}{2}[2 a+(n-1) d]] } \\ & =12[5+\frac{23 \times 5}{6}]=2[30+115]=2 \times 145=\mathbf{2 9 0 .} \end{aligned} $

8. Let the first term of the A.P. be $a$ and let the common difference be $d$.

Then, $\quad t _{12}+t _{22}=100 \Rightarrow(a+11 d)+(a+21 d)=100$

$ \Rightarrow \quad 2 a+32 d=100 $

Now sum of first 33 terms of the A.P

$ \begin{aligned} & =\frac{33}{2}(2 a+32 d)(\because S_n=\frac{n}{2}(2 a+(n-1) d). \\ & =\frac{33}{2} \times 100 \quad \quad(\text{ From }(i)) \\ & =1650 . \end{aligned} $

9. Given that,

$ \begin{aligned} & S_1=a_2+a_4+a_6+\ldots . .+ \text{ upto } 100 \text{ terms } \\ & S_2=a_1+a_3+a_5+\ldots . .+ \text{ upto } 100 \text{ terms } \end{aligned} $

Let $d$ be the common difference of the given A.P. Then,

$ d=a_2-a_1=a_3-a_2=a_4-a_3= $

Subtracting eqn (ii) from eqn (i), we have

$S_1-S_2=(a_1-a_2)+(a_4-a_3)+(a_6-a_5)+\ldots .+$ upto 100 terms

$=d+d+d+\ldots .+$ upto 100 terms

$=100 d$

$\Rightarrow d=\frac{S_1-S_2}{100}$.

10. $S_n=3 n^{2}+2 n$

$ \begin{aligned} \therefore \quad S _{n-1} & =3(n-1)^{2}+2(n-1) \\ & =3 n^{2}-6 n+3+2 n-2=3 n^{2}-4 n+1 \\ \therefore \quad T_n & =S_n-S _{n-1} \\ & =(3 n^{2}+2 n)-(3 n^{2}-4 n+1)=6 n-1 . \end{aligned} $

Given, $T_n=299$

$\Rightarrow \quad 6 n-1=299$

$ \Rightarrow \quad 6 n=300 \Rightarrow \boldsymbol{n}=\mathbf{5 0} \text{. } $

$ \frac{S_m}{S_n}=\frac{m / 2[2 a_1+(m-1) d]}{n / 2[2 a_1+(n-1) d]}=\frac{m^{2}}{n^{2}} $

$(d \to$ common difference of A.P.)

$\Rightarrow n[2 a_1+(m-1) d]=m[2 a_1+(n-1) d]$

$\Rightarrow \quad 2 a_1(n-m)=d(n-m) \Rightarrow d=2 a_1$

$\therefore \quad \frac{a_m}{a_n}=\frac{m \text{ th term }}{n \text{th term }}=\frac{a_1+(m-1) d}{a_1+(n-1) d}$

$=\frac{a_1+(m-1) \cdot 2 a_1}{a_1+(n-1) \cdot 2 a_1}=\frac{-a_1+2 a_1 m}{-a_1+2 a_1 n}$

$=\frac{a_1(2 m-1)}{a_1(2 n-1)}=\frac{\mathbf{2 m}-\mathbf{1}}{\mathbf{2 n}-\mathbf{1}}$.

12. Let $a$ and $d$ be the first term and common difference respectively of the A.P.

$ \begin{gathered} S _{10}=\frac{10}{2}[2 a+9 d] \\ S_5=\frac{5}{2}[2 a+4 d] \end{gathered} $

Given, $S _{10}=4 S_5$

$ \begin{aligned} \Rightarrow & & 5(2 a+9 d) & =4 \times \frac{5}{2}[2 a+4 d] \\ \Rightarrow & & 10 a+45 d & =20 a+40 d \\ \Rightarrow & & 5 d & =10 a \\ \Rightarrow & & \frac{a}{d} & =\frac{5}{10}=\frac{1}{2} \Rightarrow \boldsymbol{a}: \boldsymbol{d}=\mathbf{1}: \mathbf{2} . \end{aligned} $

13. The numbers between 107 and 253 divisible by 5 are

110,115,120, …., 245, 250.

This is an A.P with first term $(a)=110$ and common difference $(d)=5$. Let the last term be the $n$th term.

$ \therefore T_n =a+(n-1) d $

$\Rightarrow 110+(n-1) \times 5 =250 $

$ \Rightarrow 5 n =250-105=145 $

$ \Rightarrow n =29 .$

$ \therefore \text{ Required sum } =\frac{n}{2}(a+T_n)=\frac{29}{2}(110+250) $

$ =\frac{29}{2} \times 360=\mathbf{5 2 2 0 .}$

14. Let the two A.P.’s be $a, a+d, a+2 d, \ldots \ldots$. and $A, A+D$, $A+2 D, \ldots \ldots$.

Given, $\quad \frac{n / 2[2 a+(n-1) d]}{n / 2[2 A+(n-1) D]}=\frac{7 n+1}{4 n+27}$ $\Rightarrow \quad \frac{2 a+(n-1) d}{2 A+(n-1) D}=\frac{7 n+1}{4 n+27}$

Now, we have to find the ratio $\frac{t _{11}}{T _{11}}=\frac{a+10 d}{A+10 D}=\frac{2 a+20 d}{2 A+20 D}$

Putting $n=21$ in $(i)$, we get

$ \begin{aligned} & \frac{2 a+20 d}{2 A+20 D} & =\frac{7 \times 21+1}{4 \times 21+27}=\frac{147+1}{84+27}=\frac{148}{111} \\ \therefore \quad & t _{11}: T _{11} & =\mathbf{1 4 8 : 1 1 1 .} \end{aligned} $

15. Let $a$ be the first term and $d$ the common difference of the given A.P.

Then,

$ \begin{aligned} S_n & =\frac{n}{2}[2 a+(n-1) d] \\ S _{2 n} & =\frac{2 n}{2}[2 a+(2 n-1) d] \\ S _{3 n} & =\frac{3 n}{2}[2 a+(3 n-1) d] \end{aligned} $

Given, $\quad S _{2 n}=3 S_n$

$\Rightarrow \quad \frac{2 n}{2}[2 a+(2 n-1) d]=\frac{3 n}{2}[2 a+(n-1) d]$

$\Rightarrow 4 a+4 n d-2 d=6 a+3 n d-3 d$

$\Rightarrow \quad d+n d=2 a$

$\Rightarrow \quad a=\frac{d(n+1)}{2}$

Now,

$ \begin{aligned} \frac{S _{3 n}}{S_n} & =\frac{\frac{3 n}{2}[2 a+(3 n-1) d]}{\frac{n}{2}[2 a+(n-1) d]} \\ & =\frac{3[(n+1) d+(3 n-1) d]}{[(n+1) d+(n-1) d]} \\ & =\frac{3 n d+3 d+9 n d-3 d}{n d+d+n d-d} \\ & =\frac{12 n d}{2 n d}=\mathbf{6} . \end{aligned} $

$ =\frac{3[(n+1) d+(3 n-1) d]}{[(n+1) d+(n-1) d]} \quad(\text{ From }(i)) $

16. For the Ist A.P., $2,5,8,11, \ldots \ldots$. ,

First term $(a_1)=2$, common difference $(d_1)=3$

$\therefore$ The sum of this A.P. to $2 n$ terms

$ \begin{aligned} & =\frac{2 n}{2}[2 a_1+(2 n-1) d_1] \\ & =n[4+(2 n-1) 3] \\ & =4 n+6 n^{2}-3 n=6 n^{2}+n=n(6 n+1) \end{aligned} $

For the second A.P., 57, 59, 61, 63, …….

First term $(b_1)=57$, common difference $(d_2)=2$

$\therefore \quad$ The sum of this A.P. to $n$ term is

$ \begin{aligned} Sum_n & =\frac{n}{2}[2 b_1+(n-1) d_2] \\ & =\frac{n}{2}[114+(n-1) 2] \\ & =57 n+n^{2}-n=n^{2}+56 n=n(n+56) \end{aligned} $

$ \begin{aligned} \text{ Given, } & & S _{2 n} & =Sum_n \\ \Rightarrow & & n(6 n+1) & =n(n+56) \\ \Rightarrow & & 6 n+1 & =n+56 \Rightarrow 5 n=55 \Rightarrow n=11 . \end{aligned} $

17. Let the first term of the given A.P. be $a$ and the common difference be $d$.

Then, $S_1=\frac{n_1}{2}[2 a+(n_1-1) d]$

$ S_2=\frac{n_2}{2}[2 a+(n_2-1) d]$

$S_3=\frac{n_3}{2}[2 a+(n_3-1) d]$

$ \therefore \quad \frac{S_1}{n_1}(n_2-n_3)+\frac{S_2}{n_2}(n_3-n_1)+\frac{S_3}{n_3}(n_1-n_2) $

$ =\frac{\frac{n_1}{2}}{n_1}[2a + (n_1 - 1) d](n_2 - n_3)+\frac{\frac{n_2}{2}}{n_2}[2 a+(n_2 - 1) d $

$ (n_3-n_1)+\frac{\frac{n_3}{2}}{n_3}[2 a+(n_3 - d)](n_1 - n_2) $

$=\frac{1}{2}[2 a(n_2-n_3)+n_1(n_2-n_3)-d(n_2-n_3)]$

$\quad +\frac{1}{2}[2 a(n_3-n_1)+n_2(n_3-n_1)-d(n_3-n_1)]$

$\quad +\frac{1}{2}[2 a(n_1-n_2)+n_3(n_1-n_2)-d(n_1-n_2)]$

$ =\frac{1}{2}[2 a n_2-2 a n_3+2 a n_3-2 a n_1+2 a n_1-2 a n_2)+(n_1 n_2 - n_1 n_3 + n_2 n_3 - n_1 n_2 + n_3 n_1 - n_3 n_2) - d(n_2 - n_3 + n_3 - n_1 + n_1 - n_2)]$

$=\mathbf{0}$.

18. $n$th term of an A.P. is given by

$ T_n=\text{ Sum of } n \text{ terms }- \text{ Sum of }(n-1) \text{ terms } $

$ \begin{aligned} & =S_n-S _{n-1} \\ & =n P+\frac{n}{2}(n-1) Q-[(n-1) P+\frac{(n-1)}{2}(n-2) Q] \\ & =n P-n P+P+\frac{Q}{2}[n^{2}-n-n^{2}+3 n-2] \\ & =P+\frac{Q}{2}(2 n-2) \\ & =P+(n-1) Q . \end{aligned} $

$\therefore$ Common difference $(d)=T_n-T _{n-1}$

$ \begin{aligned} & =P+(n-1) Q-(P+(n-2) Q) \\ & =\boldsymbol{Q} . \end{aligned} $

19. Let $a_1, a_2, a_3, \ldots ., a_n$ be the $n$ arithmetic means between 3 and 17. Then,

$3, a_1, a_2, a_3, \ldots ., a_n, 17$ form an A.P. Let $d$ be the common difference of this A.P. Then,

$ a_1=3+d \quad \text{ and } \quad a_n=17-d $

Given, $\quad \frac{a_n}{a_1}=\frac{3}{1} \Rightarrow \frac{17-d}{3+d}=\frac{3}{1}$

$ \begin{matrix} \Rightarrow & & 17-d & =9+3 d \\ \Rightarrow & 4 d & =8 \quad \Rightarrow \quad d=2 . \end{matrix} $

Also 17 is the $(n+2)$ th term of the given A.P.

$ \begin{aligned} & \therefore \quad 17=3+(n+2-1) 2 \\ & \Rightarrow \quad 17=3+(n+1) 2 \\ & \Rightarrow \quad 14=(n+1) 2 \Rightarrow n+1=7 \Rightarrow n=\mathbf{6} \text{. } \end{aligned} $

20. Given, $a_1, a_2, a_3, \ldots ., a_n$ are in A.P and $a_1=0$.

Let $d$ be the common difference of the the A.P. So,

$ a_2=a_1+d=d$

$\therefore \quad a_3=a_1+2 d=2 d, a_4=3 d, a_5=4 d, \ldots, a_n=(n-1) d $

$ \therefore \quad(\frac{a_3}{a_2} + \frac{a_4}{a_3} + \ldots + \frac{a_n}{a _{n-1}}) - a_2(\frac{1}{a_2} + \frac{1}{a_3} + \ldots + \frac{1}{a _{n-2}}) $

$=[\frac{2 d}{d}+\frac{3 d}{2 d}+\ldots .+\frac{(n-1) d}{(n-2) d}] $

$=[2+\frac{3}{2}+\ldots .+\frac{n-1}{n-2}]-[1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n-3}] $

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

$-{1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n-3}}$

$=1+1+1 \ldots$ to $(n-2)$ term

$ \begin{matrix} -(1+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{(n-3)}+\frac{1}{(n-2)}) \\ -(1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n-3}) \end{matrix} $

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

21. Let $a_1=a-3 d, a_2=a-d, a_3=a+d, a_4=a+3 d$

Given, $\quad a_1+a_4=10 \Rightarrow a-3 d+a+3 d=10$

$ \Rightarrow \quad 2 a=10 \Rightarrow a=5 $

Also, $\quad a_2 a_3=24 \Rightarrow(a-d)(a+d)=24$

$ \Rightarrow \quad a^{2}-d^{2}=24 $

$ \Rightarrow \quad 25-d^{2}=24 \Rightarrow d^{2}=1 \quad \Rightarrow \quad \boldsymbol{d}=\mathbf{1} $

$\therefore \quad$ The least term is $a-3 d=5-3=\mathbf{2}$.

22. Sum of first 10 terms $=\frac{1}{2}$ (Sum of next 10 terms)

$ \begin{aligned} & \Rightarrow \quad S _{10}=\frac{1}{2}(S _{20}-S _{10}) \Rightarrow 3 S _{10}=S _{20} \\ & \Rightarrow \quad 3 \times \frac{10}{2}[2 a+9 d]=\frac{20}{2}[2 a+19 d] \\ & \Rightarrow \quad 3(2 a+9 d)=2(2 a+19 d) \\ & \Rightarrow \quad 6 a+27 d=4 a+38 d \\ & \Rightarrow \quad 2 a S_n=\frac{n}{2}[2 a+(n-1) d]] \\ & \Rightarrow \quad 2(13+d)=11 d(\because \text{ Second term }=13 \text{ and } 13-a=d) \\ & \Rightarrow \quad 26-2 d=11 d \\ & \Rightarrow \quad 13 d=26 \Rightarrow \quad d=\mathbf{2} . \end{aligned} $

23. Let the sides of the triangle be $a-d, a, a+d$.

Perimeter of the triangle $=a-d+a+a+d=3 a$

$\therefore \quad$ Each side of the equilateral triangle $=\frac{3 a}{3}=a$

$\therefore \quad$ Area of equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$.

Area of the given $\Delta=\sqrt{s(s-(a-d)(s-a)(s-(a+d))}$

where $s=\frac{3 a}{2}$

$ =\sqrt{\frac{3 a}{2}(\frac{3 a}{2}-a+d)(\frac{3 a}{2}-a)(\frac{3 a}{2}-a-d)} $

$ \begin{aligned} & =\sqrt{\frac{3 a}{2}(\frac{1}{2} a+d)(\frac{1}{2} a)(\frac{1}{2} a-d)} \\ & =\sqrt{\frac{3}{4} a^{2}(\frac{1}{4} a^{2}-d^{2})} \end{aligned} $

Given, $\sqrt{\frac{3}{4} a^{2}(\frac{1}{4} a^{2}-d^{2})}=\frac{3}{5} \times \frac{\sqrt{3}}{4} a^{2}$

$\Rightarrow \quad \frac{3}{16} a^{4}-\frac{3}{4} a^{2} d^{2}=(\frac{3 \sqrt{3}}{20} a^{2})^{2}$

$\Rightarrow \quad \frac{3}{16} a^{4}-\frac{3}{4} a^{2} d^{2}=\frac{27}{400} a^{4}$

$\Rightarrow \quad \frac{3}{16} a^{4}-\frac{27}{400} a^{4}=\frac{3}{4} a^{2} d^{2}$

$\Rightarrow \quad \frac{1}{16} a^{2}-\frac{9}{400} a^{2}=\frac{1}{4} d^{2}$

$ \Rightarrow \frac{25 a^{2}-9 a^{2}}{400} =\frac{1}{4} d^{2} $

$ \Rightarrow \frac{16 a^{2}}{400} =\frac{1}{4} d^{2}$

$\Rightarrow \frac{a^{2}}{d^{2}} =\frac{400}{64} \Rightarrow \frac{a}{d}=\frac{20}{8}=k \text{ (say) } $

$ \therefore (a-d): a:(a+d) =(20 k-8 k): 20 k:(20 k+8 k) $

$ 12 k: 20 k: 28 k = 3:5:7 $

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

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

$(\because a, b, c$ in A.P $\Rightarrow 2 b=a+c)$

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

$(\because a \log b=\log b^{a}.$ and $.\log a+\log b=\log a b)$

$ \Rightarrow \quad(2^{x}-5)^{2}=2^{x+1}-7 $

Let $\quad 2^{x}=y$. Then,

$ (y-5)^{2} =2 y-7 $

$\Rightarrow y^{2}-10 y+25 =2 y-7 $

$\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 $

$ 2^{x} =8 \Rightarrow \boldsymbol{x}=\mathbf{3}$

(because $2^{x}=4$ shall make the term $\log _3(2^{x}-5)$ negative which is not possible)

25. $d=a_2-a_1=a_3-a_2=a_4-a_3=\ldots . .=a_n-a _{n-1}$.

Now, sin d(cosec $a_1$ cosec $a_2$ + cosec $a_2$ cosec $a_3$ + $\ldots ..$ + cosec $a _{n-1}$ cosec $a_n$)

$= \frac{sin {~d}}{\sin a_1 \sin a_2} + \frac{\sin d}{\sin a_2 \sin a_3} + ….. + \frac{\sin d}{\sin a _{n-1} \sin a_n} $

$= \frac{\sin (a_2 - a_1)}{\sin a_1 \sin a_2} + \frac{\sin (a_3-a_2)}{\sin a_2 \sin a_3} + …. + \frac{\sin (a_n - a _{n-1})}{\sin a _{n-1} \sin a_n} $

$=\frac{\sin a_2 \cos a_1-\cos a_2 \sin a_1}{\sin a_1 \sin a_2} $

$\quad \quad + \frac{\sin a_3 \cos a_2 - \cos a_3 \sin a_2}{\sin a_2 \sin a_3}+ …..$

$\quad \quad + \frac{\sin a_n \cos a _{n-1} - \cos a_n \sin a _{n-1}}{\sin a _{n-1} \sin a_n} $

$ =\frac{\cos a_1}{\sin a_1}-\frac{\cos a_2}{\sin a_2}+\frac{\cos a_2}{\sin a_2}-\frac{\cos a_3}{\sin a_3}+ …..$

$ =\cot a_1 - a _{n-1} $

GEOMETRIC PROGRESSION (G.P.)

KEY FACTS

1. A Geometric Progression (G.P) is one in which the ratio of any term to its predecessor is always the same number. This ratio is called the common ratio. Thus, a sequence, $a_1, a_2, …., a_n$ is said to be in G.P.,

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

where $a_1, a_2, a_3 \ldots .$. are all non-zero terms of the G.P. and $\boldsymbol{r}$ is the common ratio.

Examples of G.P.

	Progression	Common Ratio
(i)	1, 3, 9, 27, 81, …	r = 3
(ii)	16/27, - 8/9, 4/3, - 2…	r = -3/2
(iii)	$x, x^2, x^3, x^4, …$	r = x

If $\boldsymbol{a}$ denotes the first term of a G.P. and $\boldsymbol{r}$, the common ratio, then a standard G.P. is $a, a r, a r^{2}, \ldots .$.

2. The $\boldsymbol{n}$ th term of a G.P.

For a standard G.P with first term $=a$, common ratio $=r$ and number of terms $=\boldsymbol{n}$, we know that:

First term	Second term	Third term	Fourth term
$t_1=a$	$t_2=a r$	$t_3=a r^{2}$	$t_4=a r^{3} \quad$ and so on.

So,

$ n \text{th term }=t_n=a r^{n-1} $

Ex. The sixth of the G.P. $-3,6,-12,24, \ldots$. is:

Here $ a =-3, r=\frac{6}{-3}=-2 $

$\therefore t_6 =a r^{6-1}=-3 \times(-2)^{5} $

$ =-3 \times-32 = 9 6$

3. Geometric Mean: When three real numbers form a geometric progression, then the middle one is called the geometric mean of the other two quantities.

Thus, if $\boldsymbol{G}$ is the geometric mean of two non-zero numbers $a$ and $b$, then $a, G, b$ forms a geometric progression, so that $\frac{G}{a}=\frac{b}{G}$

$ \therefore \quad G^{2}=a b \Rightarrow G= \pm \sqrt{a b} $

Maths Alert. $G=\sqrt{a b}$ if $a$ and $b$ are positive numbers.

$ G=-\sqrt{a b} \text{ if } a \text{ and } b \text{ are negative numbers. } $

Also, if there are $n$ positive integers $a_1, a_2, \ldots, a_n$, then their geometric mean is defined to be equal to

$ (a_1 \cdot a_2, a_3, \ldots . a_n)^{1 / n} . $

Ex. (i) G.M. of 4 and $36=\sqrt{4 \times 36}=\sqrt{144}=\mathbf{1 2}$.

(ii) G.M. of $\frac{1}{3}, 1,3,9,27,81 \text{ and } 243 =(\frac{1}{3} \times 1 \times 3 \times 9 \times 27 \times 81 \times 243)^{1 / 7} $

$ =(3^{2} \times 3^{3} \times 3^{4} \times 3^{5})^{1 / 7}=(3^{14})^{1 / 7}=3^{2}=\mathbf{9}$

4. Sum of $\boldsymbol{n}$ terms of a Geometric Progression:

Consider the geometric progression $a, a r, a r^{2}, \ldots a r^{n-1}$ of $n$ terms.

Then, the sum of $n$ terms $S_n=\frac{a(1-r^{n})}{1-r}(r \neq 1)$ or $\frac{a(r^{n}-1)}{(r-1)}(r \neq 1)$ where, $a=$ first term, $r=$ common ratio, $n=$ number of terms

Also, $S_n=\frac{\boldsymbol{a}-\boldsymbol{l} \boldsymbol{r}}{\mathbf{1}-\boldsymbol{r}}$ or $\frac{\boldsymbol{l} \boldsymbol{r}-\boldsymbol{a}}{\boldsymbol{r}-\mathbf{1}}$, where $l$ is the last term.

Ex. The sum of the G.P. $4+2+1+\frac{1}{2}+\frac{1}{4}+\ldots$ to 10 term is:

Here $\quad a=4, r=\frac{1}{2}, n=10$

$\therefore \quad S _{10}=\frac{4[1-(\frac{1}{2})^{10}]}{1-\frac{1}{2}}=\frac{4(1-\frac{1}{1024})}{\frac{1}{2}}=\frac{8 \times 1023}{1024}=\mathbf{8}$ (approx.)

5. Sum of an infinite G.P.

$ S _{\infty}=\frac{a}{1-r}, \text{ where } \begin{aligned} & a=\text{ first term } \\ & r=\text{ common ratio. } \end{aligned} $

Note: Sum of an infinite series exists only when $r$ is numerically less than 1 , i.e., $|r|<1$.

Ex. Sum of the G.P. $16,-8,4, \ldots$. to infinity is:

$ \begin{aligned} & \text{ Here } \quad a=16, r=-\frac{8}{16}=-\frac{1}{2} \\ & \therefore \quad S _{\infty}=\frac{a}{1-r}=\frac{16}{1-(\frac{1}{-2})}=\frac{16}{1+\frac{1}{2}}=\frac{16}{\frac{3}{2}}=\frac{\mathbf{3 2}}{\mathbf{3}} . \end{aligned} $

6. Properties of a G.P.

(a) If each term of a given geometric progression is multiplied or divided by the same number, then the resulting progression is also a G.P.

$ \begin{aligned} & \text{ If } a_1, a_2, a_3, a_4, \ldots \ldots . . . \text{ be a G.P. with common ratio } r \text{, then } \\ & a_1 c, a_2 c, a_3 c, a_4 c, \ldots \ldots . . . \text{ and } \frac{a_1}{c}, \frac{a_2}{c}, \frac{a_3}{c}, \frac{a_4}{c}, \ldots . \text{ are also G.P.s with common ratio } r . c \text{ is the constant number. } \end{aligned} $

(b) The reciprocals of the terms of a G.P. also form a G.P.

If $a_1, a_2, a_3, a_4, \ldots \ldots .$. be a G.P. with common ratio $r$, then $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4}, \ldots .$. is also a G.P. with common ratio $\frac{1}{r}$.

(c) If each term of a G.P. be raised to the same power, then the resulting progression is also a G.P.

If $a_1, a_2, a_3, a_4, \ldots \ldots . .$. be a G.P. with common ratio $r$, then $a_1{ }^{k}, a_2{ }^{k}, a_3{ }^{k}, a_4{ }^{k}, \ldots .$. is also a G.P. with common ratio $r$.

(d) If $a_1, a_2, a_3, a_4, \ldots \ldots . .$. and $b_1, b_2, b_3, b_4, \ldots$ be two G.P.s with common ratio $r_1$ and $r_2$ respectively, then the progressions $a_1 b_1, a_2 b_2, a_3 b_3, \ldots$ is a G.P. with common ratio $r_1 r_2$, and $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, \frac{a_4}{b_4}, \ldots$ is a G.P. with common ratio $\frac{r_1}{r_2}$.

(e) If $a_1, a_2, a_3, \ldots \ldots . .$. be a G.P. with common ratio $r$, such that each term of the progression is a positive number, then $\log a_1, \log a_2, \log a_3, \ldots$ is an A.P. with common difference $\log r$.

Conversely, if $\log a_1, \log a_2, \log a_3, \ldots .$. are terms of an A.P., then $a_1, a_2, a_3, \ldots$ are terms of a G.P.

(f) The Arithmetic mean between two positive numbers is always greater than equal to their Geometric Mean.

A.M. $\geq$ G.M.

(g) If $\boldsymbol{A}$ and $\boldsymbol{G}$ be the arithmetic and geometric mean respectively between two numbers, then the numbers are $A \pm \sqrt{A^{2}-G^{2}}$.

7. Useful note: When product is known, suppose the numbers in G.P. as under:

(i) If the odd number of terms are to be considered, suppose them as follows by taking $a$ as the mid-term and common ratio $r$.

$ \frac{a}{r}, a, a r ; \quad \frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2} $

(ii) If even number of terms are to be considered, suppose them as under by taking $\frac{a}{r}$, ar as the two mid-terms and $r^{2}$ as the common ratio.

$ \begin{aligned} & \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3} \quad[4 \text{ terms] } \\ & \frac{a}{r^{5}}, \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}, a r^{5} \quad[6 \text{ terms] } \end{aligned} $

SOLVED EXAMPLES

Ex.1 . Write the G.P. whose 4th term is 54 and 7 th term is 1458.

Sol.

$ \begin{aligned} & t_4=a r^{3}=54 \\ & t_7=a r^{6}=1458 \end{aligned} $

$\therefore$ Dividing (ii) by $(i)$, we get

$ \begin{aligned} & \frac{t_7}{t_4}=r^{3}=\frac{1458}{54}=27 \\ & \Rightarrow \quad r=3 \end{aligned} $

$\therefore \quad$ The G.P. is $2,2 \times 3,2 \times 3^{2}, \ldots . .$, i.e. $2,6,18,54, \ldots$.

Ex. 2. If the 10th term of a G.P. is 9 and the 4th term is 4, then what is its 7 th term?

Sol. Let the first term and common ratio of the given G.P. be $a$ and $r$ respectively.

$ \begin{aligned} T _{10} & =a r^{9}=9 \\ T_4 & =a r^{3}=4 \\ \frac{T _{10}}{T_4} & =\frac{a r^{9}}{a r^{3}}=\frac{9}{4} \\ r^{6} & =\frac{9}{4} \Rightarrow(r^{3})^{2}=\frac{9}{4} \Rightarrow r^{3}=\frac{3}{2} \end{aligned} $

From $(i), a r^{9}=9 \Rightarrow a(r^{3})^{3}=9$

$\therefore$ Substituting the value of $r^{3}$ from (iii), we get

$ a \cdot(\frac{3}{2})^{3} =9 \quad \Rightarrow \quad a=\frac{9 \times 8}{27}=\frac{8}{3} . $

$\therefore \quad T_7 =a r^{6} =\frac{8}{3} \times \frac{9}{4}=6 $

Ex. 3. What is the sum of 10 terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ldots$. ?

Sol. $\quad S_n=\frac{a(r^{n}-1)}{(r-1)}$

Here $\quad a=\sqrt{2}, r=\frac{\sqrt{6}}{\sqrt{2}}=\sqrt{3}, n=10$

$ \begin{aligned} \therefore \quad S _{10} & =\frac{\sqrt{2}((\sqrt{3})^{10}-1)}{\sqrt{3}-1}=\frac{\sqrt{2}(3^{5}-1)}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{\sqrt{2}(243-1)(\sqrt{3}+1)}{(3-1)} \\ & =\frac{\sqrt{2}(242)(\sqrt{3}+1)}{2}=\mathbf{1 2 1}(\sqrt{\mathbf{6}}+\sqrt{\mathbf{2}}) . \end{aligned} $

Ex. 4. What is the sum to infinity of the series $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots \ldots$ ?

(NDA/NA 2012)

Sol. Given series is an infinite G.P. with $a=1, r=-\frac{1}{2} / 1=-\frac{1}{2}$.

$ \begin{aligned} \therefore \quad S _{\infty} & =\frac{a}{1-r} \\ & =\frac{1}{1-(-\frac{1}{2})}=\frac{1}{\frac{3}{2}}=\frac{\mathbf{2}}{\mathbf{3}} . \end{aligned} $

Ex. 5. If $a, b, c$ are the three consecutive terms of an A.P. and $x, y, z$ are three consecutive terms of a G.P., then prove that $x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1$.

Sol. $a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$

$x, y, z$ are in G.P. $\Rightarrow y=x r, z=x r^{2}$, where $r$ is the common ratio

$ \begin{aligned} \therefore \quad x^{b-c} \cdot y^{c-a} \cdot z^{a-b} & =x^{b-c} \cdot(x r)^{c-a} \cdot(x r^{2})^{a-b} \\ & =x^{b-c} \cdot x^{c-a} \cdot r^{c-a} \cdot x^{a-b} \cdot r^{2 a-2 b}=x^{b-c+c-a+a-b} \cdot r^{c-a+2 a-2 b} \\ & =x^{0} \cdot r^{c+a-2 b}=x^{0} \cdot r^{2 b-2 b}=x^{0} \cdot r^{0}=1 \end{aligned} $

Ex. 6. Let two numbers have arithmetic mean 9 and geometric mean 4 . Then find the equation which has these two numbers as its roots.

(AIEEE 2004)

Sol. Let the two numbers be $a$ and $b$.

Then $\quad \frac{a+b}{2}=9 \Rightarrow a+b=18$

$ \sqrt{a b}=4 \Rightarrow a b=16 $

$\therefore$ Required eqn is $x^{2}-$ Sum of roots $x+$ Product of roots $=0$

$ \Rightarrow x^{2}-18 x+16=0 $

Ex. 7. Insert 3 geometric means between 16 and 256.

Sol. Let $G_1, G_2, G_3$ be the required means.

Then $16, G_1, G_2, G_3, 256$ form a G.P.

Let $r$ be the common ratio.

$ \begin{matrix} \Rightarrow & 256=5 \text{ th term }=a r^{4}=16 \times r^{4} \\ \Rightarrow & 16 r^{4}=256 \Rightarrow r r^{4}=16 \Rightarrow r=2 \\ \therefore & G_1=a r=16 \times 2=32 \end{matrix} $

$ \begin{aligned} & G_2=a r^{2}=16 \times 4=64 \\ & G_3=a r^{3}=16 \times 8=128 \end{aligned} $

Hence 32, 64 and 128 are the required G.M’s between 16 and 256.

Ex. 8. If one AM ’ $A$ ’ and two GM $p$ and $q$ are inserted between two given numbers, then find the value of $\frac{p^{2}}{q}+\frac{q^{2}}{p}$ in terms of $A$ ?

(VITEEE 2011)

Sol. Let $a$ and $b$ be the two given numbers.

Then, $\quad$ A.M. $=\frac{a+b}{2} \Rightarrow A=\frac{a+b}{2} \Rightarrow 2 A=a+b$

As, $p$ and $q$ are G.M’s between $a$ and $b, a, p, q, b$ forms a G.P.

$\therefore \quad \frac{p}{a}=\frac{q}{p}=\frac{b}{q} \Rightarrow \frac{p^{2}}{q}=a$ and $\frac{q^{2}}{p}=b$

Now, $\frac{p^{2}}{q}+\frac{q^{2}}{p}=a+b=\mathbf{2} \boldsymbol{A}$.

Ex. 9. What is the sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+$ to $n$ terms?

(KCET 2001)

Sol. The series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\ldots$. to $n$ terms can be written as

$ \begin{aligned} & (1-\frac{1}{3})+(1-\frac{1}{9})+(1-\frac{1}{27})+(1-\frac{1}{81})+\ldots . . \text{ to } n \text{ terms } \\ & =(1+1+1+1+\ldots . \text{ to } n \text{ terms })-(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\ldots . \text{ to } n \text{ terms }) \\ & =n-\frac{\frac{1}{3}(1-(\frac{1}{3})^{n})}{1-\frac{1}{3}} \\ & =n-\frac{\frac{1}{3}(1-\frac{1}{3^{n}})}{2 / 3}=n-\frac{1}{\mathbf{2}}(\mathbf{1}-\mathbf{3}^{-n}) . \end{aligned} $

Ex. 10. If $1+\sin x+\sin ^{2} x+\sin ^{3} x+\ldots . . \infty=4+2 \sqrt{3}, 0<x<\pi$, then find the value of $x . \quad$ (Kerala PET 2004)

Sol. Given, $1+\sin x+\sin ^{2} x+\sin ^{3} x+\ldots . \infty=4+2 \sqrt{3}$

$ \Rightarrow \frac{1}{1-\sin x} =4+2 \sqrt{3} $

$\Rightarrow \quad 1-\sin x =\frac{1}{4+2 \sqrt{3}} $

$\Rightarrow \quad \sin x =1-\frac{1}{4+2 \sqrt{3}}=\frac{4+2 \sqrt{3}-1}{4+2 \sqrt{3}}=\frac{3+2 \sqrt{3}}{4+2 \sqrt{3}} \times \frac{4-2 \sqrt{3}}{4-2 \sqrt{3}} $

$ =\frac{12+8 \sqrt{3}-6 \sqrt{3}-12}{16-12}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}$

$\Rightarrow \quad x =\frac{\pi}{\mathbf{3}} \text{ or } \frac{\mathbf{2 \pi}}{\mathbf{3}} \text{ as } 0<x<\pi$

$\because S _{\infty}=\frac{a}{1-r}$. Here $a=1, r=\sin x$ Also $|\sin x|<1$ as $|\sin x|=1$ will make the the sum of infinte G.P. infinite)

Ex. 11. If $a, b, c$ are in G.P. and $4 a, 5 b, 4 c$ are in A.P. such that $a+b+c=70$, then what is the value of the smallest of the numbers $a, b$ and $c$ ?

Sol. $a, b, c$ are in G.P. $\Rightarrow b^{2}=a c$

$4 a, 5 b, 4 c$ are in A.P. $\Rightarrow 2 \times 5 b=4 a+4 c \quad \Rightarrow \quad 10 b=4 a+4 c \Rightarrow 5 b=2 a+2 c$

Also, given $a+b+c=70$

$\Rightarrow \quad 2 a+2 b+2 c=140 \quad \Rightarrow \quad 5 b+2 b=140$

$ \Rightarrow \quad 7 b=140 \Rightarrow b=20 $

Now, from $(i), 400=a c$.

Also, from (ii), $a+20+c=70 \Rightarrow a+c=50$

$ \therefore \quad(a-c)^{2}=(a+c)^{2}-4 a c $

$ =2500-1600=900 $

$ \Rightarrow \quad a-c= \pm 30 $

$a+c =50$

$a-c = \pm 30$

$ \Rightarrow a=40, c=10 \text{ or } a=10, c=40 $

$\therefore$ The least value out of $a, b$ and $c$ is $\mathbf{1 0}$.

Ex. 12. Find the sum of $n$ terms of the series $1+(1+x)+(1+x+x^{2})+\ldots .$. ?

(AMU 2003)

Sol. $1+(1+x)+(1+x+x^{2})+$ $n$ terms

$ \begin{aligned} \Rightarrow \quad \text{ Required sum } & =\frac{1}{(1-x)}[(1-x)+(1-x)(1+x)+(1-x)(1+x+x^{2})+\ldots . n \text{ terms }] \\ & =\frac{1}{(1-x)}[(1-x)+(1-x^{2})+(1-x^{3})+\ldots . . n \text{ terms }] \\ & =\frac{1}{(1-x)}[(1+1+1+\ldots n \text{ terms })-(x+x^{2}+x^{3}+\ldots . n \text{ terms }]. \\ & =\frac{1}{(1-x)}[n-\frac{x(1-x^{n})}{(1-x)}] \quad(\because S_n=\frac{a(1-r^{n})}{(1-r)}, \text{ Here } a=x, r=x) \\ & =\frac{\boldsymbol{n}(\mathbf{1}-\boldsymbol{x})-\boldsymbol{x}(\mathbf{1}-\boldsymbol{x}^{n})}{(1-\boldsymbol{x})^{2}} . \end{aligned} $

Ex. 13. Five numbers are in A.P. with common difference $\neq \mathbf{0}$. If 1 st, 3 rd and 4 th terms are in G.P., then which term is always zero?

(WBJEE 2013)

Sol. Let the five numbers in A.P. be $a-2 d, a-d, a, a+d, a+2 d$.

Since 1st, 3rd and 4th terms are in G.P.

$ (t_3)^{2} =t_1 \cdot t_4 $

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

$\Rightarrow a^{2} =a^{2}-a d-2 d^{2} $

$\Rightarrow -a d =2 d^{2} $

$\Rightarrow a =-2 d $

$\Rightarrow a+2 d =0 \quad \Rightarrow \quad t_5=0 $

$\therefore$ 5th term is always zero.

Ex. 14. Find the sum to $\boldsymbol{n}$ terms of the series, $7+77+777+$

Sol.

$ \begin{aligned} S_n & =7+77+777+\ldots . . \text{ to } n \text{ terms } \\ & =7[1+11+111+\ldots . . \text{ to } n \text{ terms }] \\ & =\frac{7}{9}[9+99+999+\ldots . . \text{ to } n \text{ terms }] \end{aligned} $

$ \begin{aligned} & =\frac{7}{9}[(10-1)+(100-1)+(1000-1)+\ldots . . \text{ to } n \text{ terms }] \\ & =\frac{7}{9}[{10+10^{2}+10^{3}+\ldots . . \text{ to } n \text{ terms }}-{1+1+1+\ldots . . \text{ to } n \text{ terms }}] \\ & =\frac{7}{9}[\frac{10(10^{n}-1)}{10-1}-n]=\frac{\mathbf{7}(\mathbf{1 0}^{\boldsymbol{n}+\mathbf{1}}-\mathbf{1 0})}{\mathbf{8 1}}-\frac{\mathbf{7}}{\mathbf{9}} \boldsymbol{n} . \end{aligned} $

Ex. 15. What is the sum of $\boldsymbol{n}$ terms of the series $0.2+0.22+0.222+\ldots . . ?$

(WBJEE 2009)

Sol. $0.2+0.22+0.222+\ldots$. to $n$ terms

$ \begin{aligned} & =2[0.1+0.11+0.111+\ldots . . \text{ to } n \text{ terms }] \\ & =\frac{2}{9}[0.9+0.99+0.999+\ldots . . \text{ to } n \text{ terms }] \\ & =\frac{2}{9}[(1-0.1)+(1-0.01)+(1+0.001)+\ldots . . \text{ to } n \text{ terms }] \\ & =\frac{2}{9}[(1+1+1+\ldots . \text{ to } n \text{ terms })-(0.1+0.01+0.001+\ldots . \text{ to } n \text{ terms })] \\ & =\frac{2}{9}[n-\frac{0.1(1-(0.1)^{n}.}{(1-0.1)}]=\frac{2}{9}[n-\frac{\frac{1}{10}(1-\frac{1}{10^{n}})}{1-\frac{1}{10}}]=\frac{\mathbf{2}}{\mathbf{9}}[\boldsymbol{n}-\frac{\mathbf{1}}{\mathbf{9}}(\mathbf{1}-\frac{\mathbf{1}}{\mathbf{1 0}^{n}})] . \end{aligned} $

Ex. 16. Find the value of $0.2 \overline{34}$ regarding it as a geometric series.

Sol.

$ \begin{aligned} 0.2 \overline{34} & =0.2343434 \ldots \ldots . \\ & =0.2+0.034+0.00034+0.0000034+\ldots \ldots+\infty \\ & =\frac{2}{10}+\frac{34}{1000}+\frac{34}{100000}+\frac{34}{10000000}+\ldots \ldots+\infty \\ & =\frac{2}{10}+\frac{34}{10^{3}}[1+\frac{1}{10^{2}}+\frac{1}{10^{4}}+\ldots . . \infty] \\ & =\frac{2}{10}+\frac{34}{10^{3}} \times(\frac{1}{1-\frac{1}{10^{2}}}) \quad \quad \quad \quad(\because S _{\infty}=\frac{a}{1-r} . \text{ Here } a=1, r=\frac{1}{10^{2}}) \\ & =\frac{2}{10}+\frac{34}{1000} \times \frac{100}{99}=\frac{198+34}{990}=\frac{232}{990}=\frac{\mathbf{1 1 6}}{\mathbf{4 9 5}} . \end{aligned} $

Ex. 17. Six positive numbers are in G.P. such that their product is 1000 . If the fourth term is 1 , then find the last term.

(WBJEE 2013)

Sol. Let the six numbers in G.P. be $\frac{a}{r^{5}}, \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}, a r^{5}$.

Then, $\quad$ Product $=1000$

$ \begin{aligned} \Rightarrow & & \frac{a}{r^{5}} \times \frac{a}{r^{3}} \times \frac{a}{r} \times a r \times a r^{3} \times a r^{5} & =1000 \\ \Rightarrow & & a^{6}=1000 \Rightarrow & a=\sqrt{10} \end{aligned} $

Given, Fourth term $=t_4=a r=1$

$ \Rightarrow \quad \sqrt{10} \times r=1 \Rightarrow r=\frac{1}{\sqrt{10}} $

$\therefore \quad$ Last term $=a r^{5}=\sqrt{10} \times(\frac{1}{\sqrt{10}})^{5}=\frac{\mathbf{1}}{\mathbf{1 0 0}}$.

Ex. 18. If $a, b, c, d$ are in G.P., prove that $(a^{2}+b^{2}+c^{2}),(a b+b c+c d),(b^{2}+c^{2}+d^{2})$ are in G.P.

Sol. $a, b, c, d$ are in G.P $\Rightarrow b=a r, c=a r^{2}, d=a r^{3}$, where $\boldsymbol{r}=$ common ratio

$ \therefore \begin{aligned} (a b+b c+c d)^{2} & =(a \cdot a r+a r \cdot a r^{2}+a r^{2} \cdot a r^{3})^{2}=[a^{2} r(1+r^{2}+r^{4})]^{2} \\ (a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2}) & =(a^{2}+a^{2} r^{2}+a^{2} r^{4})(a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}) \\ & =a^{2}(1+r^{2}+r^{4}) \cdot a^{2} r^{2}(1+r^{2}+r^{4})=a^{4} r^{2}(1+r^{2}+r^{4})^{2} \\ & =[a^{2} r(1+r^{2}+r^{4})]^{2}=(a b+b c+c d)^{2} \end{aligned} $

$\therefore \quad(a^{2}+b^{2}+c^{2}),(a b+b c+c d),(b^{2}+c^{2}+d^{2})$ are in G.P.

Ex. 19. If $x, y, z$ are all positive and are the $p$ th, $q$ th and $r$ th terms of a geometric progression respectively, then find the value of the determinant $ \begin{vmatrix} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1\end{vmatrix} $.

(VITEEE 2009)

Sol. Let $a$ be the first term and $R$ the common ratio of the G.P.

Then,

$ \begin{aligned} T_p & =a R^{p-1}=x \\ T_q & =a R^{q-1}=y \\ T_r & =a R^{r-1}=z \\ \log x & =\log (a R^{p-1})=\log a+(p-1) \log R \\ \log y & =\log (a R^{q-1})=\log a+(q-1) \log R \\ \log z & =\log (a R^{r-1})=\log a+(r-1) \log R \end{aligned} $

$ \Rightarrow \quad \log x=\log (a R^{p-1})=\log a+(p-1) \log R $

$= \begin{vmatrix} \log a & p & 1 \\ \log a & q & 1 \\ \log a & r & 1\end{vmatrix} + \begin{vmatrix} (p-1) \log R & p & 1 \\ (q-1) \log R & q & 1 \\ (r-1) \log R & r & 1\end{vmatrix} $

$=\log a \begin{vmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1\end{vmatrix} +\log R \begin{vmatrix} (p-1) & p & 1 \\ (q-1) & q & 1 \\ (r-1) & r & 1\end{vmatrix} $

$=\log a \times 0+\log R \begin{vmatrix} p & p & 1 \\ q & q & 1 \\ r & r & 1\end{vmatrix} -\log R \begin{vmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1\end{vmatrix} $

$=0+\log R \times 0-\log R \times 0=0 \quad(\because$ The value of a determinants with two identical rows or columns is zero).

Ex. 20. If $A=1+r^{a}+r^{2 a}+r^{3 a}+\ldots . ., B=1+r^{b}+r^{2 b}+r^{3 b}+\ldots . . \infty$, then show that $\frac{a}{b}=\log (\frac{B-1}{B})(\frac{A-1}{A})$.

Sol.

$ A=1+r^{a}+r^{2 a}+r^{3 a}+\ldots \ldots \infty $

$ \begin{matrix} \Rightarrow \quad A & =\frac{1}{1-r^{a}} \\ \Rightarrow \quad 1-r^{a} & =\frac{1}{A} \Rightarrow r^{a}=1-\frac{1}{A}=\frac{A-1}{A} \Rightarrow a \log r=\log (\frac{A-1}{A}) \\ \Rightarrow \quad & a & =\frac{\log (\frac{A-1}{A})}{\log r} \end{matrix} $

$ \begin{aligned} & \text{ Now, } \quad B=1+r^{b}+r^{2 b}+r^{3 b}+\ldots \infty \\ & \Rightarrow \quad B=\frac{1}{1-r^{b}} \Rightarrow 1-r^{b}=\frac{1}{B} \Rightarrow r^{b}=1-\frac{1}{B}=\frac{B-1}{B} \Rightarrow \log r^{b}=\log (\frac{B-1}{B}) \\ & \Rightarrow \quad b \log r=\log (\frac{B-1}{B}) \Rightarrow b=\frac{\log (\frac{B-1}{B})}{\log r} \\ & \therefore \quad \frac{a}{b}=\frac{\log (\frac{A-1}{A})}{\log r} \times \frac{\log r}{\log (\frac{B-1}{B})} \\ & \Rightarrow \quad \frac{a}{b}=\frac{\log (\frac{A-1}{A})}{\log (\frac{B-1}{B})}=\log _{(\frac{B-1}{B})}(\frac{A-1}{A}) . \quad[\because \frac{\log a}{\log b}=\log _{b} a] . \end{aligned} $

Ex. 21 . $S_1, S_2, S_3, \ldots . ., S_n$ are the sums of $n$ infinite geometric progressions. The first term $S$ of these progressions are $1,2^{2}-1,2^{3}-1,2^{4}-1, \ldots ., 2^{n}-1$ and the common ratios are $\frac{1}{2}, \frac{1}{2^{2}}, \frac{1}{2^{3}}, \ldots \ldots . ., \frac{1}{2^{n}}$. Calculate the sum $S_1+S_2+S_3+\ldots . .+S_n$

Sol. Since $S _{\infty}=\frac{a}{1-r}$,

$ \begin{aligned} & S_1=\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2 \\ & S_2=\frac{2^{2}-1}{1-\frac{1}{2^{2}}}=\frac{2^{2}-1}{\frac{2^{2}-1}{2^{2}}}=2^{2} \\ & S_3=\frac{2^{3}-1}{1-\frac{1}{2^{3}}}=\frac{2^{3}-1}{\frac{2^{3}-1}{2^{3}}}=2^{3} \end{aligned} $

$ \begin{gathered} S_n=\frac{2^{n}-1}{1-\frac{1}{2^{n}}}=\frac{2^{n}-1}{\frac{2^{n}-1}{2^{n}}}=2^{n} \\ \therefore S_1+S_2+S_3+\ldots . .+S_n=2+2^{2}+2^{3}+\ldots . .+2^{n} \\ =\frac{2(2^{n}-1)}{2-1}=\mathbf{2}(2^{n}-\mathbf{1}) \quad(\because S_n=\frac{a(r^{n}-1)}{r-1}, \text{ Here } a=2, r=2) . \end{gathered} $

PRACTICE SHEET

Level-1

Which term of the G.P. $3,3 \sqrt{3}, 9, \ldots \ldots$ is 2187 ? (a) 13 (b) 14 (c) 15 (d) 16

(Kerala 2004)

If $n !, 3 \times(n !)$ and $(n+1)$ ! are in G.P, then the value of $n$ will be (a) 3 (b) 4 (c) 8 (d) 10
If $1, x, y, z, 16$ are in G.P., then what is the value of $x+y+z$ ? (a) 8 (b) 12 (c) 14 (d) 16

(NDA/NA 2008)

The geometric mean of $1,2,2^{2}, \ldots \ldots ., 2^{n}$ is (a) $2^{n / 2}$ (b) $n^{(n+1) / 2}$ (c) $2^{(n-1) / 2}$ (d) $2^{n(n+1) / 2}$

(MPPET 2009)

If the first term of a G.P. is 729 and its 7 th term is 64 , then the sum of the first seven terms is (a) 2187 (b) 2059 (c) 1458 (d) 2123

(Kerala 2013)

If the third term of a G.P. is 3 , then the product of its first 5 terms is: (a) 15 (b) 81 (c) 243 (d) Cannot be determined

( $J$ & K CET 2009)

In a G.P, $t_2+t_5=216$ and $t_4: t_6=1: 4$ and all the terms are integers, then its first term is (a) 16 (b) 14 (c) 12 (d) None of these
The sum of the first $n$ terms of the series

(AMU 2010) $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots .$. is (a) $2^{n}-n-1$ (b) $1-2^{-n}$ (c) $n+2^{-n}-1$ (d) $2^{n}-1$

(AMU 2003)

In a geometric progression consisting of positive terms, each term equals the sum of next two terms. Then, the common ratio of the progression equals (a) $\frac{\sqrt{5}}{2}$ (b) $\sqrt{5}$ (c) $\frac{\sqrt{5}-1}{2}$ (d) $\frac{\sqrt{5}+1}{2}$

(AIEEE 2007)

If in an infinite G.P, the first term equal to twice the sum of all the successive terms, then the common of this G.P. is (a) $\frac{1}{3}$ (b) $\frac{1}{4}$ (c) $\frac{2}{5}$ (d) $\frac{2}{3}$

Level-2

(Rajasthan 2002)

Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is 4 and the second term is $\frac{3}{4}$, then (a) $a=2, r=\frac{1}{2}$ (b) $a=2, r=\frac{3}{8}$ (c) $a=1, r=\frac{3}{4}$ (d) None of these

(AMU 2013)

If $x, 2 x+2,3 x+3$, are the first three terms of a G.P, then the fourth term is (a) $\frac{-27}{2}$ (b) $\frac{27}{2}$ (c) $\frac{-33}{2}$ (d) $\frac{33}{2}$

(NDA/NA 2009, Rajasthan PET 2005)

Six positive numbers are in G.P, such that their product is 1000. If the fourth term is 1 , then the last term is (a) 1000 (b) 100 (c) $\frac{1}{100}$ (d) $\frac{1}{1000}$

(WBJEE 2013) 14. If $a$ is the A.M. of $b$ and $c$ and the two geometric means are $G_1$ and $G_2$, then $G_1{ }^{3}+G_2{ }^{3}$ is equal to (a) $\frac{a b c}{2}$ (b) $a b c$ (c) $2 a b c$ (d) $\frac{3}{2} a b c$

(BCECE 2009, IIT 1997)

If $a, b, c$ are unequal numbers such that $a, b, c$ are in A.P. and $b-a, c-b, a$ are in G.P, then $a: b: c$ is (a) $1: 2: 3$ (b) $1: 2: 4$ (c) $1: 3: 4$ (d) $2: 3: 4$

(AMU 2005)

What is the sum of the 100 terms of the series $9+99+999+\ldots$ ? (a) $\frac{10}{9}(10^{100}-1)-100$ (b) $\frac{10}{9}(10^{99}-1)-100$ (c) $100(100^{10}-1)$ (d) $\frac{9}{100}(10^{100}-1)$

(NDA/NA2008)

The sum of the first 20 terms of the sequence

$0.7,0.77,0.777, \ldots$. is (a) $\frac{7}{81}(179-10^{-20})$ (b) $\frac{7}{9}(99-10^{-20})$ (c) $\frac{7}{81}(179+10^{-20})$ (d) $\frac{7}{9}(99+10^{-20})$

(IIT JEE 2013)

If $\frac{2}{3}=(x-\frac{1}{y})+(x^{2}-\frac{1}{y^{2}})+\ldots$. to $\infty$ and $x y=2$, then the value of $x$ and $y$ under the condition $x<1$ are (a) $x=\frac{1}{3}, y=6$ (b) $x=\frac{1}{2}, y=4$ (c) $x=\frac{1}{4}, y=8$ (d) $x=\frac{1}{6}, y=12$
The first two terms of a geometric progression add upto 12. The sum of the third and fourth terms is 48 . If the terms of the geometric progression are alternately positive and negative, then the first term is (a) -4 (b) -12 (c) 12 (d) 4

(AIEEE 2008)

If 64, 27 and 36 are the $P$ th, $Q$ th and $R$ th terms of a G.P, then $P+2 Q$ is equal to (a) $R$ (b) $2 R$ (c) $3 R$ (d) $4 R$

Level-3

(WBJEE 2012)

In a sequence of 21 terms, the first 11 terms are in A.P. with common difference 2 and the last 11 terms are in G.P with common ratio 2 . If the middle term of A.P. be equal to the middle term of G.P., then the middle term of the entire sequence is (a) $\frac{-10}{31}$ (b) $\frac{10}{31}$ (c) $\frac{32}{31}$ (d) $\frac{-31}{32}$
If $x=1+a+a^{2}+\ldots . . \infty$ and $y=1+b+b^{2}+\ldots \ldots \infty$, where $a$ and $b$ are proper fractions, then $1+a b+a^{2} b^{2}+\ldots \infty$ equals (a) $\frac{x+y}{x-y}$ (b) $\frac{x^{2}+y^{2}}{x-y}$ (c) $\frac{x^{2}-y^{2}}{x+y-1}(d) \frac{x y}{x+y-1}$

(Punjab CET 2007)

The first term of an infinite G.P. is $x$ and its sum is 5. Then, (a) $-10<x<0$ (b) $0<x<10$ (c) $0 \leq x \leq 10$ (d) $x>10$

(IIT 2004) 24. If $x>0$ and $\log _3 x+\log _3 \sqrt{x}+\log _3(\sqrt[4]{x})+\log _3(\sqrt[8]{x})$ $+\log _3 \sqrt[16]{x}+\ldots . .=4$, then $x$ equals (a) 1 (b) 9 (c) 27

(d) 81

(VITEEE 2007)

If $a, b, c, d$ are in G.P, then $(a+b+c+d)^{2}$ is equal to

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

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

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

(Kerala PET 2012)

ANSWERS

(a)
(c)
(c)
(a)
(b)
(c)
(c)
(c)
(c)
(a)
(c)
(a)
(c)
(c)
(a)
$($ a)
(c)
(b)
(b)
(c)
(a)
(d)
(b)
(b)
(a)

HINTS AND SOLUTIONS

Here first term $a=3$, common ratio $r=\frac{3 \sqrt{3}}{3}=\sqrt{3}$

Let the $n$th term be 2187 . Then,

$ \begin{matrix} T_n & =a r^{n-1}=2187 \\ & \Rightarrow \quad 3 \times(\sqrt{3})^{n-1} & =2187 \\ \Rightarrow \quad(\sqrt{3})^{n-1} & =729 \\ \Rightarrow \quad(\sqrt{3})^{n-1} & =3^{6}=(\sqrt{3})^{12} \\ \Rightarrow \quad n-1 & =12 \quad \Rightarrow \quad n=\mathbf{1 3 .} \end{matrix} $

Given, $n$ !, $3 \times n$ ! and $(n+1)$ ! are in G.P

$ \begin{aligned} & \Rightarrow \quad(3 \times n !)^{2}=n ! \times(n+1) ! \\ & (\because a, b, c \text{ in G.P. } \Rightarrow b^{2}=a c) \\ & \Rightarrow \quad 9 \times(n !)^{2}=n ! \times(n+1) ! \\ & \Rightarrow \quad 9 \times n !=(n+1) ! \\ & \Rightarrow \quad 9 n !=(n+1) . n ! \quad(\because n !=1.2 .3 \ldots \ldots . n) \\ & \Rightarrow \quad n+1=9 \Rightarrow n=\mathbf{8} . \end{aligned} $

$1, x, y, z, 16$ are in G.P.

$ \begin{aligned} & \therefore \quad a=1, n=5, T_5=a r^{4}=16 \\ & \Rightarrow \quad r^{4}=16 \Rightarrow r=2 \\ & \therefore \quad x=a r=2, y=a r^{2}=4, z=a r^{3}=8 \\ & \therefore \quad x+y+z=2+4+8=\mathbf{1 4} \text{. } \end{aligned} $

In the G.P. $1,2,2^{2}, \ldots . ., 2^{n}$, there are $(n+1)$ terms.

$\therefore \quad$ Geometric mean of this G.P. $=(1 \times 2 \times 2^{2} \times \ldots . \times 2^{n})^{\frac{1}{n+1}}$

$ \begin{aligned} & =(2^{1+2+3+\ldots . .+n})^{\frac{1}{n+1}} \\ & =(2^{\frac{n}{2}(n+1)})^{\frac{1}{n+1}}=\mathbf{2}^{\mathbf{n / 2}} . \end{aligned} $

Let the first term and common ratio of the G.P. be $a$ and $r$ respectively. Then, $T_1=a=729, T_7=a r^{6}=64$

$ \begin{aligned} & \therefore \frac{a r^{6}}{a}=\frac{64}{729} \Rightarrow r^{6}=\frac{64}{729}=\frac{2^{6}}{3^{6}} \Rightarrow r=\frac{2}{3} . \\ & \therefore \quad S_7=\frac{a(1-r^{n})}{1-r}=\frac{729(1-(\frac{2}{3})^{7})}{1-2 / 3} \\ & \quad=\frac{729(1-\frac{128}{2187})}{\frac{1}{3}}=2187(\frac{2187-128}{2187})=\mathbf{2 0 5 9 .} \end{aligned} $

Let $a$ and $r$ be the first term and common ratio respectively of the given G.P. Then,

$ \begin{aligned} T_3 & =a r^{2}=3 \\ \text{ Required product } & =a \cdot a r \cdot a r^{2} \cdot a r^{3} \cdot a r^{4} \\ & =a^{5} r^{15}=(a r^{3})^{5}=3^{5}=\mathbf{2 4 3 .} . \end{aligned} $

Let $a$ and $r$ be the first term and common ratio respectively of the given G.P. Then, $t_2=a r, t_4=a r^{3}, t_5=a r^{4}, t_6=a r^{5}$

Given, $t_2+t_5=a r+a r^{4}=216$

and

$ \frac{t_4}{t_6}=\frac{a r^{3}}{a r^{5}}=\frac{1}{4} \Rightarrow \frac{1}{r^{2}}=\frac{1}{4} \Rightarrow \frac{1}{r}=\frac{1}{2} $

Now putting $r=2$, in $(i)$ we get

$ 2 a+16 a=216 \quad \Rightarrow \quad 18 a=216 \quad \Rightarrow \quad a=12 $

Let $S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$. to $n$ terms

$ \begin{aligned} & =(1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})(1-\frac{1}{16})+\ldots \text{ to } n \text{ terms } \\ & =(1+1+1+1+\ldots \ldots \text{ to } n \text{ terms }) \\ & -{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} \ldots . . \text{ to } n \text{ terms }} \end{aligned} $

$ =n-\frac{\frac{1}{2}(1-(\frac{1}{2})^{n})}{1-\frac{1}{2}}=n-(1-2^{-n})=\boldsymbol{n}+\mathbf{2}^{-n}=\mathbf{1} \text{. } $

Let the G.P. be $a, a r, a r^{2}, \ldots .$.

As all the terms of the given G.P. are positive, $a>0, r>0$.

Given, $\quad a=a r+a r^{2}$

$\Rightarrow a r^{2}+a r-a=0$

$\Rightarrow \quad r^{2}+r-1=0$.

$\therefore \quad r=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}$

Let the given G.P. be $a, a r, a r^{2}, a r^{3}, \ldots ., \infty$.

As the given infinite G.P. has a finite sum, $|r|<1$

Also, given $\quad a=2(a r+a r^{2}+a r^{3}+\ldots .+\infty)$

$\Rightarrow \quad a=2(\frac{a r}{1-r})(\because S _{\infty}=\frac{a}{1-r}.$. Here $.a=a r, r=r)$

$\Rightarrow a-a r=2 a r$

$\Rightarrow 1-r=2 r \Rightarrow 3 r=1 \Rightarrow r=\frac{1}{\mathbf{3}}$.

Sum of an infinite G.P $=\frac{a}{1-r}$, where first term $=a$, common ratio $=r$ and $|r|<1$

Given, $\quad \frac{a}{1-r}=4$ and $a r=\frac{3}{4}$

$\Rightarrow \quad a=4-4 r$ and $\quad a=\frac{3}{4 r}$

$\Rightarrow \quad \frac{3}{4 r}=4-4 r \Rightarrow 3=16 r-16 r^{2}$

$\Rightarrow 16 r^{2}-16 r+3=0$

$\Rightarrow(4 r-3)(4 r-1)=0$

$\Rightarrow \quad 4 r=3$ or $4 r=1 \Rightarrow r=\frac{3}{4}$ or $\frac{1}{4}$

Now when $\quad r=\frac{3}{4}, a=\frac{3}{4 \times \frac{3}{4}}=1$

$ \begin{gathered} r=\frac{1}{4}, a=\frac{3}{4 \times \frac{1}{4}}=3 \\ \therefore \quad(a, r)=(\mathbf{1}, \frac{\mathbf{3}}{\mathbf{4}}) \text{ or }(3, \frac{\mathbf{1}}{\mathbf{4}}) . \end{gathered} $

$x, 2 x+2,3 x+3$, are in G.P.

$ \begin{aligned} \Rightarrow & & (2 x+2)^{2} & =x(3 x+3) \\ \Rightarrow & & 4 x^{2}+8 x+4 & =3 x^{2}+3 x \\ \Rightarrow & & x^{2}+5 x+4 & =0 \\ \Rightarrow & & (x+4)(x+1) & =0 \\ \Rightarrow & & x & =-1 \text{ or }-4 \end{aligned} $

Let the fourth term of the given G.P. be $a$. Then,

$ \begin{aligned} r & =\frac{a}{3 x+3}=\frac{2 x+2}{x} \\ \Rightarrow \quad a & =\frac{(2 x+2)(3 x+3)}{x} \end{aligned} $

When $x=-4$, Fourth term $a=\frac{(-8+2)(-24+3)}{-4}$

$ =\frac{-6 \times-21}{-4}=-\frac{27}{2} $

When

$ x=-1, a=\frac{(-2+2)(-3+3)}{-1}=\mathbf{0} \text{. } $

$\therefore \quad$ Fourth term $=-\frac{\mathbf{2 7}}{\mathbf{2}}$.

Let the six numbers in G.P. be $\frac{a}{r^{5}}, \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}, a r^{5}$

Given, $\frac{a}{r^{5}} \times \frac{a}{r^{3}} \times \frac{a}{r} \times a r \times a r^{3} \times a r^{5}=1000$

$\Rightarrow \quad a^{6}=1000 \Rightarrow a=\sqrt{10}$

Given, $\quad T_4=a r=1 \Rightarrow \sqrt{10} r=1 \Rightarrow r=\frac{1}{\sqrt{10}}$

$\therefore \quad$ Last term of G.P. $=a r^{5}=\sqrt{10} \times \frac{1}{(\sqrt{10})^{5}}=\frac{\mathbf{1}}{\mathbf{1 0 0}}$.

$a$ is the A.M of $b$ and $c \Rightarrow 2 a=b+c$

Given, $G_1$ and $G_2$ are G.Ms between $b$ and $c$

$\Rightarrow b, G_1, G_2, c$ are in G.P.

Let $r$ be the common ratio of the G.P.

$\Rightarrow \quad G_1=b r, G_2=b r^{2}, c=b r^{3}$

Now $\quad c=b r^{3} \Rightarrow r^{3}=\frac{c}{b} \Rightarrow r=(\frac{c}{b})^{\frac{1}{3}}$

$\therefore \quad G_1=b \times \frac{c^{1 / 3}}{b^{1 / 3}}=b^{2 / 3} c^{1 / 3}=(b^{2} c)^{1 / 3}$

$ G_2=b \times(\frac{c}{b})^{\frac{2}{3}}=\frac{b \times c^{2 / 3}}{b^{2 / 3}}=b^{1 / 3} c^{2 / 3}=(b c^{2})^{1 / 3} $

Now $G_1{ }^{3}+G_2{ }^{3}=((b^{2} c)^{1 / 3})^{3}+((b c^{2})^{1 / 3})^{3}$

$ =b^{2} c+b c^{2}=b c(b+c)=b c .2 a=\mathbf{2} \boldsymbol{a} \boldsymbol{c} c $

$a, b, c$ are in A.P. $\Rightarrow b-a=c-b$

$(b-a),(c-b), a$ are in G.P $\Rightarrow(c-b)^{2}=(b-a) a \quad \ldots(i i)$

$\therefore \quad$ From $(i)$ and (ii)

$ \begin{aligned} & \Rightarrow \quad(b-a)^{2}=(b-a) a \\ & \Rightarrow \quad(b-a)[(b-a)-a]=0 \\ & \Rightarrow \quad b-a=0 \quad \text{ or } \quad b-2 a=0 \\ & \Rightarrow \quad b=2 a \end{aligned} $

$(\because a$ and $b$ are distinct and $b-a \neq 0)$

$a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$

$ \Rightarrow \quad 4 a=a+c \Rightarrow c=3 a . $

$\therefore \quad a: b: c=a: 2 a: 3 a=\mathbf{1}: \mathbf{2}: \mathbf{3}$.

Let $S _{100}=9+99+999+\ldots \ldots$ upto 100 terms

$ =(10-1)+(100-1)+(1000-1) $

$+\ldots . .+$ upto 100 terms

$=(10+10^{2}+10^{3}+\ldots.$ upto 100 terms $)$

$-(1+1+1+\ldots$. upto 100 terms $)$

$=\frac{10(10^{100}-1)}{10-1}-100$

$(\because S_n=\frac{a(r^{n}-1)}{r-1}.$ when $.r>1)$

$=\frac{10}{9}(10^{100}-1)-100$.

Let $S _{20}=0.7+0.77+0.777+\ldots$. upto 20 terms

$ =7(0.1+0.11+0.111+\ldots . . \text{ upto } 20 \text{ terms }) $

$ \begin{aligned} & =\frac{7}{9}(0.9+0.99+0.999+\ldots . . \text{ upto } 20 \text{ terms }) \\ & =\frac{7}{9}[(1-0.1)+(1-0.01)+(1-0.001)+\text{ upto } \end{aligned} $

20 terms)

$=\frac{7}{9}[(1+1+1+\ldots .$. upto 20 terms $)-(0.1$

$+0.01+0.001+\ldots$. upto 20 terms) $]$

$=\frac{7}{9}[20-\frac{0.1{1-(0.1)^{2}}}{(1-0.1)}]$

$(\because S_n=\frac{a(1-r^{n})}{1-r}.$, when $.r<1)$

$=\frac{7}{9}[20-\frac{1}{9}(1-(\frac{1}{10})^{20})]=\frac{7}{9}[20-\frac{1}{9}+\frac{10}9^{-20}]$

$=\frac{7}{9}[\frac{179+10^{-20}}{9}]=\frac{\mathbf{7}}{\mathbf{8 1}}(\mathbf{1 7 9}+\mathbf{1 0}^{-\mathbf{2 0}})$.

Given $(x-\frac{1}{y})+(x^{2}-\frac{1}{y^{2}})+\ldots .$. to $\infty=\frac{2}{3}$

$\Rightarrow \quad(x+x^{2}+x^{3}+\ldots.$. to $.\infty)-(\frac{1}{y}+\frac{1}{y^{2}}+\ldots ..$. to $.\infty)=\frac{2}{3}$

$\Rightarrow \quad \frac{x}{1-x}-\frac{\frac{1}{y}}{1-\frac{1}{y}}=\frac{2}{3} \Rightarrow \frac{x}{1-x}-\frac{1}{y-1}=\frac{2}{3}$

$\Rightarrow \quad \frac{\frac{2}{y}}{1-\frac{2}{y}}-\frac{1}{y-1}=\frac{2}{3} \quad(\because x y=2 \Rightarrow x=2 / y)$

$\Rightarrow \quad \frac{2}{y-2}-\frac{1}{y-1}=\frac{2}{3} \Rightarrow \frac{2 y-2-y+2}{y^{2}-3 y+2}=\frac{2}{3}$

$\Rightarrow \quad 3 y=2(y^{2}-3 y+2)$

$\Rightarrow \quad 2 y^{2}-9 y+4=0 \quad \Rightarrow \quad 2 y^{2}-8 y-y+4=0$

$\Rightarrow \quad(2 y-1)(y-4)=0$.

$ \Rightarrow \quad y=\frac{1}{2} \text{ or } y=4 $

When

$ y=\frac{1}{2}, x=4 \text{ and } y=4, x=\frac{1}{2} $

$ \because \quad x<1 \quad \therefore \quad y=4, x=\frac{1}{2} \text{ is true. } $

Let $a$ and $r$ be the first term and common ratio respectively of the given G.P.

Then,

$a+a r =12$ … (i)

$a r^{2}+a r^{3} =48$ … (ii)

$ \Rightarrow \frac{ar^2(1+r)}{a(1+r)}=\frac{48}{12} \quad \quad$ (Dividing (ii) by (i))

$\Rightarrow \quad r^{2}=4 \Rightarrow r \pm 2 \Rightarrow r=-2$

as the terms of the G.P. are alternately positive and negative.

Now $a(1+r)=12 \Rightarrow a(1-2)=12 \Rightarrow \boldsymbol{a}=\mathbf{- 1 2}$.

Let $a$ and $r$ be the first term and common ratio respectively of the given G.P.

$ \begin{aligned} & P \text{ th term of G.P }=a r^{P-1}=64=2^{6} \\ & Q \text{ th term of G.P }=a r^{Q-1}=27=3^{3} \\ & R \text{ th term of G.P }=a r^{R-1}=36=2^{2} \cdot 3^{2} \end{aligned} $

Now from $(i)$,

$ 2=a^{1 / 6} r^{(P-1) / 6} $

From (ii),

$ 3=a^{1 / 3} r^{(Q-1) / 3} $

From (iii),

$ 2.3=a^{1 / 2} r^{(R-1) / 2} $

$\therefore$ From (iv), (v) and (vi) we have

$ \begin{matrix} \Rightarrow & a^{\frac{1}{6}} r^{\frac{P-1}{6}} \cdot a^{\frac{1}{3}} \cdot r^{\frac{Q-1}{3}} & =a^{\frac{1}{2}} r^{\frac{R-1}{2}} \\ \Rightarrow & a^{\frac{1}{6}+\frac{1}{3}} r^{\frac{P-1}{6}+\frac{Q-1}{3}} & =a^{\frac{1}{2}} r^{\frac{R-1}{2}} \\ \Rightarrow & a^{\frac{1}{2}} r^{\frac{P-1+2 Q-2}{6}} & =a^{\frac{1}{2}} r^{\frac{R-1}{2}} \\ \Rightarrow & \frac{P-1+2 Q-2}{6} & =\frac{R-1}{2} \\ \Rightarrow & P+2 Q-3 & =3 R-3 \\ \Rightarrow & P+2 Q & =\mathbf{3 R} . \end{matrix} $

Let the first term of the A.P. be $a$ and common difference $d$. Given $\quad d=2 \Rightarrow T _{11}$ of A.P. $=a+10 d=a+20$

Let the first term of the G.P. be $b$ and common ratio $r$.

Given $\quad r=2$.

Now, the middle term of A.P = middle term of G.P

$\Rightarrow T_6$ of A.P $=a+5 d=T_6$ of G.P $=b r^{5}$

$\Rightarrow \quad a+5 d=b r^{5}$

$\Rightarrow \quad a+10=32 b \quad(\because r=2)$

Also the last term of A.P. is the first term of G.P.

$ \therefore \quad b=T _{11} \text{ of A.P }=a+20 $

$\therefore \quad$ From (i) and (ii)

$ \begin{aligned} a+10 & =32 \cdot(a+20) \\ \Rightarrow \quad 31 a & =-630 \Rightarrow a=\frac{-630}{31} \end{aligned} $

$\therefore \quad$ Middle term of the entire sequence of 21 terms $=11$ th term $=a+10 d$

$=\frac{-630}{31}+20=\frac{-630+620}{31}=\frac{-\mathbf{1 0}}{\mathbf{3 1}}$.

Since $a$ and $b$ are proper fractions, $|a|<1,|b|<1$

$ \begin{aligned} & \therefore \quad x=1+a+a^{2}+\ldots . . \infty=\frac{1}{1-a}(\because S _{\infty}=\frac{a}{1-r}) \\ & \text{ and } \quad y=1+b+b^{2}+\ldots . . \infty=\frac{1}{1-b} \\ & \text{ Also, } 1+a b+a^{2} b^{2}+\ldots . . \infty=\frac{1}{1-a b} \end{aligned} $

Now $\quad x=\frac{1}{1-a} \Rightarrow x-x a=1$

$\Rightarrow \quad x a=x-1 \quad \Rightarrow \quad a=\frac{x-1}{x}$

$ y=\frac{1}{1-b} \quad \Rightarrow y-y b=1 $

$\Rightarrow \quad y b=y-1 \quad \Rightarrow \quad b=\frac{y-1}{y}$

$\therefore \quad$ Putting the values of a & b from (ii) and (iii) in (i), we get

$ \begin{aligned} \text{ Reqd. sum } & =\frac{1}{1-a b}=\frac{1}{1-(\frac{x-1}{x})(\frac{y-1}{y})} \\ & =\frac{x y}{x y-(x y-x-y+1)}=\frac{\boldsymbol{x y}}{\boldsymbol{x}+\boldsymbol{y}-\mathbf{1}} . \end{aligned} $

Let the common ratio of the given G.P be $r$. Then,

$ S _{\infty}=\frac{x}{1-r} \Rightarrow 5=\frac{x}{1-r} \Rightarrow 5-5 r=x $

$ \Rightarrow \quad r=\frac{5-x}{5}=1-\frac{x}{5} $

$\because$ Sum to infinity of the given series is a finite quantity, $|r|<1$.

$ \begin{matrix} \therefore & |1-\frac{x}{5}|<1 \Rightarrow-1<(1-\frac{x}{5})<1 \\ \Rightarrow & -1<(x / 5-1)<1 \end{matrix} $

(Multiplying the inequality by $(-1)$ )

$ \Rightarrow \quad 0<\frac{x}{5}<2 \Rightarrow 0<x<10 . $

$\log _3 x+\log _3 \sqrt{x}+\log _3 \sqrt[4]{x}++\log _3 \sqrt[8]{x}+\ldots \ldots \infty=4$

$ \begin{aligned} & \Rightarrow \quad \log _3 x+\log _3 x^{1 / 2}+\log _3 x^{1 / 4}+\log _3(x^{1 / 8})+\ldots \ldots \infty=4 \\ & \Rightarrow \quad \log _3(x \cdot x^{1 / 2} \cdot x^{1 / 4} \cdot x^{1 / 8} \ldots \ldots \infty)=4 \\ & \Rightarrow \quad \log _3(x^{1+1 / 2+1 / 4+1 / 8} \ldots \ldots \infty)=4 \\ & \Rightarrow \quad \log _3 x^{\frac{1}{1-\frac{1}{2}}}=4 \\ & (\because S _{\infty}=a / 1-r) \\ & \Rightarrow \log _3 x^{2}=4 \Rightarrow x^{2}=3^{4}=(3^{2})^{2} \Rightarrow x=9 \end{aligned} $

$a, b, c, d$ are in G.P.

$\Rightarrow b=a r, c=a r^{2}, d=a r^{3}$, where $r$ is the common ratio of the G.P.

$ \begin{aligned} \text{ LHS } & =(a+b+c+d)^{2}=(a+a r+a r^{2}+a r^{3})^{2} \\ & ={a(1+r+r^{2}+r^{3})}^{2}={a((1+r)+r^{2}(1+r))}^{2} \\ & ={a(1+r)(1+r^{2})}^{2}=a^{2}(1+r)^{2}(1+r^{2})^{2} \\ & =a^{2}(1+r)^{2}(r^{4}+2 r^{2}+1) \\ & =a^{2}(r^{4}(1+r)^{2}+2 r^{2}(1+r)^{2}+(1+r)^{2}) \\ & =a^{2} r^{4}(1+r)^{2}+2 a^{2} r^{2}(1+r)^{2}+a^{2}(1+r)^{2} \\ & =(a r^{2}+a r^{3})^{2}+2(a r+a r^{2})^{2}+(a+a r)^{2} \\ & =(\boldsymbol{c}+\boldsymbol{d})^{2}+\mathbf{2}(\boldsymbol{b}+\boldsymbol{c})^{2}+(\boldsymbol{a}+\boldsymbol{b})^{\mathbf{2} .} \end{aligned} $

HARMONIC PROGRESSION (H.P.)

KEY FACTS

A sequence of numbers is said to form a Harmonic Progression, when the reciprocals of the numbers form an Arithmetic progression.

Ex. (i) $1+2+3+4+\ldots$. is an A.P

$ \begin{aligned} & \Rightarrow 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots . . \text{ is an H.P. } \\ \text{ (ii) } & a+(a+d)+(a+2 d)+(a+3 d) \ldots . . \text{ is an A.P. } \\ & \Rightarrow \frac{1}{a}+\frac{1}{(a+d)}+\frac{1}{(a+2 d)}+\frac{1}{(a+3 d)}+\ldots . . \text{ is an H.P. } \end{aligned} $

2. $n$th term of an H.P.

The $n$th term of an H.P. is the reciprocal of the $n$th term of the A.P. formed by the reciprocals of the terms of the H.P.

If the given H.P. is $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \ldots .$. , then its $n$th term is $\frac{1}{\boldsymbol{a}+(\boldsymbol{n}-\mathbf{1}) \boldsymbol{d}}$.

Ex. Find the 7th term of the H.P. $\frac{2}{13}, \frac{1}{6}, \frac{2}{11}, \ldots . .$.

The reciprocals of the terms of the given H.P, i.e., $\frac{13}{2}, 6, \frac{11}{2}, \ldots$. form an A.P with first term $=\frac{13}{2}$ and common difference $=6-\frac{13}{2}=-\frac{1}{2}$.

$\therefore \quad 7$ th term of this A.P. $=a+6 d=\frac{13}{2}+(-\frac{1}{2}) \times 6=\frac{7}{2}$.

Hence 7 th of the given H.P. $=\frac{1}{7 / 2}=\frac{\mathbf{2}}{7}$.

3. Harmonic Mean

If $a, H, b$ are three quantities in H.P., then $\mathbf{H}$ is said to be the Harmonic Mean between $a$ and $b$. $a, H, b$ are in H.P

$ \begin{aligned} & \Rightarrow \frac{1}{a}, \frac{1}{H}, \frac{1}{b} \text{ are in A.P. } \\ & \Rightarrow \frac{1}{H}-\frac{1}{a}=\frac{1}{b}-\frac{1}{H} \Rightarrow \frac{2}{H}=\frac{1}{b}+\frac{1}{a} \Rightarrow \frac{2}{H}=\frac{a+b}{a b} \Rightarrow \boldsymbol{H}=\frac{\mathbf{2} \boldsymbol{a} \boldsymbol{b}}{\boldsymbol{a}+\boldsymbol{b}} . \end{aligned} $

Ex. The Harmonic Mean between 3 and -5 is:

$ H=\frac{2 \times 3 \times(-5)}{3+(-5)}=\frac{-30}{-2}=\mathbf{1 5} $

If $a_1, a_2, a_3, a_4, \ldots . a_n$ are $n$ non-zero numbers in H.P., then their

$ \text{ Harmonic Mean }=\frac{(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\ldots .+\frac{1}{a_n})}{n} $

Relations between the three series, i.e., A.P, G.P and H.P.

(a) Three numbers $a, b, c$ will be in A.P, G.P or in H.P. according as $\frac{a-b}{b-c}=\frac{a}{a}, \frac{a-b}{b-c}=\frac{a}{b}, \frac{a-b}{b-c}=\frac{a}{c}$ respectively.

(i) $\frac{\boldsymbol{a}-\boldsymbol{b}}{\boldsymbol{b}-\boldsymbol{c}}=\frac{\boldsymbol{a}}{\boldsymbol{a}} \Rightarrow a^{2}-b a=a b-a c \Rightarrow 2 a b=a^{2}+a c \Rightarrow \boldsymbol{2} \boldsymbol{b}=\boldsymbol{a}+\boldsymbol{c} \Rightarrow \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are in A.P.

(ii) $\frac{\boldsymbol{a}-\boldsymbol{b}}{\boldsymbol{b}-\boldsymbol{c}}=\frac{\boldsymbol{a}}{\boldsymbol{b}} \Rightarrow a b-b^{2}=a b-c a \Rightarrow \boldsymbol{b}^{\mathbf{2}}=\boldsymbol{a} \boldsymbol{c} \Rightarrow \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are in G.P.

(iii) $\frac{\boldsymbol{a}-\boldsymbol{b}}{\boldsymbol{b}-\boldsymbol{c}}=\frac{\boldsymbol{a}}{\boldsymbol{c}} \Rightarrow a c-b c=a b-a c \Rightarrow 2 a c=a b+b c \Rightarrow \boldsymbol{b}=\frac{\mathbf{2} \boldsymbol{a} \boldsymbol{c}}{\boldsymbol{a}+\boldsymbol{c}} \Rightarrow \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ are H.P.

(b) If $A, G, H$ are respectively the Arithmetic, the Geometric and the Harmonic means between any two unequal positive numbers, then

(i) $\boldsymbol{A}, \boldsymbol{G}, \boldsymbol{H}$ are in G.P

(ii) $\boldsymbol{A}>\boldsymbol{G}>\boldsymbol{H}$.

(i) Let the two positive numbers be $a$ and $b$. Then,

$ A=\frac{a+b}{2}, G=\sqrt{a b}, H=\frac{2 a b}{a+b} $

Then

$ A \times H=(\frac{a+b}{2}) \times \frac{2 a b}{a+b}=a b=G^{2} $

$ \Rightarrow \quad G=\sqrt{A H} \Rightarrow A, \boldsymbol{G}, \boldsymbol{H} \text{ are in G.P. } $

(ii) $A-G=\frac{a+b}{2}-\sqrt{a b}=\frac{a+b-2 \sqrt{a b}}{2}=\frac{(\sqrt{a}-\sqrt{b})^{2}}{2}>0 \Rightarrow A-G>0 \Rightarrow A>G$

Also $\quad G-H=\sqrt{a b}-\frac{2 a b}{a+b}=\frac{(a+b) \sqrt{a b}-2 a b}{(a+b)}=\frac{\sqrt{a b}(a+b-2 \sqrt{a b})}{a+b}=\frac{\sqrt{a b}}{a+b}(\sqrt{a}-\sqrt{b})^{2}>0$

$\Rightarrow \quad G-H>0 \Rightarrow G>H$

$\therefore \quad \boldsymbol{A}>\boldsymbol{G}>\boldsymbol{H}$.

SOLVED EXAMPLES

Ex. 1 . The third and seventh terms of an H.P are $\frac{7}{11}$ and $\frac{7}{31}$. Find the first term and the eighth term.

Sol. Third and seventh terms of an H.P are $\frac{7}{11}$ and $\frac{7}{31}$

$\Rightarrow$ Third and seventh terms of an A.P are $\frac{11}{7}$ and $\frac{31}{7}$

Let the first term of the A.P. be $a$ and common difference $d$.

Then,

$ \begin{aligned} & T_3=a+2 d=\frac{11}{7} \\ & T_7=a+6 d=\frac{31}{7} \end{aligned} $

(ii) $-(i) \Rightarrow 4 d=\frac{20}{7} \Rightarrow d=\frac{5}{7}$

$\therefore$ From $(i)$

$ \begin{aligned} a+\frac{10}{7} & =\frac{11}{7} \Rightarrow a=\frac{1}{7} \\ T_8 & =a+7 d=\frac{1}{7}+7 \times \frac{5}{7}=\frac{1}{7}+5=\frac{36}{7} \end{aligned} $

$\therefore \quad$ The first and eighth terms of the H.P. are respectively the reciprocals of the first and eighth terms of the A.P, i.e. 7 and $\frac{7}{36}$.

Ex. 2 . Insert three harmonic means between 5 and 6.

Sol. 3 harmonic means between 5 and 6

$\Rightarrow 3$ arithmetic means between $\frac{1}{5}$ and $\frac{1}{6}$

Let $A_1, A_2, A_3$ be the arithmetic means between $\frac{1}{5}$ and $\frac{1}{6}$.

Then, $\frac{1}{5}, A_1, A_2 A_3, \frac{1}{6}$ form an A.P,

where $t_1=a=\frac{1}{5}, t_5=a+4 d=\frac{1}{6}$

$ \begin{aligned} & \therefore \quad \frac{1}{5}+4 d=\frac{1}{6} \Rightarrow 4 d=\frac{1}{6}-\frac{1}{5}=-\frac{1}{30} \Rightarrow d=-\frac{1}{120} \\ & \therefore \quad A_1=a+d=\frac{1}{5}+(-\frac{1}{120})=\frac{23}{120} \\ & A_2=a+2 d=\frac{1}{5}+2 \times(-\frac{1}{120})=\frac{1}{5}-\frac{1}{60}=\frac{11}{60} \end{aligned} $

$ A_3=a+3 d=\frac{1}{5}+3 \times(-\frac{1}{120})=\frac{1}{5}-\frac{1}{40}=\frac{7}{40} $

$\therefore$ Required harmonic means are $\frac{\mathbf{1 2 0}}{\mathbf{2 3}}, \frac{\mathbf{6 0}}{\mathbf{1 1}}, \frac{\mathbf{4 0}}{7}$.

Ex. 3 . If the $m$ th term of an H.P. is $n$ and $n$th term is $m$, show that the $r$ th term is $\frac{m n}{r}$.

Sol. Let the corresponding A.P. be $a, a+d, a+2 d, \ldots$.

Since the $m$ th term and $n$th term of the H.P. are $n$ and $m$ respectively, then for the A.P.,

$ \begin{aligned} & m \text{ th term }=a+(m-1) d=\frac{1}{n} \\ & n \text{th term }=a+(n-1) d=\frac{1}{m} \end{aligned} $

(ii) $-(i) \Rightarrow(n-1) d-(m-1) d=\frac{1}{m}-\frac{1}{n}$

$ \Rightarrow \quad(n-m) d=\frac{n-m}{m n} \Rightarrow d=\frac{1}{m n} $

Putting in $(i), a+(m-1) \times \frac{1}{m n}=\frac{1}{n}$

$ \begin{matrix} \Rightarrow & a=\frac{1}{n}-\frac{1}{n}+\frac{1}{m n}=\frac{1}{m n} \\ \therefore & t_r=a+(r-1) d=\frac{1}{m n}+(r-1) \frac{1}{m n}=\frac{r}{m n} \end{matrix} $

$\therefore \quad r$ th term of H.P. $=\frac{\boldsymbol{m} \boldsymbol{n}}{\boldsymbol{r}}$.

Ex. 4 . If $H$ be the harmonic mean between $x$ and $y$, then prove that $\frac{H+x}{H-x}+\frac{H+y}{H-y}=2$.

Sol. $H$ being the H.M. between $x$ and $y$

$ \begin{matrix} \Rightarrow & H & =\frac{2 x y}{x+y} \Rightarrow \frac{H}{x}=\frac{2 y}{x+y} \text{ and } \frac{H}{y}=\frac{2 x}{x+y} \\ \Rightarrow & & \frac{H+x}{H-x}=\frac{2 y+x+y}{2 y-(x+y)} \text{ and } \frac{H+y}{H-y}=\frac{2 x+x+y}{2 x-(x+y)} \end{matrix} $

$ \begin{matrix} \Rightarrow & \frac{H+x}{H-x}=\frac{3 y+x}{y-x} \text{ and } \frac{H+y}{H-y}=\frac{3 x+y}{x-y} \\ \therefore & \frac{H+x}{H-x}+\frac{H+y}{H-y}=\frac{3 y+x}{y-x}+\frac{3 x+y}{x-y}=\frac{3 y+x-3 x-y}{y-x}=\frac{2(y-x)}{y-x}=\mathbf{2} . \end{matrix} $

(Using Componendo and Dividendo)

Ex. 5 . In an H.P., $p$ th term is $q r$ and $q$ th term is $p r$, show that $r$ th term is $p q$.

Sol. $T_p$ of H.P $=q r \Rightarrow T_p$ of A.P. $=\frac{1}{q r} \Rightarrow a+(p-1) d=\frac{1}{q r}$

Where $a$ and $d$ are the first term and common difference respectively of the A.P.

Also $\quad T_q$ of H.P $=p r \Rightarrow T_q$ of A.P $=\frac{1}{p r} \Rightarrow a+(q-1) d=\frac{1}{p r}$ $Eqn(i i)-Eqn(i) \Rightarrow(q-1) d-(p-1) d=\frac{1}{p r}-\frac{1}{q r} \Rightarrow(q-p) d=\frac{q-p}{p q r} \Rightarrow d=\frac{1}{p q r}$

$\Rightarrow \quad a+(p-1) \frac{1}{p q r}=\frac{1}{q r} \Rightarrow a=\frac{1}{q r}-(p-1) \frac{1}{p q r}=\frac{1}{q r}-\frac{1}{q r}+\frac{1}{p q r}=\frac{1}{p q r}$.

$\therefore \quad T_r$ of A.P $=a+(r-1) d=\frac{1}{p q r}+(r-1) \frac{1}{p q r}=\frac{1}{p q r}+\frac{1}{p q}-\frac{1}{p q r}=\frac{1}{p q}$

$\Rightarrow \quad T_r$ of H.P $=\boldsymbol{p q}$.

Ex. 6 . Let $a, b, c$ be in A.P and $|a|<1,|b|<1$ and $|c|<1$. If

$ \begin{aligned} & x=1+a+a^{2}+\ldots . . \text{ to } \infty \\ & y=1+b+b^{2}+\ldots . . \text{ to } \infty \\ & z=1+c+c^{2}+\ldots \ldots \text{ to } \infty, \text{ then show that } x, y, z \text{ are in H.P. } \end{aligned} $

Sol.

$ \begin{aligned} & x=1+a+a^{2}+\ldots . . \text{ to } \infty=\frac{1}{1-a} \\ & y=1+b+b^{2}+\ldots . . \text{ to } \infty=\frac{1}{1-b} \\ & z=1+c+c^{2}+\ldots . . \text{ to } \infty=\frac{1}{1-c} . \end{aligned} $

Now, $a, b, c$ are in A.P.

$\Rightarrow 1-a, 1-b, 1-c$ are in A.P.

$\Rightarrow \frac{1}{1-a}, \frac{1}{1-b}, \frac{1}{1-c}$ are in H.P.

$\Rightarrow x, y, z$ are in H.P.

Ex. 7 . If $a_1, a_2, a_3, \ldots . ., a_n$ are in H.P. then what will $(a_1 a_2+a_2 a_3+\ldots . .+a _{n-1} a_n)$ equal to?

(AMU 2003)

Sol. $a_1, a_2, a_3 \ldots ., a_n$ are in H.P.

$\Rightarrow \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3} \ldots . ., \frac{1}{a_n}$ are in A.P

If $d$ is the common difference of the A.P., then,

$ \begin{aligned} & \frac{1}{a_2}-\frac{1}{a_1}=d, \frac{1}{a_3}-\frac{1}{a_2}=d, \ldots \ldots, \frac{1}{a_n}-\frac{1}{a _{n-1}}=d \Rightarrow \frac{a_1-a_2}{a_1 a_2}=d, \frac{a_2-a_3}{a_2 a_3}=d, \ldots \ldots, \frac{a _{n-1}-a_n}{a _{n-1} a_n}=d \\ \Rightarrow & a_1-a_2=a_1 a_2 d, a_2-a_3=a_2 a_3 d, \ldots . ., a _{n-1}-a_n=a _{n-1} a_n d \\ \Rightarrow & (a_1-a_2+a_2-a_3+\ldots . .+a _{n-1}-a_n)=a_1 a_2 d+a_2 a_3 d+\ldots . .+a _{n-1} a_n d \\ \Rightarrow & (a_1-a_n)=(a_1 a_2+a_2 a_3+\ldots . .+a _{n-1} a_n) d \end{aligned} $

Also, $\frac{1}{a_1}, \frac{1}{a_2}, \ldots . . \frac{1}{a_n}$ is an A.P with common difference $d$

$\Rightarrow \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \Rightarrow(n-1) d=\frac{1}{a_n}-\frac{1}{a_1} \Rightarrow(n-1) d=\frac{a_1-a_n}{a_1 a_n}$

$\Rightarrow a_1-a_n=(n-1) d a_1 a_n$

$\therefore$ From (i) and (ii)

$ \begin{aligned} (n-1) d a_1 a_n & =(a_1 a_2+a_2 a_3+\ldots .+a _{n-1} a_n) d \\ \Rightarrow \quad(a_1 a_2+a_2 a_3+\ldots .+a _{n-1} a_n) & =(\boldsymbol{n}-\mathbf{1}) \boldsymbol{a} _{\mathbf{1}} \boldsymbol{a} _{\boldsymbol{n}} \end{aligned} $

Ex. 8 . If $\frac{1}{a^{2}}, \frac{1}{b^{2}}, \frac{1}{c^{2}}$ are in H.P, show that $b+c, c+a, a+b$ are in H.P.

Sol. We know by the property of H.P, if $x, y, z$ are H.P, $\frac{x-y}{y-z}=\frac{x}{z}$

$ \begin{aligned} & \therefore \frac{1}{a^{2}}, \frac{1}{b^{2}}, \frac{1}{c^{2}} \text{ are in H.P. } \Rightarrow \frac{\frac{1}{a^{2}}-\frac{1}{b^{2}}}{\frac{1}{b^{2}}-\frac{1}{c^{2}}}=\frac{\frac{1}{a^{2}}}{\frac{1}{c^{2}}} \Rightarrow \frac{\frac{b^{2}-a^{2}}{a^{2} b^{2}}}{\frac{c^{2}-b^{2}}{b^{2} c^{2}}}=\frac{c^{2}}{a^{2}} \Rightarrow \frac{c^{2}(b^{2}-a^{2})}{a^{2}(c^{2}-b^{2})}=\frac{c^{2}}{a^{2}} \\ & \Rightarrow \quad b^{2}-a^{2}=c^{2}-b^{2} \Rightarrow 2 b^{2}=c^{2}+a^{2} \end{aligned} $

Now, for $b+c, c+a, a+b$ to be in H.P

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

$\Rightarrow (b-a)(b+a) =(c+b)(c-b)$

Ex. 9 . The sum of the reciprocals of three numbers in H.P. is 15 and the product of the numbers is $\frac{1}{105}$. Find the numbers.

Sol. Let the three numbers in H.P. be $\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}$

$\therefore \quad(a-d)+a+(a+d)=15$

$\Rightarrow 3 a=15 \Rightarrow a=5$

Also $\quad \frac{1}{(a-d)} \cdot \frac{1}{a} \cdot \frac{1}{(a+d)}=\frac{1}{105} \Rightarrow \frac{1}{a(a^{2}-d^{2})}=\frac{1}{105} \Rightarrow \frac{1}{5(25-d^{2})}=\frac{1}{105}$

$\Rightarrow \quad 25-d^{2}=21$

$\Rightarrow \quad d^{2}=4 \Rightarrow d= \pm 2$

Hence the numbers are $\frac{1}{3}, \frac{1}{5}, \frac{1}{7}$ when $d=2$

or $\frac{1}{7}, \frac{1}{5}, \frac{1}{3}$ when $d=-2$.

Ex. 10 . Insert between $\frac{1}{6}$ and $\frac{1}{16}$ two numbers such that the first three may be in H.P. and the last three in G.P.

Sol. Let $a$ and $b$ be the two numbers between $\frac{1}{6}$ and $\frac{1}{16}$. Then $\frac{1}{6}, a, b$ are in H.P.

$ \begin{matrix} \Rightarrow \quad a=\frac{2 \times b \times \frac{1}{6}}{\frac{1}{6}+b} \\ \Rightarrow \quad \frac{a}{6}+a b=\frac{b}{3} \Rightarrow a+6 a b=2 b \end{matrix} $

$ (\because a, b, c \text{ in H.P } \Rightarrow b=\frac{2 a c}{a+c}) $

Also, $a, b, \frac{1}{16}$ are in G.P. $\Rightarrow \quad b^{2}=a \times \frac{1}{16} \Rightarrow b^{2}=\frac{a}{16} \Rightarrow a=16 b^{2}$

$\therefore$ From (i) and (ii)

$ \begin{matrix} \Rightarrow & 16 b^{2}+6 \times 16 b^{2} \times b=2 b \Rightarrow 16 b^{2}+96 b^{3}-2 b=0 \Rightarrow b(16 b+96 b^{2}-2)=0 \\ \Rightarrow & b(96 b^{2}+16 b-2)=0 \Rightarrow b(48 b^{2}+8 b-1)=0 \\ \Rightarrow & b(12 b-1)(4 b+1)=0 \\ \Rightarrow & b=\frac{1}{12} \text{ or }-\frac{1}{4} \text{ or } 0 \end{matrix} $

Since $-\frac{1}{4}$ and 0 do not lie between $\frac{1}{6}$ and $\frac{1}{16}$, therefore, $b=\frac{\mathbf{1}}{\mathbf{1 2}}$

Hence $\quad a=16 b^{2}=16 \times \frac{1}{144}=\frac{1}{9}$.

$\therefore$ The required numbers are $\frac{1}{4}$ and $\frac{1}{12}$.

Ex. 11 . If A.M. between two numbers is to their G.M. as $5: 4$ and the difference of their G.M. and H.M. is 16/5, find the numbers.

Sol. Let the two numbers be $a$ and $b$. Then,

$ \begin{aligned} \text{ A.M } & =\frac{a+b}{2} \\ \text{ G.M. } & =\sqrt{a b} \\ \text{ H.M. } & =\frac{2 a b}{a+b} \end{aligned} $

Given,

$ \text{ A.M : G.M }=5: 4 $

$\Rightarrow \quad \frac{(a+b) / 2}{\sqrt{a b}}=\frac{5}{4} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4}$

Also,

$ \text{ G.M. }- \text{ H.M. }=16 / 5 $

$\Rightarrow \quad \sqrt{a b}-\frac{2 a b}{a+b}=\frac{16}{5}$

From $(i), a+b=\frac{5 \times 2}{4} \sqrt{a b} \Rightarrow a+b=\frac{5}{2} \sqrt{a b}$.

Putting this value of $(a+b)$ in (ii), we have

$ \begin{aligned} & \sqrt{a b}-\frac{2 a b}{\sqrt{a b}} \times \frac{2}{5}=\frac{16}{5} \Rightarrow \sqrt{a b}-\frac{4}{5} \sqrt{a b}=\frac{16}{5} \\ & \Rightarrow \quad \frac{1}{5} \sqrt{a b}=\frac{16}{5} \Rightarrow \sqrt{a b}=16 \Rightarrow a b=256 \\ & \therefore \text{ From (iii), } \quad a+b=\frac{5}{2} \times 16=40 \text{. } \\ & \therefore \quad(a-b)^{2}=(a+b)^{2}-4 a b=40^{2}-4 \times 256=1600-1024=576 \\ & \Rightarrow \quad a-b= \pm 24 \end{aligned} $

Now solving $a+b=40$ and $a-b= \pm 24$, we get

$ a=32, b=8 \quad \text{ or } \quad a=8, b=32 \text{. } $

$\therefore \quad$ The numbers are 8 and 32 .

Ex. 12 . If $\frac{a-x}{p x}=\frac{a-y}{q y}=\frac{a-z}{r z}$ and $p, q, r$ are in A.P., show that $x, y, z$ are in H.P.

Sol. Let

$ \frac{a-x}{p x}=\frac{a-y}{q y}=\frac{a-z}{r z}=k $

Then,

$ \frac{a-x}{p x}=k \Rightarrow \frac{a-x}{k x}=p \Rightarrow p=\frac{1}{k}(\frac{a}{x}-1) $

Similarly,

$ q=\frac{1}{k}(\frac{a}{y}-1), r=\frac{1}{k}(\frac{a}{z}-1) $

Now, $p, q, r$ are in A.P.

$\Rightarrow \frac{1}{k}(\frac{a}{x}-1), \frac{1}{k}(\frac{a}{y}-1), \frac{1}{k}(\frac{a}{z}-1)$ are in A.P.

$\Rightarrow \frac{a}{x}-1, \frac{a}{y}-1, \frac{a}{z}-1$ are in A.P.

(Multiplying each term by $k$ )

$\Rightarrow \frac{a}{x}, \frac{a}{y}, \frac{a}{z}$ are in A.P.

(Adding 1 to each term)

$\Rightarrow \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.

(Dividing each term by $a$ )

$\Rightarrow x, y, z$ are in H.P.

Ex. 13 . Let $a, b$ be two positive real numbers. If $a, A_1, A_2, b$ are in arithmetic progression, $a, G_1, G_2, b$ are in geometric progression and $a, H_1, H_2, b$ are in harmonic progression, then show that

$ \frac{G_1 G_2}{H_1 H_2}=\frac{A_1+A_2}{H_1+H_2}=\frac{(2 a+b)(a+2 b)}{9 a b} $

Sol. $a, A_1, A_2, b$ are in A.P. $\Rightarrow A_1-a=b-A_2 \Rightarrow A_1+A_2=a+b$

$a, G_1, G_2, b$ are in G.P. $\Rightarrow \frac{G_1}{a}=\frac{b}{G_2} \Rightarrow G_1 G_2=a b$

$a, H_1, H_2, b$ are in H.P.

$\Rightarrow \frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{b}$ are in A.P.

$\Rightarrow \quad \frac{1}{H_1}-\frac{1}{a}=\frac{1}{b}-\frac{1}{H_2}$

$\Rightarrow \quad \frac{1}{H_1}+\frac{1}{H_2}=\frac{1}{a}+\frac{1}{b}$

$\Rightarrow \quad \frac{H_1+H_2}{H_1 H_2}=\frac{a+b}{a b}=\frac{A_1+A_2}{G_1 G_2}$

(From (i) and (ii))

$\Rightarrow \quad \frac{G_1 G_2}{H_1 H_2}=\frac{A_1+A_2}{H_1+H_2}$. Hence proved.

Now $\frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2}$ are in A.P

$(\because a, H_1, H_2.$, are in H.P. $)$

$\Rightarrow \quad \frac{1}{H_1}-\frac{1}{a}=\frac{1}{H_2}-\frac{1}{H_1}$ $\Rightarrow \quad \frac{2}{H_1}-\frac{1}{H_2}=\frac{1}{a}$

Also, $\frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{b}$ are in A.P

$(\because H_1, H_2, b.$ are in H.P.)

$\Rightarrow \quad \frac{1}{H_2}-\frac{1}{H_1}=\frac{1}{b}-\frac{1}{H_2} \Rightarrow \frac{2}{H_2}-\frac{1}{H_1}=\frac{1}{b}$

Eq. (iii) $+2 \times$ Eqn. (iv) $\Rightarrow(\frac{2}{H_1}-\frac{1}{H_2})+(\frac{4}{H_2}-\frac{2}{H_1})=\frac{1}{a}+\frac{2}{b}$

$\Rightarrow \frac{3}{H_2}=\frac{b+2 a}{a b} \Rightarrow H_2=\frac{3 a b}{2 a+b}$

Eq. (iv) $+2 \times$ Eq. (iii) $\Rightarrow(\frac{2}{H_2}-\frac{1}{H_1})+(\frac{4}{H_1}-\frac{2}{H_2})=\frac{1}{b}+\frac{2}{a}$

$\Rightarrow \frac{3}{H_1}=\frac{a+2 b}{a b} \Rightarrow H_1=\frac{3 a b}{a+2 b}$

$\therefore \quad \frac{G_1 G_2}{H_1 H_2}=\frac{a b}{\frac{3 a b}{(a+2 b)} \times \frac{3 a b}{(2 a+b)}}=\frac{(\boldsymbol{a}+\mathbf{2 b})(\mathbf{2} \boldsymbol{a}+\boldsymbol{b})}{\mathbf{9} \boldsymbol{a} \boldsymbol{b}}$ Hence proved.

Ex. 14 . If $x>1, y>1, z>1$ are in G.P., then show that $\frac{1}{1+\log x}, \frac{1}{1+\log y}, \frac{1}{1+\log z}$ are in H.P.

(IIT 1999, Manipal 2012)

Sol. $x, y, z$ are in G.P.

$ \begin{aligned} & \Rightarrow y^{2}=x z \Rightarrow 2 \log y=\log x+\log z \\ & \Rightarrow \log x, \log y, \log z \text{ are in A.P. } \\ & \Rightarrow 1+\log x, 1+\log y, 1+\log z \text{ are in A.P. } \\ & \Rightarrow \quad \frac{1}{1+\log x}, \frac{1}{1+\log y}, \frac{1}{1+\log z} \text{ are in H.P. } \end{aligned} $

(Adding 1 to each term)

Ex. 15 . Find the harmonic mean of the roots of the equation

$ (5+\sqrt{2}) x^{2}-(4+\sqrt{5}) x+(8+2 \sqrt{5})=0 . $

(IIT 1999, MPPET 2010, EAMCET 2013)

Sol. Let the roots of the equation be $\alpha$ and $\beta$. Then,

$ \begin{aligned} & \text{ Sum of roots }=\alpha+\beta=\frac{(4+\sqrt{5})}{5+\sqrt{2}} \\ & \text{ Product of roots }=\alpha \beta=\frac{8+2 \sqrt{5}}{5+\sqrt{2}} \\ & \text{ Now, Harmonic Mean of the roots, } \alpha \text{ and } \beta=\frac{2 \alpha \beta}{\alpha+\beta}=\frac{2(\frac{8+2 \sqrt{5}}{5+\sqrt{2}})}{\frac{(4+\sqrt{5})}{5+\sqrt{2}}}=\frac{4(4+\sqrt{5})}{(4+\sqrt{5)}}=4 \text{. } \end{aligned} $

PRACTICE SHEET

The $n$th term of the H.P. $4+4 \frac{2}{7}+4 \frac{8}{13}+5+\ldots .$. is (a) $\frac{7}{26-2 n}$ (b) $\frac{3}{3 n-2}$ (c) $\frac{60}{16-n}$ (d) None of these
The harmonic mean of $\frac{a}{1-a b}$ and $\frac{a}{1+a b}$ is (a) $a$ (b) $\frac{1}{1-a^{2} b^{2}}$ (c) $\frac{a}{1-a^{2} b^{2}}$ (d) $\frac{a}{\sqrt{1-a^{2} b^{2}}}$

(AMU 2002)

The 5 th and 11 th term of an H.P. are $\frac{1}{45}$ and $\frac{1}{69}$ respectively. Then, its 16 th term will be (a) $\frac{1}{77}$ (b) $\frac{1}{81}$ (c) $\frac{1}{85}$ (d) $\frac{1}{89}$

(Rajasthan PET 2003)

If the A.M. and G.M. of two numbers be 27 and 18 respectively, then what is their H.M. equal to? (a) 24 (b) 12 (c) 16 (d) 28
If the first two terms of an H.P. are $\frac{2}{5}$ and $\frac{12}{13}$ respectively, then the largest term is (a) 2nd term (b) 3rd term (c) 4th term (d) 6th term

(AMU 2007)

If $a$ and $b$ are two real numbers such that $0<a<b$ and the arithmetic mean between $a$ and $b$ is $\frac{4}{3}$ times the harmonic mean between them, then $b / a$ is equal to (a) $\frac{2}{\sqrt{3}}$ (b) $\frac{3}{2}$ (c) 3 (d) $\frac{8}{3}$
G.M. and H.M. of two numbers are 10 and 8 respectively. The numbers are (a) 1,100 (b) 2,50 (c) 4,25 (d) 5,20

(WBJEE 2010)

Five numbers are in H.P. The middle term is 1 and the ratio of the second and fourth terms is $2: 1$. Then, the sum of the first three terms is (a) $11 / 2$ (b) 5 (c) 2 (d) $14 / 2$

(WBJEE 2013)

If for two numbers the ratio of their H.M. to G.M. is $20: 29$, then the numbers are in the ratio (a) $3: 40$ (b) $4: 25$ (c) $1: 22$ (d) $2: 27$

(Type IIT) 10. If $2(y-a)$ is the H.M. between $y-x$ and $y-z$, then $(x-a)$, $(y-a),(z-a)$ are in (a) A.P. (b) G.P. (c) H.P. (d) None of these

(Rajasthan PET 2001)

If $a, b, c$ are in H.P., then $(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})$ is equal to (a) $\frac{4}{b^{2}}-\frac{3}{a c}$ (b) $\frac{3}{b^{2}}-\frac{4}{a c}$ (c) $\frac{4}{a c}-\frac{3}{b^{2}}$ (d) $\frac{3}{b^{2}}+\frac{4}{a c}$

(Kerala PET)

If the $l$ th, $m$ th and $n$th terms of an H.P. are in H.P. then $l, m, n$, are in (a) H.P. (b) A.P. (c) G.P. (d) None of these
If $\log (a+c), \log (c-a)$ and $\log (a-2 b+c)$ are in A.P, then (a) $a, b, c$ are in A.P. (b) $a^{2}, b^{2}, c^{2}$ are in A.P. (c) $a, b, c$ are in G.P. (d) $a, b, c$ are in H.P.

(DCE 2002)

If the sum of the roots of the equation $a x^{2}+b x+c=0$ is equal to sum of their squares, then $\frac{c}{a}, \frac{b}{a}, \frac{c}{b}$ are in (a) A.P. (b) G.P (c) H.P (d) None of these

(Punjab CET 2008)

If $a^{x}=b^{y}=c^{z}$ and $x, y, z$ are in H.P., then $a, b, c$ are in (a) G.P. (b) A.P. (c) H.P. (d) None of these
If $x^{2}+9 y^{2}+25 z^{2}=x y z(\frac{15}{x}+\frac{5}{y}+\frac{3}{z})$, then $x, y, z$ are in (a) G.P. (b) H.P. (c) A.P. (d) None of these

(DCE 2004)

If $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in H.P., then which of the following statement can be true? (a) $a, b,-\frac{c}{2}$ are in G.P. (b) $a=b=c$ (c) Any of these (d) None of these

(IIT 2003)

The H.M. of two numbers is 4. Their A.M. is $A$ and G.M. is $G$. If $2 A+G^{2}=27$, then $A$ is equal is (a) $\frac{9}{2}$ (b) 18 (c) $\frac{27}{2}$ (d) 27

(WBJEE 2011)

Let the positive numbers $a, b, c, d$ be in A.P. Then, $a b c, a b d$, $a c d, b c d$ are (a) NOT in A.P./G.P./H.P (b) In A.P. (c) In G.P. (d) In H.P.

(IIT 2001)

If $a^{x}=b^{y}=c^{z}=d^{u}$ and $a, b, c, d$ are in GP, then $x, y, z, u$ are in (a) A.P. (b) G.P. (c) H.P. (d) None of these

(VITEEE 2010)

If $a, a_1, a_2, \ldots ., a _{2 n}, b$ are in arithmetic progression and $a$, $g_1, g_2, \ldots . g _{2 n}, b$ are in geometric progression, and $h$ is the harmonic mean of $a$ and $b$, then $\frac{a_1+a _{2 n}}{g_1 g _{2 n}}+\frac{a_2+a _{2 n-1}}{g_2 g _{2 n-1}}+\ldots .$. $+\frac{a_n+a _{n+1}}{g_n g _{n+1}}$ is equal to (a) $2 n h$ (b) $n / h$ (c) $n h$ (d) $2 n / h$

(DCE 2009)

Find the value of $n$ for which $(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}})$ is the harmonic mean between $a$ and $b$. (a) -2 (b) $-\frac{3}{2}$ (c) -1 (d) $-\frac{1}{2}$
If $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$ and $b \neq(a+c)$, then $a, b, c$ are in (a) A.P (b) G.P (c) H.P (d) None of these
If $a, b, c$ are in H.P. and half the middle term be subtracted from the three terms, then the resulting series will be in (a) A.P. (b) G.P (c) H.P (d) None of these
If $p, q, r$ are in H.P and the $(p+1)$ th, $(q+1)$ th and $(r+1)$ th terms of an A.P. are in G.P., then the ratio of the first term to the common difference of the A.P. is equal to (a) $\frac{-q}{2}$ (b) $\frac{-p r}{q}$ (c) $\frac{-p r}{q^{2}}$ (d) $\frac{-2 q}{p r}$

ANSWERS

(c)
(a)
$(d)$
$(b)$
(a)
(c)
$($ d)
(a)
(b)
(b)
(c)
(b)
$($ d $)$
(c)
(a)
(b)
(c)
(a)
(d)
(c)
$(d)$
$(c)$
(c)
(b)
(a)

HINTS AND SOLUTIONS

$4,4 \frac{2}{7}, 4 \frac{8}{13}, 5$, . are in H.P.

$\Rightarrow \frac{1}{4}, \frac{7}{30}, \frac{13}{60}, \frac{1}{5}, \ldots .$. are in A.P.

Now, here $\quad a=\frac{1}{4}, d=\frac{7}{30}-\frac{1}{4}=-\frac{1}{60}$

$\therefore \quad T_n=a+(n-1) d=\frac{1}{4}+(n-1)(-\frac{1}{60})$

$=\frac{1}{4}-\frac{n}{60}+\frac{1}{60}=\frac{15+1}{60}-\frac{n}{60}=\frac{16-n}{60}$

$\therefore \quad n$th term of the H.P. $=\frac{\mathbf{6 0}}{\mathbf{1 6 - n}}$.

$\because \quad$ Harmonic mean of two quantities $a$ and $b$ is $\frac{2 a b}{a+b}$

$ \begin{aligned} \therefore \text{ Reqd. H.M. } & =\frac{2 \times(\frac{a}{1-a b}) \times(\frac{a}{1+a b})}{\frac{a}{1-a b}+\frac{a}{1+a b}} \\ & =\frac{\frac{2 a^{2}}{(1-a^{2} b^{2})}}{\frac{a+a^{2} b+a-a^{2} b}{(1-a^{2} b^{2})}}=\frac{2 a^{2}}{a}=a . \end{aligned} $

Given, 5 th term of an H.P. $=\frac{1}{45}$

$\Rightarrow 5$ th term of corresponding A.P. $=45$

11th term of an H.P. $=\frac{1}{69}$ $\Rightarrow \quad 11$ th term of corresponding A.P. $=69$

Let $a$ and $d$ be the first term and common difference of this A.P.

So $\quad a+4 d=45$

$ a+10 d=69 $

(ii) $-(i) \Rightarrow 6 d=24 \quad \Rightarrow \quad d=4$

From (i) $a+16=45 \Rightarrow a=29$.

$\therefore \quad T _{16}=a+15 d=29+15 \times 4=29+69=89$

$\therefore \quad$ 16th term of corresponding H.P. $=\frac{\mathbf{1}}{\mathbf{8 9}}$.

$\because \quad(G . M)^{2}=$ (A.M) . (H.M.)

$ \therefore \quad(18)^{2}=27 \times \text{ H.M. } \Rightarrow \quad \text{ H.M. }=\frac{18 \times 18}{27}=\mathbf{1 2} . $

First term of H.P $=\frac{2}{5} \Rightarrow$ First term of A.P $=\frac{5}{2}=\frac{30}{12}$

Second term of H.P $=\frac{12}{13} \Rightarrow$ Second term of A.P $=\frac{13}{12}$

$\therefore \quad$ The first and second terms of the corresponding A.P. are $\frac{30}{12}$ and $\frac{13}{12}$ respectively. The common difference $d=\frac{13}{12}-\frac{30}{12}=-\frac{17}{12}$

$\therefore \quad$ The corresponding A.P. is $\frac{30}{12}, \frac{13}{12},-\frac{4}{12},-\frac{21}{12}, \ldots \ldots$

$\therefore \quad$ The corresponding H.P is $\frac{12}{30}, \frac{12}{13},-\frac{12}{4},-\frac{12}{21}, \ldots$.

All the terms after the second term are negative, so the largest term is out of the first two terms, i.e., between $\frac{12}{30}$ and $\frac{12}{13}$. $\because \quad \frac{12}{13}>\frac{12}{30}$, so, $\frac{12}{13}$ i.e., second term is the largest term of the H.P.

Arithmetic mean between $a$ and $b$ is $A=\frac{a+b}{2}$

Harmonic mean betwen $a$ and $b$ is $H=\frac{2 a b}{a+b}$

Given, $\quad A=\frac{4}{3} H \Rightarrow \frac{a+b}{2}=\frac{4}{3}(\frac{2 a b}{a+b})$

$\Rightarrow 3(a+b)^{2}=16 a b \Rightarrow 3 b^{2}-10 a b+3 a^{2}=0$

$\Rightarrow 3(\frac{b}{a})^{2}-10(\frac{b}{a})+3=0$ (Dividing throughout by $a^{2}$ )

$.\Rightarrow(\frac{3 b}{a}-1))(\frac{b}{a}-3)=0$

$\Rightarrow \frac{b}{a}=\frac{1}{3}$ or $\frac{b}{a}=3 \quad[\because 0<a<b \Rightarrow \frac{b}{a} \neq \frac{1}{3}]$

$\Rightarrow \frac{b}{a}=3$.

Let the two required numbers be $a$ and $b$.

Then, $\quad$ G.M $=\sqrt{a b}=10 \Rightarrow a b=100$

H.M. $=\frac{2 a b}{a+b}=8 \Rightarrow a+b=\frac{a b}{4}=\frac{100}{4}=25$

$\therefore \quad(a-b)^{2}=(a+b)^{2}-4 a b$

$=625-400=225$

$\Rightarrow \quad a-b= \pm 15$

Solving ( $i$ ) and (ii), we get $a=20, b=5$ or $a=5, b=20$

$\therefore \quad$ The required numbers are $\mathbf{2 0}$ and $\mathbf{5}$.

Let the terms of the corresponding A.P. be

$a-2 d, a-d, a, a+d, a+2 d$.

Given, $\quad a=1$ and $\frac{a-d}{a+d}=\frac{1}{2}$

$\Rightarrow 2 a-2 d=a+d$

$\Rightarrow \quad a=3 d \Rightarrow d=\frac{1}{3} a=\frac{1}{3}$.

$(\because a=1)$

$\therefore \quad$ First three terms of A.P are $1-\frac{2}{3}, 1-\frac{1}{3}, 1$, i.e., $\frac{1}{3}, \frac{2}{3}, 1$

$\Rightarrow$ First three terms of corresponding H.P are $3, \frac{3}{2}, 1$

$\therefore \quad$ Required sum $=3+\frac{3}{2}+1=5 \frac{1}{2}=\frac{\mathbf{1 1}}{\mathbf{2}}$.

Let the two numbers be $a$ and $b$.

Given, $\frac{\text{ H.M }}{\text{ G.M }}=\frac{20}{29} \Rightarrow \frac{2 a b / a+b}{\sqrt{a b}}=\frac{20}{29}$

$\Rightarrow \quad \frac{2 \sqrt{a b}}{a+b}=\frac{20}{29} \quad \Rightarrow \quad 58 \sqrt{a b}=20(a+b)$

$\Rightarrow \quad 20 a-58 \sqrt{a b}+20 b=0$

$ \begin{aligned} & .\Rightarrow \quad 20 \frac{a}{b}-58 \sqrt{\frac{a}{b}}+20=0 \quad \text{ (Dividing all terms by } b) \\ & \Rightarrow \quad 20 x^{2}-58 x+20=0(\text{ where } x=\sqrt{\frac{a}{b}}) \\ & \Rightarrow \quad 20 x^{2}-50 x-8 x+20=0 \\ & \Rightarrow 10 x(2 x-5)-4(2 x-5)=0 \\ & \Rightarrow \quad(10 x-4)(2 x-5)=0 \\ & \Rightarrow \quad x=\frac{2}{5} \text{ or } \frac{5}{2} \\ & \Rightarrow \quad \sqrt{\frac{a}{b}}=\frac{2}{5} \text{ or } \frac{5}{2} \\ & \Rightarrow \quad \frac{a}{b}=\frac{4}{25} \text{ or } \frac{25}{4} \end{aligned} $

Thus from the given options, the two numbers are in the ratio $4: 25$.

Given, $2(y-a)$ is the H.M. between $(y-x)$ and $(y-z)$

$\Rightarrow \quad(y-x), 2(y-a),(y-z)$ are in H.P.

$\Rightarrow \frac{1}{y-x}, \frac{1}{2(y-a)}, \frac{1}{y-z}$ are in A.P.

$\Rightarrow \quad \frac{1}{2(y-a)}-\frac{1}{(y-x)}=\frac{1}{(y-z)}-\frac{1}{2(y-a)}$

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

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

$\Rightarrow \quad \frac{x+y-2 a}{x-y}=\frac{y+z-2 a}{y-z}$

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

Now applying componendo and dividendo, we have

$ \begin{aligned} & \quad(\text{ By comp. and div., } \frac{x}{y}=\frac{a}{b} \Rightarrow \frac{x+y}{x-y}=\frac{a+b}{a-b}) \\ \Rightarrow & \frac{2(x-a)}{2(y-a)}=\frac{2(y-a)}{2(z-a)} \\ \Rightarrow & (x-a)(z-a)=(y-a)^{2} \\ \Rightarrow & (x-a),(y-a),(z-a) \text{ are in G.P. } \end{aligned} $

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

$ \begin{aligned} \therefore(\frac{1}{a}. & .+\frac{1}{b}-\frac{1}{c})(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}) \\ & =(\frac{1}{a}+\frac{a+c}{2 a c}-\frac{1}{c})(\frac{a+c}{2 a c}+\frac{1}{c}-\frac{1}{a}) \\ & =(\frac{2 c+a+c-2 a}{2 a c})(\frac{a+c+2 a-2 c}{2 a c}) \\ & =(\frac{3 c-a}{2 a c})(\frac{3 a-c}{2 a c})=\frac{10 a c-3 a^{2}-3 c^{2}}{4 a^{2} c^{2}} \end{aligned} $

$ \begin{aligned} & =\frac{16 a c-(3 a^{2}+3 c^{2}+6 a c)}{4 a^{2} c^{2}}=\frac{16 a c-3(a+c)^{2}}{4 a^{2} c^{2}} \\ & =\frac{4}{a c}-3(\frac{a+c}{2 a c})^{2}=\frac{4}{a c}-\frac{3}{b^{2}} \end{aligned} $

Let the $l$ th, $m$ th and $n$th terms of the corresponding A.P. be

$ \begin{aligned} & T_l=a+(l-1) d \Rightarrow \quad \Rightarrow 2 T_m=T_l+T_n \\ & T_m=a+(m-1) d \quad \Rightarrow \quad 2[a+(m-1) d] \\ & =a+(l-1) d+a+(n-1) d \\ & .T_n=a+(n-1) d] \Rightarrow(2 m-2) d=(l+n-2) d \\ & \Rightarrow 2 m=l+n \\ & \Rightarrow l, m, n \text{ are in A.P } \end{aligned} $

$\log (a+c), \log (c-a)$ and $\log (a-2 b+c)$ are in A.P.

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

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

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

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

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

Let $\alpha, \beta$ be the roots of the equation $a x^{2}+b x+c=0$.

Then, $\alpha+\beta=-\frac{b}{a}, \quad \alpha \beta=\frac{c}{a}$

Also, given $\alpha+\beta=\alpha^{2}+\beta^{2} \Rightarrow \alpha+\beta=(\alpha+\beta)^{2}-2 \alpha \beta$

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

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

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

$\Rightarrow \frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in A.P.

$x, y, z$ are in H.P $\Rightarrow y=\frac{2 x z}{x+z}$

Given, $a^{x}=b^{y}=c^{z}=k$ (say)

Then $x \log a=\log k, y \log b=\log k, z \log c=\log k$

$\Rightarrow x=\frac{\log k}{\log a}, y=\frac{\log k}{\log b}, \quad z=\frac{\log k}{\log c}$

$\therefore \quad$ Putting these values of $x, y, z$ in (i), we get

$ \begin{aligned} & \frac{\log k}{\log b}=\frac{2(\frac{\log k}{\log a})(\frac{\log k}{\log c})}{\frac{\log k}{\log a}+\frac{\log k}{\log c}} \\ & \Rightarrow \quad \frac{\log k}{\log b}=\frac{2(\log k)^{2}}{\log k \log c+\log a \log k} \\ & \Rightarrow \quad \frac{(\log k)^{2}}{\log b}=\frac{2(\log k)^{2}}{\log c+\log a} \\ & \Rightarrow \quad 2 \log b=\log c+\log a \Rightarrow \log b^{2}=\log (a c) \\ & \Rightarrow \quad b^{2}=a c \Rightarrow a, b \text{ are in G.P. } \end{aligned} $

Given, $x^{2}+9 y^{2}+25 z^{2}=x y z(\frac{15}{x}+\frac{5}{y}+\frac{3}{z})$

$\Rightarrow x^{2}+9 y^{2}+25 z^{2}-15 y z-5 x z-3 x y=0$

$\Rightarrow 2 x^{2}+18 y^{2}+50 z^{2}-30 y z-10 x z-6 x y=0$

$\Rightarrow \quad(x-3 y)^{2}+(3 y-5 z)^{2}+(x-5 z)^{2}=0$

$\Rightarrow x=3 y, 3 y=5 z, 5 z=x \quad \Rightarrow \quad x=3 y=5 z$

$\Rightarrow \frac{x}{1}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}} \Rightarrow \frac{x}{15}=\frac{y}{5}=\frac{z}{3}$

$\Rightarrow x, y, z$ are in H.P. $\quad(\begin{matrix} \because y & =\frac{2 x z}{x+z} \\ y & =\frac{2 \times 15 \times 3}{15+3}=\mathbf{5}\end{matrix} )$

$a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$

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

From eqn (ii)

$ \begin{aligned} & b^{2}(a^{2}+c^{2})=2 a^{2} c^{2} \Rightarrow b^{2}{(a+c)^{2}-2 a c}=2 a^{2} c^{2} \\ \Rightarrow & b^{2}{4 b^{2}-2 a c}=2 a^{2} c^{2} \quad(\text{ From }(i) 2 b=a+c) \\ \Rightarrow & 2 b^{4}-b^{2} a c-a^{2} c^{2}=0 \\ \Rightarrow & (b^{2}-a c)(2 b^{2}+a c)=0 \\ \Rightarrow & b^{2}-a c=0 \quad \text{ or } \quad 2 b^{2}+a c=0 \\ \Rightarrow & (\frac{a+c}{2})^{2}-a c=0 \quad b^{2}=-\frac{a c}{2} \\ \Rightarrow & (a-c)^{2}=0 \quad a, b,-\frac{c}{2} \text{ are in G.P. } \\ \Rightarrow & 2 b=2 c \quad(\text{ From }(i)) \\ \Rightarrow & \boldsymbol{b}=c . \end{aligned} $

Let the two numbers be $a$ and $b$. Then,

$ H \cdot M=\frac{2 a b}{a+b},=4 \Rightarrow a+b=\frac{a b}{2} $

Also given, $\quad A=\frac{a+b}{2}, G=\sqrt{a b}$

$ \begin{aligned} & \therefore \quad 2 A+G^{2}=27 \Rightarrow 2 \times \frac{a+b}{2}+a b=27 \\ & \Rightarrow \quad a+b+a b=27 \Rightarrow(a+b)+2(a+b)=27 \\ & \Rightarrow \quad 3(a+b)=27 \Rightarrow a+b=9 \Rightarrow \frac{a+b}{2}=\frac{9}{2}=A \end{aligned} $

If $a, b, c, d$ are in A.P.

$\Rightarrow d, c, b, a$ are in A.P.

$\Rightarrow \frac{d}{a b c d}, \frac{c}{a b c d}, \frac{b}{a b c d}, \frac{a}{a b c d}$ are in A.P.

(Dividing all terms by $a b c d$ )

$\Rightarrow \frac{1}{a b c}, \frac{1}{a b d}, \frac{1}{a c d}, \frac{1}{b c d}$ are in A.P.

$\Rightarrow a b c, a b d, a c d, b c d$ are in H.P.

Given,

$ a^{x}=b^{y}=c^{z}=d^{u}=k \text{ (say) } $

Then, $\quad a=k^{1 / x}, b=k^{1 / y}, c=k^{1 / z}, d=k^{1 / u}$

Since $a, b, c, d$ are in G.P.

$ \frac{b}{a}=\frac{c}{b}=\frac{d}{c} $

$ \Rightarrow \quad \frac{k^{1 / y}}{k^{1 / x}}=\frac{k^{1 / z}}{k^{1 / y}}=\frac{k^{1 / u}}{k^{1 / z}} $

(Using (i)

$\Rightarrow \quad k^{1 / y-1 / x}=k^{1 / z-1 / y}=k^{1 / u-1 / z}$

$\Rightarrow \quad \frac{1}{y}-\frac{1}{x}=\frac{1}{z}-\frac{1}{y}=\frac{1}{u}-\frac{1}{z}$

$\Rightarrow \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{u}$ are in A.P $\Rightarrow x, y, z, u$ are in H.P.

Given, $a, a_1, a_2, a_3, \ldots \ldots, a _{2 n}, b$ are in A.P.

$\Rightarrow \quad a_1+a _{2 n}=a_2+a _{2 n-1}=\ldots \ldots=a+b$

$ (\because a_1-a=b-a _{2 n}, a_2-a=b-a _{2 n-1}) $

Also, $a, g_1, g_2, \ldots . g _{2 n}, b$ are in G.P.

$ \begin{aligned} & \Rightarrow \quad g_1 g _{2 n}=g_2 g _{2 n-1}=\ldots \ldots=a b \\ & \therefore \quad \frac{a_1+a _{2 n}}{g_1 g _{2 n}}+\frac{a_2+a _{2 n-1}}{g_2 g _{2 n-1}}+\ldots . .+\frac{a_n+a _{n+1}}{g_n g _{n+1}} \\ & =\frac{a+b}{a b}+\frac{a+b}{a b}+\ldots . .+\frac{a+b}{a b} \quad(n \text{ times) } \\ & =\frac{n(a+b)}{a b}=\frac{\mathbf{2 n}}{\boldsymbol{h}}, \text{ where, } \end{aligned} $

[Given, $h=\frac{2 a b}{a+b}$ is the harmonic mean between $a$ and $b$ ]

If $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ is the harmonic mean between $a$ and $b$, then

$ \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\frac{2 a b}{a+b} $

$\Rightarrow a^{n+1} \cdot a+a^{n+1} b+b^{n+1} a+b^{n+1} \cdot b=2 a b \cdot a^{n}+2 a b b^{n}$

$\Rightarrow a^{n+2}+a^{n+1} b+b^{n+1} a+b^{n+2}=2 a^{n+1} b+2 a b^{n+1}$

$\Rightarrow a^{n+2}+b^{n+2}=a^{n+1} b+a b^{n+1}$

$\Rightarrow a^{n+2}-b a^{n+1}=a b^{n+1}-b^{n+2}$

$\Rightarrow a^{n+1}(a-b)=b^{n+1}(a-b)$

$\Rightarrow a^{n+1}=b^{n+1} \Rightarrow(\frac{a}{b})^{n+1}=1 \Rightarrow(\frac{a}{b})^{n+1}=(\frac{a}{b})^{0}$

$\Rightarrow n+1=0 \Rightarrow n=-\mathbf{1}$.

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

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

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

$\Rightarrow b^{2}(a+c)-b(a+c)^{2}+2 a c(a+c)-2 a b c=0$

$ \Rightarrow b(a+c)(b-(a+c))+2 a c(a+c-b)=0 $

$ \Rightarrow b(a+c)(b-(a+c))-2 a c(b-(a+c))=0$

$ \Rightarrow [b(a+c) - 2ac] (b - (a+c)) = 0$

$ \Rightarrow b(a+c)-2 a c=0(\because b \neq a+c \Rightarrow b-(a+c) \neq 0)$

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

Let the three numbers in H.P. be $a, b, c$.

$ \therefore \quad b=\frac{2 a c}{a+c} $

The new numbers after subtracting half the middle term from each term are:

$ \begin{aligned} & a-\frac{b}{2}=a-\frac{a c}{a+c}=\frac{a^{2}}{a+c} \\ & b-\frac{b}{2}=\frac{b}{2}=\frac{a c}{a+c} \\ & c-\frac{b}{2}=c-\frac{a c}{a+c}=\frac{c^{2}}{a+c} \end{aligned} $

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

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

$\Rightarrow(a-\frac{b}{2}), \frac{b}{2},(c-\frac{b}{2})$ are in G.P.

$p, q, r$ are in H.P. $\Rightarrow q=\frac{2 p r}{p+r}$

Let $a$ and $d$ be the first term and common difference respectively of the A.P.

$ \therefore \quad \begin{aligned} T _{p+1} & =a+p d \\ T _{q+1} & =a+q d \\ T _{r+1} & =a+r d \end{aligned} $

Given, $T _{p+1}, T _{q+1}, T _{r+1}$ are in G.P.

$ \begin{aligned} \Rightarrow & (T _{q+1})^{2}=T _{p+1} \cdot T _{r+1} \\ & (a+q d)^{2}=(a+p d)(a+r d) \\ \Rightarrow & a^{2}+2 a q d+q^{2} d^{2}=a^{2}+a d(p+r)+p r d^{2} \\ \Rightarrow & a d(2 q-p-r)=d^{2}(p r-q^{2}) \\ \Rightarrow & \frac{a}{d}=\frac{p r-q^{2}}{2 q-p-r}=\frac{\frac{1}{2}(p+r) q-q^{2}}{2 q-(p+r)} \\ & \frac{-\frac{q}{2}[2 q-(p+r)]}{2 q-(p+r)}=-\frac{\boldsymbol{q}}{\mathbf{2}} . \end{aligned} $

(Using (i))

ARITHMETICO-GEOMETRIC SERIES

KEY FACTS

A series in which each term is the product of corresponding terms of an A.P. and a G.P. is called an Arithmetico-Geometric series.

The general or standard form of such a series is $a+(a+d) r+(a+2 d) r^{2}+(a+3 d) r^{3}+\ldots . .+{a+(n-1) d} r^{n-1}+\ldots$. where each term is formed by multiplying the corresponding terms of the two series.

A.P. : $a+(a+d)+(a+2 d)+\ldots . .+(a+(n-1) d)+\ldots$. and

G.P. : $1+r+r^{2}+\ldots . .+r^{n-1}+\ldots .$. 2. $n$th term of an Arithmetico-Geometric series is

$ T_n={a+(n-1) d} r^{n-1} . $

3. Sum of $\boldsymbol{n}$ terms of an Arithmetico-Geometric Series

Let $S_n$ be the sum of the $n$ terms of the series $a+(a+d) r+(a+2 d) r^{2}+\ldots . .+(a+(n-1) d) r^{n-1}$

Then,

$ \begin{aligned} S_n & =a+(a+d) r+(a+2 d) r^{2}+\ldots .+(a+(n-1) d) r^{n-1} \\ r S_n & =\quad a r+(a+d) r^{2}+\ldots \ldots+(a+(n-2) d) r^{n-1}+(a+(n-1) d) r^{n} \end{aligned} $

Subtracting eqn (ii) from eqn $(i)$, we get

$ \begin{aligned} (1-r) S_n & =(a+d r+d r^{2}+\ldots \ldots+d r^{n-1})-(a+(n-1) d) r^{n} \\ & =a+\frac{d r(1-r^{n-1})}{(1-r)}-(a+(n-1) d) r^{n} \\ \Rightarrow \quad S_n & =\frac{\boldsymbol{a}}{\mathbf{1}-\boldsymbol{r}}+\frac{\boldsymbol{d r}(\mathbf{1}-\boldsymbol{r}^{n-1})}{(\mathbf{1}-\boldsymbol{r})^{2}}-\frac{(\boldsymbol{a}+(\boldsymbol{n}-\mathbf{1}) \boldsymbol{d}) \boldsymbol{r}^{n}}{(\mathbf{1}-\boldsymbol{r})} . \end{aligned} $

4. Sum of an infinite Arithmetico-Geometric Series

Let the infinite Arithmetico-Geometric series be

$ a+(a+d) r+(a+2 d) r^{2}+\ldots \ldots+(a+(n-1) d) r^{n-1}+\ldots \ldots \infty $

If $|r|<1$, then $r^{n} \to 0$ and $r^{n-1} \to 0$ as $n \to \infty$, then

$ S _{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^{2}} $

SOLVED EXAMPLES

Ex. 1 . Find the $n$th term of the given arithmetico-geometric series: (i) $\frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\ldots$. (ii) $1-2 x+3 x^{2}-4 x^{3}+\ldots$.

Sol. (i) The A.P. and G.P. corresponding to the given series

$ \begin{aligned} & \frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\ldots . \text{ are respectively } \\ & 1,3,5,7, \ldots \ldots \ldots . . \text{ and } \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, \ldots \ldots . \\ & \quad n \text{th term of A.P. }=(1+(n-1) 2)=(2 n-1) \\ & \quad n \text{th term of G.P. }=\frac{1}{3} \cdot(\frac{1}{3})^{n-1}=(\frac{1}{3})^{n} \\ & \therefore n \text{th term of the given series }=(2 n-1)(\frac{\mathbf{1}}{\mathbf{3}})^{n} . \end{aligned} $

(ii) The given arithmetico-geometric series is

$1-2 x+3 x^{2}-4 x^{3}+\ldots$.

where corresponding A.P. and G.P. are respectively $1,2,3,4, \ldots$ and $1,-x,(-x)^{2},(-x)^{3}, \ldots$.

$n$th term of A.P $=(1+(n-1) 1)=n$

$n$th term of G.P $=1 .(-x)^{n-1}=(-\mathbf{1})^{n-1}(x)^{n-1}$.

$\therefore \quad n$th term of the given series $=n \cdot(-1)^{n-1} x^{n-1}=(-\mathbf{1})^{\boldsymbol{n}-\mathbf{1}} \boldsymbol{n} \boldsymbol{x}^{\boldsymbol{n}-\mathbf{1}}$.

Ex. 2 . Find the sum to $n$ terms of the series $1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\ldots .$.

Sol. $1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\ldots \ldots$ is an arithmetico-geometric series with corresponding A.P. and G.P as:

A.P. : $1+3+5+7+\ldots \ldots$

G.P. : $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .$.

$n$th term of A.P. $=(1+(n-1) 2)=(2 n-1)$

$n$th term of G.P. $=1 \cdot(\frac{1}{2})^{n-1}=\frac{1}{2^{n-1}}$

$\therefore n$th term of the given A.G.P. $=\frac{2 n-1}{2^{n-1}}$

$\therefore \quad S_n=1+\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\ldots \ldots \ldots+\frac{2 n-1}{2^{n-1}}$

$\Rightarrow \quad \frac{1}{2} S_n=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\ldots \ldots+\ldots \ldots+\frac{2 n-3}{2^{n-1}}+\frac{2 n-1}{2^{n}}$

On subtraction, we get

$ \Rightarrow (1-\frac{1}{2}) S_n =1+\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\ldots \ldots \ldots+\frac{2}{2^{n-1}}-\frac{2 n-1}{2^{n}} $

$\frac{1}{2} S_n =1+2[{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \ldots+\frac{1}{2^{n-1}}}]-\frac{2 n-1}{2^{n}} $

$ \quad \frac{1}{2} S_n =1+2[{\frac{(\frac{1}{2}(1-(\frac{1}{2})^{n})}{1-\frac{1}{2}}}]-\frac{2 n-1}{2^{n}}=1+2-\frac{4}{2^{n}}-\frac{2 n-1}{2^{n}}=3-\frac{2 n+3}{2^{n}}$

$\quad S_n =6-\frac{2 n+3}{2^{n-1}}$

Ex. 3 . Find the sum of the series $1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\ldots . \infty$

Sol. Let

$ \begin{aligned} S _{\infty} & =1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\ldots \infty \\ \frac{1}{5} S _{\infty} & =\frac{1}{5}+\frac{4}{5^{2}}+\frac{7}{5^{3}}+\ldots \infty \end{aligned} $

Subtracting eqn (ii) from eqn (i), we get

$ (1-\frac{1}{5}) S _{\infty}=1+\frac{3}{5}+\frac{3}{5^{2}}+\frac{3}{5^{3}}+\ldots \infty $

$ \begin{aligned} \frac{4}{5} S _{\infty} & =1+3(\frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}+\ldots \infty) \\ & =1+3(\frac{\frac{1}{5}}{1-\frac{1}{5}})(\because \text{ Sum of infinite G.P }=\frac{a}{1-r})=1+\frac{3 / 5}{4 / 5}=1+\frac{3}{4}=\frac{7}{4} \\ \therefore \quad S _{\infty} & =\frac{7}{4} \times \frac{5}{4}=\frac{\mathbf{3 5}}{\mathbf{1 6}} . \end{aligned} $

Ex. 4 . If the sum to infinity of the series $3+5 r+7 r^{2}+\ldots \ldots \infty$ is $\frac{44}{9}$, find the value of $r$.

Sol. $3+5 r+7 r^{2}+\ldots \ldots \infty$ is an infinite arithmetico-geometric series, where

$ a=3, d=2 \text{, common ratio }(r)=r \text{. } $

Sum to infinity of an A.G.P., with first term of A.P, as $a$, common difference $d$ and common ratio $r$ is

$s_{\infty} =\frac{a}{1-r}+\frac{d r}{(1-r)^{2}}$

$\therefore \frac{44}{9} =\frac{3}{1-r}+\frac{2 r}{(1-r)^{2}} \Rightarrow \frac{44}{9}=\frac{3(1-r)+2 r}{(1-r)^{2}} $

$\Rightarrow 44(1-r)^{2} =9(3-r) \Rightarrow 44(1-2 r+r^{2})=27-9 r $

$\Rightarrow 44-88 r+44 r^{2} =27-9 r \Rightarrow 44 r^{2}-79 r+17=0 $

$\Rightarrow \quad(4 r-1)(11 r-17) =0 \Rightarrow r=\frac{1}{4} \text{ or } \frac{17}{11} $

$r \neq \frac{17}{11} \text{ as it is not possible to find the sum of an infinite G.P. with }|r|>1$

Ex. 5 . Show that $2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times 8^{\frac{1}{16}} \times 16^{\frac{1}{32}} \times \ldots . . \infty=2$

Sol. Let

$ x=2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times 8^{\frac{1}{16}} \times 16^{\frac{1}{32}} \times \ldots . . \infty $

$ \begin{aligned} \therefore \quad \log x & =\frac{1}{4} \log 2+\frac{1}{8} \log 4+\frac{1}{16} \log 8+\frac{1}{32} \log 16+\ldots . . \infty \\ & =\frac{1}{4} \log 2+\frac{1}{8} \log 2^{2}+\frac{1}{16} \log 2^{3}+\frac{1}{32} \log 2^{4}+\ldots . . \infty \\ & =\frac{1}{4} \log 2+\frac{2}{8} \log 2+\frac{3}{16} \log 2+\frac{4}{32} \log 2+\ldots \ldots \infty \\ & =(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\ldots . . \infty) \log 2 \end{aligned} $

Now, $\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\ldots . . \infty$ is an Arithmetico-Geometric series.

Let

$ \begin{aligned} S & =\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\ldots . . \infty \\ \frac{1}{2} S & =\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\ldots . \infty \end{aligned} $

On subtracting eqn (iii) from eqn (ii), we get

$ \begin{matrix} \therefore & \log x=1 \times \log 2 \\ \Rightarrow & \log x=\log 2 \Rightarrow x=\mathbf{2} . \end{matrix} $

$ \begin{aligned} & \frac{1}{2} S=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots . . \infty=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2} \Rightarrow S=1 \\ & \log x=1 \times \log 2 \end{aligned} $

PRACTICE SHEET

The sum to $n$ terms of the series $1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\ldots .$. is (a) $(\frac{25}{14}-\frac{7 n+10}{14 \times 5^{n-1}})$ (b) $(\frac{17}{12}-\frac{4 n+7}{12 \times 5^{n-1}})$ (c) $(\frac{35}{16}-\frac{12 n+7}{16 \times 5^{n-1}})$ (d) $(\frac{15}{11}-\frac{10 n+2}{11 \times 5^{n-1}})$
Sum the series: $1+2.2+3.2^{2}+\ldots . .+100.2^{99}$ (a) $99.2^{100}$ (b) $100.2^{100}$ (c) $99.2^{100}+1$ (d) $1000.2^{100}$

(Kerala PET 2001) 3. The sum to infinity of the series $1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\frac{14}{3^{4}}+\ldots$. (a) 6 (b) 2 (c) 3 (d) 4

(AIEEE 2009)

Given $\cot \theta=2 \sqrt{2}$, the sum of the infinite series $1+2(1-\sin \theta)+3(1-\sin \theta)^{2}+4(1-\sin \theta)^{3}+\ldots .$. is (a) $6 \sqrt{2}$ (b) 8 (c) 9 (d) $8 \sqrt{2}$

(AMU 2006)

If $4+\frac{4+d}{5}+\frac{4+2 d}{5^{2}}+\ldots . . \infty=10$, then $d$ is equal to (a) 5 (b) 8 (c) 10 (d) 16

(DCE 2007)

ANSWERS

(c)
(c)
(b)
(c)
$(d)$

HINTS AND SOLUTIONS

$1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+$ is an A.G.P. with A.P. : $1+4+7+10+$

G.P. : $1+\frac{1}{5}+\frac{1}{5^{2}}+\frac{1}{5^{3}}+\ldots .$.

$n$th term of A.P. $=1+3(n-1)=3 n-2$

$n$th term of G.P. $=1 \times(\frac{1}{5})^{n-1}=\frac{1}{5^{n-1}}$

$\therefore \quad n$th term of the A.G.P. $=\frac{3 n-2}{5^{n-1}}$

$ (n-1) \text{ th term of the A.G.P. }=\frac{3(n-1)-2}{5^{n-2}}=\frac{3 n-5}{5^{n-2}} $

Now let $\quad S_n=1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+\ldots . .+\frac{3 n-2}{5^{n-1}}$

$\therefore \quad \frac{1}{5} S_n=\frac{1}{5}+\frac{4}{5^{2}}+\frac{7}{5^{3}}+\ldots . .+\frac{3 n-5}{5^{n-1}}+\frac{3 n-2}{5^{n}}$

$\therefore S_n-\frac{1}{5} S_n=1+\frac{3}{5}+\frac{3}{5^{2}}+\frac{3}{5^{3}}+\ldots . .+\frac{3}{5^{n-1}}-\frac{3 n-2}{5^{n}}$

$\Rightarrow(1-\frac{1}{5}) S_n=1+\frac{3}{5}(1+\frac{1}{5}+\frac{1}{5^{2}}+\ldots . .+\frac{1}{5^{n-2}})-\frac{3 n-2}{5^{n}}$

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

$ \because S_n=\frac{a(1-r^{n})}{1-r}, r<1 $

$ \begin{aligned} & =1+\frac{3}{5} \times \frac{5}{4}[1-\frac{1}{5^{n-1}}]-\frac{3 n-2}{5^{n}} \\ & =1+\frac{3}{4}-\frac{3}{4 \cdot 5^{n-1}}-\frac{3 n-2}{5^{n}}=\frac{7}{4}-\frac{3}{4.5^{n-1}}-\frac{3 n-2}{5^{n}} \\ \Rightarrow S_n & =\frac{35}{16}-(\frac{15}{16 \times 5^{n-1}}+\frac{3 n-2}{4.5^{n-1}}) \\ & =\frac{35}{16}-(\frac{15+12 n-8}{16 \times 5^{n-1}})=\frac{\mathbf{3 5}}{\mathbf{1 6}}-\frac{\mathbf{1 2 n - 7}}{\mathbf{1 6} \times \mathbf{5}^{n-1}} . \end{aligned} $

$1+2.2+3.2^{2}+4.2^{3}+\ldots . .+100.2^{99}$ is

clearly an AGP with A.P. : $1+2+3+\ldots . .100$

and G.P. : $1+2+2^{2}+\ldots . .+2^{99}$.

Let $\quad S=1+2.2+3.2^{2}+4.2^{3}+\ldots . .+100.2^{99}$

$\therefore \quad 2 S=2+2.2^{2}+3.2^{3}+\ldots . .+99.2^{99}+100.2^{100}$

$\therefore \quad S-2 S=(1+2+2^{2}+2^{3}+\ldots . .2^{99})-100.2^{100}$

$\Rightarrow \quad-S=\frac{(2^{100}-1)}{2-1}-100.2^{100}$

$=2^{100}-1-100.2^{100}$

$ S=2^{100} \cdot 100-2^{100}+1=2^{100} \cdot(100-1)+1 $

$ =99.2^{100}+1 . $

$ S=1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\frac{14}{3^{4}}+\ldots . . \text{ upto } \infty $

$\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^{2}}+\frac{6}{3^{3}}+\frac{10}{3^{4}}+\ldots .$. upto $\infty$ $\Rightarrow \quad S-\frac{1}{3} S=1+\frac{1}{3}+\frac{4}{3^{2}}+\frac{4}{3^{3}}+\frac{4}{3^{4}}+\ldots .$. upto $\infty$

$ =\frac{4}{3}+4[\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\ldots . . \infty] $

$ =\frac{4}{3}+4[\frac{\frac{1}{3^{2}}}{1-\frac{1}{3}}] \quad(\because S _{\infty}=\frac{a}{1-r}) $

$=\frac{4}{3}+4[\frac{\frac{1}{9}}{\frac{2}{3}}]=\frac{4}{3}+\frac{4}{6}=\frac{12}{6}=\mathbf{2}$.

$ S _{\infty}=1+2(1-\sin \theta)+3(1-\sin \theta)^{2} $

$\Rightarrow \quad S _{\infty}=1+2 a+3 a^{2}+4 a^{3}+\ldots . . \infty$

(where $a=1-\sin \theta$ )

$\therefore \quad a S _{\infty}=a+2 a^{2}+3 a^{3}+\ldots . . \infty$

$\Rightarrow S _{\infty}-a S _{\infty}=1+a+a^{2}+a^{3}+\ldots \ldots \infty$

$(1-a) S _{\infty}=\frac{1}{1-a}$

$\Rightarrow \quad S _{\infty}=\frac{1}{(1-a)^{2}}=\frac{1}{(1-1+\sin \theta)^{2}}$

$ \begin{gathered} =\frac{1}{\sin ^{2} \theta}=cosec^{2} \theta \\ \Rightarrow \quad S _{\infty}=cosec^{2} \theta=1+\cot ^{2} \theta=1+(2 \sqrt{2})^{2}=1+8=\mathbf{9} \end{gathered} $

Let $S=4+\frac{4+d}{5}+\frac{4+2 d}{5^{2}}+\frac{4+3 d}{5^{3}}+\ldots . . \infty$

$ \therefore \quad \frac{1}{5} S=\frac{4}{5}+\frac{4+d}{5^{2}}+\frac{4+2 d}{5^{3}}+\ldots . . \infty $

$ \begin{aligned} & \Rightarrow S-\frac{1}{5} S=4+\frac{d}{5}+\frac{d}{5^{2}}+\frac{d}{5^{3}}+\ldots . . \infty \\ & \Rightarrow \quad \frac{4}{5} S=4+\frac{d}{5}[1+\frac{1}{5}+\frac{1}{5^{2}}+\ldots . . \infty] \\ & \Rightarrow \quad \frac{4}{5} S=4+\frac{d}{5} \times \frac{1}{1-\frac{1}{5}} \Rightarrow \frac{4}{5} S=4+\frac{d}{5} \times \frac{5}{4} \\ & \Rightarrow \quad \frac{4}{5} S=4+\frac{d}{4} \Rightarrow S=5+\frac{5 d}{16} \end{aligned} $

Given $\quad S=10$

$\therefore \quad 5+\frac{5 d}{16}=10 \Rightarrow \frac{5 d}{16}=5 \Rightarrow \boldsymbol{d}=\mathbf{1 6}$.

SOME SPECIAL SERIES

KEY FACTS

1. Sum of first $\boldsymbol{n}$ natural numbers

Let

$ S=1+2+3+4+\ldots \ldots+n $

Clearly, this is an A.P. with $a=1, l=n, n=n$.

$ \therefore \quad S=\frac{n}{2}(a+l)=\frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})}{2} \quad \text{ or } \Sigma \boldsymbol{n}=\frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})}{2}, \text{ where } \Sigma \text{ stands for summation. } $

2. Sum of the squares of first $\boldsymbol{n}$ natural numbers

Let

$ S=1^{2}+2^{2}+3^{2}+4^{2}+\ldots . .+n^{2} . $

Then,

$ S=\Sigma n^{2}=\frac{n(n+1)(2 n+1)}{6} $

3. Sum of the cubes of first $\boldsymbol{n}$ natural numbers

Let

$ S=1^{3}+2^{3}+3^{3}+4^{3}+\ldots \ldots+n^{3} $

Then,

$ S=\Sigma n^{3}={\frac{n(n+1)}{2}}^{2}=(\Sigma n)^{2} $

$ \begin{aligned} T_n & =a n^{3}+b n^{2}+c n+d, \text{ then } \\ S_n & =a \Sigma n^{3}+b \Sigma n^{2}+c \Sigma n+d n \\ & =a[\frac{n(n+1)}{2}]^{2}+b[\frac{n(n+1)(2 n+1)}{6}]+c[\frac{n(n+1)}{2}]+d n . \end{aligned} $

SOLVED EXAMPLES

Ex. 1 . Find the $\boldsymbol{n}$ th term and then sum to $\boldsymbol{n}$ terms of the following series.

(i) $1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+\ldots$

(ii) $3.5+4.7+5.9+\ldots .$.

Sol. (i) $n$th term of the given series

$T_n =1^{2}+2^{2}+3^{2}+\ldots . .+n^{2}=\Sigma n^{2}=\frac{n(n+1)(2 n+1)}{6}=\frac{2 n^{3}+3 n^{2}+n}{6}=\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{n}{6}$

$\therefore S_n =\frac{1}{3} \cdot \Sigma n^{3}+\frac{1}{2} \cdot \Sigma n^{2}+\frac{1}{6} \cdot \Sigma n=\frac{1}{3} \cdot \frac{n^{2}(n+1)^{2}}{4}+\frac{1}{2} \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{1}{6} \cdot \frac{n(n+1)}{2} $

$ =\frac{n(n+1)}{12}{n(n+1)+(2 n+1)+1}=\frac{n(n+1)(n^{2}+3 n+2)}{12}=\frac{n(n+1)(n+1)(n+2)}{12}=\frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})^{2}(\boldsymbol{n}+\mathbf{2})}{\mathbf{1 2}} $

(ii) In the given series each term is the product of two factors. The factors $3,4,5, \ldots .$. are in A.P. having 3 as the first term and 1 as the common difference, therefore the $n$th term of this A.P. $=3+(n-1) 1=2+n$. Also, the factors $5,7,9, \ldots$ are in A.P. having 5 as the first term and 2 as the common difference, therefore the $n$th term of this A.P. $=5+(n-1)^{2}=2 n+3$

$\therefore n$th term of the series $=(2+n)(2 n+3)$

$ \begin{aligned} & \Rightarrow \quad T_n=2 n^{2}+7 n+6 \\ & \therefore S_n=2 \cdot \Sigma n^{2}+7 \Sigma n+6 n=\frac{2 n(n+1)(2 n+1)}{6}+\frac{7 n(n+1)}{2}+6 n \\ & =\frac{2 n(n+1)(2 n+1)+21 n(n+1)+36 n}{6}=\frac{\mathbf{4} \boldsymbol{n}^{3}+\mathbf{2 7} \boldsymbol{n}^{2}+\mathbf{5 9 n}}{\mathbf{6}} . \end{aligned} $

Ex. 2 . What is the sum of the series $15^{2}+16^{2}+17^{2}+\ldots . .+30^{2}$ equal to?

Sol. $15^{2}+16^{2}+17^{2}+\ldots . .+30^{2}=(1^{2}+2^{2}+3^{2}+\ldots .+30^{2})-(1^{2}+2^{2}+3^{2}+\ldots . .+14^{2})$

$ \begin{aligned} & =\sum _{k=1}^{30} k^{2}-\sum _{k=1}^{14} k^{2}=\frac{1}{6} \times 30 \times 31 \times(2 \times 30+1)-\frac{1}{6} \times 14 \times 15 \times(2 \times 14+1) \\ & (\because \sum _{k=1}^{n} k^{2}=\frac{1}{6} \times n(n+1)(2 n+1)) \\ & =\frac{30 \times 31 \times 61}{6}-\frac{14 \times 15 \times 29}{6}=9455-1015=\mathbf{8 4 4 0} . \end{aligned} $

Ex. 3 . What is the sum of the series $\mathbf{1}^{3}-2^{3}+3^{3}-4^{3}+\ldots . . . .+9^{3}$ equal to?

(AIEEE 2002)

Sol. Reqd. Sum $=(1^{3}+2^{3}+3^{3}+4^{3}+\ldots . .+9^{3})-2(2^{3}+4^{3}+6^{3}+8^{3})$

$ \begin{aligned} & =(1^{3}+2^{3}+3^{3}+\ldots .+9^{3})-2^{4}(1^{3}+2^{3}+3^{3}+4^{3}) \\ & =\sum _{k=1}^{9} k^{3}-2^{4} \sum _{k=1}^{4} k^{3}=\frac{9^{2}(9+1)^{2}}{4}-2^{4} \times \frac{4^{2} \times(4+1)^{2}}{4} \\ & =\frac{81 \times 100}{4}-\frac{16 \times 16 \times 25}{4}=2025-1600=\mathbf{4 2 5 .} \end{aligned} $

(Kerala PET 2004)

$ \begin{aligned} & t_4=4^{2}+3^{4} \\ & t_n=n^{2}+3^{n} \end{aligned} $

$\therefore \quad$ Adding columnwise, we get

$ \begin{aligned} t_1+t_2+t_3+\ldots \ldots+t_n & =(1^{2}+2^{2}+3^{2}+\ldots . .+n^{2})+(3^{1}+3^{2}+\ldots . .+3^{n}) \\ & =\sum _{k=1}^{n} k^{2}+\frac{3 \cdot(3^{n}-1)}{(3-1)}=\frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})(\mathbf{2} \boldsymbol{n}+\mathbf{1})}{\mathbf{6}}+\frac{\mathbf{3}}{\mathbf{2}}(\mathbf{3}^{\boldsymbol{n}}-\mathbf{1}) . \end{aligned} $

Ex. 5 . If $S_1, S_2$ and $S_3$ are the sums of the first $n$ natural numbers, their squares, their cubes respectively, then show that $9 S_2{ }^{2}=S_3(1+8 S_1)$

Sol.

$ \begin{aligned} S_1 & =\frac{n(n+1)}{2}, S_2=\frac{n(n+1)(2 n+1)}{6}, S_3=\frac{n^{2}(n+1)^{2}}{4} \\ S_2^{2} & =9{\frac{1}{6} n(n+1)(2 n+1)}^{2}=9{\frac{1}{36} n^{2}(n+1)^{2}(2 n+1)^{2}} \\ & =\frac{1}{4} n^{2}(n+1)^{2}(4 n^{2}+4 n+1)=\frac{1}{4} n^{2}(n+1)^{2}(4 n(n+1)+1) \\ & =\frac{1}{4} n^{2}(n+1)^{2}(1+8(\frac{1}{2} n(n+1))=S_3(\mathbf{1}+\mathbf{8} S_1) .. \end{aligned} $

$ \therefore \quad 9 S_2^{2}=9{\frac{1}{6} n(n+1)(2 n+1)}^{2}=9{\frac{1}{36} n^{2}(n+1)^{2}(2 n+1)^{2}} $

PRACTICE SHEET

What is the $n$th term of the series $1+\frac{(1+2)}{2}+\frac{(1+2+3)}{3}$ $+\ldots \ldots$ ? (a) $\frac{n+1}{2}$ (b) $\frac{n(n+1)}{2}$ (c) $n^{2}-(n+1)$ (d) $\frac{(n+1)(2 n+3)}{2}$
For $n \in N$, the sum of the series $2.3+3.5+4.7+\ldots$. $+(n+1)(2 n+1)$ is equal to (a) $\frac{n}{3}(2 n^{2}+3 n+1)$ (b) $\frac{1}{6} n(n^{2}+n-1)$ $\begin{matrix} \text{ (c) } \frac{n}{3}(3 n^{2}+5 n+11) & \text{ (d) } \frac{n}{6}(4 n^{2}+15 n+17)\end{matrix} $

(AMU 2004)

The value of $1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+11^{2}$ is equal to (a) 55 (b) 66 (c) 77 (d) 88

(Kerala PET 2011)

Sum of $n$ terms of the series $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$. is (a) $n^{2}(2 n^{2}-1)$ (b) $2 n^{2}+3 n^{2}$ (c) $n^{3}(n-1)$ (d) $n^{3}+8 n+4$

(WBJEE 2010)

If $S_n=1^{3}+2^{3}+\ldots . .+n^{3}$ and $T_n=1+2+3+\ldots . .+n$, then (a) $S_n=T_n^{2}$ (b) $S_n=T_n^{3}$ (c) $S_n^{2}=T_n$ (d) $S_3^{2}=T_n$

(EAMCET 2007)

ANSWERS

HINTS AND SOLUTIONS

Reqd. $n$th term $=\frac{1+2+3+\ldots . .+n}{n}$

$ =\frac{n(n+1)}{2} \times \frac{1}{n}=\frac{\boldsymbol{n}+\mathbf{1}}{\mathbf{2}} . $

Let $S=2 \cdot 3+3 \cdot 5+4 \cdot 7+\ldots . .+(n+1)(2 n+1)$

$ \begin{aligned} & =\sum _{k=1}^{n}(k+1)(2 k+1)=\sum _{k=1}^{n}(2 k^{2}+3 k+1) \\ & =2 \sum _{k=1}^{n} k^{2}+3 \sum _{k=1}^{n} k+n \end{aligned} $

$ \begin{aligned} & =2 \times \frac{n(n+1)(2 n+1)}{6}+3 \times \frac{n(n+1)}{2}+n \\ & =\frac{n}{6}[2(n+1)(2 n+1)+9(n+1)+6] \\ & =\frac{n}{6}[4 n^{2}+6 n+2+9 n+9+6] \\ & =\frac{n}{6}(4 n^{2}+\mathbf{1 5} \boldsymbol{n}+\mathbf{1 7}) . \end{aligned} $

Let $S=1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+11^{2}$

$=(1^{2}+2^{2}+3^{2}+4^{2}+\ldots . .+11^{2})-2(2^{2}+4^{2}+6^{2}+8^{2}+11^{2})$

$=(1^{2}+2^{2}+3^{2}+4^{2}+\ldots . .+11^{2})-2^{3}(1^{2}+2^{2}+3^{2}+4^{2}+5^{2})$

$=\frac{11 \times(11+1) \times(2 \times 11+1)}{6}-\frac{8 \times 5 \times(5+1) \times(2 \times 5+1)}{6}$

$=\frac{11 \times 12 \times 23}{6}-\frac{8 \times 5 \times 6 \times 11}{6}=22 \times 23-40 \times 11$

$=506-440=\mathbf{6 6}$.

Let $S_n=1^{3}+3^{3}+5^{3}+7^{3}+\ldots$. upto $n$ terms

$n$th term of the series $=(2 n-1)^{3}$

$ \begin{aligned} & \therefore S_n=\sum _{k=1}^{n}(2 k-1)^{3}=\sum _{k=1}^{n}[8 k^{3}-1-12 k^{2}+6 k] \\ & =8 \sum _{k=i}^{n} k^{3}-12 \sum _{k=1}^{n} k^{2}+6 \sum _{k=1}^{n} k-n \\ & =8 \frac{n^{2}(n+1)^{2}}{4}-12 \times \frac{n(n+1)(2 n+1)}{6} \\ & +6 \times \frac{n(n+1)}{2}-n \\ & =2 n^{2}(n^{2}+2 n+1)-2 n(2 n^{2}+3 n+1)+3 n(n+1)-n \\ & =2 n^{4}+4 n^{3}+2 n^{2}-4 n^{3}-6 n^{2}-2 n+3 n^{2}+3 n-n \\ & =2 n^{4}+n^{2}=n^{2}(2 n^{2}-\mathbf{1}) . \\ & \text{ 5. } S_n=\Sigma n^{3} \\ & =\frac{n^{2}(n+1)^{2}}{4}=(\frac{n(n+1)}{2})^{2} \\ & T_n=\frac{n(n+1)}{2} \\ & \therefore \text{ From }(i) \text{ and }(i i) \\ & S_n=T_n{ }^{2} \end{aligned} $

SELF ASSESSMENT SHEET

If the first term of an A.P. be ’ $a$ ‘, second be ’ $b$ ‘, and $n$th be ’ $2 a$ ‘, then the sum of $n$ terms is (a) $\frac{3 a b}{2(b-a)}$ (b) $\frac{2 a b}{5(b-a)}$ (c) $\frac{a b}{2(b-a)}(d) \frac{3 a b}{(b-a)}$

(MPPET 2010)

If $S_n$ denotes the sum of $n$ terms of an A.P., then $S _{n+3}-3 S _{n+2}+3 S _{n+1}-S_n$ is equal to (a) 0 (b) 1 (c) 2 (d) 3

(Kerala PET 2000)

The interior angles of a polygon are in A.P. The smallest angles is $120^{\circ}$ and the common difference is $5^{\circ}$. The number of sides of the polygon is (a) 8 (b) 9 (c) 12 (d) 19

(AMU 2010)

Divide 20 into four parts which are in A.P. and such that the product of the first and fourth is to the product of the second and third in the ratio $2: 3$. Find the product of the first and fourth term of the A.P. (a) 12 (b) 16 (c) 20 (d) 25
The sum of the first $p$ terms of an A.P. is $q$ and the sum of the first $q$ terms is $p$. Find the sum of the first $(p+q)$ terms. (a) $p q$ (b) $p-q$ $(c)-(p+q)$ (d) 0
In $a$ G.P, the ratio of the sum of first 3 terms to that of the first 6 terms is $125: 152$. Find the common ratio of the G.P. (a) $\frac{2}{3}$ (b) $\frac{3}{4}$ (c) $\frac{4}{3}$ (d) $\frac{3}{5}$
The sum of $n$ terms of the following series $1+(1+x)$ $+(1+x+x^{2})+\ldots$ will be (a) $\frac{1-x^{n}}{1-x}$ (b) $\frac{x(1-x^{n})}{1-x}$ (c) $\frac{n(1-x)-x(1-x^{n})}{(1-x)^{2}}$ (d) None of the above

(AMU 2003)

If $a_1, a_2, \ldots ., a _{50}$ are in G.P., then $\frac{a_1-a_3+a_5-\ldots . .+a _{49}}{a_2-a_4+a_5-\ldots . .+a _{50}}$ is equal to (a) 0 (b) 1 (c) $\frac{a_1}{a_2}$ (d) $\frac{a_1}{a _{50}}$

(Kerala PET 2006)

If $a, b, c$ are in G.P and $x, y$ are the arithmetic means of $a$, $b$ and $b, c$ respectively, then $\frac{1}{x}+\frac{1}{y}$ is equal to (a) $\frac{b}{2}$ (b) $\frac{b}{3}$ (c) $\frac{2}{b}$ (d) $\frac{3}{b}$
If $S$ is the sum to infinity of a G.P. whose first term is 1 , then the sum of its first $n$ terms is (a) $S(1-\frac{1}{S})^{n-1}$ (b) $S(1-\frac{1}{S})^{n}$ (c) $S{1-(1-\frac{1}{S})^{n-1}}$ (d) $S{1-(1-\frac{1}{S})^{n}}$
If $H$ is the harmonic mean between $P$ and $Q$, then $\frac{H}{P}+\frac{H}{Q}=$ (a) 2 (b) $\frac{1}{2}$ (c) $\frac{P Q}{P+Q}$ (d) $\frac{P+Q}{P Q}$

(VITEEE 2007)

If $a, b, c$ are in G.P, then $\log _{a} 10, \log _{b} 10$ and $\log _{c} 10$ are in (a) A.P (b) G.P (c) H.P. (d) None of these
If the sum of the roots of the quadratic equation $a x^{2}+b x$ $+c=0$ be equal to the sum of the reciprocals of their squares, then $b c^{2}, c a^{2}, a b^{2}$ will be in (a) A.P. (b) G.P. (c) H.P. (d) None of these
Let $a_1, a_2, \ldots, a _{10}$ be in A.P. and $h_1, h_2, \ldots, h _{10}$ be in H.P. If $a_1=h_1=2$ and $a _{10}=h _{10}=3$, then $a_4 h_7$ is (a) 2 (b) 3 (c) 5 (d) 6

(IIT)

If $a, b, c$ are in H.P, then $\frac{a}{b+c}, \frac{b}{c+a}+\frac{c}{a+b}$ will be in (a) A.P. (b) G.P. (c) H.P. (d) None of these

(VITEEE 2011)

If $|x|<1$, then the square root of the sum $1+2 x+3 x^{2}$ $+4 x^{3}+\ldots \infty$ (a) $(1-x)$ (b) $(1+x)$ (c) $(1-x)^{-1}$ (d) $(1+x)^{-1}$

(Rajasthan PET 2007) 17. The sum to $n$ terms of the series $1+2(1+\frac{1}{n})+3(1+\frac{1}{n})^{2}$ $+\ldots$ is (a) $n^{2}$ (b) $n(n-1)$ (c) $(n+1)^{2} \quad(d) n(n+1)$

(EAMCET 2000)

Find the sum of $n$ terms of the series whose $n$th term is $2 n^{2}+3 n$ (a) $\frac{1}{2} n(n+1)(2 n+5)$ (b) $\frac{1}{6} n(n+1)(4 n+1)$ (c) $\frac{1}{3}(n^{2}-6 n+3)$ (d) $\frac{1}{6} n(n-1)(n^{2}+1)$
Find the sum of the series $1.2^{2}+3.3^{2}+5.4^{2}+\ldots$. to $n$ terms (a) $\frac{n}{2}(n^{3}+4 n^{2}+4 n-1)$ (b) $\frac{n}{2}(n+1)^{2}(2 n+1)$ (c) $\frac{n^{2}}{3}(n+1)^{2}(2 n+1)$ (d) $\frac{n}{3}(n^{3}+4 n^{2}-2 n+1)$
If the sum of first $n$ terms of an A.P. is $c n^{2}$, then the sum of the squares of these $n$ terms is (a) $\frac{n(4 n^{2}-1) c^{2}}{6}$ (b) $\frac{n(4 n^{2}+1) c^{2}}{6}$ (c) $\frac{n(4 n^{2}-1) c^{2}}{3}$ (d) $\frac{n(4 n^{2}+1) c^{2}}{3}$

(IIT 2009)

ANSWERS

(a)
(a)
$(b)$
$(b)$
(c)
$(d)$
(c)
(c)
(c)
$(d)$
(a)
(c)
(a)
$(d)$
(c)
$(c)$
(a)
(b)
(a)
$(c)$

HINTS AND SOLUTIONS

Let the common difference of the A.P. be ’ $d$ ‘. Then

$ \begin{matrix} d & =b-a \\ \text{ and } \quad T_n & =2 a=a+(n-1) d \\ \Rightarrow \quad 2 a & =a+(n-1)(b-a) \\ \Rightarrow \quad 2 a & =a+b n-b-n a+a \\ \Rightarrow \quad & b & =n(b-a) \Rightarrow \quad n=\frac{b}{b-a} \\ \therefore \quad & S_n & =\frac{n}{2}{a+T_n}=\frac{b}{2(b-a)}(a+2 a)=\frac{\mathbf{3 a b}}{\mathbf{2}(\boldsymbol{b}-\boldsymbol{a})} . \end{matrix} $

$S _{n+3}-3 S _{n+2}+3 S _{n+1}-S_n$

$ \begin{aligned} & =(S _{n+3}-S _{n+2})-2(S _{n+2}-S _{n+1})+(S _{n+1}-S_n) \\ & =t _{n+3}-2 t _{n+2}+t _{n+1} \end{aligned} $

$ (\because t_n=S_n-S _{n-1}) $

$\because t _{n+1}, t _{n+2}, t _{n+3}$ are consecutive terms of an A.P,

$ 2 t _{n+2}=t _{n+1}+t _{n+3} $

$\therefore$ From $(i)$

$ \begin{aligned} \text{ Reqd. Sum } & =(t _{n+3}+t _{n+1})-2 t _{n+2} \\ & =2 t _{n+2}-2 t _{n+2}=0 \end{aligned} $

(Using (ii))

Let the polygon have $n$ sides.

Then, sum of its interior $\angle s(S_n)=(2 n-4)$ rt. $\angle s$

$ =(n-2) \times 180^{\circ} $

Number of sides of the polygon $=$ number of interior angles of the polygon $=n$.

The interior angles form an A.P. with first term $=120^{\circ}$ and common difference $5^{\circ}$.

$ \begin{aligned} \therefore \quad S_n & =\frac{n}{2}[2 \times 120^{\circ}+(n-1) \times 5^{\circ}] \\ & =\frac{n}{2}[240^{\circ}+5 n-5^{\circ}] \end{aligned} $

From (i) and (ii),

$ (n-2) \times 180^{\circ}=\frac{n}{2}[240^{\circ}+5 n-5^{\circ}] $

$\Rightarrow(n-2) \times 360=5 n^{2}+235 n$

$\Rightarrow 5 n^{2}+235 n-360 n+720=0$

$\Rightarrow 5 n^{2}-125 n+720=0 \Rightarrow n^{2}-25 n+144=0$

$\Rightarrow(n-16)(n-9)=0 \quad \Rightarrow n=16$ or 9 .

when $n=16$, the last angle $a_n=a+(n-1) d$

$ =120^{\circ}+(16-1) \times 5^{\circ}=195^{\circ} $

which is not possible.

Hence, $\quad \boldsymbol{n}=\mathbf{9}$.

Let the required four parts be $(a-3 d),(a-d),(a+d)$, $(a+3 d)$

Given, $\quad a-3 d+a-d+a+d+a+3 d=20$

$\Rightarrow \quad 4 a=20 \Rightarrow a=5$

Also, $\frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}=\frac{2}{3}$

$\Rightarrow \quad 3(a^{2}-9 d^{2})=2(a^{2}-d^{2})$

$\Rightarrow \quad 3(25-9 d^{2})=2(25-d^{2})$

$\Rightarrow \quad 75-27 d^{2}=50-2 d^{2} \Rightarrow 25=25 d^{2}$

$\Rightarrow \quad d= \pm 1$

Taking $d=1$, the numbers are $5-3,5-1,5+1,5+3$, i.e., 2, $4,6,8$

Taking $d=-1$, the numbers are $5+3,5+1,5-1,5-3$, i.e., $8,6,4,2$

$\therefore \quad$ Required product in either case $=2 \times 8=\mathbf{1 6}$.

Since $S_n=\frac{n}{2}[2 a+(n-1) d]$ for an A.P. whose first term $=a$, common difference $=d$, number of terms $=n$.

$ \begin{aligned} & \therefore & S_p & =q=\frac{p}{2}(2 a+(p-1) d) \\ & \Rightarrow & 2 q & =2 a p+p(p-1) d \\ & & S_q & =p=\frac{q}{2}(2 a+(q-1) d) \\ & & 2 p & =2 a q+q(q-1) d \end{aligned} $

Subtracting eqn (ii) from eqn (i), we get

$ \begin{aligned} 2(q-p) & =2 a(p-q)+(p^{2}-q^{2}) d-(p-q) d \\ \Rightarrow-2(p-q) & =2 a(p-q)+(p-q)(p+q) d-(p-q) d \\ -2 & =2 a+[(p+q)-1] d \end{aligned} $

Now $\quad S _{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]$

$ \begin{aligned} & =\frac{p+q}{2} \times-2 \\ & =-(\boldsymbol{p}+\boldsymbol{q}) . \end{aligned} $

(From (iii))

Let $a$ be the first term and $r$ the common ratio of the G.P.

$ \begin{aligned} & \text{ Then, } \quad \begin{aligned} \quad \frac{S_3}{S_6} & =\frac{125}{152} \Rightarrow \frac{\frac{a(r^{3}-1)}{(r-1)}}{\frac{a(r^{6}-1)}{(r-1)}}=\frac{125}{152} \\ \frac{(r^{3}-1)}{(r^{6}-1)} & =\frac{125}{152} \Rightarrow \frac{(r^{3}-1)}{(r^{3}+1)(r^{3}-1)}=\frac{125}{152} \\ \Rightarrow \quad r^{3}+1 & =\frac{152}{125} \Rightarrow r^{3}=\frac{152}{125}-1=\frac{27}{125} \end{aligned} \end{aligned} $

$ \Rightarrow \quad r^{3}=(\frac{3}{5})^{3} \Rightarrow r=\frac{3}{5} . $

$S_n=1+(1+x)+(1+x+x^{2})+\ldots . n$ terms

$ \begin{aligned} \Rightarrow S_n & =\frac{1}{1-x}[(1-x)+(1-x)(1+x)+(1-x)(1+x+x^{2}). \\ & +\ldots . \text{ to } n \text{ terms }] \\ & =\frac{1}{1-x}[(1-x)+(1-x^{2})+(1-x^{3})+\ldots \text{ to } n \text{ terms }] \\ & =\frac{1}{1-x}[n-(x+x^{2}+x^{4}+\ldots . \text{ to } n \text{ terms }]. \\ & =\frac{1}{1-x}[n-\frac{x(1-x^{n})}{1-x}] \quad(\because S_n=\frac{a(1-r^{n})}{1-r}) \\ & =\frac{\boldsymbol{n}(\mathbf{1}-\boldsymbol{x})-\boldsymbol{x}(\mathbf{1}-\boldsymbol{x}^{\boldsymbol{n}})}{(1-\boldsymbol{x})^{\mathbf{2}}} . \end{aligned} $

Let the first term of the G.P. be $a$ and common ratio $r$. Then,

$ \begin{aligned} \frac{a_1-a_3+a_5-\ldots .+a _{49}}{a_2-a_4+a_6-\ldots .+a _{50}} & =\frac{a-a r^{2}+a r^{4}-\ldots .+a r^{48}}{a r-a r^{3}+a r^{5}-\ldots .+a r^{49}} \\ & =\frac{a(1-r^{2}+r^{4}-\ldots .+r^{48})}{a r(1-r^{2}+r^{4}-\ldots .+r^{48})} \\ & =\frac{\boldsymbol{a}}{\boldsymbol{a r}}=\frac{\boldsymbol{a} _{\mathbf{1}}}{\boldsymbol{a} _{\mathbf{2}}} . \end{aligned} $

$a, b, c$ are in G.P $\Rightarrow b^{2}=a c$

$x$ is the A.M. of $a, b \quad \Rightarrow \quad x=\frac{a+b}{2}$

$y$ is the A.M. of $b, c \Rightarrow y=\frac{b+c}{2}$

Now $\frac{1}{x}+\frac{1}{y}=\frac{2}{a+b}+\frac{2}{b+c}=\frac{2(b+c)+2(a+b)}{(a+b)(b+c)}$

$ \begin{aligned} & =\frac{2(a+2 b+c)}{a b+b^{2}+a c+b c}=\frac{2(a+2 b+c)}{a b+b^{2}+b^{2}+b c} \\ & =\frac{2(a+2 b+c)}{b(a+2 b+c)}=\frac{\mathbf{2}}{\boldsymbol{b}} . \end{aligned} $

Let the first term of the G.P. be $a$ and common ratio $r$.

$ \begin{aligned} & \text{ Given } \quad S=\frac{a}{1-r} \Rightarrow S=\frac{1}{1-r} \\ & \Rightarrow \quad 1-r=\frac{1}{S} \Rightarrow r=1-\frac{1}{S} \end{aligned} $

Let $S_n$ be the sum of first $n$ terms of the series. Then,

$ \begin{aligned} S_n & =\frac{1(1-r^{n})}{1-r} \\ & =\frac{1(1-(1-\frac{1}{S})^{n})}{(1-(1-\frac{1}{S}))}=S{1-(1-\frac{1}{S})^{n}} . \end{aligned} $

$H$ being the harmonic mean between $P$ and $Q$.

$ \begin{aligned} H & =\frac{2 P Q}{P+Q} \\ \therefore \quad \frac{H}{P}+\frac{H}{Q} & =\frac{2 Q}{P+Q}+\frac{2 P}{P+Q}=\frac{2(P+Q)}{P+Q}=\mathbf{2} . \end{aligned} $

$a, b, c$ are in G.P $\Rightarrow b^{2}=a c$

$ \begin{aligned} & \Rightarrow \frac{1}{\log _{a} 10}+\frac{1}{\log _{c} 10}=\log _{10} a+\log _{10} c=\log _{10}(a c) \\ & =\log _{10} b^{2}=2 \log _{10} b=\frac{2}{\log _{b} 10} \\ & \Rightarrow \frac{1}{\log _{a} 10}, \frac{1}{\log _{b} 10}, \frac{1}{\log _{c} 10} \text{ are in A.P. } \\ & \Rightarrow \log _{a} 10, \log _{b} 10, \log _{c} 10 \text{ are in H.P. } \end{aligned} $

Let $\alpha, \beta$, be the roots of the equation $a x^{2}+b x+c=0$. Then

$ \begin{aligned} & \text{ Sum of roots }=\alpha+\beta=-\frac{b}{a} \\ & \text{ Product of roots }=\alpha \beta=\frac{c}{a} \end{aligned} $

Let $d$ be the common difference of the given A.P, $a_1, a_2, \ldots, a _{10}$.

Then, given, $a_1=2$ and $a _{10}=a_1+9 d=3$

$\Rightarrow \quad 2+9 d=3 \Rightarrow d=\frac{1}{9}$

$\Rightarrow \quad a_4=a_1+3 d=2+3 \times \frac{1}{9}=2 \frac{1}{3}=\frac{7}{3}$

Now, $h_1, h_2, \ldots, h _{10}$ are in H.P.

$\Rightarrow \frac{1}{h_1}, \frac{1}{h_2}, \ldots ., \frac{1}{h _{10}}$ are in A.P.

Let $d_1$ be the common difference of this A.P.

Given, $h_1=2 \Rightarrow$ First term of the A.P $=\frac{1}{h_1}=\frac{1}{2}$ Also, $h _{10}=3 \Rightarrow \frac{1}{h _{10}}=\frac{1}{h_1}+9 d_1$

$\Rightarrow \quad \frac{1}{3}=\frac{1}{2}+9 d_1 \Rightarrow 9 d_1=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$

$\Rightarrow \quad d_1=-\frac{1}{54}$

$\therefore \quad \frac{1}{h_7}=\frac{1}{h_1}+6 d_1=\frac{1}{2}+6 \times-\frac{1}{54}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}$

$\Rightarrow \quad h_7=\frac{18}{7}$

$\therefore \quad$ From (i) and (ii) $a_4 h_7=\frac{7}{3} \times \frac{18}{7}=\mathbf{6}$.

$a, b, c$ are in H.P.

$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P

$\Rightarrow \frac{a+b+c}{a}, \frac{a+b+c}{b}, \frac{a+b+c}{c}$ are in A.P

(Multiplying each term by $(a+b+c)$ )

$\Rightarrow 1+\frac{b+c}{a}, 1+\frac{a+c}{b}, 1+\frac{a+b}{c}$ are in A.P

$\Rightarrow \frac{b+c}{a}, \frac{a+c}{b}, \frac{a+b}{c}$ are in A.P.

(Subtracting 1 from each term)

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

Let $S _{\infty}=1+2 x+3 x^{2}+4 x^{3}+\ldots . . \infty$

$S _{\infty}$ is an infinite A.G.P. with A.P. : $1+2+3+\ldots \infty$

and G.P. : $1+x+x^{2}+x^{3}+\ldots \infty$

The common ratio of the A.G.P. is $x$

$\therefore \quad x S _{\infty}=x+2 x^{2}+3 x^{2}+\ldots \infty$

$Eq(i)-Eq(i i)$

$\Rightarrow \quad(1-x) S _{\infty}=1+x+x^{2}+x^{3}+\ldots \infty$

$\Rightarrow \quad(1-x) S _{\infty}=\frac{1}{1-x} \quad(\because S _{\infty}=\frac{a}{1-r})$

$\Rightarrow \quad S _{\infty}=\frac{1}{(1-x)^{2}}=(1-x)^{-2}$

$\therefore \quad$ Square root of $S _{\infty}=(\mathbf{1}-\boldsymbol{x})^{-1}$.

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

$\therefore(1+\frac{1}{n}) S_n=(1+\frac{1}{n})+2(1+\frac{1}{n})^{2}+\ldots$

$+(n-1)(1+\frac{1}{n})^{n-1}+n(1+\frac{1}{n})^{n}$

$(\because S_n.$ is an A.G.P with common ratio $.(1+\frac{1}{n}))$ $\Rightarrow S_n[1-(1+\frac{1}{n})]=1+(1+\frac{1}{n})+(1+\frac{1}{n})^{2}+\ldots$

$ \begin{aligned} & \Rightarrow-\frac{1}{n} S_n=\frac{1((1+\frac{1}{n})^{n}-1)}{(1+\frac{1}{n})-1}-n(1+\frac{1}{n})^{n-1}-n(1+\frac{1}{n})^{n} \\ & \Rightarrow-\frac{1}{n} S_n=n[(1+\frac{1}{n})^{n}-1]-n(1+\frac{1}{n})^{n} \\ & \Rightarrow-\frac{1}{n} S_n=-n \Rightarrow S_n=\boldsymbol{n}^{2} . \end{aligned} $

$T_n=2 n^{2}+3 n \quad \therefore \quad S_n=2 \sum _{k=1}^{n} k^{2}+3 \sum _{k=1}^{n} k$

$ \begin{aligned} & =2 \times \frac{1}{6} \times n(n+1)(2 n+1)+3 \times \frac{n(n+1)}{2} \\ & =\frac{1}{3} \times n(n+1)(2 n+1)+\frac{3}{2} n(n+1) \\ & =n(n+1)[\frac{2 n+1}{3}+\frac{3}{2}]=n(n+1)[\frac{4 n+2+9}{6}] \\ & =\frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})(\mathbf{4 n + 1 )}}{\mathbf{6}} . \end{aligned} $

$n$th term $(T_n)$ of the given series

$ \begin{aligned} & =(2 n-1)(n+1)^{2}=(2 n-1)(n^{2}+2 n+1) \\ & =2 n^{3}-n^{2}+4 n^{2}-2 n+2 n-1=2 n^{3}+3 n^{2}-1 \\ & \therefore S_n=2 \sum _{n=1}^{n} k^{3}+3 \sum _{k=1}^{n} k^{2}-n \end{aligned} $

$ =2 \times \frac{n^{2}(n+1)^{2}}{4}+3 \times \frac{1}{6} \times n(n+1)(2 n+1)-n $

$ =\frac{1}{2}[n^{2}(n^{2}+2 n+1)+(n^{2}+n)(2 n+1)-2 n]$

$ =\frac{1}{2}[n^{4}+2 n^{3}+n^{2}+2 n^{3}+2 n^{2}+n^{2}+n-2 n] $

$ =\frac{1}{2}[n^{4} + 4n^{3} + 4n^{2} - n]=\frac{n}{2}[n^3+4n^2 + 4n - 1] $

Let the sum of first $n$ terms of the A.P, $S_n=c n^{2}$

$ \therefore \quad S _{n-1}=c(n-1)^{2}=c(n^{2}-2 n+1) $

$\therefore n$th term of the A.P. $=S_n-S _{n-1}$

$ =c n^{2}-c(n^{2}-2 n+1)=c(2 n-1) $

Let $S_n^{2}$ be the sum of the squares of these $n$ terms. Then,

$ S_n^2 =\sum_{k=1}^n t_k^2 \quad \text{ where } t_k=c(2 k-1) $

$ =\sum_{k=1}^{n}[c^{2}(2 k-1)^{2}]=\sum_{k=1}^{n}[c^{2} \cdot 4 k^{2}-c^{2} \cdot 4 k+c^{2}] $

$ =4 c^{2} \sum_{k=1}^{n} k^{2}-4 c^{2} \sum_{k=1}^{n} k+c^{2} n $

$ =4 c^{2} \times \frac{n(n+1)(2 n+1)}{6}-4 c^{2} \times \frac{n(n+1)}{2}+c^{2} n $

$ =\frac{c^{2} n}{6}[4(n+1)(2 n+1)-12(n+1)+6] $

$ =\frac{c^{2} n}{6}[8 n^{2}+12 n+4-12 n-12+6] $

$ =\frac{c^{2} n}{6}[8 n^{2}-2]=\frac{\boldsymbol{n}(\mathbf{4} n^{2}-\mathbf{1}) c^{2}}{\mathbf{3}} $

CHAPTER 2 Functions

KEY FACTS

Definition: Let $A$ and $B$ be two non-empty sets, then if by some rule or correspondence we can associate with every element $x$ of set $A$ one and only one element $y=f(x)$ of set $B$, then this rule or correspondence denoted by $f: A \to B$ is called a function from set $A$ to set $B$.

The element $y=\boldsymbol{f}(\boldsymbol{x})$ of $B$ is called the image of $\boldsymbol{x}$ and $\boldsymbol{x}$ is called the pre-image of $\boldsymbol{f}(\boldsymbol{x})$.

Here

Set $\mathbf{A}$ is called the domain of the function
Set B is called the co-domain of the function
Set $\mathbf{C}$ of all members of $B$ assigned to members of $A$ by the function is called the range of the function.

Note: (1) An empty set is not considered a function

(2) A function is usually denoted by $f, g, h, F, G, H \ldots$

(3) Range of $\boldsymbol{f} \subseteq$ (subset) codomain of $\boldsymbol{f}$.

(4) A function is also thought of as a mapping of its domain into its range.

In terms of figures, for a mapping to be a function, each \to should start from a different point in the domain, whether they terminate at the same point in the co-domain is immaterial.

Therefore, the essential requirements for a function are:

The function $f: A \to B$ is defined under following conditions:

(a) Every $x \in A$ is associated with some $y \in B$, i.e., a function is defined only when the domain is entirely used up. The set $B$ (codomain) need not be entirely used up.

(b) The function may associate more than one $x$ to same $y$.

Ex. (i) The relation ${(1,-2),(3,7),(4,-6),(8,11)}$ is a function as no two ordered pairs have the same first component.

Domain $={1,3,4,8}$, Range $={-2,7,-6,11}$ (ii) ${(1,0),(1,-2),(2,3),(4,-7)}$ is not a function as two ordered pairs $(1,0)$ and $(1,-2)$ have the same first component.

(iii) For $x>3, f(x)=3 x-2$ and for $-2 \leq x \leq 2, f(x)=x^{2}-2$. Then, $f(0)+f(4)=((0)^{2}-2)+(3 \times 4-2)=-2+12-2=\mathbf{8}$.

2. Types of Functions

(a) One-one function (Injective)

A function $\boldsymbol{f}: A \to B$ is called a one-one or injective function if an element of $A$ is associated to an unique element of set $B$, i.e., distinct elements of set $A$ have distinct images in set $B$.

$f: A \to B$ is one-one if and only if, for all $a_1, a_2 \in A, f(a_1)=f(a_2) \Rightarrow a_1=a_2$.

Ex. If $A={1,2,3}$ and $B={a, b, c, d}$ and $f: A \to B={(1, a),(2, b),(3, d)}$, then $f$ is one-one, where as $f: R \to R$ such that $f(x)=x^{2}$ is not one-one as $f(-2)=$ 4 and $f(2)=4$, i.e., two distinct elements -2 and 2 have the same image 4.

Method to check whether a function is one-one or not

Take two arbitrary elements $a_1$ and $a_2$ in the domain of $f$.
Put $f(a_1)=f(a_2)$ and solve the equation
If $f(a_1)=f(a_2) \Rightarrow a_1=a_2$, then $f: A \to B$ is a one-one function otherwise not.

The function $f: R \to R$ defined by $f(x)=x^{3}, x \in R$ can be shown one-one as follows:

Let $a_1, a_2 \in R$. Then $f(a_1)=f(a_2) \Rightarrow a_1{ }^{3} \Rightarrow a_2{ }^{3} \Rightarrow a_1=a_3 \Rightarrow f$ is one-one.

(b) Many-one function.

If a function $f: A \to B$ is such that two or more elements $a_1, a_2, \ldots$ of $A$ have the same $f$-image in $B$, then the function is called many-one function, i.e., two or more distinct elements of $A$ have the same image in $B$.

Ex. Thus, $f(x)=x^{2}$ is many-one since two elements in domain $A$ namely -1 and 1

have same image, i.e., 1 .

Method to check whether a function is many-one or not.

Consider any two arbitrary elements $a_1, a_2 \in A$
Put $f(a_1)=f(a_2)$ and solve the equation
If $f(a_1)=f(a_2) \Rightarrow a_1 \neq a_2$, then $f$ is many-one

Therefore, the function $f: R \to R$ defined by $f(x)=x^{2}+5$ is many-one

Since, $f(-1)=(-1)^{2}+5=6$ and $f(1)=1^{2}+5=6$

$\therefore \quad f(-1)=f(1) \Rightarrow(-1) \neq 1$

The mapping $f: A \to B$ is called an onto function if the set $B$ is entirely used up, i.e., if every element of $B$ is the image of at least one element of $A$. $\Rightarrow$ For every $b \in B$, there exists at least one element $a \in A$ such that $f(a)=b$.

$\Rightarrow$ Range of $f$ is the co-domain of $f$. (Range $=$ co-domain)

Ex. Let $A={-2,2,-3,3}, B={4,9}$ and $f: A \to B$ be a function defined by $f(x)=x^{2}$, then $\boldsymbol{f}$ is onto, because $f(-2)=4, f(2)=4, f(-3)=9, f(3)=9$,

i.e., $f(A)={4,9}=B$. Here, range $={4,9}=B=$ co-domain

Method to check whether a function is onto or not.

Suppose $f: A \to B$ is a given function

Step 1 : Choose any arbitrary element $y$ in $B$.

Step 2 : Putf $(x)=y$

Step 3 : Solve the equation $f(x)=y$ for $x$ and obtain $x$ in terms of $y$. Let $x=g(y)$

Step 4 : If for all values of $y \in B$, the values of $x$ obtained from $x=g(y)$ are in $A$, then $f$ is onto. If there are some $y \in B$ for which $x$, given by $x=g(y)$ is not in $A$, then $f$ is not onto.

Ex. Let $f: Z \to Z$ defined by $f(x)=3 x+2$ for all $x \in Z$. Let $y$ be an arbitrary element of $Z$ (codomain). Then

$ f(x)=y \Rightarrow y=3 x+2 \Rightarrow x=\frac{y-2}{3} $

Now if $y=0$, then $x=-\frac{2}{3} \notin Z$. Thus, $y=0$ in $Z$ (codomain) does not have its pre-image in $Z$ (domain).

Hence, $f$ in not an onto function.

(d) Into function: The mapping $f: A \to B$ is called an into function if the set $B$ is not entirely used up, i.e., there exists at least one element in $B$ having no pre-image in $A$.

(e) One-one onto function (Bijection): If a function $\boldsymbol{f}$ is both one-one and onto, then it is called a one-one onto function, i.e., a function $f: A \to B$

is one-one onto if

(i) it is one-one, i.e., $f(x)=f(y) \Rightarrow x=y$ for all $x, y \in A$

(ii) it is onto, i.e., for all $y \in B$, there exists an $x \in A$ such that $f(x)=y$.

Ex. If $\boldsymbol{R}$ is a set of real numbers and $\boldsymbol{Z}$ is the set of integers, then $f: R \to R$ defined by $f(x)=x^{3}$ is a one-one onto function while $f: Z \to Z$ defined by $f(x)=x^{3}$ is not a one-one onto function as elements $-3,-2$, in co-domain $Z$ have no preimage in domain $Z$.

Bijection

(f) Real valued functions: If $R$ be a set of real numbers and $A, B$ are subsets of $R$. Then the function $f: A \to B$ is called a real valued function or real function.

(g) Piece functions: Although functions are generally described by single formulas, sometimes a function is defined in two or more parts as:

$ g(x)=\begin{cases} x ; x \geq 0 \\ -x ; x<0 \end{cases}$

or,

$ h(x) = \begin{cases} -1 ; x<0 \\ 0 \quad x=0 \\ 1 \quad x>0 \end{cases} $

Modulus function

As we can see, for function $g$ domain $=$ set of all real numbers Range $=$ set of non-negative real numbers

Signum function

For function $\boldsymbol{h}$ Domain $=$ set of all real numbers Range $={-1,0,1}$

Thus, modulus function and signum functions are examples of piece functions.

(h) Inverse function: If $f: A \to B$ is a one-one onto function (bijection), then for every $x \in A$, we have a $y \in B$, such that $y=f(x)$. A new function $f^{-1}$ from $B$ to $A$ which associates each element $y \in B$ to its pre-image $x=f^{-1}(y) \in A$ can be defined and this function $f^{-1}$ is called the inverse of function $\boldsymbol{f}$.

Thus, $f: A \to B$ is a bijection and $f: a \to b$, then $f^{-1}: b \to a$.

Method to find $f^{-1}$ if f is invertible

Step 1. Put $y=f(x)$.

Step 2. Solve $y=f(x)$ and express $x$ in terms of $y$.

Step 3. The value of $x$ obtained in step 2 gives $f^{-1}(y)$

Step 4. Replace y by $x$ in $f^{-1}(y)$ to obtain $f^{-1}(x)$.

Ex. To find the inverse of the function $f(x)=5 x-8$, where $x \in R$.

$ f(x)=5 x-8, x \in R . $

For a function to be invertible, it should be a bijection, i.e., one-one onto function.

For all $x_1, x_2 \in R, f(x_1)=f(x_2) \Rightarrow 5 x_1-8=5 x_2-8 \Rightarrow x_1=x_2 \Rightarrow f(x)$ is one-one.

Let $y=f(x)=5 x-8 \Rightarrow 5 x=y+8 \Rightarrow x=\frac{y+8}{5}$

Thus, for all $y \in R, x=\frac{y+8}{5} \in R \Rightarrow f(x)$ is onto.

$f$ being one-one onto $\Rightarrow f$ is invertible.

$ \begin{aligned} & \text{ Let } \quad y=f(x)=5 x-8 \Rightarrow x=\frac{y+8}{5} \\ & \Rightarrow \quad f^{-1}(y)=\frac{y+8}{5} \Rightarrow f^{-1}(x)=\frac{x+8}{5} . \end{aligned} $

(i) Even and odd functions

If $f(x)$ is a function of $x$ such that $f(-x)=f(x)$ for every $x \in$ domain, then $f(x)$ is an even function of $x$.

Ex. (i) $f(x)=\cos x$ is an even function as $f(-x)=\cos (-x)=\cos x=f(x)$

(ii) $f(x)=x^{4}+2$ is an even function as $f(-x)=(-x)^{4}+2=x^{4}+2=f(x)$.

If $f(x)$ is a function of $x$ such that $f(-x)=-f(x)$ for every $x \in$ domain, then $f(x)$ is an odd function of $x$.

Ex. (i) $f(x)=\sin x$ is an odd function as $f(-x)=\sin (-x)=-\sin x=-f(x)$

(ii) $f(x)=x^{3}+6 x$ is an odd function as $f(-x)=(-x)^{3}+6(-x)=-x^{3}-6 x=-(x^{3}+6 x)=-f(x)$

Some functions as $y=x^{2}-5 x+3$ are neither even nor odd.

(j) Constant function: If $f: A \to B$ is such that for all $x \in A f(x)=c$ (a constant value) then $\boldsymbol{f}$ is a constant function. Its domain is the set $A$ and range is a singleton set containing a single value $\boldsymbol{c}$.

Ex. $f: R \to R$ such that $f(x)=5$ for all $x \in R$ is a constant function.

(k) Equal functions: Two functions $\boldsymbol{f}$ and $\boldsymbol{g}$ are said to be equal if and only if

(i) domain of $f=$ domain of $g$

(ii) co-domain of $f=$ co-domain of $g$

(iii) $f(x)=g(x)$ for every $x$ belonging to their common domain.

Ex. Let $A={1,2}$ and $B={3,6}$. Then $f: A \to B$ given by $f(x)=x^{2}+2$ and $g: A \to B$ given by $g(x)=3 x$ are equal as:

$f(1)=3=g(1), f(2)=6=g(2)$
Both $f$ and $g$ have the same domain and co-domain.

(l) Identity Function: The identity function maps each member of the domain onto itself, i.e., if $I$ is the identity function $I: R \to R$ then for all $x \in R, \boldsymbol{I}(\boldsymbol{x})=\boldsymbol{x}$.

Classification of Functions: Functions can be broadly categorized into two groups. (a) Algebraic Functions (b) Transcendental Functions

(a) Algebraic Functions: A function which consists of a finite number of terms involving powers and roots of the independent variable $\boldsymbol{x}$ and the four fundamental operations of addition, subtraction, multiplication and division is called an algebraic function, e.g. $5 x^{3}-7 x^{2}+6,4 \sqrt{7 x-12}, 2 x^{\frac{1}{2}}+4 x-7, \frac{2 x+1}{2 x-3}$, etc.

The particular cases of algebraic functions are:

(i) Polynomial Functions: A function $f(x)=a_0+a_1 x+a_2 x^{2}+\ldots . .+a_n x^{n}$, where $n \in N$ and $a_1, a_2, a_3, \ldots . ., a_n \in R$ is called a polynomial function.

Its Domain is the set of real numbers

Range is the set of real numbers. (ii) Rational Function: A function $f: A \to R, f(x)=\frac{P(x)}{Q(x)}$, where $Q(x) \neq 0$ is called a rational function. Here $A={x: x \in R}$ such that $Q(x) \neq 0$, and $P(x)$ and $Q(x)$ are polynomial functions of $x$.

Its domain is the set of all real numbers except the real roots of $Q(x)$.

Ex. (i) $\frac{x^{2}+5 x+6}{x^{2}-3 x+2}$ is a polynomial function with domain $=R-{1,2}$

$(\because.$ The roots of $x^{2}-3 x+2$ are $.x=1, x=2)$

(ii) $f(x)=\frac{1}{x} \quad$ Domain $=R-{0}$, Range $=R-{0}$

(iii) $f(x)=\frac{1}{x^{2}} \quad$ : Domain $=R-{0}$, Range $=$ Set of positive real numbers.

(iv) $f(x)=\frac{1}{x^{3}} \quad$ : Domain $=R-{0}$, Range $=R-{0}$.

(iii) Irrational Functions: Functions as $\sqrt{3 x^{2}-7 x+4}, 4 x^{2}+\sqrt{3 x}, \frac{1}{\sqrt[3]{5+3 x}}$, i.e., involving radicals or non-integral powers of $x$ are called irrational functions.

(b) Transcendental Functions: A function that is not algebraic is called transcendental. Particular cases are:

(i) Trigonometric functions as $\sin x, \cos x, \tan 3 x, cosec 2 x$, etc.

(ii) Inverse Trigonometric functions as $\sin ^{-1} x, \tan ^{-1} 2 x$, etc

(iii) Exponential functions as $y=a^{x}$

(iv) Logarithmic functions as the correspondence $x \to \log x$.

Algebraic Operations on Functions: If $\boldsymbol{f}$ and $\boldsymbol{g}$ are real valued functions of $\boldsymbol{x}$ with domain set $A$ and $B$ respectively, then both $f$ and $g$ are defined in $A \cap B$. Then,

(a) $(f+g) x=f(x)+g(x) \quad:$ Domain $A \cap B$

(b) $(f-g) x=f(x)-g(x) \quad:$ Domain $A \cap B$

(d) $[\frac{f}{g}] x=\frac{f(x)}{g(x)} \quad:$ Domain $A \cap B$

(e) $(f+k) x=f(x)+k \quad: k$ is constant, so domain $=A$

(f) $(k f) x=k f(x)$

(g) $f^{n}(x)=[f(x)]^{n}$

SOLVED EXAMPLES

Ex. 1 . Find whether the following functions are one-one or many-one?

(i) $f: R-{3} \to R$ defined by $f(x)=\frac{5 x+7}{x-3}, x \in R-{3}$

(ii) Modulus function $f: R \to R$ defined by $f(x)=\begin{cases} x ; x \geq 0 \\ -x ; x<0\end{cases}$

(iii) Greatest integer function $f: R \to R$ defined by $f(x)=[x]=n$ (an integer) for all $n \leq x \leq n+1$.

Sol. (i) Given : $f(x)=\frac{5 x+7}{x-3}, x \in R-{3}$

Let $a_1, a_2$ be two arbitrary elements $\in R-{3}$ such that $f(a_1)=f(a_2)$.

Then, $\quad f(a_1)=f(a_2) \Rightarrow \frac{5 a_1+7}{a_1-3}=\frac{5 a_2+7}{a_2-3}$

$ \begin{aligned} & \Rightarrow 5 a_1 a_2+7 a_2-15 a_1-21=5 a_2 a_1-21+7 a_1-15 a_2 \\ & \Rightarrow-22 a_1=-22 a_2 \Rightarrow a_1=a_2 \Rightarrow f \text{ is one-one. } \end{aligned} $

(ii) Given: $\quad f(x)=|x| \Rightarrow f(1)=|1|=1$ and $f(-1)=|-1|=1$

Thus, $\quad f(1)=f(-1)=1$ but $1 \neq-1$

$\therefore f$ is many-one.

(iii) $\because$

$f(x)=[x]=n$ for all $n \leq x<n+1$, so, $1 \leq x<2 \quad \Rightarrow \quad f(x)=1$

$\therefore$ The elements $1.1,1.25,1.4,1.9 \ldots \in$ domain are mapped onto to the same value 1 in the range. Hence the function $\boldsymbol{f}$ is many-one.

Ex. 2 . If $A=R-{3}$ and $B=R-{1}$ and $f: A \to B$ is a mapping defined by $f(x)=\frac{x-2}{x-3}$, Show that $f$ is one-one onto.

Sol. One-One

Let $x, y$ be any two elements of $A$, then

$ \begin{matrix} f(x)=f(y) \Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3} \Rightarrow(x-2)(y-3)=(y-2)(x-3) \\ \Rightarrow \quad x y-2 y-3 x+6=x y-2 x-3 y+6 \Rightarrow-x=-y \Rightarrow x=y \Rightarrow f \text{ is one-one. } \end{matrix} $

Onto

Let $y$ be any element of $B$. Then

$ \begin{aligned} f(x)=y & \Rightarrow y=\frac{(x-2)}{(x-3)} \Rightarrow(x-3) y=(x-2) \Rightarrow x y-3 y=x-2 \\ & \Rightarrow x y-x=3 y-2 \Rightarrow x(y-1)=3 y-2 \Rightarrow x=\frac{3 y-2}{y-1} \end{aligned} $

For $y \neq 1, x=\frac{3 y-2}{y-1}$ is a real number.

Also, $A=R-{3} \Rightarrow \frac{3 y-2}{y-1} \neq 3$, because if we take $\frac{3 y-2}{y-1}=3$, then $.3 y-2=3 y-3)$

$\Rightarrow \quad 2=3$ which is not true. Hence, $\boldsymbol{f}$ is onto.

$\Rightarrow f$ is both one-one and onto.

Ex. 3 . Prove that function $f: R \to R, f(x)=x^{2}+x$ is a many-one into function?

Sol. Many-one

Let $a_1, a_2$ be any two arbitrary elements of $R$, then

$ \begin{aligned} f(a_1)=f(a_2) & \Rightarrow a_1^{2}+a_1=a_2^{2}+a_2 \Rightarrow a_1^{2}-a_2^{2}+a_1+a_2=0 \\ & \Rightarrow(a_1-a_2)(a_1+a_2)+(a_1-a_2)=0 \Rightarrow(a_1-a_2)(a_1+a_2+1)=0 \\ & \Rightarrow a_1-a_2=0 \text{ or } a_1+a_2+1=0 \Rightarrow a_1=a_2 \text{ or } a_1+a_2=-1 \in R \\ & \Rightarrow \text{ Both the inferences can be true. } \end{aligned} $

So, $f(a_1)=f(a_2)$ does not necessarily imply $a_1=a_2 \Rightarrow \boldsymbol{f}$ is many-one.

Into

Let $y=x^{2}+x$, then for all $y \in R$, there does not exist all $x \in R$, as for $y=-1,-2, \ldots$, etc. There is no pre-image in $R$. Hence $\boldsymbol{f}$ is an into function.

$\Rightarrow f$ is many-one into function.

Ex. 4 . Let $f: N \to N$ be defined by

$ f(n)=\begin{cases} \frac{n+1}{2}, \text{ if } n \text{ is odd } \\ \frac{n}{2} , \text{ if } n \text{ is even } \end{cases} \text{ for all } n \in N. $

Find whether the function is bijective or not. Give reasons.

Sol. $f(3)=\frac{3+1}{2}=2$ and $f(4)=\frac{4}{2}=2$

So, although $3 \neq 4, f(3)=f(4) \Rightarrow \boldsymbol{f}$ is not one-one but many-one. For onto, consider any $n \in N$.

When $n$ is odd, then $(2 n-1)$ is odd, $so f(2 n-1)=\frac{2 n-1+1}{2}=n$

when $n$ is even, then $2 n$ is even, so $f(2 n)=\frac{2 n}{2}=n$

Thus for every $n \in N$, whether even or odd, there exists a pre-image $n \in N$, so the function $\boldsymbol{f}$ is onto.

Hence the function $\boldsymbol{f}$ is many-one onto.

Ex. 5 . If $f(x)=\frac{x-|x|}{|x|}$, then find $f(-1)$.

Sol. $f(-1)=\frac{-1-|-1|}{|-1|}=\frac{-1-1}{1}=-\mathbf{2}$.

Ex. 6 . If $f(x)=\log (\frac{1-x}{1+x})$, show that $f(a)+f(b)=f(\frac{a+b}{1+a b})$.

Sol.

$ \begin{aligned} f(a) & =\log (\frac{1-a}{1+a}), f(b)=\log (\frac{1-b}{1+b}) \\ f(\frac{a+b}{1+a b}) & =\log (\frac{1-\frac{a+b}{1+a b}}{1+\frac{a+b}{1+a b}}) \\ \Rightarrow f(\frac{a+b}{1+a b}) & =\log (\frac{\frac{1+a b-a-b}{1+a b}}{\frac{1+a b+a+b}{1+a b}})=\log (\frac{(1-a)+b(a-1)}{(1+a)+b(a+1)})=\log (\frac{(1-a)(1-b)}{(1+a)(1+b)}) \\ & =\log (\frac{1-a}{1+a})+\log (\frac{1-b}{1+b}) \quad(\because \log a b=\log a+\log b) \\ & =\boldsymbol{f}(\boldsymbol{a})+\boldsymbol{f}(\boldsymbol{b}) . \end{aligned} $

Ex. 7 . If $f(x)=\begin{cases} 2 x-1, \text{ when } x \leq 0 \\ x^{2}, \text{ when } x>0\end{cases}$ , then find $f(\frac{3}{4})$ and $f(-\frac{3}{4})$.

Sol. $\quad f(\frac{3}{4})=(\frac{3}{4})^{2}=\frac{\mathbf{9}}{\mathbf{1 6}} \quad$ as $f(x)=x^{2}$ when $x>0$

$ f(-\frac{3}{4})=2 \times(\frac{-3}{4})-1=-\frac{3}{2}-1=-\frac{\mathbf{5}}{\mathbf{2}} \quad \text{ as } f(x)=2 x-1 \text{ when } x \leq 0 . $

Ex. 8 . Consider $f:{4,5,6} \to{p, q, r}$ given by $f(4)=p, f(5)=q, f(6)=r$. Find the inverse of $f$ i.e., $f^{-1}$ and show that $(f^{-1})^{-1}=f$.

Sol. Given, $f:{4,5,6} \to{p, q, r}$ such that

$ f(4)=p, f(5)=q, f(6)=r \Rightarrow f={4, p),(5, q),(6, r)} \quad(f \text{ defined by ordered pairs }) $

$\Rightarrow f$ is one-one and onto $\Rightarrow f^{-1}$ exists

$ \Rightarrow \quad f^{-1}={(p, 4),(q, 5),(r, 6)} $

( $\because$ components of ordered pains are interchanged in case of inverse functions)

$\Rightarrow (f^{-1})^{-1}=[(4, p),(5, q),(6, r)]=f$.

Ex. 9 . Consider $f: R^{+} \to(4, \infty)$ given by $f(x)=x^{2}+4$. Show that $f$ is invertible with $f^{-1}=\sqrt{y-4}$ where $R^{+}$is the set of all non-negative real numbers.

Sol. For the function $f$ to be invertible, $\boldsymbol{f}$ has to be one-one onto.

One-One

Let $x_1, x_2 \in R^{+}$such that $f(x_1)=f(x_2)$

$\Rightarrow x_1^{2}+4=x_2^{2}+4 \Rightarrow x_1^{2}=x_2^{2} \Rightarrow(x_1-x_2)(x_1+x_2)=0$

$\Rightarrow x_1-x_2=0 \quad \Rightarrow x_1=x_2 \quad(\because x_1+x_2 \neq 0.$ as $.x_1, x_2 \in R^{+})$

$\Rightarrow f$ is one-one

Onto

Let $y=f(x)=x^{2}+4 \Rightarrow x^{2}=y-4 \Rightarrow x= \pm \sqrt{y-4} \Rightarrow x=f^{-1}(y)=\sqrt{y-4}$

$\Rightarrow \quad f^{-1}(x)=\sqrt{x-4} \quad(\because x \in R^{+}.$so we ignore-ve value $)$

For every element $y \in[4, \infty]$, there exists a pre image $f^{-1}(x) \in R^{+}$. So $\boldsymbol{f}$ is onto.

Hence $\boldsymbol{f}$ being one-one onto is invertible.

Ex. 10 . If $f(x)=[4-(x-7)^{3}]^{1 / 5}$ is a real valued function, then show that $f$ is invertible and find $f^{-1}$.

Sol. For $\boldsymbol{f}$ to be invertible, $\boldsymbol{f}$ should be bijective.

One-One

Let $x_1, x_2$ be any arbitrary value $\in R$, then

$ \begin{aligned} f(x_1)=f(x_2) & \Rightarrow[4-(x_1-7)^{3}]^{1 / 5}=[4-(x_2-7)^{3}]^{1 / 5} \\ & \Rightarrow 4-(x_1-7)^{3}=4-(x_2-7)^{3} \\ & \Rightarrow(x_1-7)^{3}=(x_2-7)^{3} \\ & \Rightarrow x_1-7=x_2-7 \\ & \Rightarrow x_1=x_2 \Rightarrow f \text{ is one-one. } \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 4-(x_1-7)^{3}=4-(x_2-7)^{3} \quad \text{ (Putting both the sides to power 5) } \\ & \Rightarrow \quad x_1-7=x_2-7 \quad \text{ (Taking cube root of both the sides) } \end{aligned} $

Onto Let $y=f(x)=[4-(x-7)^{3}]^{1 / 5}$

$ \begin{aligned} & \Rightarrow y^{5}=4-(x-7)^{3} \Rightarrow(x-7)^{3}=4-y^{5} \\ & \Rightarrow \quad x-7=\sqrt[3]{4-y^{5}} \Rightarrow x=\sqrt[3]{4-y^{5}}+7 \end{aligned} $

Thus for every $y \in R$, there exists an $x=(\sqrt[3]{4-y^{5}}+7) \in R \quad \Rightarrow \quad \boldsymbol{f}$ is onto.

$\therefore$ From (i) $x=f^{-1}(y)=\sqrt[3]{4-y^{5}}+7 \quad$ or $f^{-1}(x)=\sqrt[3]{4-y^{5}}+7$ (Substituting $x$ for $y$ ).

Ex. 11 . Examine whether $x[\frac{a^{x}+1}{a^{x}-1}]$ is an odd or even function.

Sol. Let $\quad f(x)=x(\frac{a^{x}+1}{a^{x}-1})$

Then

$f(-x)=(-x)(\frac{a^{-x}+1}{a^{-x}-1})=(-x)[\frac{\frac{1}{a^{x}}+1}{\frac{1}{a^{x}}-1}]=(-x)[\frac{1+a^{x}}{1-a^{x}}]=(-x)[\frac{a^{x}+1}{-(a^{x}-1)}]=x[\frac{a^{x}+1}{a^{x}-1}]=f(x)$

$\therefore f(x)$ is an even function.

PRACTICE SHEET-1

If A= {1,2,3,4}, B={1,2,3,4,5,6} are two sets and the function $\boldsymbol{f}: A \to B$ is defined by $f(x)=x+2 \forall x \in A$, then the function $f$ is (a) bijective (b) one-one (c) onto (d) many-one

(WBJEE 2010)

Define $g: R \to R$ by $g(x)=x^{2}+1$. Then $g$ is

(a) neither injective nor surjective

(b) injective but not surjective

(d) bijective

Let $f: R \to R$ be a function defined by $f(x)=\frac{x-m}{x-n}$, when $m \neq n$, then (a) $f$ is one-one onto (b) $f$ is many-one onto (c) $f$ is one-one into (d) $f$ is many-one into

(UPSEAT 2004)

If $f:(0, \infty) \to(0, \infty)$ and $f(x)=\frac{x}{1+x}$, then the function $f$ is (a) one-one and onto (b) one-one but not onto (c) onto but not one-one (d) neither one-one nor onto.

(J&K CET 2011)

If $f: R \to R$ be a function defined as $f(x)=x^{4}$, then (a) $f$ is one-one onto (b) $f$ is many-one onto (c) $f$ is one-one but not onto (d) $f$ is neither one-one nor onto
If $\quad f(x)=\begin{cases} x, \text{ if } x \text{ is rational } \\ 0, \text{ if } x \text{ is irrational }\end{cases} .$

and $g(x)=\begin{cases} 0, \text{ if } x \text{ is rational } \\ x, \text{ if } x \text{ is irrational }\end{cases} $

then $f-g$ is (a) neither one-one nor onto (b) one-one and onto (c) one-one and into (d) many-one and onto

(IIT 2005)

The function $f: R \to R$ defined by $f(x)=(x-1)(x-2)$ $(x-3)$ is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto

(VITEEE 2012)

Which one of the following is a bijective function on the set of real numbers? (a) $2 x-5$ (b) $|x|$ (c) $x^{2}+1$ (d) $x^{4}-x^{2}+1$

(Kerala PET 2002)

A mapping $f: R \to R$ defined by $f(x)=e^{x}$ where $x \in R$ is (a) one-one onto (b) one-one into (c) many-one into (d) many-one into

(Odisha JEE 2012)

If $f(x)=\frac{x}{x-1}$, then $\frac{f(a)}{f(a+1)}$ equals (a) $f(\frac{1}{a})$ (b) $f(a)^{2}$ (c) $f(-a)$ (d) $f(0)$
If $f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}$ for $x>2$, then $f(11)$ equals (a) $\frac{7}{6}$ (b) $\frac{5}{6}$ (c) $\frac{6}{7}$ (d) $\frac{5}{7}$

(EAMCET 2003)

Let $\phi(x)=\frac{b(x-a)}{b-a}+\frac{a(x-b)}{a-b}$, where $x \in R$ and $a$ and $\boldsymbol{b}$ are fixed numbers with $a \neq b$. Then $\phi(a+b)$ is equal to (a) $\phi(0)$ (b) $\phi(a-b)$ (c) $\phi(a b)$ (d) $\phi(a)+\phi(b)$

(Kerala PET 2012)

If $f(x)=\frac{4^{x}}{4^{x}+2}$, then $f(x)+f(1-x)$ is equal to (a) -1 (b) 0 (c) 1 (d) None of these
If $f(x)=\frac{1}{1+\frac{1}{x}} ; g(x)=\frac{1}{1+\frac{1}{f(x)}}$, then $g(2)$ equals (a) $\frac{1}{5}$ (b) $\frac{1}{25}$ (c) $\frac{2}{5}$ (d) $\frac{1}{16}$
If $f(x)=\cos (\log x)$, then $f(x^{2}) . f(y^{2})$

$-\frac{1}{2}[f(x^{2} \cdot y^{2})+f(\frac{x^{2}}{y^{2}})]$ equals (a) -2 (b) -1 (c) $\frac{1}{2}$ (d) 0

The function $f(x)=\sin x+\cos x$ will be (a) an even function (b) an odd function (c) a constant function (d) none of these
Let $A=[-2,2]-{0}$. Define $f: A \to R$ and $g: A \to R$ by $f(x)=\frac{1}{x^{3}+2 x|x|}$ and $g(x)=\sqrt{x^{4}+\frac{3}{x^{2}}}$, then (a) $f$ and $g$ are odd (b) $f$ is odd, $g$ is even (c) $f$ is even, $g$ is odd (d) $f$ and $g$ are even
The function $f(x)=\sin [\log (x+\sqrt{x^{2}+1})]$ is (a) an odd function (b) an even function (c) neither even nor odd (d) None of these.
If $f(x)=x^{2}+k x+1$ for all $x$ and $f$ is an even function, find $k$ where $k \in R$ (a) 1 (b) -2 (c) 0 $(d)-1$
Let $F(x)=[\frac{g(x)-g(-x)}{f(x)+f(-x)}]^{m}$ such that $m=2 n, n \in N$ and $f(-x) \neq-f(x)$. Then, $F(x)$ is (a) an odd function (b) an even function (c) constant function (d) None of these
Let $f: R \to R$ be defined as $f(x)=x^{2}+1$. Then find $f^{-1}(-5)$ (a) ${0}$ (b) $\phi$ (c) ${5}$ (d) ${-5,5}$

(VITEEE 2011)

Let the function $\boldsymbol{f}$ be defined by $f(x)=\frac{2 x+1}{1-3 x}$, then $f^{-1}(x)$ is (a) $\frac{1-3 x}{2 x+1}$ (b) $\frac{x-1}{3 x+2}$ (c) $\frac{x+1}{2 x-1}$ (d) $\frac{3-x}{2+x}$

(Kerala PET 2002)

If $f(x)=x-x^{2}+x^{3}-x^{4}+\ldots . \infty$ for $|x|<1$, then $f^{-1}(x)$ is equal to (a) $\frac{1+x}{x}$ (b) $\frac{x}{1+x}$ (c) $\frac{1-x}{x}$ (d) $\frac{x}{1-x}$
If $f:[1, \infty) \to[2, \infty)$ is given by $f(x)=x+\frac{1}{x}$, then $f^{-1}(x)$ equals (a) $\frac{x+\sqrt{x^{2}-4}}{2}$ (b) $\frac{x}{1+x^{2}}$ (c) $\frac{x-\sqrt{x^{2}-4}}{2}$ (d) $1+\sqrt{x^{2}-4}$

(IIT 2001)

The inverse of $f(x)=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}$ is (a) $\frac{1}{2} \log _{10}(\frac{1+x}{1-x})$ (b) $\log _{10}(2-x)$ (c) $\frac{1}{2} \log _{10}(2 x-1)$ (d) $\frac{1}{4} \log _{10}(\frac{2 x}{2-x})$

(Odisha JEE, WBJEE 2007)

ANSWERS

1. $(b)$	2. (a)	3. $(c)$	4. $(b)$	5. $(d)$	6. $(b)$	7. $(b)$	8. (a)	9. $(b)$	10. $(b)$
11. (c)	12. $(d)$	13. (c)	14. (c)	15. $(d)$	16. $(d)$	17. $(b)$	18. (a)	19. $(c)$	20. (b)
21. $(b)$	22. (b)	23. (d)	24. (a)	25. (a)

HINTS AND SOLUTIONS

$f(x)=x+2$ and $f: A \to B$

$ \begin{aligned} \therefore \quad f(1) & =1+2=3, f(2)=2+2=4 \\ f(3) & =3+2=5, f(4)=4+2=6 \end{aligned} $

Thus each element in $A$ has a unique image in $B$ and no two elements in $B$ have the same pre-image in $A$. So $f$ is one-one. But not all the elements of $B$, i.e., 1 and 2 have a pre-image in $A$. So, the function $\boldsymbol{f}$ is not onto.

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

$\Rightarrow$ Both -1 and $1 \in R$ have the same image $2 \in R$

$\Rightarrow g$ is not one-one or injective

Let $z=g(x)=x^{2}+1 \Rightarrow x= \pm \sqrt{z-1}$ $x$ is not defined when $z-1<0$, i.e., $z<1$

$\therefore$ for $0 \in R$, there exists no pre-image in $R$

$\Rightarrow g(x)$ is not onto or surjective.

$\therefore \quad g$ is neither injective nor surjective.

$ \text{ 3. } \begin{aligned} \forall(x, y) \in R, f(x)=f(y) \Rightarrow \frac{x-m}{x-n}=\frac{y-m}{y-n} \\ \Rightarrow(x-m)(y-n)=(y-m)(x-n) \\ \Rightarrow x y-m y-n x+m n=y x-m x-n y+m n \\ \Rightarrow m x-n x=m y-n y \Rightarrow(m-n) x=(m-n) y \\ \Rightarrow x=y \Rightarrow f \text{ is one-one. } \end{aligned} $

Let $\quad z=f(x)=\frac{x-m}{x-n} \Rightarrow z x-z n=x-m$

$\Rightarrow z x-x=z n-m \Rightarrow x(z-1)=z n-m$

$\Rightarrow x=\frac{z n-m}{z-1}=\frac{m-z n}{1-z}$

$x$ is not defined for $z=1 \Rightarrow$ for $z=1$, there exists no pre-image in $R$

$\Rightarrow f$ is not onto.

$\therefore \boldsymbol{f}$ is one-one, into function.

$\forall(x, y) \in(0, \infty) \quad f(x)=f(y) \Rightarrow \frac{x}{1+x}=\frac{y}{1+y}$

$\Rightarrow x+x y=y+x y \Rightarrow x=y \Rightarrow f$ is one-one.

Let $z=f(x)=\frac{x}{1+x} \Rightarrow z+z x=x$

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

$\Rightarrow x$ is not defined for $z=1$ and $1 \in(0, \infty)$

$\therefore 1 \in(0, \infty)$ has no pre-image in $(0, \infty)$

$\Rightarrow f$ is not onto.

For all $x, y \in R$,

$ \begin{aligned} & f(x)=f(y) \Rightarrow x^{4}=y^{4} \Rightarrow x^{4}-y^{4}=0 \\ \Rightarrow & (x^{2}-y^{2})(x^{2}+y^{2})=0 \Rightarrow(x-y)(x+y)(x^{2}+y^{2})=0 \\ \therefore & (x-y)=0 \text{ or }(x+y)=0 \text{ or }(x^{2}+y^{2})=0 \\ \Rightarrow & x=y \text{ or } x=-y \text{ or } x= \pm y \\ \therefore & x=-y \Rightarrow f \text{ is a many-one function } \end{aligned} $

Also let $z=f(x)=x^{4} \Rightarrow x=(z)^{1 / 4}$

Now $z=-1 \in R$ has no pre-image in $R$ as $(-1)^{1 / 4} \notin R$.

$\therefore f: R \to R$ is not an onto function.

$(f-g)(x)=f(x)-g(x)=\begin{cases} x, \text{ if } x \text{ is rational } \\ -x, \text{ if } x \text{ is irrational }\end{cases} $

Distinct elements of $R$ have distinct images in $(f-g)$. Hence $(f-g)$ is one-one.

Also Range of $(f-g)=R \Rightarrow f$ is onto. $\Rightarrow(f-g) R \to R$ is a one-one onto function.

Given $f(x)=(x-1)(x-2)(x-3)$

$\Rightarrow f(1)=f(2)=f(3)=0$

$\Rightarrow f: R \to R$ is not one-one

For each $y \in R, y=(x-1)(x-2)$

$(x-3)$, there exists an $x \in R$, such

that $f(x)=y \quad \Rightarrow \quad f$ is onto.

Let $f(x)=2 x-5 \quad f: R \to R$

Then, for all $x, y \in R, f(x)=f(y)$

$\Rightarrow 2 x-5=2 y-5 \Rightarrow x=y \Rightarrow f$ is one-one

Also, let $z=f(x)=2 x-5 \quad \Rightarrow \quad x=\frac{z+5}{2}$

$\therefore$ For all $z \in R$, there exists an $x \in R$ such that

$z=f(x)=2 x+5 \Rightarrow f$ is onto.

$\therefore f$ is one-one onto $\Rightarrow f$ is bijective.

Now, check for other functions:

Let $g(x)=|x|, g: R \to R$ Then $g(-1)$ and $g(1)=1$

$\Rightarrow$ Both 1 , and -1 have the same image in $R$

$\Rightarrow g$ is not one-one

Similarly, for the functions $x^{2}+1$ and $x^{2}-x^{2}+1 ;-1$ and 1 have the same image in $R$, so both the functions are not one-one, hence bijective.

For all $x_1, x_2 \in R$,

$ f(x_1)=f(x_2) $

$\Rightarrow e^{x_1}=e^{x_2} \Rightarrow x_1=x_2 \Rightarrow f$ is one-one.

Let $y=f(x)=e^{x} \Rightarrow x=\log _{e} y$

Now $x$ is not defined for $y<0$

$\therefore \quad y \in R$ does not have a pre-image in $R$ when $y$ is negative.

$\therefore \boldsymbol{f}$ is into.

Given, $f(x)=\frac{x}{x-1} \Rightarrow \frac{f(a)}{f(a+1)}=\frac{\frac{a}{a-1}}{\frac{a+1}{a+1-1}}$

$ =\frac{a}{a-1} \times \frac{a}{a+1}=\frac{a^{2}}{a^{2}-1}=f(a^{2}) . $

Given, $f(x)=\frac{1}{\sqrt{x+2 \sqrt{2 x-4}}}+\frac{1}{\sqrt{x-2 \sqrt{2 x-4}}}$

$\Rightarrow \quad f(11)=\frac{1}{\sqrt{11+2 \sqrt{22-4}}}+\frac{1}{\sqrt{11-2 \sqrt{22-4}}}$

$=\frac{1}{\sqrt{11+2 \sqrt{18}}}+\frac{1}{\sqrt{11-2 \sqrt{18}}}$

$=\frac{1}{\sqrt{11+6 \sqrt{2}}}+\frac{1}{\sqrt{11-6 \sqrt{2}}}$

$=\frac{1}{\sqrt{(9+2 \times 3 \times \sqrt{2}+(\sqrt{2})^{2}.}}$

$+\frac{1}{\sqrt{(9-2 \times 3 \times \sqrt{2}+(\sqrt{2})^{2}.}}$

$=\frac{1}{\sqrt{(3+\sqrt{2})^{2}}}+\frac{1}{\sqrt{(3-\sqrt{2})^{2}}}$

$=\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{3-\sqrt{2}+3+\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}$

$=\frac{6}{(9-2)}=\frac{6}{7}$.

$ \phi(x)=\frac{b(x-a)}{b-a}+\frac{a(x-b)}{a-b} $

Then, $\phi(a+b)=\frac{b(a+b-a)}{b-a}+\frac{a(a+b-b)}{a-b}$

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

Now

$ \phi(a)=\frac{b \times 0}{b-a}+\frac{a(a-b)}{a-b}=a $

$ \phi(b)=\frac{b(b-a)}{b-a}+\frac{a \times 0}{a-b}=b $

(i), (ii) and (iii) $\Rightarrow \phi(a+b)=\phi(a)+\phi(b)$.

Given, $f(x)=\frac{4^{x}}{4^{x}+2}$

$ \begin{aligned} f(1-x) & =\frac{4^{1-x}}{4^{1-x}+2}=\frac{4 \cdot 4^{-x}}{4^{1} \cdot 4^{-x}+2}=\frac{4 \cdot \frac{1}{4^{x}}}{4 \cdot \frac{1}{4^{x}}+2} \\ & =\frac{\frac{4}{4^{x}}}{\frac{4+2 \cdot 4^{x}}{4^{x}}}=\frac{4}{4+2 \cdot 4^{x}}=\frac{2}{2+4^{x}} \end{aligned} $

Now $f(x)+f(1-x)=\frac{4^{x}}{4^{x}+2}+\frac{2}{2+4^{x}}=\frac{4^{x}+2}{4^{x}+2}=\mathbf{1}$.

Given, $f(x)=\frac{1}{1+\frac{1}{x}}$ and $g(x)=\frac{1}{1+\frac{1}{f(x)}}$

$\therefore \quad f(x)=\frac{1}{\frac{x+1}{x}}=\frac{x}{x+1}$

$\Rightarrow \quad g(x)=\frac{1}{1+\frac{1}{\frac{x}{x+1}}}=\frac{1}{1+\frac{x+1}{x}}=\frac{x}{2 x+1}$

$\Rightarrow \quad g(2)=\frac{2}{2 \times 2+1}=\frac{\mathbf{2}}{\mathbf{5}}$.

Given, $f(x)=\cos (\log x)$

$ \begin{aligned} & \Rightarrow \quad f(x^{2})=\cos (\log x^{2})=\cos (2 \log x) \\ & f(y^{2})=\cos (\log y^{2})=\cos (2 \log y) \\ & f(\frac{x^{2}}{y^{2}})=\cos (\log \frac{x^{2}}{y^{2}})=\cos (\log x^{2}-\log y^{2}) \\ & =\cos (2 \log x-2 \log y) \\ & f(x^{2} y^{2})=\cos (\log x^{2} y^{2})=\cos (\log x^{2}+\log y^{2}) \\ & =\cos (2 \log x+2 \log y) \end{aligned} $

Now, $f(x^{2}) \cdot f(y^{2})-\frac{1}{2}[f(\frac{x^{2}}{y^{2}})+f(x^{2} y^{2})]$

$ \begin{matrix} =\cos (2 \log x) \cdot \cos (2 \log y)-\frac{1}{2}[\cos (2 \log x-2 \log y) \\ +\cos (2 \log x+2 \log y)] \\ =\cos (2 \log x) \cdot \cos (2 \log y) \quad \\ -\frac{1}{2}[2 \cos (\frac{2 \log x-2 \log y+2 \log x+2 \log y}{2}). \\ .\cos (\frac{2 \log x+2 \log y-2 \log x+2 \log y}{2})] \end{matrix} $

$ (\because \cos C+\cos D=2 \cos (\frac{C+D}{2}) \cos (\frac{D-C}{2})) $

$=\cos (2 \log x) \cdot \cos (2 \log y)$

$ -\frac{1}{2}[2 \cos (2 \log x) \cdot \cos (2 \log y)]=\mathbf{0} . $

$f(x)=\sin x+\cos x$

$f(-x)=\sin (-x)+\cos (-x)=-\sin x+\cos x \neq f(x)$ or $-f(x)$

$\therefore \quad f(x)$ is neither even nor odd nor constant.

$f(x)=\frac{1}{x^{3}+2 x|x|}$

$ \begin{aligned} \therefore f(-x) & =\frac{1}{(-x)^{3}+2(-x)|-x|} \\ & =\frac{1}{-x^{3}-2 x|x|} \quad(\because|-x|=|x|) \\ & =-\frac{1}{x^{3}+2 x|x|}=-f(x) \Rightarrow f \text{ is odd. } \end{aligned} $

$ g(x)=\sqrt{x^{4}+\frac{3}{x^{2}}} $

$\therefore g(-x)=\sqrt{(-x)^{4}+\frac{3}{(-x)^{2}}}=\sqrt{x^{4}+\frac{3}{x^{2}}}=g(x)$

$\Rightarrow g$ is even.

$f(x)=\sin [\log (x+\sqrt{x^{2}+1})]$

$ \begin{aligned} \therefore f(-x) & =\sin [\log (-x+\sqrt{(-x)^{2}+1})] \\ & =\sin [\log (\sqrt{1+x^{2}}-x)] \\ & =\sin [\log (\sqrt{1+x^{2}}-x) \frac{(\sqrt{1+x^{2}}+x)}{(\sqrt{1+x^{2}}+x)}] \\ & =\sin [\log (\frac{(1+x^{2})-x^{2}}{\sqrt{1+x^{2}}+x})] \\ & =\sin [\log (\sqrt{1+x^{2}}+x)^{-1}] \\ & =\sin [-\log (\sqrt{1+x^{2}}+x)](\because \log a^{-1}=-\log a) \\ & =-\sin [\log (\sqrt{1+x^{2}}+x)](\because \sin (-x)=-\sin x) \\ & =-f(x) \Rightarrow f \text{ is an odd function. } \end{aligned} $

$f(x)=x^{2}+k x+1$

Given, $f(x)$ is an even function

$ \begin{aligned} & \Rightarrow f(x)=f(-x) \Rightarrow x^{2}+k x+1=(-x)^{2}+k(-x)+1 \\ & \Rightarrow x^{2}+k x+1=x^{2}-k x+1 \Rightarrow 2 k x=0 \Rightarrow k=\mathbf{0} . \end{aligned} $

$F(x)=[\frac{g(x)-g(-x)}{f(x)+f(-x)}]^{m}$

$ \begin{aligned} \therefore F(-x) & =[\frac{g(-x)-g(x)}{f(-x)+f(x)}]^{m}=[\frac{-(g(x)-g(-x)}{f(-x)+f(x)}]^{m} \\ & =[-(\frac{g(x)-g(-x)}{f(-x)+f(x)})]^{2 n} \\ & =[\frac{g(x)-g(-x)}{f(-x)+f(x)}] \quad(\because 2 n \Rightarrow \text{ even power }) \\ & =F(x) \end{aligned} $

$\Rightarrow \quad F$ is an even function.

$f(x)=x^{2}+1$

Let $y=f(x)=x^{2}+1$

$\Rightarrow y-1=x^{2} \Rightarrow x= \pm \sqrt{y-1} \Rightarrow f^{-1}(x)= \pm \sqrt{x-1}$

$\therefore \quad f^{-1}(-5)= \pm \sqrt{-5-1}= \pm \sqrt{-6} \notin R$

$\therefore f^{-1}(-5)$ is the null set.

$f(x)=\frac{2 x+1}{1-3 x}$

Let $y=f(x)=\frac{2 x+1}{1-3 x} \Rightarrow y(1-3 x)=2 x+1$

$\Rightarrow y-3 x y=2 x+1 \Rightarrow x(-3 y-2)=1-y$

$\Rightarrow x=\frac{1-y}{-3 y-2}=\frac{y-1}{3 y+2}$ or $f^{-1}(x)=\frac{\boldsymbol{x}-\mathbf{1}}{\mathbf{3} \boldsymbol{x}+\mathbf{2}}$.

$f(x)=x-x^{2}+x^{3}-x^{4}+\ldots . . \infty,|x|<1$

This is an infinite G.P. with $a=x, r=-x$.

$\therefore f(x)=S _{\infty}=\frac{a}{1-r}=\frac{x}{1+x}$

$ \begin{aligned} & \text{ Let } y=f(x)=\frac{x}{1+x} \Rightarrow y(1+x)=x \Rightarrow y+x y=x \\ & \Rightarrow y=x(1-y) \Rightarrow x=\frac{y}{1-y} \Rightarrow f^{-1}(x)=\frac{\boldsymbol{x}}{\mathbf{1 - x}} \end{aligned} $

Given, $f(x)=x+\frac{1}{x}$

$ \begin{aligned} \text{ Let } & y & =f(x)=x+\frac{1}{x} \Rightarrow y=\frac{x^{2}+1}{x} \\ \Rightarrow & x y & =x^{2}+1 \Rightarrow x^{2}-x y+1=0 \\ \therefore & & x=\frac{y \pm \sqrt{y^{2}-4}}{2} \end{aligned} $

Since $y \in[2, \infty)$, so $f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}$.

Given, $y=\frac{10^{x}-10^{-x}}{10^{x}+10^{-x}}$

$ \begin{aligned} \Rightarrow \frac{y+1}{y-1} & =\frac{10^{x}-10^{-x}+10^{x}+10^{-x}}{10^{x}-10^{-x}-10^{x}-10^{-x}} \\ & =\frac{10^{x}}{-10^{-x}} \text{ (Using componendo and dividendo) } \end{aligned} $

$\Rightarrow-10^{2 x}=\frac{y+1}{y-1} \Rightarrow 10^{2 x}=\frac{1+y}{1-y}$

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

$\Rightarrow f^{-1}(x)=\frac{1}{2} \log _{10}(\frac{1+x}{1-x})$.

Domain and Range of a Function

KEY FACTS

The set of numbers $x$ for which the function $f(x)$ is defined is called the domain of the function.

Ex. (i) $f(x)=\frac{1}{x}$; Domain $=R-{0}$ as the function is defind for all real numbers except 0 .

(ii) $f(x)=\sqrt{x}$; Domain $=$ All non-negative real numbers as $\sqrt{x}$ is not defined when $x$ is a negative real numbers.

To find the domain of a function, following points should be kept in mind.

(i) For algebraic functions (a) denominator should be non-zero

(b) expression under an even root should be non-negative.

(ii) For trigonometric functions (a) $\sin x$ and $\cos x$ are defined for all values of $x$

(b) tan $x$ and sec $x$ are defined for all real values of $x$ except $x=(2 n+1) \frac{\pi}{2}$ where $n \in I$.

(iii) Logarithmic functions: $\log _{\boldsymbol{b}} \boldsymbol{a}$ is defined when $a>0, b>1$ and $b \neq 1$.

(iv) Exponential functions: $a^{x}$ is defined for all real values of $x$, where $a>0$.

Once the domain of a function $f(x)$ is known, the set of all the values that $\boldsymbol{f}(\boldsymbol{x})$ can take is called the range of the function.

To find the range of a function $\boldsymbol{y}=\boldsymbol{f}(\boldsymbol{x})$ :

Step 1. Find the domain of the function $y=f(x)$

Step 2. Solve the equation $y=f(x)$ to find $x$ in terms of $y$.

Step 3. Find the real values of $y$ for which $x$ is real. The set of values of $y$ so obtained makes up the range off.

SOLVED EXAMPLES

Ex. 1 . Write the domain of the following real functions? (a) $\sqrt{9-x^{2}}$ (b) $10^{x}$ (c) $\frac{2}{4 x+7}$ (d) $\log (2-3 x)$

Sol. (i) Since $f$ is a real function, $9-x^{2} \geq 0 \Rightarrow(3+x)(3-x) \geq 0 \Rightarrow-a \leq x \leq a$

$\therefore$ Domain of $f(x)=\sqrt{9-x^{2}}$ is ${x \mid-\mathbf{3} \leq \boldsymbol{x} \leq \mathbf{3}}$.

(ii) Let $f(x)=10^{x}$

$a^{x}$ is defined for all real values of $x$ when $a>0$

Here $a=10, \therefore$ Domain of $f(x)=10^{x}$ is $\boldsymbol{R}$.

(iii) $f(x)=\frac{2}{4 x+7}$

Since $\boldsymbol{f}$ is real, $x$ can take all real values except the value for which $4 x+7=0$, i.e., $x=-\frac{7}{4} \Rightarrow x \neq-\frac{7}{4}$

$\therefore \quad$ Domain of $f(x)=\frac{2}{4 x+7}=\boldsymbol{R}-{-\frac{7}{4}}$. (iv) $f(x)=\log (2-3 x)$

For $f(x)$ to be defined $(2-3 x)>0 \Rightarrow 2>3 x \Rightarrow x<\frac{2}{3}$

$\therefore$ Domain of $f(x)=\log (2-3 x)=-\infty<x<\frac{2}{3}$ or $x \in(-\infty \frac{\mathbf{2}}{\mathbf{3}})$

Ex. 2 . Find the domain of $\sqrt{x}+\sqrt{3 x-2}$

Sol. Let $f(x)=\sqrt{x}$ and $g(x)=\sqrt{3 x-2}$

Let the domain of $f(x)$ be $A$ and that of $g(x)$ be $B$

Then, $\quad$ domain of $f(x)=A=[0, \infty)$

$(\because \sqrt{x}$ is defined when $x>0$

domain of $g(x)=B=[\frac{2}{3}, \infty) \quad(\because \sqrt{3 x-2}.$ is defined when $.3 x-2 \geq 0 \Rightarrow x \geq \frac{2}{3})$

$\therefore$ Domain of $\sqrt{x}+\sqrt{3 x-2}=A \cap B=[0, \infty) \cap[\frac{2}{3}, \infty)=[\frac{\mathbf{2}}{\mathbf{3}}, \infty)$.

Ex. 3 . Find the domain of the function $f(x)=\frac{\log _2(x+3)}{x^{2}+3 x+2}$.

(Kerala PET 2013)

Sol. The function is defined when $x+3>0$ and $x^{2}+3 x+2 \neq 0$

$ \begin{aligned} & \Rightarrow \quad x>-3 \quad \text{ and } \quad(x+2)(x+1) \neq 0 \\ & \Rightarrow \quad x \in(-3, \infty) \text{ and } \quad x \neq-2 ; x \neq-1 \end{aligned} $

$\therefore \quad$ Domain of $f(x)=\frac{\log _2(x+3)}{x^{2}+3 x+2}$ is $(-\mathbf{3}, \infty)-{-\mathbf{1},-\mathbf{2}}$.

Ex. 4 . What is the domain of the function $f(x)=\log _4(\log _5(\log _3(18 x-x^{2}-77)))$.

Sol. $\log _{b} a$ is defined for all $a>0, b>0$ and $b \neq 1$

$\therefore f(x)$ will be defined if $\log _5(\log _3(18 x-x^{2}-77))>0 \Rightarrow \log _3(18 x-x^{2}-77)>5^{0}$, i.e., 1

$ \begin{aligned} & \Rightarrow \quad 18 x-x^{2}-77>3^{1} \Rightarrow 18 x-x^{2}-80>0 \quad \Rightarrow \quad x^{2}-18 x+80<0 \\ & \Rightarrow \quad(x-8)(x-10)<0 \Rightarrow 8<x<10 \Rightarrow \boldsymbol{x} \in(\mathbf{8}, \mathbf{1 0}) . \end{aligned} $

$\therefore \quad$ Domain of $f(x)$ is $\boldsymbol{x} \in(\mathbf{8}, \mathbf{1 0})$.

Ex. 5 . What is domain of the real function $f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$ ?

Sol.

$ f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}=\log _{(1-x)} 10+\sqrt{x+2} \quad(\because \frac{1}{\log _{b} a}=\log _{a} b) $

Let

$ F(x)=\log _{(1-x)} 10 \text{ and } G(x)=\sqrt{x+2} $

$F(x)$ is defined when $(1-x)>0$ and $1-x \neq 1$

$\Rightarrow x<1$ and $x \neq 0 \Rightarrow x \in(-\infty, 1)-{0}$

$G(x)$ is defined when $x+2 \geq 0 \Rightarrow x \geq-2 \Rightarrow x \in[-2, \infty)$

From (i) and (ii), the domain of $f(x)=F(x)+G(x)$ is common domain of $F(x)$ and $G(x)$, i.e.,

$ (-\infty, 1)-{0} \cap[-2, \infty)=[-\mathbf{2}, \mathbf{1})-{\mathbf{0}} $

Ex. 6 . Find the range of each of the following functions:

(i) $|x-3|$ (ii) $\sqrt{x-5}$ (iii) $\sqrt{3 x^{2}-4 x+5}$ (iv) $\frac{x}{1+x^{2}}$

(DCE 2009)

Sol. (i) $f(x)=|x-3|$ is defined for all $x \in R$, so domain of $f(x)=R$.

Now $|x-3| \geq 0$ for all $x \in R \quad \Rightarrow \quad 0 \leq|x-3|<\infty$ for all $x \in R$

$\Rightarrow f(x) \in[0, \infty)$ for all $x \in R \Rightarrow$ Range of $f(x)=|x-3|$ is $[0, \infty)$

(ii) Let $y=f(x)=\sqrt{x-5}$

$f(x)$ is defined for $x-5 \geq 0 \Rightarrow x \geq 5 \Rightarrow x \in[5, \infty)$

Now $y=\sqrt{x-5} \Rightarrow x-5=y^{2} \Rightarrow x=y^{2}+5 \Rightarrow x>0$ for all value of $y$

For $x$ to be real and $x \in[5, \infty), \boldsymbol{y} \in[0, \infty)$.

$\therefore$ Range of $f$ is $[0, \infty)$

(iii) Let $y=f(x)=\sqrt{3 x^{2}-4 x+5}$

$f(x)$ is defined if $3 x^{2}-4 x+5 \geq 0 \Rightarrow 3(x^{2}-\frac{4}{3} x+\frac{5}{3}) \geq 0 \Rightarrow 3(x^{2}-\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{5}{3}) \geq 0$

$\Rightarrow 3((x-\frac{2}{3})^{2}+\frac{11}{9}) \geq 0$, which is true for all real numbers, i.e., Domain of $f$ is $(-\infty, \infty)$

$y=\sqrt{3 x^{2}-4 x+5} \Rightarrow 3 x^{2}-4 x+5=y^{2} \Rightarrow 3 x^{2}-4 x+(5-y^{2})=0$

$\Rightarrow x=\frac{4 \pm \sqrt{(-4)^{2}-4 \times 3 \times(5-y^{2})}}{2 \times 3}$

For $x$ to be real, $(-4)^{2}-4 \times 3 \times(5-y^{2}) \geq 0 \quad \Rightarrow \quad 16-60+12 y^{2} \geq 0$

$\Rightarrow-44+12 y^{2} \geq 0 \Rightarrow 12 y^{2} \geq 44 \Rightarrow y^{2} \geq \frac{11}{3} \Rightarrow y \geq \sqrt{\frac{11}{3}}$

$\therefore \quad$ Range of $y=[\sqrt{\frac{11}{3}}, \infty)$

(iv) Let $y=f(x)=\frac{x}{1+x^{2}}$

The given function $f(x)=\frac{x}{1+x^{2}}$ is defined for all real numbers, so domain of $\boldsymbol{f}$ is $\boldsymbol{R}$.

$ y=\frac{x}{1+x^{2}} \Rightarrow x^{2} y-x+y=0 \Rightarrow x=\frac{-(-1) \pm \sqrt{1-4 y^{2}}}{2 y} $

Now, $\frac{1 \pm \sqrt{1-4 y^{2}}}{2 y}$ will be a real number if and only if

$ \begin{aligned} & \quad 1-4 y^{2} \geq 0 \text{ and } y \neq 0 \Rightarrow 4 y^{2}-1 \leq 0 \text{ and } y \neq 0 \Rightarrow(y^{2}-\frac{1}{4}) \leq 0 \text{ and } y \neq 0 \\ & \Rightarrow \quad(y-\frac{1}{2})(y+\frac{1}{2}) \leq 0 \text{ and } y \neq 0 \Rightarrow-\frac{1}{2} \leq y \leq \frac{1}{2} \text{ and } y \neq 0 \Rightarrow y \in[-\frac{1}{2}, 0) \cup(0, \frac{1}{2}] \\ & \text{ For } x=0, y=0 \quad \therefore \quad \text{ range is }[-\frac{\mathbf{1}}{\mathbf{2}}, \frac{\mathbf{1}}{\mathbf{2}}] . \end{aligned} $

Ex. 7 . If the domain of the function $f(x)=x^{2}-6 x+7$ is $(-\infty, \infty)$ then find the range of the function?

(MP PET)

Sol. Let $\quad f(x)=y=x^{2}-6 x+7 \Rightarrow x^{2}-6 x+7-y=0 \Rightarrow x=\frac{6 \pm \sqrt{36-4(7-y)}}{2}=3 \pm \sqrt{8+4 y}$

For real values of $x, 8+4 y \geq 0 \Rightarrow y \geq-2$

$\therefore$ For $f(x)=x^{2}-6 x+7$, the range is $[-2, \infty)$

Ex. 8 . Find the range of the function $f(x)=\frac{x^{2}-2}{x^{2}-3}$.

Sol. Let

$ \begin{aligned} & y=\frac{x^{2}-2}{x^{2}-3} \\ & y=\frac{x^{2}-2}{x^{2}-3} \text{ is not defined for } x^{2}-3=0, \text{ i.e., } x= \pm \sqrt{3} \end{aligned} $

$\therefore$ Domain of $y=R-{-\sqrt{3},+\sqrt{3}}$.

Now $\quad y=\frac{x^{2}-2}{x^{2}-3} \Rightarrow y x^{2}-3 y=x^{2}-2 \Rightarrow x^{2}(y-1)=3 y-2 \Rightarrow x^{2}=\frac{3 y-2}{y-1}$

$\because \quad$ LHS is a perfect square $\Rightarrow \frac{3 y-2}{y-1} \geq 0$

$ \begin{aligned} & \Rightarrow \quad(3 y-2) \geq 0 \text{ and }(y-1) \geq 0 \text{ or }(3 y-2) \leq 0,(y-1) \leq 0 \text{ and } y \neq 1 \text{ (Note) } \\ & \Rightarrow \quad y \geq \frac{2}{3} \text{ and } y>1 \text{ or } y \leq \frac{2}{3}, y<1 \Rightarrow y \in(1, \infty) \text{ or } y \in(-\infty, \frac{2}{3}] \\ & \Rightarrow \quad y \in(-\infty, \frac{2}{3}] \cup(1, \infty) \end{aligned} $

For $x= \pm \sqrt{3}, y$ does not exist.

Ex. 9 . Find the domain and range of the function $\frac{x^{2}-4}{x-2}$.

Sol. Let $\quad y=\frac{x^{2}-4}{x-2} \Rightarrow y=x+2$ when $x \neq 2$

Now $\frac{x^{2}-4}{x-2}$ is not defined only when $x=2$, therefore

Domain of $y=\frac{x^{2}-4}{x-2}$ is $\boldsymbol{R}-{\mathbf{2}}$.

But $y=x+2$ will take up all real values, but $y=4$, when $x=2$, (not possible)

$\therefore$ Range of function is $\boldsymbol{R}-{\mathbf{4}}$.

PRACTICE SHEET-2

The domain of the function $f(x)=\sqrt{|x|}$ is (a) $x \in(-\infty, \infty)$ (b) $x \in(0, \infty)$ (c) $x \in(-\infty, 0) \cup(0, \infty)$ (d) $x \in[0, \infty)$
The domain of the function $f(x)=(\sqrt{x})^{2}$ is (a) $-\infty<x<\infty$ (b) $0<x<\infty$ (c) $0 \leq x<\infty$ (d) None of these
The domain of the function $f(x)=\log x^{2}$ is (a) $R$ (b) $R^{+}$ (c) $R-{0}$ (d) $R^{+} \cup{0}$
The domain of the function is $f(x)=\frac{1}{\log _{10} x}$ (a) $R$ (b) $R^{+} \cup{0}$ (c) $R^{+}-{1} \quad(d) R-{0}$
The domain of the function $f(x)=\sqrt{6 x-4}+\sqrt{2 x-4}$ is (a) $[0, \infty)$ (b) $[\frac{2}{3}, \infty)$ (c) $[\frac{3}{2}, \infty)$ (d) $[2, \infty)$
The domain of $y=\frac{1}{\sqrt{|x|-x}}$ is: (a) $[0, \infty)$ (b) $(0, \infty)$ (c) $(-\infty, 0)$ (d) $(-\infty, 0)$

(Odisha JEE 2010)

The domain of $f(x)=\frac{x}{\sqrt{x^{2}-5 x+6}}$ is (a) $(-\infty, \infty)$ (b) $[0, \infty)$ (c) $2<x<3$ (d) $(-\infty, 2] \cup[3, \infty)$
The domain of $y=\log _{10}(\sqrt{7-x}+\sqrt{x-3})$ is (a) $R$ (b) $R-{0}$ (c) $[7, \infty)$ (d) $(3,7)$
The domain of the function $f(x)=\frac{1}{\sqrt{9-x^{2}}}$ is (a) $-3 \leq x \leq 3$ (b) $-3<x<3$ (c) $-9 \leq x \leq 9$ (d) $-\infty<x<\infty$

(Kerala PET 2011)

The domain of the function $f(x)=\log _2(\log _3(\log _4 x))$ is (a) $(-\infty, 4)$ (b) $(0,4)$ (c) $(4, \infty)$ (d) $(1, \infty)$

(Kerala PET 2009)

The domain of the definition of the function $y=f(x)$ given by the equation $y=f(x): 2^{x}+2^{y}=2$ is (a) $0<x<1$ (b) $0<x \leq 1$ (c) $0 \leq x \leq 1$ (d) None of these

(DCE 2001)

Domain and range are equal for the (a) constant function (b) one-one function (c) identity function (d) onto function.

(MP PET 2013)

The domain of the function $f(x)=\exp \sqrt{5 x-3-2 x^{2}}$ is: (a) $[\frac{3}{2}, \infty)$ (b) $[1, \frac{3}{2}]$ (c) $[-\infty, 1]$ (d) $(1, \frac{3}{2})$

(MP PET 2004)

The domain of the real valued function

$ f(x)=\sqrt{5-4 x-x^{2}}+x^{2} \log (x+4) \text{ is } $

(a) $-5 \leq x \leq 1$ (b) $-5 \leq x$ and $x \geq 1$ (c) $-4<x \leq 1$ (d) $0 \leq x \leq 1$

(Kerela CEE 2006) 15. The domain of the function $f(x)=\log _{3+x}(x^{2}-1)$ is

(a) $(-3,-1) \cup(1, \infty)$

(b) $[-3,-1] \cup[1, \infty)$

(d) $[-3,-2) \cup(-2,-1) \cup(1, \infty)$

(Odisha JEE 2003)

The range of the function $f(x)=-|x|$ is (a) $(-\infty, \infty)$ (b) $(-\infty, 0)$ (c) $(-\infty, 0]$ (d) $(0, \infty)$
The range of the real function $f(x)=\frac{1}{1-x^{2}}$ is (a) $R-{-1,1}$ (b) $(0, \infty)$ (c) $[1, \infty)$ (d) $(-\infty, 0)$
The domain and range of the function $f(x)=\frac{x^{2}-9}{x-3}$ respectively are (a) $R, R-{3}$ (b) $R-{3}, R$ (c) $R-{3}, R-{6}$ (d) $R, R$
Let $f:{x, \frac{x^{2}}{1+x^{2}}: x \in R}$ be function from $R \to R$, the range of $\boldsymbol{f}$ is (a) $[0,1]$ (b) $[0,1)$ (c) $(0,1]$ (d) $(-\infty, \infty)$

(DCE 2007)

The range of the function $f(x)=\sqrt{(x-1)(3-x)}$ is (a) $[0,1]$ (b) $(-1,1)$ (c) $(-3,3)$ (d) $(-3,1)$

(AMU 2005)

The range of the function $f(x)=x^{2}+\frac{1}{x^{2}+1}$ (a) $[1, \infty)$ (b) $[2, \infty)$ (c) $[\frac{3}{2}, \infty)$ (d) None of these

(Manipal Engg. 2008)

If $f: R \to R$ and is defined by $f(x)=\frac{1}{2-\cos 3 x}$ for each $x \in R$, then the range of $\boldsymbol{f}$ is (a) $(\frac{1}{3}, 1)$ (b) $[\frac{1}{3}, 1]$ (c) $(1,2)$ (d) $[1,2]$

(EAMCET 2007)

ANSWERS

(a)
(c)
(c)
(c)
$(d)$
(c)
(d)
(d)
(b)
(c)
(a)
(c)
$(d)$
(c)
(d)
(c)
(c)
(c)
(b)
(a)
(a)
(b)

HINTS AND ANSWERS

$\because|x| \geq 0 \forall x \in R$

$\therefore \sqrt{|x|}$ is defined for all real values of $x$.

Hence the domain of $\sqrt{|x|}$ is $R$ or $(-\infty, \infty)$

Since $\sqrt{x}$ is defined for only non-negative real numbers, therefore domain of $f(x)=(\sqrt{x})^{2}$ is $[0, \infty)$
Since $\log _{b} a$ is defined only when $a>0, b>0$ and $b \neq 1$, $\log x^{2}$ is defined only when $x^{2}>0$, which is true for all real values of $x$ except $x=0$.

$\therefore \quad$ Domain of $\log x^{2}$ is $R-{0}$.

$\frac{1}{\log _{10} x}=\log _{x} 10$ which is defined only when $x>0, x \neq 1$

$\therefore$ Domain of the given function is the set of all non-zero positive real number except 1, i.e., $R^{+}-$ {1}.

Note : Similarly the domain of $\frac{1}{\log |x|}$ is $R-{0,1,-1}$

$f(x)=\sqrt{6 x-4}+\sqrt{2 x-4}$

$f(x)$ will be defined if $6 x-4 \geq 0$ and $2 x-4 \geq 0$

$\Rightarrow \quad x \geq \frac{2}{3}$ and $x \geq 2 \Rightarrow x \geq 2$

(For $x<2, \sqrt{2 x-4}$ will not be defined)

$\therefore$ Domain of $f(x)=\sqrt{6 x-4}+\sqrt{2 x-4}$ is $[2, \infty)$.

For the function $y=\frac{1}{\sqrt{|x|-x}}$ to be defined,

$|x|-x>0 \quad$ as $\quad|x|-x \neq 0$

$\Rightarrow|x|>x$ which is only possible when $x<0$ as $|x|=x$ for all $x \geq 0$.

$\therefore \quad$ Domain of $y=\frac{1}{\sqrt{|x|-x}}$ is $x<0$ or $(-\infty, 0)$

For $f(x)$ to be defined, $x^{2}-5 x+6>0$ as $x^{2}-5 x+6 \neq 0$ and $+ve$.

$\Rightarrow(x-2)(x-3)>0$

$\Rightarrow x<2$ or $x>3$

$\Rightarrow x \in(-\infty, 2)$ or $(3, \infty)$

$\therefore \quad$ Domain of $f(x)=\frac{x}{\sqrt{x^{2}-5 x+6}}$ is $(-\infty, \mathbf{2}) \cup(\mathbf{3}, \infty)$.

The function $\boldsymbol{f}$ will be defined if

$ (\sqrt{7-x}+\sqrt{x-3})>0 $

i.e., $7-x>0$ and $x-3>0 \Rightarrow 7>x$ and $x>3$

$\Rightarrow 3<x<7$

$\therefore$ Domain of the given function is $(3,7)$.

$f(x)=\frac{1}{\sqrt{9-x^{2}}}$ will be defined if $9-x^{2}>0$

$\Rightarrow x^{2}-9<0 \Rightarrow(x+3)(x-3)<0 \Rightarrow \boldsymbol{x} \in(-3,3)$ 10. $f(x)=\log _2(\log _3(\log _4 x))$ is defined if

$\log _3(\log _4 x)>0 \Rightarrow \log _4 x>3^{0}$

$\Rightarrow \log _4 x>1 \quad \Rightarrow \quad x>4^{1}=4$

$\therefore$ Domain of the given function is $(4, \infty)$.

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

$\Rightarrow$ The function will be defined only when $2-2^{x}>0$

$\Rightarrow 2^{x}<2^{1} \Rightarrow x<1$

Also, $x>0, y<0$

$\therefore \boldsymbol{x} \in \mathbf{( 0 , 1 )}$.

$\because$ Identity functions maps each element onto itself, i.e. $I(x)=x$, i.e., the domain and range are equal.
$Given f(x)=\exp (\sqrt{5 x-3-2 x^{2}}=\exp (\sqrt{-(2 x^{2}-5 x+3)}..$

For $f(x)$ to be defined

$-(2 x^{2}-5 x+3) \geq 0 \Rightarrow 2 x^{2}-5 x+3 \leq 0$

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

$\Rightarrow(2 x-3)(x-1) \leq 0 \quad(\begin{matrix} \because(x-a)(x-b) \leq 0 \\ \Rightarrow a \leq x \leq b \text{ where } a<b\end{matrix} )$

$\Rightarrow x \leq \frac{3}{2}$ and $x \geq 1 \Rightarrow \mathbf{1} \leq x \leq \frac{\mathbf{3}}{\mathbf{2}}$

For $f(x)$ to be defined, $5-4 x-x^{2} \geq 0$ and $x+4>0$

$\Rightarrow x^{2}+4 x-5 \leq 0$ and $x>-4$

$\Rightarrow(x+5)(x-1) \leq 0$ and $x>-4$

$\Rightarrow-5 \leq x \leq 1$ and $x>-4$

$\Rightarrow-4<x \leq 1$.

The function is defined, when $x^{2}-1>0,3+x>0$ and $3+x \neq 1$

$\Rightarrow x<-1$ or $x>1$ and $x>-3$ and $x \neq-2$

$\therefore$ Domain of the function $=(-3,-2) \cup(-2,-1) \cup(1, \infty)$.

$f(x)=-|x|$

$f(x)$ is defined for each $x \in R$, so domain of $f(x)=R$.

Let $y=-|x|$, then $y$ is either 0 or a negative real number

$\therefore$ Range of $\boldsymbol{f}=$ set of non-positive real numbers.

$\therefore$ Range of $\boldsymbol{f}=(-\infty, \mathbf{0}]$.

$f(x)=\frac{1}{1-x^{2}}$ is not defined when $1-x^{2}=0$, i.e. $x= \pm 1$

$\therefore \quad$ Domain of $f(x)=R-{-1,1}$

Let $y=\frac{1}{1-x^{2}} \Rightarrow(1-x^{2})=\frac{1}{y}$

$\Rightarrow x^{2}=1-\frac{1}{y} \Rightarrow x \pm \sqrt{1-\frac{1}{y}}$

This shows that $x$ will not be defined when $1-\frac{1}{y}<0$, i.e., $y<1$

$\therefore \quad$ Range of $f: y \geq 1$, i.e. $\boldsymbol{y} \in[\mathbf{1}, \infty)$.

$f(x)=\frac{x^{2}-9}{x-3}$ is not defined for $x-3=0$, i.e., $x=3$.

$\therefore$ Domain of $\boldsymbol{f}$ is $\boldsymbol{R}-{\mathbf{3}}$.

Let $y=\frac{x^{2}-9}{x-3} \Rightarrow y=x+3($ for $x \neq 3)$

$\therefore y$ will take all real values of $x$, except 6 , when $x \in R-{3}$.

$\therefore$ Range of $\boldsymbol{f}=\boldsymbol{R}-{\mathbf{6}}$.

Let $y=\frac{x^{2}}{1+x^{2}} \geq 0$ for all $x \in R$

$\Rightarrow y(1+x)^{2} \geq x^{2} \Rightarrow y+y x^{2}-x^{2} \geq 0$

$\Rightarrow(y-1) x^{2}+y \geq 0$

Let $y=\frac{x^{2}}{x^{2}+1}$ and $y \geq 0$ for all $x \in R$.

$\Rightarrow \frac{x^{2}}{x^{2}+1} \geq 0$

Also $x^{2}<x^{2}+1 \Rightarrow \frac{x^{2}}{x^{2}+1}<1$

$\therefore \quad$ Required range $=[\mathbf{0}, \mathbf{1})$.

Let $y=f(x)=\sqrt{(x-1)(3-x)}$

$\Rightarrow y^{2}=(x-1)(3-x)=-x^{2}+4 x-3$

$\Rightarrow x^{2}-4 x+(3+y^{2})=0$

This is a quadratic in $x$, so

$ x=\frac{+4 \pm \sqrt{16-4(3+y^{2})}}{2}=\frac{4 \pm 2 \sqrt{1-y^{2}}}{2} $

Since $x$ is real, $1-y^{2} \geq 0 \Rightarrow y^{2}-1 \leq 0 \Rightarrow-1 \leq y \leq 1$

But $f(x)$ attains only non-negative values, so range of $f=[0,1]$

Given $f(x)=x^{2}+\frac{1}{x^{2}+1}=\frac{x^{4}+x^{2}+1}{x^{2}+1}$

$ \begin{aligned} & =\frac{x^{4}+2 x^{2}+1-x^{2}}{x^{2}+1}=\frac{(x^{2}+1)^{2}-x^{2}}{x^{2}+1} \\ & =(x^{2}+1)-\frac{x^{2}}{x^{2}+1} \\ & =1+x^{2}(1-\frac{1}{x^{2}+1}) \geq 1 \forall x \in R \end{aligned} $

$\therefore \quad$ Range of $f(x)=[1, \infty)$.

$f(x)=\frac{1}{2-\cos 3 x}$

$\because \quad-1 \leq \cos 3 x \leq 1$

$\Rightarrow-1 \leq-\cos 3 x \leq 1 \quad$ (Multiplying all terms by -1 )

$\Rightarrow 2-1 \leq 2-\cos 3 x \leq 3 \quad$ (Adding 2 to each term)

$\Rightarrow 1 \leq 2-\cos 3 x \leq 3 \Rightarrow 1 \geq \frac{1}{2-3 \cos 3 x} \geq \frac{1}{3}$

$\therefore \quad$ Range of $f$ is $[\frac{1}{\mathbf{3}}, \mathbf{1}]$.

Composition of Functions

KEY FACTS

Let $\boldsymbol{f}: A \to B$ and $\boldsymbol{g}: B \to C$ be two functions. Then a function $g \circ f: A \to C$ defined by $(g \circ f)(x)=g(f(x)$, for all $x \in A$ is called the composition of $\boldsymbol{f}$ and $g$.

$\because f: A \to B \Rightarrow$ for each $x \in A$ there exists an unique element $f(x) \in B$.

Also $g: B \to C \Rightarrow$ for each $f(x) \in B$ there exists an unique element $g(f(x)) \in C$.

Ex. Given that $f: x \to 2 x-5$ and $g: x \to x^{4}$ for $x \in R$. Then,

(i) $\quad(g \circ f)(x)=g(f(x))=g(2 x-5)=(2 x-5)^{4}$

(ii) $(f \circ g)(x)=f(g(x))=f(x^{4})=2 x^{4}-5$

(iii) $(f \circ f)(x)=f(f(x))=f(2 x-5)=2(2 x-5)-5=4 x-15$

(iv) $(gog)(x)=g(g(x))=g(x^{4})=(x^{4})^{4}=x^{16}$.

SOLVED EXAMPLES

~~ Ex.1. . Given $f$ and $g$ defined by $f(x)=x^{2}+3$ and $g(x)=1-\frac{1}{x}$, find $(g \circ f)(x)$ and $(f \circ g)(x)$.

Sol. $(gof)(x)=g(f(x))=g(x^{2}+3)=1-\frac{1}{x^{2}+3}=\frac{x^{2}+2}{x^{2}+3}$.

$(f \circ g)(x)=f(g(x))=f(1-\frac{1}{x})=(1-\frac{1}{x})^{2}+3=\frac{(x-1)^{2}+3 x^{2}}{x^{2}}=\frac{x^{2}-2 x+1+3 x^{2}}{x^{2}}=\frac{\mathbf{4} \boldsymbol{x}^{2}-\mathbf{2} \boldsymbol{x}+\mathbf{1}}{\boldsymbol{x}^{2}}$.

~~ Ex. 2. . If $f(x)=\frac{2 x+1}{3 x-2}$, then what is ( $.f \circ f)(2)$ equal to?

Sol. $(f \circ f)(2)=f(f(2))=f(\frac{2 \times 2+1}{3 \times 2-2})=f(\frac{5}{4})=\frac{2 \times \frac{5}{4}+1}{3 \times \frac{5}{4}-2}=\frac{\frac{14}{4}}{\frac{7}{4}}=\frac{14}{7}=\mathbf{2}$.

~~ Ex. 3. . If $h(x)=\sec x$, find $(h o h)^{2}$ at $x=\frac{\pi}{4}$.

Sol. $\quad(h o h)(x)=h(h(x))=h(\sec x)=\sec (\sec x)$

$\Rightarrow \quad(h o h)^{2}=\sec ^{2}(\sec x) \therefore(h o h)^{2}$ at $x=\frac{\pi}{4}=\sec ^{2}(\sec \frac{\pi}{4})=\sec ^{2}(\sqrt{2})$

~~ Ex. 4. . If $f(x)=\frac{\alpha x}{x+1}, x \neq-1$, for what value of $\alpha$ is $f(f(x))=x$ ?

(Kerala CEE)

Sol.

$ f(f(x))=f(\frac{\alpha x}{x+1})=\frac{\alpha \cdot(\frac{\alpha x}{x+1})}{(\frac{\alpha x}{x+1})+1}=\frac{\frac{\alpha^{2} x}{x+1}}{\frac{\alpha x+x+1}{x+1}}=\frac{\alpha^{2} x}{\alpha x+x+1} $

Given,

$ f(f(x))=x \Rightarrow \frac{\alpha^{2} x}{\alpha x+x+1}=x \Rightarrow \alpha^{2} x=\alpha x^{2}+x^{2}+x \Rightarrow \alpha^{2}-1=(\alpha+1) x $

$\Rightarrow \quad(\alpha-1)(\alpha+1)-(\alpha+1) x=0 \Rightarrow(\alpha+1)(\alpha-1-x)=0 \Rightarrow \alpha+1=0 \Rightarrow \alpha=-1 \quad[\because \alpha-1-x \neq 0]$

~~ Ex. 5. . If $f: R \to R$ and $g: R \to R$ are defined by $f(x)=x-3$ and $g(x)=x^{2}+1$, then find the values of $x$ for which $g{f(x)}=10$.

Sol. $\quad g{f(x)}=g(x-3)=(x-3)^{2}+1=x^{2}-6 x+9+1=x^{2}-6 x+10$.

Given, $g{f(x)}=10 \Rightarrow x^{2}-6 x+10=10 \Rightarrow x^{2}-6 x=0 \Rightarrow x(x-6)=0 \Rightarrow \boldsymbol{x}=\mathbf{0}, \mathbf{6}$.

~~ Ex. 6 . If $f(y)=\frac{y}{\sqrt{1+y^{2}}}$ and $g(y)=\frac{y}{\sqrt{1-y^{2}}}$, then find $(f \circ g)(y)$.

(EAMCET)

Sol.

$ (f \circ g)(y)=f{g(y)}=f{\frac{y}{\sqrt{1-y^{2}}}}=\frac{\frac{y}{\sqrt{1-y^{2}}}}{\sqrt{1+(\frac{y}{\sqrt{1-y^{2}}})^{2}}}=\frac{\frac{y}{\sqrt{1-y^{2}}}}{\sqrt{\frac{(1-y^{2}+y^{2})}{(1-y^{2})}}}=\frac{\frac{y}{\sqrt{1-y^{2}}}}{\frac{1}{\sqrt{1-y^{2}}}}=\boldsymbol{y} \text{. } $

~~ Ex. 7. . Let $f: R \to R$ be a function defined by

$ f(x)=\begin{cases} 1, \text{ if } x \text{ is a rational number. } \\ 0, \text{ if } x \text{ is an irrational number. } \end{cases} . $

Find (fof) $(1-\sqrt{3})$.

Sol. (fof) $(1-\sqrt{3})=f{f(1-\sqrt{3})}=f(0)$

$(\because 1-\sqrt{3}$ is an irrational number)

$=\mathbf{1}$

$(\because 0$ is rational)

~~ Ex. 8 . Let $g(x)=1+x-[x]$ and $f(x)=\begin{cases}-1, & x<0 \\ 0, & x=0, \\ 1, & x>0\end{cases}.$ then for all $x$, what is $f(g(x))$ equal to?

Sol.

$ f(g(x))=f(1+x-[x]) . $

Now $1+x-[x]>1$ since $x-[x] \geq 0$

$\therefore f(1+x-[x])=\mathbf{1}$ as $f(x)=1$ when $x>0$.

PRACTICE SHEET-3

~~ 1. If $f(x)=x+1$ and $g(x)=2 x$, then $f(g(x)$ is equal to

(a) $2(x+1)$

(b) $2 x(x+1)$

(d) $2 x+1$

(Kerala PET 2012)

~~ 2. If $f(x)=\frac{1+x}{1-x}$, then the value of $f[f(x)]$ is

(a) $x$

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

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

(MP PET 2013)

~~ 3. If $f: R \to R$ is such that $f(x)=\sin x$ and $g: R \to R$ is such that $g(x)=x^{2}$, then composite function $f o g$ is

(a) $\sin x$

(b) $x^{2}$

(d) $\sin x^{2}$

(Odisha JEE 2012)

~~ 4. Let $R$ be a set of real numbers and the mapping $f: R \to R$ and $g: R \to R$ be defined by $f(x)=5-x^{2}$ and $g(x)=3 x-4$, then the value of $f \circ g(-1)$ is

(a) -44

(b) -54

(d) -64

(WBJEE 2010)

~~ 5. If $f(x)=\sin x+\cos x+1$ and $g(x)=x^{2}+x, x \in R$, then $f \circ g(x)$ at $x=0$ is

(a) 0

(b) 1

(d) 3

(Rajasthan PET 2009)

~~ 6. Two functions $f: R \to R, g: R \to R$ one defined as follows:

$\begin{matrix} & f(x)= \begin{cases}0, & x \text{ is rational } \\ 1, & x \text{ is irrational }\end{cases} \\ & g(x)=\begin{cases} -1, & x \text{ is rational } \\ 0, & x \text{ is irrational } \end{cases}. \end{matrix}$

Then $f \circ g(\pi)+g \circ f(e)$ is equal to

(a) 0

(b) -1

(d) 1

(EAMCET 2001)

~~ 7. If $f(x)=(20-x^{4})^{1 / 4}$ for $0<x<\sqrt{5}$, then $f(f(\frac{1}{2}))$ is equal to

(a) $2^{-4}$

(b) $2^{-3}$

(d) $2^{-1}$

(EAMCET)

~~ 8. If $f(x)=e^{x}$ and $g(x)=\log _{e} x$, then which of the following is true?

(a) $f{g(x)} \neq g{f(x)}$

(b) $f{g(x)}=g{f(x)}$

(d) $f{g(x)}-g{f(x)}=1$

(MP PET 2008)

~~ 9. If $x \neq 1$ and $f(x)=\frac{x+1}{x-1}$ is a real function, then $f f f(2)$ is

(a) 1

(b) 2

(d) 4

(Kerala PET 2001)

~~ 10 If $f(x)=a x+b$ and $g(x)=c x+d$, then $f{g(x)}=g{f(x)}$ is equivalent to

(a) $f(a)=g(c)$

(b) $f(b)=g(b)$

(d) $f(c)=g(a)$

(AMU 2009)

~~ 11 Let $R$ be a set of real numbers and the functions $f: R \to R$ and $g: R \to R$ be defined by $f(x)=x^{2}+2 x-3$ and $g(x)=x+1$, then the value of $x$ for which $f \lbrace g(x)\rbrace =g \lbrace f(x) \rbrace$ is

(a) -1

(b) 0

(d) 2

(WBJEE 2012)

~~ 12 Let $f(x)=\frac{1}{x+1}, g(x)=\cos x$ and $a=\frac{4-\pi}{\pi}$.

Then $(g \circ f)(a)$ is equal to

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

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

(d) 1

~~ 13 If $f(x)=2 x+1$ and $g(x)=\frac{x-1}{2}$ for all real $x$, then $(f \circ g)^{-1}(\frac{1}{x})$ is equal to

(a) $-x$

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

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

~~ 14 If $f(x)=\log _{e}(\frac{1+x}{1-x})$ and $g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$, then $(f \circ g)(x)$ is equal to

(a) $3 f(x)$

(b) $(f(x))^{3}$

(d) $3 g(x)$

(MP PET 2012)

~~ 15 Functions $f$ and $g$ are defined over the domain of real numbers as $f: x \to 2 x-1$ and $g: x \to x^{3}$. The values of $x$ for which $(f \circ g)(x)=(g \circ f)(x)$ holds true are:

(a) $-\frac{1}{2}, 0,1$

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

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

ANSWERS

1. (d) 2. $(d)$ 3. $(d)$ 4. (a) 5. (c) 6. (b) 7. $(d)$ 8. (b) 9. (c) 10 $(c)$ 11 (a) 12 $(c)$ 13 $(d)$ 14 (a) 15 $(c)$

HINTS AND SOLUTIONS

~~ 1. $f(g(x)=f(2 x)=2 x+1$

$(\because f(x)=x+1)$

~~ 2. Given, $f(x)=\frac{1+x}{1-x}$

$ \begin{aligned} \therefore f(f(x) & =f(\frac{1+x}{1-x})=\frac{1+(\frac{1+x}{1-x})}{1-(\frac{1+x}{1-x})}=\frac{\frac{1-x+1+x}{1-x}}{\frac{1-x-1-x}{1-x}} \\ & =\frac{2}{-2 x}=-\frac{1}{x} . \end{aligned} $

~~ 3. $f \circ g(x)=f(g(x))=f(x^{2})=\sin x^{2}$.

~~ 4. $f \circ g(-1)=f(g(-1))=f(3 \times-1-4)=f(-7)$

$=5-(-7)^{2}=5-49=-44$.

~~ 5. $f \circ g(x)=f(g(x)=f(x^{2}+x)=\sin (x^{2}+x)+\cos (x^{2}+x)+1.$

$\therefore \quad f \circ g(0)=\sin 0+\cos 0+1=0+1+1=\mathbf{2}$.

~~ 6. $f \circ g(\pi)=(g(\pi))=f(-1)=0$

( $\because \pi$ is irrational and 0 is rational)

$ g o f(e)=g(f(e))=g(1)=-1 $

( $\because e$ is irrational and 1 is rational)

$ \therefore \quad f \circ g(\pi)+g \circ f(e)=0+(-1)=-\mathbf{1} . $

~~ 7. $\quad f(x)=(20-x^{4})^{\frac{1}{4}}$

$ \begin{aligned} \therefore \quad f(\frac{1}{2}) & =(20-\frac{1}{16})^{\frac{1}{4}}=(\frac{319}{16})^{\frac{1}{4}} \\ f(f(\frac{1}{2})) & =f(\frac{319}{16})^{\frac{1}{4}}=(20-((\frac{319}{16})^{\frac{1}{4}})^{4})^{\frac{1}{4}} \\ & =(20-\frac{319}{16})^{\frac{1}{4}}=(\frac{1}{16})^{\frac{1}{4}}=\frac{1}{2}=\mathbf{2}^{-\mathbf{1}} . \end{aligned} $

~~ 8.

$ \begin{aligned} & \therefore \quad f{g(x)}=f(\log _{e} x)=e^{\log _{e} x}=x \\ & g{f(x)}=g(e^{x})=\log _{e} e^{x}=x \\ & \therefore \quad f{g(x)}=g{f(x)} . \end{aligned} $

~~ 9. $f(x)=\frac{x+1}{x-1}$

$ \begin{aligned} f f f(2) & =f[f{f(2)}]=f[f{\frac{2+1}{2-1}}]=f[f(3)] \\ & =f[\frac{3+1}{3-1}]=f(2)=\frac{2+1}{2-1}=\mathbf{3} \end{aligned} $

~~ 10 Given,

$f(x)=a x+b, g(x)=c x+d $

$\because \quad f\lbrace g(x)=g{f(x)} \rbrace$

$\Rightarrow \quad f(c x+d)=g(a x+b)$

$\Rightarrow a(c x+d)+b=c(a x+b)+d$

$\Rightarrow a c x+a d+b=c a x+b c+d$

$\Rightarrow \quad a d+b=b c+d \quad \Rightarrow \quad \boldsymbol{f}(\boldsymbol{d})=\boldsymbol{g}(\boldsymbol{b})$.

~~ 11 Given,

$f(x)=x^{2}+2 x-3$ and $g(x)=x+1$. Then,

$ \begin{aligned} f(g(x))=f(x+1) & =(x+1)^{2}+2(x+1)-3 \\ & =x^{2}+2 x+1+2 x+2-3 \\ & =x^{2}+4 x \\ g(f(x))=g(x^{2}+2 x-3) & =x^{2}+2 x-3+1 \\ & =x^{2}+2 x-2 \end{aligned} $

$ \Rightarrow \quad 2 x=-2 \Rightarrow x=-1 . $

~~ 12 $(g \circ f)(a)=g(f(a))=g(\frac{1}{a+1})=\cos \frac{1}{a+1}=\cos \frac{1}{\frac{4-\pi}{\pi}+1}$

$ =\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} . $

~~ 13 $f(x)=2 x+1$ and $g(x)=\frac{x-1}{2}$

$ \begin{aligned} \therefore & f \circ g(x)=f(g(x))=f(\frac{x-1}{2})=2(\frac{x-1}{2})+1=x \\ & \quad f \circ g(x)=x \Rightarrow(f \circ g)^{-1}(x)=x \Rightarrow(f \circ g)^{-1}(\frac{1}{x})=\frac{\mathbf{1}}{\boldsymbol{x}} . \end{aligned} $

~~ 14 Given, $f(x)=\log _{e}(\frac{1+x}{1-x})$ and $g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$

$ \therefore f \circ g(x)=f(g(x))=f(\frac{3 x+x^{3}}{1+3 x^{2}})=\log _{e}(\frac{1+\frac{3 x+x^{3}}{1+3 x^{2}}}{1-\frac{3 x+x^{3}}{1+3 x^{2}}}) $

$ \begin{aligned} & =\log _{e}(\frac{1+3 x^{2}+3 x+x^{3}}{1+3 x^{2}-3 x-x^{3}})=\log _{e}(\frac{(1+x)^{3}}{(1-x)^{3}}) \\ & =\log _{e}(\frac{1+x}{1-x})^{3}=3 \log _{e}(\frac{1+x}{1-x})=\mathbf{3} \boldsymbol{f}(\boldsymbol{x}) \end{aligned} $

~~ 15 $f(x)=2 x-1, g(x)=x^{3}$

$ \begin{aligned} \therefore & (f \circ g)(x)=f{g(x)}=f(x^{3})=2 x^{3}-1 \\ & (g \circ f)(x)=g{f(x)}=g(2 x-1)=(2 x-1)^{3} \\ & (f \circ g)(x)=(g \circ f)(x) \\ \Rightarrow & 2 x^{3}-1=(2 x-1)^{3} \Rightarrow 2 x^{3}-1=8 x^{3}-1-12 x^{2}+6 x \\ \Rightarrow & 6 x^{3}-12 x^{2}+6 x=0 \Rightarrow 6 x(x^{2}-2 x+1)=0 \\ \Rightarrow & x(x-1)^{2}=0 \Rightarrow x=0,1 \end{aligned} $

$\therefore$ The values of $x$ for which $(f \circ g)(x)=(g \circ f)(x)$ are $\mathbf{0}$ and 1 .

SELF ASSESSMENT SHEET

~~ 1. The function $f:[0, \infty) \to[0, \infty)$ defined by $f(x)=\frac{2 x}{1+2 x}$ is

(a) one-one and onto

(b) one-one but not onto

(d) neither one-one nor onto

(J&K CET 2011)

~~ 2. Let $A=[-1,1]$ and $f: A \to A$ be defined as $f(x)=x|x|$ for all $x \in A$, then $f(x)$ is

(a) many-one into function

(b) one-one into function

(d) one-one onto function.

(BITSAT 2007)

~~ 3. The quotient of the identity function by the reciprocal function is given by

(a) $x^{2} \forall x \in R$ (set of real numbers)

(b) $\frac{1}{x^{2}} \forall x \in R-{0}$

(d) None of these

(J&K CET 2013)

~~ 4. Which of the following functions is an even function?

(a) $f(x)=\frac{a^{x}+1}{a^{x}-1}$

(b) $f(x)=\frac{a^{x}-a^{-x}}{a^{x}+a^{-x}}$

(d) $f(x)=\sin x$

~~ 5. The function $f(x)=\sec [\log (x+\sqrt{1+x^{2}})]$ is

(a) odd

(b) even

(d) constant

(WBJEE 2010)

~~ 6. If the function $f: N \to N$ is defined by $f(x)=\sqrt{x}$, then $\frac{f(25)}{f(16)+f(1)}$ is equal to

(a) $\frac{5}{6}$

(b) $\frac{5}{7}$

(d) 1

(MP PET 2009)

~~ 7. If $f: R \to R$ is defined by $f(x)=(1-x)^{1 / 3}$, the $f^{-1}(x)$ equals

(a) $(1-x)^{-1 / 3}(b)(1-x)^{3}$

(d) $1-x^{1 / 3}$

(RPET 2009)

~~ 8. Let $f(x)=(x+2)^{2}-2, x \geq-2$, then $f^{-1}(x)$ is equal to

(a) $-\sqrt{2+x}-2$

(b) $\sqrt{2+x}+2$

(d) $-\sqrt{2+x}+2$

(AMU 2013)

~~ 9. The domain of the function $\log |x^{2}-9|$ is

(a) $R$

(b) $R-[-3,3]$

(d) None of these

(Manipal Engg. 2012)

~~ 10 The domain of the function given by $f(x)=\frac{1}{\sqrt{x+[x]}}$ is given by

(a) $(0, \infty)$

(b) $(-\infty, \infty)$

(d) $[0, \infty)$

~~ 11 If $f(x)=\sqrt{\log _{10} x^{2}}$. The set of values of $x$ for which $x$ is real is

(a) $[-1,1]$

(b) $[1, \infty)$

(d) $[-\infty,-1] \cup[1, \infty)$

(VITEEE 2010)

~~ 12 If $f: R \to R$ is defined by $f(x)=2 x+|x|$, then $f(3 x)-f(-x)-4 x$ is equal to

(a) $-f(x)$

(b) $f(x)$

(d) $2 f(x)$

(EAMCET)

~~ 13 Let $f: R \to R$ be defined by

$ f(x)= \begin{cases}x+2, & x \leq 1 \\ x^{2}, & -1<x<1, \text{ then the } \\ 2-x, & x \geq 1\end{cases} $

value of $f(-1.75)+f(0.5)+f(1.5)$ is

(a) 0

(b) 1

(d) -1

~~ 14 Let $A= \lbrace x \in R, x \neq 0,-4 \leq x \leq 4)$ and $f: A \to R$ defined by $f(x)=\frac{|x|}{x}$ for $x \in A$. Then the range of $f$ is

(a) ${1,-1}$

(b) ${x: 0 \leq x \leq 1}$

(d) ${x:-4 \leq x \leq 0}$

~~ 15 If $f(x)=\frac{1}{\sqrt{-x}}$, then domain of fof is

(a) $(0, \infty)$

(b) $(-\infty, 0)$

(d) {}

(J&K CET)

~~ 16 The range of the function $f(x)=\log _{e}(3 x^{2}+4)$ is equal to

(a) $[\log _{e} 2, \infty)$

(b) $[\log _{e} 3, \infty)$

(d) $[0, \infty)$

(Kerala PET 2011)

~~ 17 If $f: R \to R$ and $g: R \to R$ are defined by $f(x)=2 x+3$ and $g(x)=x^{2}+7$, then the values of $x$ such that $g{f(x)}=8$ are

(a) 1,2

(b) $-1,2$

(d) $1,-2$

~~ 18 If $f(x)=\sin ^{2} x$ and the composite function $g(f(x)=|\sin x|$, then function $g(x)$ is equal to

(a) $\sqrt{x-1}$

(b) $\sqrt{x}$

(d) $-\sqrt{x}$

(Odisha JEE 2003)

ANSWERS

1. (b) 2. (d) 3. (c) 4. $(c)$ 5. (b) 6. $($ d) 7. (c) 8. (c) 9. (c) 10 (a) 11 (d) 12 (d) 13 (b) 14 (a) 15 $(d)$ 16 (c) 17 (c) 18 (b)

HINTS AND SOLUTIONS

~~ 1. For all $(x_1, x_2) \in[0, \infty)$

$ \begin{aligned} f(x_1)=f(x_2) & \Rightarrow \frac{2 x_1}{1+2 x_1}=\frac{2 x_2}{1+2 x_2} \\ & \Rightarrow 2 x_1+4 x_1 x_2=2 x_2+4 x_1 x_2 \\ & \Rightarrow 2 x_1=2 x_2 \Rightarrow x_1=x_2 \\ & \Rightarrow f \text{ is one-one. } \end{aligned} $

Let $\quad y=f(x)=\frac{2 x}{1+2 x} \Rightarrow y+2 x y=2 x$

$\Rightarrow \quad y=2 x-2 x y=2 x(1-y) \Rightarrow x=\frac{y}{2(1-y)}$

$x$ is not defined when $(1-y)=0$, i.e., $y=1 \in[0, \infty)$

$\Rightarrow f$ is not onto

$\therefore \quad f$ is one-one, not onto.

~~ 2. $f(x) = x|x|=\begin{cases} x^{2}, & x \geq 0 \\ -x^{2}, & x<0\end{cases} .$

Since $-1 \leq x \leq 1$, therefore $-1 \leq f(x) \leq 1$

$\Rightarrow$ function is one-one onto.

~~ 3. Identity function : $I(x)=x \forall x \in R$

Reciprocal function : $f(x)=\frac{1}{x} \forall x \in R, x \neq 0$

$\therefore \quad$ Required quotient $=\frac{I(x)}{f(x)}=\frac{x}{\frac{1}{x}} \forall x \in R, x \neq 0$

$ =x^{2} \forall x \in R-{0} $

~~ 4. (a)

$ \begin{aligned} f(x) & =\frac{a^{x}+1}{a^{x}-1} \\ \therefore \quad f(-x) & =\frac{a^{-x}+1}{a^{-x}-1}=\frac{\frac{1}{a^{x}}+1}{\frac{1}{a^{x}}-1}=\frac{1+a^{x}}{1-a^{x}} \\ & =-\frac{a^{x}+1}{a^{x}-1}=-f(x) \end{aligned} $

Hence $\boldsymbol{f}$ is odd.

(b) Similarly $\frac{a^{x}-a^{-x}}{a^{x}+a^{-x}}$ is an odd function

(c)

$ \begin{aligned} f(x) & =x \frac{a^{x}+1}{a^{x}-1} \\ \therefore \quad f(-x) & =(-x) \frac{a^{-x}+1}{a^{-x}-1}=(-x) \frac{1 / a^{x}+1}{1 / a^{x}-1} \\ & =(-x)(\frac{1+a^{x}}{1-a^{x}})=(-x)(-\frac{a^{x}+1}{a^{x}-1}) \\ & =x(\frac{a^{x}+1}{a^{x}-1})=f(x) \Rightarrow f \text{ is even. } \end{aligned} $

(d)

$ \begin{aligned} f(x) & =\sin x \\ f(-x) & =\sin (-x)=-\sin x=-f(x) \end{aligned} $

$\Rightarrow f$ is odd.

~~ 5. Type Q. 21 (Practice Sheet-1)

Hint: $\sec x$ is an even function.

~~ 6. $f: R \to R$ given by $f(x)=\sqrt{x}$

$\therefore \quad \frac{f(25)}{f(16)+f(1)}=\frac{5}{4+1}=\frac{5}{5}=\mathbf{1}$.

~~ 7. Let $y=f(x)=(1-x)^{1 / 3} \Rightarrow y^{3}=1-x$

$\Rightarrow \quad x=1-y^{3} \quad \Rightarrow f^{-1}(x)=1-x^{3}$

~~ 8. Let $y=f(x)=(x+2)^{2}-2$

$\Rightarrow y+2=(x+2)^{2} \Rightarrow x+2=\sqrt{y+2}$

$\Rightarrow \quad x=\sqrt{y+2}-2 \Rightarrow f^{-1}(x)=\sqrt{x+2}-2$

~~ 9. The function $\log |x^{2}-9|$ is defined when

$ |x^{2}-9|>0 $

$|x^{2}-9|$ is positive for all real values but $\log |x^{2}-9|$ is not defined when $|x^{2}-9|=0$, i.e., function $\log |x^{2}-9|$ does not exist when $x=-3,3$.

$\therefore \quad$ Domain of function is $R-{-3,3}$.

~~ 10 $f(x)=\frac{1}{\sqrt{x+[x]}}$

We know that,

$ \begin{cases}x+[x]>0 & \text{ for all } x>0 \\ x+[x]=0, & \text{ for all } x=0 \\ x+[x]<0, & \text{ for all } x<0\end{cases} $

Also, $f(x)=\frac{1}{\sqrt{x+[x]}}$ is defined for all $x$ satisfying $x+[x]>0$

$\therefore \quad$ Domain of $f(x)=(\mathbf{0}, \infty)$.

~~ 11 $f(x)=\sqrt{\log _{10} x^{2}}$ is defined when

$ \begin{aligned} & \log _{10} x^{2} \geq 0 \Rightarrow x^{2} \geq 10^{0}=1 \Rightarrow(x^{2}-1) \geq 0 \\ \Rightarrow & (x+1)(x-1) \geq 0 \Rightarrow x<-1 \text{ and } x>1 \\ \Rightarrow & \boldsymbol{x} \in(-\infty,-\mathbf{1}] \cup[\mathbf{1}, \infty) \end{aligned} $

~~ 12 $f(x)=2 x+|x|$

$ f(3 x)=2 \cdot(3 x)+|3 x|=6 x+3|x| $

$ f(-x)=-2 x+|x| $

$\therefore f(3 x)-f(-x)-4 x=6 x+3|x|+2 x-|x|-4 x$

$ =4 x+2|x|=2(2 x+|x|)=\mathbf{2} \boldsymbol{f}(\boldsymbol{x}) $

~~ 13 $f(-1.75)+f(0.5)+f(1.5)$

$ \begin{aligned} & =(-1.75+2)+(0.5)^{2}+(2-1.5) \\ & =0.25+0.25+0.5=\mathbf{1} \end{aligned} $

~~ 14 $\because x \in[-4,4]$

$ \begin{aligned} f(x)=\frac{|x|}{x} & =\frac{x}{x}=1 \text{ if } x \text{ is }+v e \\ & =\frac{x}{-x}=-1 \text{ if } x \text{ is }-v e \end{aligned} $

$\therefore$ Range of $\boldsymbol{f}={-1,1}$.

~~ 15 $f(x)=\frac{1}{\sqrt{-x}}$

$ f \circ f(x)=f(f(x))=f(\frac{1}{\sqrt{-x}})=f(\frac{1}{\sqrt{-\frac{1}{\sqrt{-x}}}}) $

Since $\sqrt{-\frac{1}{\sqrt{-x}}}$ is not real

$\therefore \quad$ no domain of $f o f(x)$ exists

Thus, domain of $f \circ f(x)$ is the empty set.

~~ 16 Given, $f(x)=\log _{e}(3 x^{2}+4)$

Let $\quad y=\log _{e}(3 x^{2}+4)$

$\Rightarrow 3 x^{2}+4=e^{y} \Rightarrow x=\sqrt{\frac{e^{y}-4}{3}}$

For $x$ to be defined $\frac{e^{y}-4}{3} \geq 0$

$ e^{y}-4 \geq 0 \Rightarrow e^{y} \geq 4 \Rightarrow y \geq \log _{e} 4 \Rightarrow y \geq 2 \log _{e} 2 $

$\therefore$ Range of the function is $[2 \log _{e} \mathbf{2}, \infty)$

~~ 17 $g{f(x)}=8 \quad \Rightarrow g{2 x+3}=8 \quad \Rightarrow(2 x+3)^{2}+7=8$

$\Rightarrow 4 x^{2}+12 x+9-1=0 \Rightarrow 4 x^{2}+12 x+8=0$

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

$\Rightarrow x=\mathbf{- 1 , - 2}$.

~~ 18 Given $f(x)=\sin ^{2} x$ and $g(f(x))=|\sin x|$

$ \begin{aligned} & \Rightarrow g(\sin ^{2} x)=|\sin x| \Rightarrow g(\sin ^{2} x)=\sqrt{\sin ^{2} x} \\ & \therefore g(x)=\sqrt{x} . \end{aligned} $

Matrices and Determinants

KEY FACTS

~~ 1. A Matrix (plural-matrices) is a rectangular array of real numbers, arranged in rows and columns.

The general form of a matrix with $m$ rows and $n$ columns is

$A _{m \times n}$ $= \begin{bmatrix} a _{11} & a _{12} & a _{13} & - & - & a _{1n} \\ a _{21} & a _{22} & a _{23} & - & - & a _{2n} \\ - & - & - & - & - & - \\ - & - & - & - & - & - \\ a _{m1} & a _{m2} & a _{m3} & - & - & a _{mn} \end{bmatrix}$

which is written in a compact form as $A _{m \times n}=[a _{i j}] _{m \times n}$

Hence $\boldsymbol{a} _{i j}$ is the element of the $\boldsymbol{i}$ th row and $j$ th column.

For example, $a _{12}$ is element in 1st row and 2nd column.

$a _{34}$ is the element in the 3rd row and 4th column.

Order of a Matrix is the ordered pair having first component as the number of rows and the second component as the number of columns in the matrix. Thus, a matrix of order $\boldsymbol{m} \times \boldsymbol{n}$ has $\boldsymbol{m}$ rows and $\boldsymbol{n}$ columns and is called an $\boldsymbol{m} \times \boldsymbol{n}$ (read " $\boldsymbol{m}$ by $\boldsymbol{n}$ “) matrix.

Thus the generalised form of a $3 \times 3$ matrix is :

~~ 2. Types of Matrices:

$ A _{3 \times 3}= \begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33} \end{bmatrix} $

(i) Rectangular Matrix. Any $m \times n$ matrix, where $m \neq n$ is called a rectangular matrix.

For example, $ \begin{bmatrix} 1 & 3 \\ 2 & 4 \\ -6 & 2\end{bmatrix} $ of the order $3 \times 2$ is a rectangular matrix.

(ii) Row Matrix. A matrix having only one row is called a row matrix.

For example, $ \begin{bmatrix} 3 & 7 & 1 & -2\end{bmatrix} _{1 \times 4}, \begin{bmatrix} 2 & -3\end{bmatrix} _{1 \times 2}$ are row matrices.

(iii) Column Matrix. A matrix which has a single column is called a column matrix.

For example, $ \begin{bmatrix} 2 \\ -3 \\ 4\end{bmatrix} _{3 \times 1} \begin{bmatrix} 5 \\ 7 \\ 1 \\ -6\end{bmatrix} _{4 \times 1}$ are column matrices. (iv) Square Matrix. A matrix in which the number of rows is equal to the number of columns is called a square matrix. An $m \times m$ matrix is termed as a square matrix of order $m$.

For example, $ \begin{bmatrix} 1 & 3 \\ 6 & 7\end{bmatrix} _{2 \times 2}$ is a square matrix of order 2 .

$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} _{3 \times 3} \text{ is a square matrix of order } 3 . $

(v) Diagonal Matrix. It is a square matrix all of whose elements except those in the leading diagonal are zero. The leading diagonal elements of square matrix $A=[a _{i j}] _{m \times n}$ are $a _{11}, a _{22}, a _{33}, \ldots, a _{m n}$ For example, $ \begin{bmatrix} a _{11} & 0 \\ 0 & a _{22}\end{bmatrix} _{2 \times 2}, \begin{bmatrix} 2 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & -1\end{bmatrix} _{3 \times 3}$ are all diagonal matrices. A diagonal matrix of order $n$, having $d_1, d_2$, , $d_n$ as diagonal elements may be denoted by diagonal $[d_1, d_2.$, ,$.d_n]$

Thus, the diagonal matrix $ \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 7\end{bmatrix} $ may be denoted by diagonal $ \begin{bmatrix} 1 & -2 & 7\end{bmatrix} $

(vi) Scalar Matrix. A square matrix in which the diagonal elements are all equal, all other elements being zeros is called a scalar matrix.

For example, $ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix} $ is a scalar matrix of order 3.

(vii) Unit Matrix or Identity Matrix. A square matrix in which each diagonal element is unity, all other elements being zeros, is called a unit matrix or an identity matrix.

Unit matrix of order $n$ is denoted by $I_n$.

For example, $I_2= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} , I_3= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} $

(viii) Zero or Null Matrix. A matrix each of whose elements is zero is called a zero matrix.

For example, $ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix} , \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix} $ are all null matrices.

(ix) Sub-Matrix. A matrix obtained by deleting the rows or columns or both of a matrix is called a sub-matrix.

For example, $A= \begin{bmatrix} 5 & 7 \\ -1 & 2\end{bmatrix} $ is a sub matrix of $B= \begin{bmatrix} 5 & 7 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 0\end{bmatrix} $ obtained by deleting the third row and third column of matrix $B$.

(x) Comparable Matrices. Two matrices $A$ and $B$ are said to be comparable if they are of the same order, i.e., they have the same number of rows and same number of columns.

For example, $ \begin{bmatrix} 1 & 0 & -3 \\ 2 & 7 & 4\end{bmatrix} $ and $ \begin{bmatrix} -1 & 2 & 3 \\ 4 & 0 & 5\end{bmatrix} $ are comparable matrices each of order $2 \times 3$.

(xi) Triangular Matrix. A square matrix of order $n$ is called a triangular matrix if its diagonal elements are all equal to zero.

For example, $ \begin{bmatrix} 0 & 1 & 2 \\ -2 & 0 & 3 \\ -4 & -3 & 0\end{bmatrix} $ is a triangular matrix of order 3

(a) Upper Triangular Matrix. A square matrix in which the elements below the principal diagonal are all zero is called upper triangular matrix, i.e., $a _{i j}=0$ for all $i>j$.

For example, $ \begin{bmatrix} 1 & -4 & 7 \\ 0 & -6 & 8 \\ 0 & 0 & 9\end{bmatrix} $ is an upper triangular matrix.

(b) Lower Triangular Matrix. A square matrix in which the elements above the principal diagonal are all zero is called lower triangular matrix, i.e., $a _{i j}=0$ for all $i<j$.

For example, $ \begin{bmatrix} 1 & 0 & 0 \\ -2 & 4 & 0 \\ 8 & 2 & 1\end{bmatrix} $ is a lower triangular matrix.

~~ 3. Equality of Matrices: Two matrices $A$ and $B$ are equal if and only if both matrices are of same order and each element of one is equal to the corresponding element of the other, i.e.,

$A=[a _{i j}] _{m \times n}$ and $B=[b _{i j}] _{m \times n}$ are said to be equal if $a _{i j}=b _{i j}$ for all $i, j$

For example, $ \begin{bmatrix} 2 & 1 \\ 3 & 0\end{bmatrix} = \begin{bmatrix} 4 / 2 & 2-1 \\ \sqrt{9} & 0\end{bmatrix} $, but $ \begin{bmatrix} 2 & 1 \\ 3 & 0\end{bmatrix} \neq \begin{bmatrix} 2 & 3 \\ 1 & 0\end{bmatrix} $.

4. Addition and Subtraction of Matrices.

(i) Addition of Matrices. The sum of two matrices $A=[a _{i j}] _{m \times n}$ and $B=[b _{i j}] _{m \times n}$ of the same order $m \times n$ is $(A+B) _{m \times n}$ in which the element at $i$ th row and $j$ th column is $a _{i j}+b _{i j}$ for all $1 \leq i \leq m$ and $1 \leq j \leq n$.

Thus, if $A=[a _{i j}] _{m \times n}$ and $B=[b _{i j}] _{m \times n}$, then

$ A+B=[a _{i j}+b _{i j}] _{m \times n} $

For example, $ \begin{bmatrix} 3 & 1 & 2 \\ 2 & 1 & 4\end{bmatrix} + \begin{bmatrix} 1 & 3 & 2 \\ -3 & 1 & -4\end{bmatrix} = \begin{bmatrix} 3+1 & 1+3 & 2+2 \\ 2+(-3) & 1+1 & 4+(-4)\end{bmatrix} = \begin{bmatrix} 4 & 4 & 4 \\ -1 & 2 & 0\end{bmatrix} $.

Caution: The sum of two matrices of different orders is not defined.

(ii) Negative of a matrix. The negative of a matrix $A _{m \times n}$ denoted by $-A _{m \times n}$ is the matrix formed by replacing each entry in the matrix $A _{m \times n}$ by its additive inverse.

Thus, if $A=[a _{i j}] _{m \times n}$ be any matrix, then its negative $-A=[-a _{i j}] _{m \times n}$.

For example if $A= \begin{bmatrix} 3 & -1 \\ 2 & -2 \\ -4 & 5\end{bmatrix} $, then $-A= \begin{bmatrix} -3 & 1 \\ -2 & 2 \\ 4 & -5\end{bmatrix} $.

(iii) Additive Inverse. For each matrix $A=[a _{i j}] _{m \times n}$, there exists a matrix $-A=[-a _{i j}] _{m \times n}$ (negative of matrix $A$ ), called the additive inverse of $A$, such that $A+(-A)=O$ (null matrix)

Thus, the additive inverse of $ \begin{bmatrix} -4 & 3 \\ 2 & 1\end{bmatrix} $ is $ \begin{bmatrix} 4 & -3 \\ -2 & -1\end{bmatrix} $ and $ \begin{bmatrix} -4 & 3 \\ 2 & 1\end{bmatrix} + \begin{bmatrix} 4 & -3 \\ -2 & -1\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix} =O$ (null matrix).

(iv) Subtraction of matrices. If $A$ and $B$ are matrices of the same order, then the sum $B+(-A)$ is called the difference or subtraction of $B$ and $A$ is denoted by $B-A$.

For example, if $L= \begin{bmatrix} 2 & 0 \\ -3 & 6\end{bmatrix} $ and $M= \begin{bmatrix} 1 & -2 \\ 0 & 4\end{bmatrix} $, then

$ L-M=L+(-M)= \begin{bmatrix} 2 & 0 \\ -3 & 6 \end{bmatrix} + \begin{bmatrix} -1 & +2 \\ 0 & -4 \end{bmatrix} = \begin{bmatrix} 2-1 & 0+2 \\ -3+0 & 6-4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ -3 & 2 \end{bmatrix} \text{. } $

(v) Properties of Matrix Addition.

If $A, B$ and $C$ belong to the set $S _{m \times n}$ of all $m \times n$ matrices with real numbers as elements, then I. $A+B \in S _{m \times n}$

Closure property of addition

II. $A+B=B+A$. Commutative law of addition

$ \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & -2 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix} $

III. $(A+B)+C=A+(B+C)$

IV. The matrix $O _{m \times n}$ has the property that for every matrix $A _{m \times n}$.

Associative law of addition

Additive - identity law

$ \begin{bmatrix} 1 & 2 & -1 \\ -3 & 4 & 5 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 2 & -1 \\ -3 & 4 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 2 & -1 \\ -3 & 4 & 5 \end{bmatrix} $

V. For every matrix $A _{m \times n}$, there exists a matrix $-A _{m \times n}$, such that

$ A+(-A)=0 \text{ and }(-A)+A=0 $

$ \begin{bmatrix} 4 & -5 \\ -1 & 2 \end{bmatrix} + \begin{bmatrix} -4 & 5 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \text{. } $

VI. If $A, B$ and $C$ are three matrices of the same order, then,

$ \begin{aligned} & A+B=A+C \Rightarrow B=C \\ & B+A=C+A \Rightarrow B=C \end{aligned} $

Left Cancellation Law

Right Cancellation Law

~~ 5. Scalar Multiplication. The product of a real number or scalar $\boldsymbol{k}$ and a matrix $A=[a _{i j}] _{m \times n}$ is a matrix whose elements are the products of $k$ and corresponding elements of $A$, i.e.

$k A=[k a _{i j}] _{m \times n} \forall i, j$

For example, if $\quad A= \begin{bmatrix} a & b \\ c & d\end{bmatrix} $, then $k A= \begin{bmatrix} k a & k b \\ k c & k d\end{bmatrix} $.

Properties of scalar multiplication of matrices

Let $A=[a _{i j}] _{m \times n}$ and $B=[b _{i j}] _{m \times n}$ be the two given matrices and $k_1, k_2, k_3$ the scalars, i.e., $k_1, k_2, k_3 \in R$, then I. $k_1(A+B)=k_1 A+k_1 B$

II. $(k_1+k_2) A=k_1 A+k_2 A$

III. $(k_1 k_2) A=k_1(k_2 A)=k_2(k_1 A)$

IV. $1 A=A$ and $(-1) A=-A$ V. $(-k) A=-k A$

VI. $O A=O$ and $k_1 O=O$

6. Multiplication of Matrices

Matrices need to be conformable or compatible for multiplication, i.e., for the product of two matrices. $A$ and $B, A B$ to exist the number of columns of matrix $A$ should be equal to the number of rows of $B$.

Then, the product matrix $A B$ has the same number of rows as $A$ and same number of columns as $B$.

$\therefore A _{m \times p} \times B _{p \times n}=C _{m \times n}$

Note: If two matrices are conformable for the matrix multiplication $A \times B$ then it does not necessarily imply that they are conformable for the order $B \times A$ also.

The matrix multiplication follows the “multiply row by column” process which can be shown diagrammatically as:

Let $A=[a _{i j}] _{2 \times 3}, B=[b _{i j}] _{3 \times 2}$, Then

No. of columns of $A=3=$ No. of rows of $B \Rightarrow A B$ exists and is of order $2 \times 2$

Similarly, No. of columns of $B=2=$ No. of rows of $A \Rightarrow B A$ exists and is of order $3 \times 3$.

$= \begin{bmatrix} 3+2+6 & -3+1+2 \\ 1+0+3 & -1+0+1\end{bmatrix} = \begin{bmatrix} 11 & 0 \\ 4 & 0\end{bmatrix} _{2 \times 2}$

$ \text{ and } \begin{aligned} { \begin{bmatrix} 1 & -1 \\ 2 & 1 \\ 3 & 1 \end{bmatrix} _{3 \times 2} \begin{bmatrix} 3 & 1 & 2 \\ 1 & 0 & 1 \end{bmatrix} _{2 \times 3} } & = \begin{bmatrix} 1 \times 3+(-1) \times 1 & 1 \times 1+(-1) \times 0 & 1 \times 2+(-1) \times 1 \\ 2 \times 3+1 \times 1 & 2 \times 1+1 \times 0 & 2 \times 2+1 \times 1 \\ 3 \times 3+1 \times 1 & 3 \times 1+1 \times 0 & 3 \times 2+1 \times 1 \end{bmatrix} _{3 \times 3} \\ & = \begin{bmatrix} 3-1 & 1-0 & 2-1 \\ 6+1 & 2+0 & 4+1 \\ 9+1 & 3+0 & 6+1 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 7 & 2 & 5 \\ 10 & 3 & 7 \end{bmatrix} _{3 \times 3} \end{aligned} $

(i) The product of matrices is not commutative

(a) Wherever $A B$ exists, $B A$ is not always defined. For example, if $A$ be a $5 \times 4$ matrix and $B$ be a $4 \times 3$ matrix, then $A B$ is defined and is of order $5 \times 3$, while $B A$ is not defined. (No. of columns of $B=3 \neq$ No. of rows of $A=5$ ).

(b) If $A B$ and $B A$ are both defined, it is not necessary that they are of the same order. For example, if $A$ be a $4 \times 3$ matrix and $B$ be a $3 \times 4$ matrix, then $A B$ is defined and is a $4 \times 4$ matrix. $B A$ is also defined but is a $3 \times 3$ matrix. $A B$ and $B A$ being of different orders $A B \neq B A$.

(ii) Matrix multiplication is associative if conformability is assured, i.e., $A(B C)=(A B) C$

(iii) Matrix multiplication is distributive with respect to matrix addition $A(B+C)=A B+A C$

Also, it can be proved that: $(B+C) A=B C+C A, A(B-C)=A B-A C,(B-C) A=B A-C A$

(iv) The product of two matrices can be zero without either factor being a zero matrix.

For example, Let $A= \begin{bmatrix} 0 & a \\ 0 & b\end{bmatrix} $ and $B= \begin{bmatrix} c & d \\ 0 & 0\end{bmatrix} $ where, $a \neq 0, b \neq 0, c \neq 0, d \neq 0$

Then, $A B= \begin{bmatrix} 0 & a \\ 0 & b\end{bmatrix} \begin{bmatrix} c & d \\ 0 & 0\end{bmatrix} = \begin{bmatrix} 0 \times c+a \times 0 & 0 \times d+a \times 0 \\ 0 \times c+b \times 0 & 0 \times d+b \times 0\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix} =O$.

(v) Existence of Identity Matrix: For each square matrix $A$ of order $n$, we have an identity matrix I of order $n$ such that $A _{n \times n} I _{n \times n}=A _{n \times n}=I _{n \times n} A _{n \times n}$.

For example, $ \begin{bmatrix} 2 & 3 \\ -1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 2 \times 1+3 \times 0 & 2 \times 0+3 \times 1 \\ -1 \times 1+1 \times 0 & -1 \times 0+1 \times 1\end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -1 & 1\end{bmatrix} $

$ \begin{aligned} & \begin{matrix} \mathbf{A} _2 & \mathbf{I} _2 & \mathbf{A} _2 \end{matrix} \\ & { \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 \times 2+0 \times-1 & 1 \times 3+0 \times 1 \\ 0 \times 2+1 \times-1 & 0 \times 3+1 \times 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix} } \\ & \begin{matrix} \mathbf{I} _2 & \mathbf{A} _2 & \mathbf{A} _2 \end{matrix} \end{aligned} $

Thus for any matrix $A$, there exists an identity matrix $I$ such that $A I=A=I A$, whenever the products are defined.

(vi) Zero Matrix: For any matrix $A$, we have a zero matrix $O$ such that, $A O=O=O A$, whenever these products are defined.

Thus, for any matrix $A _{p \times n}$, we have

$ O _{m \times p} A _{p \times n}=O _{m \times n} \text{ and } A _{p \times n} O _{n \times q}=O _{p \times q} $

(vii) Positive Integral Power of Matrices: The product $A$. $A$ is defined only when $A$ is a square matrix.

$ \begin{aligned} A^{2} & =A \cdot A \\ A^{3} & =A^{2} \cdot A=A \cdot A \cdot A \\ A^{4} & =A^{3} \cdot A=A \cdot A \cdot A \cdot A \\ \therefore \quad A^{m} & =(A \cdot A \cdot A \cdot A \text{—— } m \text{ times }) \end{aligned} $

Note: 1. Also for an identity matrix $I$ of any order $I=I^{2}=I^{3}=I^{4}$——— $=I^{n}$.

~~ 2. For a square matrix $A$ of order $n$,

$f(A)=a_0 I+a_1 A+a_2 A^{2}+——-+a_n A^{n}$ is a matrix polynomial.

If $f(A)$ is a zero matrix, then $A$ is the root or zero of polynomial $f(x)$

~~ 7. Transpose Matrix: The matrix obtained from any given matrix $A$ by interchanging its rows and columns is called the transpose of the given matrix and is denoted by $\boldsymbol{A}^{T}$ or $\boldsymbol{A}^{\prime}$.

If $A=[a _{i j}] _{m \times n} \Rightarrow A^{T}=[a _{j i}] _{n \times m}$

For example, if $A= \begin{bmatrix} 3 & 6 & 2 \\ 1 & -1 & 5\end{bmatrix} _{3 \times 2}$, then $A^{T}$ or $A^{\prime}= \begin{bmatrix} 3 & 1 \\ 6 & -1 \\ 2 & 5\end{bmatrix} _{2 \times 3}$

$ A= \begin{bmatrix} 4 & -1 & 2 \\ 7 & 6 & 5 \\ -3 & -2 & 0 \end{bmatrix} \text{, then } A^{\prime}= \begin{bmatrix} 4 & 7 & -3 \\ -1 & 6 & -2 \\ 2 & 5 & 0 \end{bmatrix} $

Note: $\bullet(i-j)$ th element of $A=(j-i)$ th element of $A^{T}$

If order of $A$ is $m \times n$, then order of $A^{T}$ is $n \times m$

Properties of Transpose Matrix

I. If $A$ is any matrix, then $(A^{\prime})^{\prime}=\boldsymbol{A}$

Let $A= \begin{bmatrix} 4 & -2 & -3 \\ 1 & 6 & 5\end{bmatrix} \Rightarrow A^{\prime}= \begin{bmatrix} 4 & 1 \\ -2 & 6 \\ -3 & 5\end{bmatrix} \Rightarrow(A^{\prime})^{\prime}= \begin{bmatrix} 4 & -2 & -3 \\ 1 & 6 & 5\end{bmatrix} $

II. If $A$ and $B$ are two matrices of the same order, then $(A+B)^{\prime}=A^{\prime}+B^{\prime}$

For example, Let $A= \begin{bmatrix} 2 & 0 \\ 1 & -3\end{bmatrix} , B= \begin{bmatrix} 6 & -5 \\ 0 & 8\end{bmatrix} $

$ \begin{aligned} A+B & = \begin{bmatrix} 2 & 0 \\ 1 & -3 \end{bmatrix} + \begin{bmatrix} 6 & -5 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 2+6 & 0+(-5) \\ 1+0 & -3+8 \end{bmatrix} = \begin{bmatrix} 8 & -5 \\ 1 & 5 \end{bmatrix} \\ (A+B)^{\prime} & = \begin{bmatrix} 8 & 1 \\ -5 & 5 \end{bmatrix} \end{aligned} $

$ A^{\prime}+B^{\prime}= \begin{bmatrix} 2 & 1 \\ 0 & -3 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ -5 & 8 \end{bmatrix} = \begin{bmatrix} 2+6 & 1+0 \\ 0+(-5) & -3+8 \end{bmatrix} = \begin{bmatrix} 8 & 1 \\ -5 & 5 \end{bmatrix} $

III. If $\boldsymbol{A}$ is $\boldsymbol{m} \times \boldsymbol{p}$ matrix and $\boldsymbol{B}$ is $\boldsymbol{p} \times \boldsymbol{n}$ matrix then $(\boldsymbol{A B})^{\prime}=\boldsymbol{B}^{\prime} \boldsymbol{A}^{\prime}$.

For example, let $A= \begin{bmatrix} 1 \\ -5 \\ 7\end{bmatrix} , B= \begin{bmatrix} 3 & 1 & -2\end{bmatrix} $, then

$A B= \begin{bmatrix} 1 \\ -5 \\ 7\end{bmatrix} _{3 \times 1} \begin{bmatrix} 3 & 1 & -2\end{bmatrix} _{1 \times 3}= \begin{bmatrix} 1 \times 3 & 1 \times 1 & 1 \times-2 \\ -5 \times 3 & -5 \times 1 & -5 \times-2 \\ 7 \times 3 & 7 \times 1 & 7 \times-2\end{bmatrix} _{3 \times 3}= \begin{bmatrix} 3 & 1 & -2 \\ -15 & -5 & 10 \\ 21 & 7 & -14\end{bmatrix} $

$\therefore(A B)^{\prime}= \begin{bmatrix} 3 & -15 & 21 \\ 1 & -5 & 7 \\ -2 & 10 & -14\end{bmatrix} $

Now $\quad A^{\prime}= \begin{bmatrix} 1 & -5 & 7\end{bmatrix} , B^{\prime}= \begin{bmatrix} 3 \\ 1 \\ -2\end{bmatrix} $

So, $\quad B^{\prime} A^{\prime}= \begin{bmatrix} 3 \\ 1 \\ -2\end{bmatrix} _{3 \times 1} \begin{bmatrix} 1 & -5 & 7\end{bmatrix} _{1 \times 3}= \begin{bmatrix} 3 \times 1 & 3 \times-5 & 3 \times 7 \\ 1 \times 1 & 1 \times-5 & 1 \times 7 \\ -2 \times 1 & -2 \times-5 & -2 \times 7\end{bmatrix} _{3 \times 3}= \begin{bmatrix} 3 & -15 & 21 \\ 1 & -5 & 7 \\ -2 & 10 & -14\end{bmatrix} $

$\Rightarrow(A B)^{\prime}=B^{\prime} A^{\prime}$

IV. If $\boldsymbol{A}$ is a matrix and $\boldsymbol{k}$ is a scalar, then $(\boldsymbol{k} \boldsymbol{A})^{\prime}=\boldsymbol{k} \boldsymbol{A}^{\prime}$.

~~ 8. (i) Symmetric Matrices: A square matrix $A=[a _{i j}]$ is said to be symmetric if its $(i-j)$ th element is equal to its $(j-i)$ th element, i. e., if $\boldsymbol{a} _{i j}=\boldsymbol{a} _{\boldsymbol{j} \boldsymbol{i}} \forall \boldsymbol{i}, \boldsymbol{j}$.

$ \begin{bmatrix} 1 & 2 \\ 2 & 3\end{bmatrix} , \begin{bmatrix} 2 & 4 & 7 \\ 4 & 5 & 0 \\ 7 & 0 & 3\end{bmatrix} $ are symmetric matrices

then $a _{12}=a _{21} \quad a _{12}=a _{21}, a _{13}=a _{31}, a _{23}=a _{32}$

Note: $\bullet$ Symmetric matrices are always square matrices.

For a matrix $A$ to be symmetric, it is necessary that the matrix is equal to its transpose, i.e., $A^{T}=A$.
Diagonal matrices are always symmetric.

(ii) Skew-Symmetric Matrices: A square matrix $A=[a _{i j}]$ is said to be skew-symmetric if the $(i-j)$ th element of $A$ is the negative of $(j-i)$ th element of $A$, i.e., if $a _{i j}=-a _{j i} \forall i, j$.

Thus, the matrix $ \begin{bmatrix} 0 & a & b \\ a & 0 & c \\ -b & c & 0\end{bmatrix} $ is a skew symmetric matrix.

Each element on the principal diagonal of a skew-symmetric matrix is zero as:

$ \begin{aligned} & a _{i j}=-a _{j i} \forall i, j \\ \Rightarrow & a _{i i}=-a _{i i} \forall i=j \Rightarrow 2 a _{i i}=0 \Rightarrow a _{i i}=0 \forall i \\ \Rightarrow & a _{11}=a _{22}=a _{33}=\ldots \ldots \ldots \ldots .=a _{n n}=0 . \end{aligned} $

Note: 1. For a skew-symmetric matrix, it is necessary that $A^{\prime}=-A$.

2. A matrix which is both symmetric and skew symmetric is a square null matrix.

A is symmetric as well as skew symmetric.

$ \begin{aligned} & \Rightarrow A^{\prime}=A \text{ and } A^{\prime}=-A . \\ & \Rightarrow A=-A \Rightarrow 2 A=0 \Rightarrow A=0 . \end{aligned} $

Properties of Symmetric and Skew-Symmetric Matrices:

I. The sum of two symmetric matrices is a symmetric matrix.

Let $A, B$ be two symmetric matrices. Then,

$A^{\prime}=A, B^{\prime}=B$

$\therefore(A+B)^{\prime}=A^{\prime}+B^{\prime}=(A+B) \Rightarrow A+B$ is symmetric.

II. The sum of two skew symmetric matrices is a skew symmetric matrix.

Let, $A, B$ be two skew symmetric matrices. Then,

$A^{\prime}=-A, B^{\prime}=-B$.

Then, $(A+B)^{\prime}=A^{\prime}+B^{\prime}=-A+(-B)=-(A+B)$

$\Rightarrow(A+B)$ is skew symmetric.

III. For a scalar $k$ and

(a) a symmetric matrix $A, k A$ is a symmetric matrix.

(b) a skew symmetric matrix $A, k A$ is a skew symmetric matrix.

IV. If $A$ be any square matrix, then $A+A^{\prime}$ is symmetric and $A-A^{\prime}$ is skew symmetric.

$(A+A^{\prime})^{\prime}=A^{\prime}+(A^{\prime})^{\prime}$

$=A^{\prime}+A$

$\Rightarrow A+A^{\prime}$ is symmetric.

$(A-A^{\prime})^{\prime}=A^{\prime}-(A^{\prime})^{\prime}$

$=A^{\prime}-A=-(A-A^{\prime})$

$\Rightarrow A-A^{\prime}$ is skew symmetric.

V. Every square matrix is uniquely expressible as the sum of a symmetric matrix and a skew-symmetric matrix.

Given, a square matrix $A$, it can be expressed as the sum of a symmetric matrix and a skew symmetric matrix as under:

$A=\frac{1}{2}(A+A^{\prime})+\frac{1}{2}(A-A^{\prime})$

where $\frac{1}{2}(A+A^{\prime})$ is a symmetric matrix and

$\frac{1}{2}(A-A^{\prime})$ is a skew symmetric matrix.

VI. For the symmetric matrices $A, B$, if $A B$ is a symmetric matrix, then $A B=B A$ and vice versa.

Given, $A^{\prime}=A, B^{\prime}=B$ and $(A B)^{\prime}=A B$

Now $(A B)^{\prime}=B^{\prime} A^{\prime}=B A$

From (i) and (ii) $A B=B A$

If $A B=B A \Rightarrow A B=B^{\prime} A^{\prime} \Rightarrow A B=(A B)^{\prime}$

$(\therefore A^{1}=A, B^{1}=B)$

$\Rightarrow A B$ is a symmetric matrix.

VII. All positive integral powers of a symmetric matrix are symmetric.

If $A^{\prime}=A$, then $(A^{n})^{\prime}=A^{n}$.

$(A^{n})^{\prime}=(\text{ A.A.A.A. …n times })^{\prime}$

$=A^{\prime} \cdot A^{\prime} \cdot A^{\prime} \cdot A^{\prime} \ldots n$ times $(\therefore(A)^{\prime}=A)$

$=A \cdot A \cdot A \cdot A \ldots n$ times $=A^{n}$.

$\Rightarrow A^{n}$ is symmetric.

DETERMINANTS

9. Determinant of a square matrix

Associated with each square matrix $A$ having real number entries is a real number called the determinant of $A$ and is denoted by $\delta A$ or $\triangle A$ or $|A|$ or $det(A)$.

$\quad$ For a square matrix $A= \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2\end{bmatrix} $ of order 2

$ det(A)=\Delta A= \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} =a_1 b_2-a_2 b_1 $

Note: The determinant of a matrix of order 1, i.e., $[a]$ is a itself.

If $A=[-3]$, then det. $A=|A|=|-3|=-3$.

Caution: The determinant $|-3|=-3$ should not be confused with the absolute value $|-3|=3$.

For a square matrix $A= \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix} $ of order 3, the value of determinant $A$ is calculated as explained below.

Some new concepts that will be used are:

Minors and Cofactors

Minors: The minor of an element in a determinant is a determinant that is left after removing the row and column which intersect at the element, and is of order one less than that of the given determinant.

$\therefore \quad$ In the determinant $ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix} $

Minor of $a_1= \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3\end{vmatrix} $, Minor of $b_1= \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3\end{vmatrix} $, Minor of $c_1= \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3\end{vmatrix} $

$\therefore$ In general, the minor $M _{i j}$ of element $a _{i j}$ is the value of the determinant obtained by deleting the ith row and jth column of the given determinant.

Cofactor. The cofactor of an element or element $a _{i j}$ is the minor of $a _{i j}$ with appropriate sign. Thus,

$ \text{ Cofactor of } a _{i j}=A _{i j}=(-1)^{i+j} M _{i j} \text{ (minor) } $

where $i$ and $j$ are respectively the row number and column number of the element.

$\therefore \quad$ In the determinant $ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix} $

Cofactor of $a_1=(-1)^{1+1} \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3\end{vmatrix} =b_2 c_3-b_3 c_2$

Cofactor of $a_2=(-1)^{2+1} \begin{vmatrix} b_1 & c_1 \\ b_3 & c_3\end{vmatrix} =-(b_1 c_3-b_3 c_1)$

Cofactor of $b_1=(-1)^{1+2} \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3\end{vmatrix} =-(a_2 c_3-a_3 c_2)$

Thus the value of a determinant can be determined by expanding it along any row or column.

Now, the value of a determinant of order 3 can be written as: Expanding along row $1(R_1)$

$ \begin{aligned} \Delta & = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} =a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} -b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} +c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \\ & =a_1(\text{ its cofactor })+b_1(\text{ its cofactor })+c_1(\text{ its cofactor })=a_1(\text{ its cofactor })+a_2(\text{ its cofactor })+a_3(\text{ its cofactor }) \\ & =a_2(\text{ its cofactor })+b_2(\text{ its cofactor })+c_2(\text{ its cofactor })=b_1(\text{ its cofactor })+b_2(\text{ its cofactor })+b_3(\text{ its cofactor }) \\ & =a_3(\text{ its cofactor })+b_3(\text{ its cofactor })+c_3(\text{ its cofactor })=c_1(\text{ its cofactor })+c_2(\text{ its cofactor })+c_3(\text{ its cofactor }) \end{aligned} $

Note: - The ordered pairs used are (an element, minor) for the same row or column.

While expanding a determinant by any row or column using minors, we may keep in mind the following scheme of Signs for a determinant of order 3 .

$ \begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}$

If a row or column of a determinant consists of all zeros, the value of the determinant is zero.

~~ 10 Singular matrix. A square matrix $A$ is said to be singular if $det[A]=0$, otherwise it is a non-singular matrix. For example,

(a) Let $A= \begin{bmatrix} 1 & 3 \\ 2 & -1\end{bmatrix} $

Then, $|A|= \begin{vmatrix} 1 & 3 \\ 2 & -1\end{vmatrix} =(1 \times(-1))-(2 \times 3)=-1-6=-7 \neq 0$

$\therefore A$ is a non-singular matrix.

(b) Let $A= \begin{bmatrix} 0 & 1 & -1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix} $ Then,

$ \begin{aligned} |A| & =0 \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} -1 \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} +(-1) \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \quad \text{ (Expanding along } R_1 \text{ ) } \\ & =0+(-1)(-1)+(-1) \times 1=1-1=0 \end{aligned} $

$\therefore|A|=0 \Rightarrow A$ is a singular matrix.

11. Properties of Determinants.

Determinants have some properties which help in simplifying the process of finding the value of the determinant. In fact in some cases using these properties, we can find the value of the determinant without ever expanding along a given row or column. We shall list the properties here and show them with the help of examples.

Property I: If each entry in any row, or each entry in any column of a determinant is 0, then the determinant

(Expanding by $R_2$ )

Property II: If the rows be changed in columns and columns into rows, the determinant remains unaltered.

$ \begin{aligned} \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \\ \begin{aligned} \text{ LHS } & =a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} -a_2 \begin{vmatrix} b_1 & c_1 \\ b_3 & c_3 \end{vmatrix} +a_3 \begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix} \quad \text{ (Expanding along } C_1 \text{ (col. 1)) } \\ & =a_1(b_2 c_3-b_3 c_2)-a_2(b_1 c_3-b_3 c_1)+a_3(b_1 c_2-b_2 c_1) \\ \text{ RHS } & =a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} -a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} +a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \quad \quad(\text{ Expanding along } R_1. \text{ (row 1)). } \\ & =a_1(b_2 c_3-b_3 c_2)-a_2(b_1 c_3-b_3 c_1)+a_3(b_1 c_2-b_2 c_1) \end{aligned} \end{aligned} $

Hence LHS $=$ RHS.

Property III: If any two rows (or columns) of a determinant are interchanged the resulting determinant is the negative of the original determinant.

$ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} =- \begin{vmatrix} b_1 & a_1 & c_1 \\ b_2 & a_2 & c_2 \\ b_3 & a_3 & c_3 \end{vmatrix} $

For example, $ \begin{vmatrix} 1 & 2 & -3 \\ 2 & 1 & 4 \\ 3 & 1 & 2\end{vmatrix} =1 \begin{vmatrix} 1 & 4 \\ 1 & 2\end{vmatrix} -2 \begin{vmatrix} 2 & 4 \\ 3 & 2\end{vmatrix} +(-3) \begin{vmatrix} 2 & 1 \\ 3 & 1\end{vmatrix} \quad$ (Expanding along $R_1$ )

$ \begin{aligned} & =1(2-4)-2(4-12)-3(2-3)=-2+16+3=17 \\ \begin{vmatrix} 2 & 1 & -3 \\ 1 & 2 & 4 \\ 1 & 3 & 2 \end{vmatrix} & =2 \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} -1 \begin{vmatrix} 1 & 4 \\ 1 & 2 \end{vmatrix} +(-3) \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \\ & =2(4-12)-1(2-4)-3(3-2)=-16+2-3=-\mathbf{1 7 .} \end{aligned} $

Note: If any line (row or column) of a determinant $\Delta$ be passed over in parallel lines, the resulting determinant $\Delta^{\prime}=(-1)^{m} \Delta$.

For example, if $\Delta= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix} $ and $\Delta^{\prime}= \begin{vmatrix} b_1 & c_1 & a_1 \\ b_2 & c_2 & a_2 \\ b_3 & c_3 & a_3\end{vmatrix} $, then

$ \Delta^{\prime}=(-1)^{2} \Delta=\Delta \text{. } $

Property IV: If any two rows (or columns) in a determinant are identical, the determinant is equal to zero.

$ \begin{aligned} \Delta & = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} =0 \\ \Delta & =a_1 \begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix} -b_1 \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} +c_1 \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} \\ & =a_1(b_1 c_2-b_2 c_1)-b_1(a_1 c_2-a_2 c_1)+c_1(a_1 b_2-b_1 a_2) \\ & =a_1 b_1 c_2-a_1 b_2 c_1-b_1 a_1 c_2+b_1 a_2 c_1+c_1 a_1 b_2-b_1 a_2 c_1=0 \end{aligned} $

Property V: If all the elements of any row (or column) be multiplied by a non-zero real number $k$, then the value of the new determinant is $k$ times the value of the original determinant.

Thus if

$ \Delta= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_2 \end{vmatrix} \text{ and } \Delta^{\prime}= \begin{vmatrix} k a_1 & k b_1 & k c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} $

Then

$ \Delta=k \Delta^{\prime} $

For example, $ \begin{vmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 6 & -3 & 15\end{vmatrix} = \begin{vmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3(2) & 3(-1) & 3(5)\end{vmatrix} =3 \begin{vmatrix} 1 & 3 & -1 \\ 2 & 4 & 1 \\ 2 & -1 & 5\end{vmatrix} $

Note: If two parallel lines (rows or columns) be such that the elements of one are equi-multiples of the elements of the other, the determinant is equal to zero.

Let $\Delta= \begin{vmatrix} 1 & 3 & 2 \\ 1 & 3 & 4 \\ 2 & 6 & 8\end{vmatrix} =2 \begin{vmatrix} 1 & 3 & 2 \\ 1 & 3 & 4 \\ 1 & 3 & 4\end{vmatrix} =2 \times 0=0$ (Two identical rows)

Here elements of row $2(R_3)$ are equimultiples of elements of $R_3$.

Property VI: If each entry in a row (or column) of a determinant is written as the sum of two or more terms, then the determinant can be written as the sum of two or more determinants.

$ \text{ i.e, } \Delta= \begin{vmatrix} a_1+x_1 & b_1 & c_1 \\ a_2+x_2 & b_2 & c_2 \\ a_3+x_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} x_1 & b_1 & c_1 \\ x_2 & b_2 & c_2 \\ x_3 & b_3 & c_3 \end{vmatrix} $

Property VII: If each entry of one row (or column) of a determinant is multiplied by a real number $k$ and the resulting product is added to the corresponding entry in another row (or column respectively) in the determinant, then the resulting determinant is equal to the original determinant.

$ \begin{aligned} \Delta & = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \text{ and } \Delta^{\prime}= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3+k a_1 & b_3+k b_1 & c_3+k c_1 \end{vmatrix} \text{ Then } \Delta=\Delta^{\prime} . \\ \Delta^{\prime} & = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3+k a_1 & b_3+k b_1 & c_3+k c_1 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_3 \\ k a_1 & k b_1 & k c_1 \end{vmatrix} \\ & = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} +k \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_1 & b_1 & c_1 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} +k \times 0 \quad \quad(\because \text{ Two rows are identical) } \\ & = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} =\Delta . \end{aligned} $

This property can be generalised as:

If to each element of a line (row or column) of a determinant be added the equi-multiples of the corresponding elements of one or more parallel lines, the determinant remains unaltered.

$ \begin{vmatrix} a_1+l a_2+m a_3 & a_2 & a_3 \\ b_1+l b_2+m b_3 & b_2 & b_3 \\ c_1+l c_2+l c_3 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} $

This property provides very powerful methods for simplifying the process for calculating the values of determinants. $R_1 \longrightarrow R_1+m R_2+n R_3$ means to the first row, we add $m$ times the second row and $n$ times the third row. Similarly,

$C_3 \longrightarrow C_3-C_1$ means subtracting corresponding elements of column 1 from elements of column 3 and placing them in place of elements of column 3.

For example, $ \begin{vmatrix} 43 & 1 & 6 \\ 35 & 7 & 4 \\ 17 & 3 & 2\end{vmatrix} = \begin{vmatrix} 43-7 \times 6 & 1 & 6 \\ 35-7 \times 4 & 7 & 4 \\ 17-7 \times 2 & 3 & 2\end{vmatrix} $ and so $C_1 \longrightarrow C_1-7 C_3= \begin{vmatrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 3 & 3 & 2\end{vmatrix} =0, C_1$ and $C_2$ being identical.

$ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = \begin{vmatrix} a+b+c & b & c \\ b+c+a & c & a \\ c+a+b & a & b \end{vmatrix} \text{ Operating } C_1 \longrightarrow C_1+C_2+C_3 $

$ \begin{aligned} & =(a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \text{ (Taking out }(a+b+c) \text{ common) } \\ & =(a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} \text{ Operating } R_2 \longrightarrow R_2-R_1, R_3 \longrightarrow R_3-R_1 \end{aligned} $

$ \begin{aligned} & =(a+b+c) .{(c-b)(b-c)-(a-b)(a-c)} \text{ expanding along column } 1 \\ & =(a+b+c){b c-b^{2}-c^{2}+c b-(a^{2}-a b-a c+b c)} \\ & =(a+b+c){a b+b c+c a-a^{2}-b^{2}-c^{2}} . \end{aligned} $

Property VIII : Product of two determinants

$ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \begin{vmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{vmatrix} = \begin{vmatrix} a_1 \alpha_1+b_1 \beta_1+c_1 \gamma_1 & a_1 \alpha_2+b_1 \beta_2+c_1 \gamma_2 & a_1 \alpha_3+b_1 \beta_3+c_1 \gamma_3 \\ a_2 \alpha_1+b_2 \beta_1+c_2 \gamma_1 & a_2 \alpha_2+b_2 \beta_2+c_2 \gamma_2 & a_2 \alpha_3+b_2 \beta_3+c_2 \gamma_3 \\ a_3 \alpha_1+b_3 \beta_1+c_3 \gamma_1 & a_3 \alpha_2+b_3 \beta_2+c_3 \gamma_2 & a_3 \alpha_3+b_3 \beta_3+c_3 \gamma_3 \end{vmatrix} $

12. Adjoint of a square matrix

Let $A=[a _{i j}] _{3 \times 3}$ be the given square matrix of order 3 .

Then,

$ A= \begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33} \end{bmatrix} $

If $A _{i j}$ be the cofactor $a _{i j}$, the adjoint of matrix $A$ denoted by adj. $A$ is defined as:

$ \text{ adj. } A= \begin{bmatrix} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33} \end{bmatrix} ^{T}= \begin{bmatrix} A _{11} & A _{21} & A _{31} \\ A _{12} & A _{22} & A _{32} \\ A _{13} & A _{23} & A _{33} \end{bmatrix} $

So, the adjoint of a square matrix $A$ is the transpose of the matrix obtained by replacing each element of $A$ by its cofactor in $|A|(det A)$.

For example,

(a) If $A= \begin{bmatrix} 3 & -1 \\ 4 & -2\end{bmatrix} $, then to find adj $A$, we find the cofactors.

$ \begin{aligned} & A _{11}=(-1)^{+1}|-2|=-2 \\ & A _{12}=(-1)^{1+2}|4|=-4 \\ & A _{21}=(-1)^{2+1}|-1|=+1 \\ & A _{22}=(-1)^{2+2}|3|=3 \\ & \therefore \text{ adj. } A= \begin{bmatrix} A _{11} & A _{12} \\ A _{21} & A _{22} \end{bmatrix} ^{T}= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ -4 & 3 \end{bmatrix} \end{aligned} $

(b) Find the adjoint of matrix $A= \begin{bmatrix} 1 & 4 & 3 \\ 4 & 2 & 1 \\ 3 & 2 & 2\end{bmatrix} $

Let $A _{i j}$ be the cofactor of $a _{i j}$. Then,

$ \begin{aligned} & A _{11}=+ \begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix} =(4-2)=2 \\ & A _{12}=- \begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix} =-(8-3)=-5 \\ & A _{13}= \begin{vmatrix} 4 & 2 \\ 3 & 2 \end{vmatrix} =(8-6)=2 \\ & A _{21}=- \begin{vmatrix} 4 & 3 \\ 2 & 2 \end{vmatrix} =-(8-6)=-2 \\ & A _{22}= \begin{vmatrix} 1 & 3 \\ 3 & 2 \end{vmatrix} =(2-9)=-7 \\ & A _{23}=- \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} =-(2-12)=10 \\ & A _{31}= \begin{vmatrix} 4 & 3 \\ 2 & 1 \end{vmatrix} =(4-6)=-2 \end{aligned} $

Remember the signs of cofactor by

$ (-\mathbf{1})^{i+j} \text{ or } \begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix} $

$ \begin{aligned} & A _{32}=- \begin{vmatrix} 1 & 3 \\ 4 & 1 \end{vmatrix} =-(1-12)=11 \quad A _{33}= \begin{vmatrix} 1 & 4 \\ 4 & 2 \end{vmatrix} =(2-16)=-14 \\ & \therefore adj A= \begin{bmatrix} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33} \end{bmatrix} ^{T}= \begin{bmatrix} A _{11} & A _{21} & A _{31} \\ A _{12} & A _{22} & A _{32} \\ A _{13} & A _{23} & A _{33} \end{bmatrix} = \begin{bmatrix} 2 & -2 & -2 \\ -5 & -7 & 11 \\ 2 & 10 & -14 \end{bmatrix} \end{aligned} $

Properties of adjoint of a square matrix

1. If $A$ be a square matrix of order $n$, then $(adj A) A=A(adj A)=|A| I_n$, where $|A|=determinant$ value of matrix $A$ and $I_n$ is the identity matrix of order $n$.

For a $3 \times 3$ square matrix, let

$ A= \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} , \text{ Then adj } A= \begin{bmatrix} A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3 \end{bmatrix} $

where $A_1, B_1, C_1 \ldots$. are the respective cofactors of $a_1, b_1, c_1 \ldots \ldots$.

$ \begin{aligned} \therefore \quad A(adj A)= & { \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \\ C_1 & C_2 & C_3 \end{bmatrix} } \\ = & { \begin{bmatrix} a_1 A_1+b_1 B_1+c_1 C_1 & a_1 A_2+b_1 B_2+c_1 C_2 & a_1 A_3+b_1 B_3+c_1 C_3 \\ a_2 A_1+b_2 B_1+c_2 C_1 & a_2 A_2+b_2 B_2+c_2 C_2 & a_2 A_3+b_2 B_3+c_2 C_3 \\ a_3 A_1+b_3 B_1+c_3 C_1 & a_3 A_2+b_3 B_2+c_3 C_2 & a_3 A_3+b_3 B_3+c_3 C_3 \end{bmatrix} = \begin{bmatrix} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{bmatrix} } \\ & (\because a_1 A_2+b_1 B_2+c_1 C_2=a_1(b_1 c_3-b_3 c_1)+b_1(a_1 c_3-a_3 c_1)+c_1(a_1 b_3-a_3 b_1)=0) \end{aligned} $

Similarly for all the entries besides the entries of the principal diagonal.

$ =|A| \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =|A| I_3 $

Also it can be shown similarly that $(adj A) A=|A| I_3$

Note: $Adj I=I \text{ and } Adj 0=0 $

II. If $A$ and $B$ are two non-singular matrices of the same order, then $adj(A B)=(adj B)(adj A)$.

13. Inverse of a square matrix

If $A$ and $B$ are square matrices such that $A B=B A=I$, then $B$ is called the inverse of $A$ and is written as $A^{-1}=B$ and $A$ is the inverse of $B$, written as $B^{-1}=A$.

Thus, $A^{-1} A=A A^{-1}=I$.

If $A$ is a non-singular square matrix of order $n$, then

$ A^{-1}=\frac{1}{|A|}(adj \boldsymbol{A}) \quad(|\boldsymbol{A}| \neq \mathbf{0}) $

Note: If $A B=B A=I$, then $B$ is inverse of $A$, i.e, $B=A^{-1}$

Also, we know by the property of adjoint of a square matrix that, $A(adj A)=(adj A) A=|A| I$

$ \begin{aligned} & \Rightarrow A \frac{(adj A)}{|A|}=(\frac{adj A}{|A|}) A=I(\text{ as }|A| \neq 0, A \text{ being singular }) \\ & \Rightarrow A^{-1}=\frac{1}{|A|}(adj(A)) \end{aligned} $

- Properties of matrices and inverses

I. If $A$ and $B$ are two non-singular matrices of order $n$, then $A B$ is also a non - singular matrix of order $n$ such that

$(A B)^{-1}=B^{-1} A^{-1}$

(Reversal law for the inverse of a product)

II. $(A^{T})^{-1}=(A^{-1})^{T}$

For example, If $A= \begin{bmatrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ -6 & 4 & 7\end{bmatrix} $ and $A B=B A=I$, find $B$.

If $A B=B A=I$, then $B=A^{-1}$.

For $A^{-1}$ to exists, $A$ should be a non-singular matrix.

We have, $\quad|A|= \begin{vmatrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ -6 & 4 & 7\end{vmatrix} =3 \begin{vmatrix} 5 & 9 \\ 4 & 7\end{vmatrix} -0 \begin{vmatrix} 1 & 9 \\ -6 & 7\end{vmatrix} +2 \begin{vmatrix} 1 & 5 \\ -6 & 4\end{vmatrix} =3(35-36)+2(4+30)=-3+68=65 \neq 0$

$\Rightarrow A$ is a non-singular matrix

Now we need to find adj $A$,

$ \begin{aligned} & \therefore \quad A _{11}= \begin{vmatrix} 5 & 9 \\ 4 & 7 \end{vmatrix} =(35-36)=-1 \quad A _{12}=- \begin{vmatrix} 1 & 9 \\ -6 & 7 \end{vmatrix} =-(7+54)=-61 \\ & A _{13}= \begin{vmatrix} 1 & 5 \\ -6 & 4 \end{vmatrix} =4+30=34 \quad A _{21}=- \begin{vmatrix} 0 & 2 \\ 4 & 7 \end{vmatrix} =-(0-8)=8 \\ & A _{22}= \begin{vmatrix} 3 & 2 \\ -6 & 7 \end{vmatrix} =21+12=33 \quad A _{23}=- \begin{vmatrix} 3 & 0 \\ -6 & 4 \end{vmatrix} =-(12-0)=-12 \\ & A _{31}= \begin{vmatrix} 0 & 2 \\ 5 & 9 \end{vmatrix} =0-10=-10 \quad A _{32}=- \begin{vmatrix} 3 & 2 \\ 1 & 9 \end{vmatrix} =-(27-2)=-25 \\ & A _{33}= \begin{vmatrix} 3 & 0 \\ 1 & 5 \end{vmatrix} =15 \\ & \therefore adj A= \begin{bmatrix} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33} \end{bmatrix} ^{T}= \begin{bmatrix} -1 & -61 & 34 \\ 8 & 33 & -12 \\ -10 & -25 & 15 \end{bmatrix} ^{T}= \begin{bmatrix} -1 & 8 & -10 \\ -61 & 33 & -25 \\ 34 & -12 & 15 \end{bmatrix} \\ & \Rightarrow \quad A^{-1}=\frac{1}{|A|} \text{ adj } A=\frac{1}{65} \begin{bmatrix} -1 & 8 & -10 \\ -61 & 33 & -25 \\ 34 & -12 & 15 \end{bmatrix} \text{. } \end{aligned} $

14. Some more special matrices :

1. Nilpotent Matrix: $A$ square matrix $A$ such that $A^{n}=0$ is called a nilpotent matrix of order $n$.

If there exists a matrix such that $A^{2}=0$, then $A$ is nilpotent of order 2 .

2. Idempotent Matrix: A square matrix $A$, such that $A^{2}=A$ is called an idempotent matrix.

If $A B=A$ and $B A=B$, then $A$ and $B$ are idempotent matrices.

Orthogonal Matrix: A square matrix $A$, such that $A A^{T}=I$ is called an orthogonal matrix.

If $A$ is an orthogonal matrix, then $\boldsymbol{A}^{T}=\boldsymbol{A}^{\mathbf{- 1}}$

Involuntary Matrix: A square matrix $A$, such that $A^{2}=I$ is called an involuntary matrix.

Note: (i) A matrix $A$ is involuntary $\Leftrightarrow(I-A)(I+A)=0$

(ii) A matrix $A$ is involuntary matrix then $A^{-1}=A$.

15. Application of matrices to the solution of linear equations

Consider the two simultaneous equations in two variables $x$ and $y$.

$ \begin{aligned} & a_1 x+b_1 y=c_1 \\ & a_2 x+b_2 y=c_2 \end{aligned} $

These can be written in the matrix form as:

$ \begin{gathered} { \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} } \\ A \quad X=B \end{gathered} $

where $A= \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2\end{bmatrix} $ is a $2 \times 2$ matrix and

$ X= \begin{bmatrix} x \\ y \end{bmatrix} \text{ and } B= \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \text{ are both } 2 \times 1 \text{ matrices and } $

Similarly the three simultaneous equations

$ \begin{aligned} & a_1 x+b_1 y+c_1 z=d_1 \\ & a_2 x+b_2 y+c_2 z=d_2 \\ & a_3 x+b_3 y+c_3 z=d_3 \text{ can be written in the matrix form as: } \end{aligned} $

$ \begin{aligned} { \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} } & = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} \\ A & =B \end{aligned} $

where $A$ is a square matrix of order $3(3 \times 3)$ and $X$ and $B$ are $(3 \times 1)$ column matrices.

$ \begin{matrix} \text{ Now, } & A X=B \\ \Rightarrow & A^{-1} A X=A^{-1} B \\ \Rightarrow & I X=A^{-1} B \Rightarrow X=A^{-1} B \\ \Rightarrow & X=\frac{1}{|A|}(adj A) \cdot B \end{matrix} $

The conditions for unique solution, no solution and infinite number of solution can be summarized as under:

Unique solution consistent

$ X=A^{-1} B $

No solution or infinite number of solutions

Consistent (dependent)

Special case: When $B= \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix} $

In this case, $|A| \neq 0 \Rightarrow x=0, y=0, z=0$ we say that the system has trivial solution.

If $|A|=0$, then the system has infinitely many solutions.

Note: We shall deal with equations in two variables only in this book.

Ex. 1 . Use matrix method to solve the system of equations :

$ 4 x-3 y=11,3 x+7 y=-1 $

Sol. The given system of equations can be written in the matrix form as:

$ \begin{aligned} & { \begin{bmatrix} 4 & -3 \\ 3 & 7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 11 \\ -1 \end{bmatrix} } \\ &|A|= \begin{vmatrix} 4 & -3 \\ 3 & 7 \end{vmatrix} =28+9=37 \neq 0 \\ & \Rightarrow A \text{ is a non-singular matrix and the system has an unique solution. } \\ & X=A^{-1} B . \\ & \therefore \text{ We need to find adj } A \text{ and hence the cofactors of } A . \\ & A _{11}=7, A _{12}=-3, A _{21}=-(-3)=3, A _{22}=4 \\ & \therefore A^{-1}=\frac{1}{|A|} \text{ adj } A=\frac{1}{|A|} \begin{bmatrix} A _{11} & A _{12} \\ A _{21} & A _{22} \end{bmatrix} \\ & \quad=\frac{1}{|A|} \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} =\frac{1}{37} \begin{bmatrix} 7 & 3 \\ -3 & 4 \end{bmatrix} \\ & \therefore { \begin{bmatrix} x \\ y \end{bmatrix} =A^{-1} B= \begin{bmatrix} 7 / 37 & 3 / 37 \\ -3 / 37 & 4 / 37 \end{bmatrix} \begin{bmatrix} 11 \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{77}{37} & -\frac{3}{37} \\ \frac{-33}{37} & -\frac{4}{37} \end{bmatrix} = \begin{bmatrix} \frac{74}{37} \\ \frac{-37}{37} \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix} } \\ & \Rightarrow \boldsymbol{x}=\mathbf{2 ,} \boldsymbol{y}=-\mathbf{1 .} \end{aligned} $

Ex. 2 . Use matrix method to examine the consistency or inconsistency of the system of equations

$ 6 x+4 y=2,9 x+6 y=3 . $

Sol. Writing the given system of equations in the matrix form, we have

$ \begin{aligned} { \begin{bmatrix} 6 & 4 \\ 9 & 6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} } & = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \\ X & =B \end{aligned} $

Now, $\quad|A|= \begin{vmatrix} 6 & 4 \\ 9 & 6\end{vmatrix} =36-36=0$

$\Rightarrow A$ is a singular matrix $\Rightarrow$ Either the system has no solution or infinite number of solutions.

To check that we find $(adj A) B$.

$ \begin{aligned} A _{11} & =6, A _{12}=-9, A _{21}=-4, A _{22}=6 \\ \therefore \quad adj A & = \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} 6 & -4 \\ -9 & 6 \end{bmatrix} \\ (adj A) B & = \begin{bmatrix} 6 & -4 \\ -9 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 12-12 \\ -18+18 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} =0 . \end{aligned} $

Since $(adj A) B=0$, the given system is consistent and has infinite number of solutions.

Let $y=k$ in the first equation. Then,

$ 6 x+4 k=2 \Rightarrow 6 x=2-4 k \Rightarrow x=\frac{2-4 k}{6}=\frac{1}{3}(1-2 k) $

Putting this value of $x$ in the second equation, we have 9. $\frac{1}{3}(1-2 k)+6 k=3 \Rightarrow 3-6 k+6 k=3 \Rightarrow 3=3$, which is true

Hence the given system has infinitely many solutions given by

$ x=\frac{1}{3}(1-2 k), y=k $

Ex. 3 . Use matrix method to examine the given system of equations for consistency or inconsistency.

$ \begin{aligned} & 3 x-2 y=5 \\ & 6 x-4 y=9 \end{aligned} $

Sol. Writing the given system of equations in the matrix form we have

$ \begin{matrix} { \begin{bmatrix} 3 & -2 \\ 6 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} } & = \begin{bmatrix} 5 \\ 9 \end{bmatrix} \\ A & X \end{matrix} $

Now, $|A|= \begin{vmatrix} 3 & -2 \\ 6 & -4\end{vmatrix} =-12-(-12)=-12+12=0$

$\Rightarrow A$ is a singular matrix

$\Rightarrow$ Either the system has no solution or infinitely many solutions.

To check that we find $(adj A) B$.

$ \begin{aligned} A _{11} & =-4, A _{12}=-6, A _{21}=-(-2)=2, A _{22}=3 \\ adj A & = \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} -4 & 2 \\ -6 & 3 \end{bmatrix} \\ (adj A) B & = \begin{bmatrix} -4 & 2 \\ -6 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ 9 \end{bmatrix} = \begin{bmatrix} -20+18 \\ -30+27 \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \end{bmatrix} \neq 0 \end{aligned} $

$\Rightarrow$ The system of equations is inconsistent and has no solution.

16. Application of determinants to the solution of linear equations.

(Cramer’s Rule)

Consider the simultaneous equations,

$ \begin{aligned} a_1 x+b_1 y & =c_1 \\ a_2 x+b_2 y & =c_2 \\ \Rightarrow \quad a_1 x+b_1 y-c_1 & =0 \\ a_2 x+b_2 y-c_2 & =0 \end{aligned} $

Solving these equations by cross-multiplication method, we have

$ \begin{matrix} \frac{x}{-b_1 c_2+b_2 c_1} & =\frac{y}{-a_2 c_1+a_1 c_2}=\frac{1}{a_1 b_2-a_2 b_1} \\ \therefore \quad x & =\frac{b_2 c_1-b_1 c_2}{a_1 b_2-a_2 b_1}, y & =\frac{a_1 c_2-a_2 c_1}{a_1 b_2-a_2 b_1} \end{matrix} $

The solutions can be expressed in the determinant form as :

$ x=\frac{ \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} }{ \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} }=\frac{D x}{D}, y=\frac{ \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} }{ \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} }=\frac{D y}{D} $

The determinant $D$ is the determinant of the coefficients of variables $x$ and $y$, while in $D x$, the coefficients of $x$, i.e., $a_1$ and $a_2$ are replaced by constant terms $c_1$ and $c_2$ and in determinant $D y$, the coefficients of $y$, i.e., $b_1$ and $b_2$ are replaced by the constant terms.

The solutions to the above given equations will exist only when $D \neq 0$. Likewise, for the system of linear equations in three variables.

$ \begin{aligned} & a_1 x+b_1 y+c_1 z=d_1 \\ & a_2 x+b_2 y+c_2 z=d_2 \\ & a_3 x+b_3 y+c_3 z=d_3 \end{aligned} $

We have the solutions for $x, y$ and $z$ in the determinant form as:

$ x=\frac{ \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix} }{ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} }=\frac{D x}{D}, y=\frac{ \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix} }{ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} }=\frac{D y}{D}, z=\frac{ \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix} }{ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} }=\frac{D z}{D} $

Here the important conditions are:

(i) For a solution to exist, $|D| \neq 0$

(ii) The constant terms are on the Right Hand Side of the given equations.

Conditions of consistency and inconsistency of linear equations.

A system of equations is said to be consistent if its solution exists whether unique or not, otherwise it is inconsistent.

The conditions for unique solution, infinitely many solutions and no-solution can be summarized as under:

(i) If $D \neq 0$, then the given system of equations is consistent and has a unique solution, namely, $x=\frac{D x}{D}, y=\frac{D y}{D}$

(ii) If $D=0$ and $D x=D y=0$, then the system may be consistent with infinitely many solutions or inconsistent.

(iii) If $D=0$ and at least one of $D x$ and $D y$ is non-zero, then the system has no solution, i.e., the system is inconsistent.

Note: We shall limit the conditions of consistency and inconsistency in this book only to equations with two variables.

Ex. 1 . Solve $7 x+2 y-25=0$ and $2 x-y-4=0$ by Cramer’s Rule.

Sol. The given equations are:

$ \begin{aligned} & 7 x+2 y=25 \\ & 2 x-y=4 \\ & \therefore \quad D= \begin{vmatrix} 7 & 2 \\ 2 & -1 \end{vmatrix} =-7-4=-11 \neq 0 \\ & \because \quad D \neq 0, \text{ therefore the solution exists. } \\ & D x= \begin{vmatrix} 25 & 2 \\ 4 & -1 \end{vmatrix} =-25-8=-33 \\ & D y= \begin{vmatrix} 7 & 25 \\ 2 & 4 \end{vmatrix} =28-50=-22 \\ & x=\frac{D x}{D}=\frac{-33}{-11}=3, y=\frac{D y}{D}=\frac{-22}{-11}=2 . \end{aligned} $

Hence, $x=3, y=2$

Ex. 2 . Check whether the given system of equations is consistent or inconsistent. (i) $x+3 y=2$ $2 x+6 y=7$ (ii) $2 x+7 y=9$ $4 x+14 y=18$

Sol. (i) $x+3 y=2$

$2 x+6 y=7$

$ \begin{aligned} D & = \begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix} =6-6=0 \\ D x & = \begin{vmatrix} 2 & 3 \\ 7 & 6 \end{vmatrix} =12-21=-9 \neq 0 \\ D y & = \begin{vmatrix} 1 & 2 \\ 2 & 7 \end{vmatrix} =7-4=3 \neq 0 \end{aligned} $

Since, $D=0$ and $D x \neq 0, D y \neq 0$, (at least one of the determinant $D x \neq 0$ ), the given system of equations is inconsistent, i.e., it has no solution,

(ii) $2 x+7 y=9$

$4 x+14 y=18$

$ \begin{aligned} D & = \begin{vmatrix} 2 & 7 \\ 4 & 14 \end{vmatrix} =28-28=0 \\ D x & = \begin{vmatrix} 9 & 7 \\ 18 & 14 \end{vmatrix} =126-126=0 \\ D y & = \begin{vmatrix} 2 & 9 \\ 4 & 18 \end{vmatrix} =36-36=0 \end{aligned} $

Since, $D=0$ and $D x=D y=0$, therefore the system has infinitely many solutions or is inconsistent.

Let $x=k$, Then from (1), $2 k+7 y=9 \Rightarrow y=\frac{9-2 k}{7}$

Substituting this value in $(2), 4(k)+4(\frac{9-2 k}{7})=18$

$\Rightarrow \quad 4 k+18-4 k=18 \quad \Rightarrow 18=18$.

$\therefore \quad$ The system has infinitely many solutions given by $x=k, y=\frac{9-2 k}{7}$

SOLVED EXAMPLES

Ex. 1 . If a matrix has 12 elements, what are the possible orders it can have?

Sol. We know that a matrix of order $m \times n$ has $m n$ elements. Hence to find all possible orders of a matrix having 12 elements, we will have to find all ordered pairs the product of whose components is 12 . The possible ordered pairs satisfying the above given condition are $(1,12),(12,1),(2,6),(6,2),(3,4),(4,3)$.

Hence, the possible orders are $1 \times 12,12 \times 1,2 \times 6,6 \times 2,3 \times 4$, and $4 \times 3$.

Ex. 2. Construct a $2 \times 2$ matrix $A=[a _{i j}]$ whose elements are given by

$a_{i j}=\frac{1}{2}|2 i-3 j|$.

Sol. Let $A=[a _{i j}]= \begin{bmatrix} a _{11} & a _{12} \\ a _{21} & a _{22}\end{bmatrix} $

$ \begin{aligned} \therefore & a _{11}=\frac{1}{2}|2 \times 1-3 \times 1|=\frac{1}{2}|2-3|=\frac{1}{2}|-1|=\frac{1}{2} \times 1=\frac{1}{2} \\ & a _{12}=\frac{1}{2}|2 \times 1-3 \times 2|=\frac{1}{2}|2-6|=\frac{1}{2}|-4|=\frac{1}{2} \times 4=2 \end{aligned} $

$ \begin{aligned} a _{21} & =\frac{1}{2}|2 \times 2-3 \times 1|=\frac{1}{2}|4-3|=\frac{1}{2} \times 1=\frac{1}{2} \\ a _{22} & =\frac{1}{2}|2 \times 2-3 \times 2|=\frac{1}{2}|4-6|=\frac{1}{2} \times 2=1 \\ \therefore A & = \begin{bmatrix} \frac{1}{\mathbf{2}} & \mathbf{2} \\ \frac{\mathbf{1}}{\mathbf{2}} & \mathbf{1} \end{bmatrix} . \end{aligned} $

Ex. 3 . Find the values of $x$ and $y$ so that the matrices $A= \begin{bmatrix} 2 x+1 & 3 y \\ 0 & y^{2}-5 y\end{bmatrix} , B= \begin{bmatrix} x+3 & y^{2}+2 \\ 0 & -6\end{bmatrix} $ may be equal ?

Sol. $A=[a_{i j}]=B=[b_{i j}] \Rightarrow a_{i j}=b_{i j}$

$\therefore 2 x+1=x+3 \Rightarrow x=2$

$3 y=y^{2}+2 \Rightarrow y^{2}-3 y+2=0 \Rightarrow(y-1)(y-2)=0 \Rightarrow y=1$ or 2

$y^{2}-5 y=-6 \Rightarrow y^{2}-5 y+6=0 \Rightarrow(y-3)(y-2)=0 \Rightarrow y=3$ or 2

Since, $3 y=y^{2}+2$ and $y^{2}-5 y=-6$ must hold simultaneously, we take the common solution of the two equations, i.e., $y=2$.

$\therefore A=B \Rightarrow \boldsymbol{x}=\mathbf{2}, \boldsymbol{y}=\mathbf{2}$

Ex. 4 . Solve the equation $-2[x+ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix} ]=3 x+ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} $, over $S _{3 \times 3}$.

Sol. Given equation becomes, $-2 x+-2 \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix} =3 x+ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} $

$ \begin{matrix} \Rightarrow \begin{bmatrix} -2 & -4 & -6 \\ 0 & -2 & -4 \\ 0 & 0 & -2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} =5 x & \Rightarrow 5 x= \begin{bmatrix} -2-1 & -4-0 & -6-0 \\ 0-0 & -2-0 & -4-0 \\ 0-0 & 0-0 & -2-1 \end{bmatrix} \\ \Rightarrow 5 x= \begin{bmatrix} -3 & -4 & -6 \\ 0 & -2 & -4 \\ 0 & 0 & -3 \end{bmatrix} \end{matrix} $

Ex. 5 . If $A=diag \begin{bmatrix} 3 & -2 & 1\end{bmatrix} $ and $B=diag \begin{bmatrix} 1 & 3 & -2\end{bmatrix} $, find $2 A-3 B$.

Sol. $A=diag \begin{bmatrix} 3 & -2 & 1\end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1\end{bmatrix} \quad B=diag \begin{bmatrix} 1 & 3 & -2\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2\end{bmatrix} $

$ \begin{aligned} \Rightarrow 2 A-3 B & =2 \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{bmatrix} -3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & -6 \end{bmatrix} \\ & = \begin{bmatrix} 6-3 & 0 & 0 \\ 0 & -4-9 & 0 \\ 0 & 0 & 2+6 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -13 & 0 \\ 0 & 0 & 8 \end{bmatrix} =diag \begin{bmatrix} \mathbf{3} & -\mathbf{1 3} & \mathbf{8} \end{bmatrix} . \end{aligned} $

Ex. 6 . Find $a$ and $b$, if ${3 \begin{bmatrix} 2 & 1 & -3 \\ 1 & 4 & 2\end{bmatrix} -2 \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & 3\end{bmatrix} } \begin{bmatrix} 2 \\ 0 \\ -1\end{bmatrix} = \begin{bmatrix} a \\ b\end{bmatrix} $.

Sol. LHS $={3 \begin{bmatrix} 2 & 1 & -3 \\ 1 & 4 & 2\end{bmatrix} -2 \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & 3\end{bmatrix} } \begin{bmatrix} 2 \\ 0 \\ -1\end{bmatrix} ={ \begin{bmatrix} 6 & 3 & -9 \\ 3 & 12 & 6\end{bmatrix} - \begin{bmatrix} 2 & -4 & 0 \\ 4 & -2 & 6\end{bmatrix} } \begin{bmatrix} 2 \\ 0 \\ -1\end{bmatrix} $

$ = \begin{bmatrix} 6-2 & 3+4 & -9-0 \\ 3-4 & 12+2 & 6-6 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} 4 & 7 & -9 \\ -1 & 14 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} 4 \times 2+7 \times 0-9 \times-1 \\ -1 \times 2+14 \times 0+0 \times-1 \end{bmatrix} = \begin{bmatrix} 17 \\ -2 \end{bmatrix} $

Given, $ \begin{bmatrix} 17 \\ -2\end{bmatrix} = \begin{bmatrix} a \\ b\end{bmatrix} \Rightarrow \boldsymbol{a}=\mathbf{1 7}, \boldsymbol{b}=\mathbf{- 2}$.

Ex. 7 . If $A, B, C$ are three matrices such that $A= \begin{bmatrix} x & y & z\end{bmatrix} , B= \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix} , C= \begin{bmatrix} x \\ y \\ z\end{bmatrix} $ evaluate $A B C$

Sol. $A B= \begin{bmatrix} x & \boldsymbol{y} & z\end{bmatrix} _{1 \times 3} \begin{bmatrix} \boldsymbol{a} & \boldsymbol{h} & \boldsymbol{g} \\ \boldsymbol{h} & \boldsymbol{b} & \boldsymbol{f} \\ \boldsymbol{g} & \boldsymbol{f} & \boldsymbol{c}\end{bmatrix} _{3 \times 3}= \begin{bmatrix} x a+y h+z g & x h+y b+z f & x g+y f+z c\end{bmatrix} _{1 \times 3}$

$ \begin{aligned} \Rightarrow A B C & = \begin{bmatrix} x a+y h+z g & x h+y b+z f & x g+y f+z c \end{bmatrix} _{1 \times 3} \begin{bmatrix} x \\ y \\ z \end{bmatrix} _{3 \times 1} \\ & = \begin{bmatrix} x(x a+y h+z g)+y(x h+y b+z f)+z(x g+y f+z c) \end{bmatrix} _{1 \times 1} \\ & = \begin{bmatrix} a x^{2}+b y^{2}+c z^{2}+2 h x y+2 g z x+2 f y z \end{bmatrix} . \end{aligned} $

Ex. 8 . If $A= \begin{bmatrix} 2 & 3 \\ -1 & 2\end{bmatrix} $ and $f(x)=x^{2}-4 x+7$, show that $f(A)=0$. Use this result to find $A^{5}$.

Sol. $f(A)=A^{2}-4 A+7 I_2$

$ \begin{aligned} & A^{2}=A \cdot A= \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 2+3 \times-1 & 2 \times 3+3 \times 2 \\ -1 \times 2+2 \times-1 & -1 \times 3+2 \times 2 \end{bmatrix} = \begin{bmatrix} 4-3 & 6+6 \\ -2-2 & -3+4 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \\ & \therefore f(A)=A^{2}-4 A+7 I_2= \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} -4 \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} +7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 1-8+7 & 12-12+0 \\ -4+4+0 & 1-8+7 \end{bmatrix} =0 . \end{aligned} $

Now $f(A)=0 \Rightarrow A^{2}-4 A+7 I_2=0$

$\Rightarrow A^{2}=4 A-7 I_2$

$\therefore A^{3}=A^{2} . A=(4 A-7 I_2) A=4 A^{2}-7 I_2 A=4 A^{2}-7 A$

$(\because I_2 A=A)$

$\Rightarrow A^{3}=4(4 A-7 I_2)-7 A=9 A-28 I_2$

(Replacing $A^{2}$ by $4 A-7 I_2$ )

$\therefore A^{4}=(9 A-28 I_2) A=9 A^{2}-28 I_2 A=9(4 A-7 I_2)-28 A=36 A-63 I_2-28 A=8 A-63 I_2$

$\therefore A^{5}=(8 A-63 I_2) A=8 A^{2}-63 I_2 A=8(4 A-7 I_2)-63 A=32 A-56 I_2-63 A=-31 A-56 I_2$

$\therefore A^{5}=-31 \begin{bmatrix} 2 & 3 \\ -1 & 2\end{bmatrix} -56 \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} -62 & -93 \\ 31 & -62\end{bmatrix} - \begin{bmatrix} 56 & 0 \\ 0 & 56\end{bmatrix} $

$ = \begin{bmatrix} -62-56 & -93-0 \\ 31-0 & -62-56 \end{bmatrix} = \begin{bmatrix} -\mathbf{1 1 8} & -\mathbf{9 3} \\ \mathbf{3 1} & -\mathbf{1 1 8} \end{bmatrix} $

Ex. 9 . If $A= \begin{bmatrix} 2 & 3 & 4 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix} , B= \begin{bmatrix} 4 & 0 & 5 \\ 1 & 2 & 0 \\ 0 & 3 & 1\end{bmatrix} $, verify that $(A B)^{T}=B^{T} A^{T}$.

Sol. $A= \begin{bmatrix} 2 & 3 & 4 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} 2 & 5 & -2 \\ 3 & 7 & 1 \\ 4 & 9 & 1\end{bmatrix} $

$ B= \begin{bmatrix} 4 & 0 & 5 \\ 1 & 2 & 0 \\ 0 & 3 & 1 \end{bmatrix} \Rightarrow B^{T}= \begin{bmatrix} 4 & 1 & 0 \\ 0 & 2 & 3 \\ 5 & 0 & 1 \end{bmatrix} $

Now, $\quad A B= \begin{bmatrix} 2 & 3 & 4 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix} \begin{bmatrix} 4 & 0 & 5 \\ 1 & 2 & 0 \\ 0 & 3 & 1\end{bmatrix} = \begin{bmatrix} 8+3+0 & 0+6+12 & 10+0+4 \\ 20+7+0 & 0+14+27 & 25+0+9 \\ -8+1+0 & 0+2+3 & -10+0+1\end{bmatrix} = \begin{bmatrix} 11 & 18 & 14 \\ 27 & 41 & 34 \\ -7 & 5 & -9\end{bmatrix} $

$ (A B)^{T}= \begin{bmatrix} 11 & 27 & -7 \\ 18 & 41 & 5 \\ 14 & 34 & -9 \end{bmatrix} $

Now, $\quad B^{T} A^{T}= \begin{bmatrix} 4 & 1 & 0 \\ 0 & 2 & 3 \\ 5 & 0 & 1\end{bmatrix} \begin{bmatrix} 2 & 5 & -2 \\ 3 & 7 & 1 \\ 4 & 9 & 1\end{bmatrix} = \begin{bmatrix} 8+3+0 & 20+7+0 & -8+1+0 \\ 0+6+12 & 0+14+27 & 0+2+3 \\ 10+0+4 & 25+0+9 & -10+0+1\end{bmatrix} = \begin{bmatrix} 11 & 27 & -7 \\ 18 & 41 & 5 \\ 14 & 34 & -9\end{bmatrix} $

$\therefore \quad(A B)^{T}=B^{T} A^{T}$.

Ex. 10 . If $A= \begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix} $, show that $A-A^{T}$ is a skew-symmetric matrix.

Sol. A square matrix $A=[a _{i j}]$ is said to be a skew-symmetric matrix if $A^{T}=-A$.

$ \begin{aligned} A & = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \\ \therefore \quad A-A^{T} & = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \\ (A-A^{T})^{T} & = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} =- \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} =-(A-A^{T}) . \end{aligned} $

Hence, $A-A^{T}$ is a skew-symmetric matrix.

Ex. 11 . Express the matrix $A= \begin{bmatrix} 2 & 1 & 3 \\ 1 & 1 & 4 \\ -1 & 6 & 2\end{bmatrix} $ as a sum of a symmetric and a skew-symmetric matrix.

Sol. We know that symmetric part of the matrix $A$ is $\frac{1}{2}(A+A^{T})$ and the skew-symmetric part is $\frac{1}{2}(A-A^{T})$.

Here $A= \begin{bmatrix} 2 & 1 & 3 \\ 1 & 1 & 4 \\ -1 & 6 & 2\end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} 2 & 1 & -1 \\ 1 & 1 & 6 \\ 3 & 4 & 2\end{bmatrix} $

Symmetric part = $\frac{1}{2}(A+A^{T})=\frac{1}{2} $ $\begin{Bmatrix}\begin{bmatrix} 2 & 1 & 3 \\ 1 & 1 & 4 \\ -1 & 6 & 2 \end{bmatrix}+ \begin{bmatrix} 2 & 1 & -1 \\ 1 & 1 & 6 \\ 3 & 4 & 2 \end{bmatrix} \end{Bmatrix}$ = $\frac{1}{2}$
$\begin{bmatrix} 4 & 2 & 2 \\ 2 & 2 & 10 \\ 2 & 10 & 4 \end{bmatrix}$ = $\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 5 \\ 1 & 5 & 2 \end{bmatrix}$

Skew-symmetric part = $\frac{1}{2}(A-A^{T})=\frac{1}{2} $ $ \begin{Bmatrix} \begin{bmatrix} 2 & 1 & 3 \\ 1 & 1 & 4 \\ -1 & 6 & 2 \end{bmatrix} -\begin{bmatrix} 2 & 1 & -1 \\ 1 & 1 & 6 \\ 3 & 4 & 2 \end{bmatrix} \end{Bmatrix}$ = $\frac{1}{2}$ $ \begin{bmatrix} 0 & 0 & 4 \\ 0 & 0 & -2 \\ -4 & 2 & 0 \end{bmatrix}$ = $\begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix}$

$\therefore A=\frac{1}{2}(A+A^{T})+\frac{1}{2}$ $(A-A^{T})$ = $\begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 5 \\ 1 & 5 & 2 \end{bmatrix}$ + $\begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix}$

Ex. 12 . Find the adjoint of A , where A = $\begin{bmatrix} \mathbf{1} & -\mathbf{1} & \mathbf{2} \\ \mathbf{2} & \mathbf{3} & \mathbf{5} \\ -\mathbf{2} & \mathbf{0} & \mathbf{1} \end{bmatrix} $.

Sol. Adj $A$ is the transpose of the matrix obtained by replacing the elements of $A$ by their corresponding cofactors.

$ \begin{aligned} & A=[a _{i j}]= \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix} \\ & \therefore \quad A _{11}=\text{ cofactor of } a _{11}(1)=(-1)^{1+1} \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} =3-0=3 \\ & A _{12}=\text{ cofactor of } a _{12}(-1)=(-1)^{1+2} \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} =-(2+10)=-12 \\ & A _{13}=\text{ cofactor of } a _{13}(2)=(-1)^{1+3} \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} =0+6=6 \\ & A _{21}=\text{ cofactor of } a _{21}(2)=(-1)^{2+1} \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} =-(-1-0)=1 \\ & A _{22}=\text{ cofactor of } a _{22}(3)=(-1)^{2+2} \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} =1+4=5 \\ & A _{23}=\text{ cofactor of } a _{23}(5)=(-1)^{2+3} \begin{vmatrix} 1 & -1 \\ 2 & 0 \end{vmatrix} =-(0-2)=2 \\ & A _{31}=\text{ cofactor of } a _{31}(-2)=(-1)^{3+1} \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} =-5-6=-11 \\ & A _{32}=\text{ cofactor of } a _{32}(0)=(-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} =-(5-4)=-1 \\ & A _{33}=\text{ cofactor of } a _{33}(1)=(-1)^{3+3} \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} =3+2=5 \\ & \therefore \text{ adj. } A= \begin{bmatrix} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33} \end{bmatrix} ^{T}= \begin{bmatrix} A _{11} & A _{21} & A _{31} \\ A _{12} & A _{22} & A _{32} \\ A _{13} & A _{23} & A _{33} \end{bmatrix} = \begin{bmatrix} \mathbf{3} & \mathbf{1} & \mathbf{- 1 1} \\ \mathbf{- 1 2} & \mathbf{5} & \mathbf{- 1} \\ \mathbf{6} & \mathbf{2} & \mathbf{5} \end{bmatrix} . \end{aligned} $

Ex. 13 . If $A= \begin{bmatrix} 1 & 3 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 3\end{bmatrix} $, find the value of $(adj . A) A$ without finding adj. $A$.

Sol. We know that $A(adj . A)=(adj . A) A=|A| I$

$\therefore$ Here $(adj . A) A=|A| I_3$

$ \begin{aligned} & .A= \begin{bmatrix} 1 & 3 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 3 \end{bmatrix} \Rightarrow|A|= \begin{vmatrix} 1 & 3 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 3 \end{vmatrix} =1 \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} -3 \begin{vmatrix} 2 & 0 \\ 3 & 3 \end{vmatrix} +1 \rvert, \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \\ & =1(3-0)-3(6-0)+1(4-3)=3-18+1=-14 \\ & \therefore \quad(adj . A) A=|A| I_3=-14 \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = \begin{vmatrix} -\mathbf{1 4} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{- 1 4} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{- 1 4} \end{vmatrix} \text{. } \end{aligned} $

Ex. 14 . Find the inverse of the matrix $ \begin{bmatrix} -2 & 5 \\ 3 & 4\end{bmatrix} $.

Sol. Let $A= \begin{bmatrix} -2 & 5 \\ 3 & 4\end{bmatrix} $

Then $|A|=-2 \times 4-5 \times 3=-23 \neq 0$

$\therefore|A| \neq 0 \Rightarrow A$ is a non-singular matrix $\Rightarrow A^{-1}$ exists.

$\therefore A _{11}=(-1)^{1+1}|4|=4 \quad A _{21}=(-1)^{2+1}|5|=-5$

$A _{12}=(-1)^{1+2}|3|=-3 \quad A _{22}=(-1)^{2+2}|-2|=-2$

$\therefore$ adj. $A= \begin{bmatrix} A _{11} & A _{12} \\ A _{21} & A _{22}\end{bmatrix} ^{T}= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22}\end{bmatrix} = \begin{bmatrix} 4 & -5 \\ -3 & -2\end{bmatrix} $

$\therefore A^{-1}=\frac{\text{ adj. } A}{|A|}=\frac{1}{-23} \begin{bmatrix} 4 & -5 \\ -3 & -2\end{bmatrix} = \begin{bmatrix} -4 / 23 & 5 / 23 \\ 3 / 23 & 2 / 23\end{bmatrix} $.

Ex. 15 . Let $F(\alpha)= \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix} $ and $G(\beta)= \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{bmatrix} $, show that $[F(\alpha) . G(\beta)]^{-1}=G(-\beta) . F(-\alpha)$.

Sol. $F(\alpha), F(-\alpha)= \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} \cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1\end{bmatrix} $

$ = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos ^{2} \alpha+\sin ^{2} \alpha & 0 & 0 \\ 0 & \sin ^{2} \alpha+\cos ^{2} \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =I $

$\therefore \quad F(\alpha) . F(-\alpha)=I \Rightarrow[F(\alpha)]^{-1}=F(-\alpha)$

Similarly, $G(\beta) \cdot G(-\beta)= \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{bmatrix} \begin{bmatrix} \cos (-\beta) & 0 & \sin (-\beta) \\ 0 & 1 & 0 \\ -\sin (-\beta) & 0 & \cos (-\beta)\end{bmatrix} = \begin{bmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta\end{bmatrix} \begin{bmatrix} \cos \beta & 0 & -\sin \beta \\ 0 & 1 & 0 \\ \sin \beta & 0 & \cos \beta\end{bmatrix} $

$ = \begin{bmatrix} \cos ^{2} \beta+0+\sin ^{2} \beta & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \sin ^{2} \beta+0+\cos ^{2} \beta \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =I $

$ \begin{aligned} & \therefore \quad G(\beta) \cdot G(-\beta)=I \Rightarrow[G(\beta)]^{-1}=G(-\beta) \\ & (\because(A B)^{-1}=B^{-1} A^{-1}) \\ & \therefore{F(\alpha) \cdot G(\beta)}^{-1}={G(\beta)}^{-1} \cdot{F(\alpha)}^{-1} \\ & =G(-\beta) \cdot F(-\alpha) \end{aligned} $

Ex. 16 . For the matrix $A= \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} $ show that $A^{2}-4 A+5 I=0$. Hence obtain $A^{-1}$.

Sol. $A^{2}-4 A+5 I= \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} -4 \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix} +5 \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 1-2 & -1-3 \\ 2+6 & -2+9\end{bmatrix} - \begin{bmatrix} 4 & -4 \\ 8 & 12\end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5\end{bmatrix} $

$ = \begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix} + \begin{bmatrix} -4 & +4 \\ -8 & -12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} -1-4+5 & -4+4+0 \\ 8-8+0 & 7-12+5 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} =0 $

Now premultiplying both the sides of $A^{2}-4 A+5 I$ by $A^{-1}$, we get

$ \begin{aligned} & A^{-1} A^{2}-4 A^{-1} A+5 A^{-1} I=0 \\ \Rightarrow & (A^{-1} A) A-4 I+5 A^{-1}=0 \\ \Rightarrow & I A-4 I+5 A^{-1}=0 \\ \Rightarrow & A-4 I+5 A^{-1}=0 \Rightarrow 5 A^{-1}=4 I-A \\ \Rightarrow & 5 A^{-1}=4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix} \\ \therefore & A^{-1}=\frac{\mathbf{1}}{\mathbf{5}} \begin{bmatrix} \mathbf{3} & \mathbf{1} \\ -\mathbf{2} & \mathbf{1} \end{bmatrix} . \end{aligned} $

Ex. 17 . Find the inverse of the matrix $A= \begin{bmatrix} a & b \\ c & \frac{1+b c}{a}\end{bmatrix} $ and show that $a A^{-1}=(a^{2}+b c+1) I_2-a A$.

Sol. For $A^{-1}$ to exist, $|A| \neq 0$.

$ \begin{aligned} & |A|= \begin{vmatrix} a & b \\ c & \frac{1+b c}{a} \end{vmatrix} =a(\frac{1+b c}{a})-b c=1+b c-b c=1 \neq 0 \\ \Rightarrow & A^{-1} \text{ exists and } A^{-1}=\frac{1}{|A|} \times a d j . A . \\ & A _{11}=\frac{1+b c}{a}, A _{12}=-c, A _{21}=-b, A _{22}=a \\ \therefore & \text{ adj. } A= \begin{bmatrix} A _{11} & A _{12} \\ A _{21} & A _{22} \end{bmatrix} ^{T}= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} \frac{1+b c}{a} & -b \\ -c & a \end{bmatrix} \\ \therefore & A^{-1}=\frac{1}{1} \begin{bmatrix} \frac{1+b c}{a} & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} \frac{1+b c}{a} & -b \\ -c & a \end{bmatrix} \\ \text{ LHS }= & a A^{-1}=a \begin{bmatrix} \frac{1+b c}{a} & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 1+b c & -a b \\ -a c & a^{2} \end{bmatrix} \end{aligned} $

$ (\because A^{-1} A=I \text{ and } A^{-1} I=A^{-1}) $

$ \begin{aligned} \text{ RHS } & =(a^{2}+b c+1) I_2-a A \\ & =(a^{2}+b c+1) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} -a \begin{bmatrix} a & b \\ c & \frac{1+b c}{a} \end{bmatrix} = \begin{bmatrix} a^{2}+b c+1 & 0 \\ 0 & a^{2}+b c+1 \end{bmatrix} - \begin{bmatrix} a^{2} & a b \\ a c & 1+b c \end{bmatrix} \\ & = \begin{bmatrix} a^{2}+b c+1-a^{2} & 0-a b \\ 0-a c & a^{2}+b c+1-1-b c \end{bmatrix} = \begin{bmatrix} b c+1 & -a b \\ -a c & a^{2} \end{bmatrix} =\text{ LHS. } \end{aligned} $

Ex. 18 . Solve the following system of equations by matrix method:

$ \begin{aligned} & 5 x+3 y+z=16 \\ & 2 x+y+3 z=19 \\ & x+2 y+4 z=25 \end{aligned} $

Sol. Writing the given equations in the matrix form $A X=B$, we have $ \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 16 \\ 19 \\ 25\end{bmatrix} $, where

$ \begin{aligned} & A= \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} , B= \begin{bmatrix} 16 \\ 19 \\ 25 \end{bmatrix} \\ & |A|= \begin{vmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} =5(4-6)-3(8-3)+1(4-1)=-10-15+3=-22 \neq 0 \end{aligned} $

$\therefore A$ is non - singular.

$\therefore$ The given system of equations has a unique solution $X=A^{-1} B$.

$ \begin{aligned} & A _{11}= \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} =4-6=-2, A _{12}=- \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} =-(8-3)=-5, A _{13}= \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} =4-1=3 \\ & A _{21}=- \begin{vmatrix} 3 & 1 \\ 2 & 4 \end{vmatrix} =-(12-2)=-10, A _{22}= \begin{vmatrix} 5 & 1 \\ 1 & 4 \end{vmatrix} =20-1=19, A _{23}=- \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} =-(10-3)=-7 \\ & A _{31}= \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} =(9-1)=8, A _{32}=- \begin{vmatrix} 5 & 1 \\ 2 & 3 \end{vmatrix} =-(15-2)=-13, A _{33}= \begin{vmatrix} 5 & 3 \\ 2 & 1 \end{vmatrix} =5-6=-1 \\ & \therefore \quad A^{-1}=\frac{a d j . A}{|A|}=\frac{1}{|A|} \begin{bmatrix} A _{11} & A _{21} & A _{31} \\ A _{12} & A _{22} & A _{32} \\ A _{13} & A _{23} & 33 \end{bmatrix} =\frac{1}{-22} \begin{bmatrix} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{bmatrix} = \begin{bmatrix} 2 / 22 & 10 / 22 & -8 / 22 \\ 5 / 22 & -19 / 22 & 13 / 22 \\ -3 / 22 & 7 / 22 & 1 / 22 \end{bmatrix} \\ & \therefore \quad X=A^{-1} B \\ & \Rightarrow \quad \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 / 22 & 10 / 22 & -8 / 22 \\ 5 / 22 & -19 / 22 & 13 / 22 \\ -3 / 22 & 7 / 22 & 1 / 22 \end{bmatrix} \begin{bmatrix} 16 \\ 19 \\ 25 \end{bmatrix} = \begin{bmatrix} \frac{32+190-200}{22} \\ \frac{80-361+325}{22} \\ \frac{-48+133+25}{22} \end{bmatrix} = \begin{bmatrix} \frac{22}{22} \\ \frac{44}{22} \\ \frac{110}{22} \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix} \end{aligned} $

$ \therefore \quad x=1, y=2, z=5 \text{. } $

Ex. 19 . Use matrix method to examine the following system of equations for consistency or inconsistency. $2 x+5 y=7,6 x+15 y=13$

Sol. Writing the given equations in the matrix form, we have $ \begin{bmatrix} 2 & 5 \\ 6 & 15\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 7 \\ 13\end{bmatrix} $ or $A X=B$, where

$ A= \begin{bmatrix} 2 & 5 \\ 6 & 15 \end{bmatrix} , X= \begin{bmatrix} x \\ y \end{bmatrix} , B \begin{bmatrix} 7 \\ 13 \end{bmatrix} $

Now

$ |A|= \begin{vmatrix} 2 & 5 \\ 6 & 15 \end{vmatrix} =30-30=0 \Rightarrow A \text{ is singular } $

$\Rightarrow$ Further the system has no solution or infinite number of solutions. So now we find (adj. $A$ ) $B$.

$A _{11}=15, A _{12}=-6, A _{21}=-5, A _{22}=2$

$\therefore$ adj. $A= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22}\end{bmatrix} = \begin{bmatrix} 15 & -5 \\ -6 & 2\end{bmatrix} $

$\therefore($ adj. $A) B= \begin{bmatrix} 15 & -5 \\ -6 & 2\end{bmatrix} \begin{bmatrix} 7 \\ 13\end{bmatrix} = \begin{bmatrix} 105-65 \\ -42+26\end{bmatrix} = \begin{bmatrix} 40 \\ -16\end{bmatrix} \neq 0$

$\therefore$ The given system has no solution and is therefore inconsistent.

Note: If $|A|=0$ and $(adj . A) B=0$, then the given system is consistent and has infinite number of solutions.

Based on Properties of Determinants

Ex. 20 . Without expanding, i.e., using properties of determinants, show that :

(a) $ \begin{vmatrix} 1 / a & a & b c \\ 1 / b & b & c a \\ 1 / b & c & a b\end{vmatrix} =0$

(b) $ \begin{vmatrix} 3 x+y & 2 x & x \\ 4 x+3 y & 3 x & 3 x \\ 5 x+6 y & 4 x & 6 x\end{vmatrix} =x^{3}$

Sol. (a) Given, $\Delta= \begin{vmatrix} 1 / a & a & b c \\ 1 / b & b & c a \\ 1 / c & c & a b\end{vmatrix} $

Multiply $R_1$ by $a, R_2$ by $b, R_3$ by $c$. Then,

$ \begin{aligned} & \Delta=\frac{1}{a b c} \begin{vmatrix} 1 & a^{2} & a b c \\ 1 & b^{2} & a b c \\ 1 & c^{2} & a b c \end{vmatrix} =\frac{1}{a b c} \times a b c \begin{vmatrix} 1 & a^{2} & 1 \\ 1 & b^{2} & 1 \\ 1 & c^{2} & 1 \end{vmatrix} \text{ (Taking out } a b c \text{ common from } C_3 \text{ ) } \\ & =1 \times 0=\mathbf{0}(\text{ Two columns being identical }) \\ & \Delta= \begin{vmatrix} 3 x+y & 2 x & x \\ 4 x+3 y & 3 x & 3 x \\ 5 x+6 y & 4 x & 6 x \end{vmatrix} = \begin{vmatrix} 3 x & 2 x & x \\ 4 x & 3 x & 3 x \\ 5 x & 4 x & 6 x \end{vmatrix} + \begin{vmatrix} y & 2 x & x \\ 3 y & 3 x & 3 x \\ 6 y & 4 x & 6 x \end{vmatrix} \\ & =x^{3} \begin{vmatrix} 3 & 2 & 1 \\ 4 & 3 & 3 \\ 5 & 4 & 6 \end{vmatrix} +x^{2} y \begin{vmatrix} 1 & 2 & 1 \\ 3 & 3 & 3 \\ 6 & 4 & 6 \end{vmatrix} =x^{3} \begin{vmatrix} 3 & 2 & 1 \\ 4 & 3 & 3 \\ 5 & 4 & 6 \end{vmatrix} +x^{2} y \times 0(\because \text{ Two columns } C_1 \text{ and } C_3 \text{ are identical }) \\ & =x^{3}[3(18-12)-2(24-15)+1(16-15)]=x^{3}(18-18+1)=x^{3} . \end{aligned} $

(b)

Ex. 21 . For positive number $x, y$ and $z$, show that the numerical value of the determinant $ \begin{vmatrix} 1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{vmatrix} =0$

Sol. Let $\Delta= \begin{vmatrix} 1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{vmatrix} = \begin{vmatrix} 1 & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & 1 & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & 1\end{vmatrix} \quad$ (Using $\log _{n} m=\frac{\log m}{\log n}$ )

$ =\frac{1}{\log x \cdot \log y \cdot \log z} \begin{vmatrix} \log x & \log y & \log z \\ \log x & \log y & \log z \\ \log x & \log y & \log z \end{vmatrix} $

(Multiplying $R_1, R_2, R_3$ by $\log x, \log y$ and $\log z$ respectively)

$ =\frac{1}{\log x \cdot \log y \cdot \log z} \times 0=0 $

Ex. 22 . Prove that $ \begin{vmatrix} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b\end{vmatrix} =2(a+b+c)^{3}$.

Sol. Given determinant $\Delta= \begin{vmatrix} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b\end{vmatrix} $

Applying $C_1 \longrightarrow C_1+(C_2+C_3)$, we get

$ \Delta= \begin{vmatrix} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2 a & b \\ 2(a+b+c) & a & c+a+2 b \end{vmatrix} =2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b \end{vmatrix} $

Applying $R_2 \longrightarrow R_2-R_1$ and $R_3 \longrightarrow R_3-R_1$

$ \begin{aligned} \Delta & =2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix} =2(a+b+c) \times 1 \begin{vmatrix} b+c+a & 0 \\ 0 & c+a+b \end{vmatrix} \\ & =2(a+b+c) \times{(b+c+a) .(c+a+b)-0 \times 0}=\mathbf{2}(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c})^{\mathbf{3} .} \end{aligned} $

Ex. 23 . Solve the following equations: $ \begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{vmatrix} =0$

Sol. Given, $ \begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{vmatrix} =0$

$\Rightarrow \begin{vmatrix} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x\end{vmatrix} =0 \quad \quad$ (Applying $R_1 \longrightarrow R_1+R_2+R_3$ )

$\Rightarrow(x+9) \begin{vmatrix} 1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x\end{vmatrix} =0$

$ \begin{matrix} \Rightarrow(x+9) \begin{vmatrix} 1 & 0 & 0 \\ 2 & x-2 & 0 \\ 7 & -1 & x-7 \end{vmatrix} =0 & \text{ (Applying } C_2 \longrightarrow C_2-C_1, C_3 \longrightarrow C_3-C_1 \text{ ) } \\ \Rightarrow(x+9) \begin{vmatrix} (x-2) & 0 \\ -1 & (x-7) \end{vmatrix} =0 & \text{ (Expanding along } .R_1) \\ \Rightarrow(x+9)(x-2)(x-7)=0 & \\ \Rightarrow x=-9 \text{ or } 2 \text{ or } 7 . \end{matrix} $

$\therefore$ The solution set is ${-9,2,7}$.

Ex. 24 . Without expanding prove that $ \begin{vmatrix} a_1 \alpha_1+b_1 \beta_1 & a_1 \alpha_2+b_1 \beta_2 & a_1 \alpha_3+b_1 \beta_3 \\ a_2 \alpha_1+b_2 \beta_1 & a_2 \alpha_2+b_2 \beta_2 & a_2 \alpha_3+b_2 \beta_3 \\ a_3 \alpha_1+b_3 \beta_1 & a_3 \alpha_2+b_3 \beta_2 & a_3 \alpha_3+b_3 \beta_3\end{vmatrix} =0$

Sol. We know that, $ \begin{vmatrix} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0\end{vmatrix} \begin{vmatrix} \alpha_1 & \beta_1 & 0 \\ \alpha_2 & \beta_2 & 0 \\ \alpha_3 & \beta_3 & 0\end{vmatrix} =0 \times 0=0$

(Using row by row multiplication of determinants)

$ \begin{aligned} & \Rightarrow \begin{vmatrix} a_1 \alpha_1+b_1 \beta_1+0 & a_1 \alpha_2+b_1 \beta_2+0 & a_1 \alpha_3+b_1 \beta_3+0 \\ a_2 \alpha_1+b_2 \beta_1+0 & a_2 \alpha_2+b_2 \beta_2+0 & a_2 \alpha_3+b_2 \beta_3+0 \\ a_3 \alpha_1+b_3 \beta_1+0 & a_3 \alpha_2+b_3 \beta_2+0 & a_3 \alpha_3+b_3 \beta_3+0 \end{vmatrix} =0 \\ & \Rightarrow \begin{vmatrix} a_1 \alpha_1+b_1 \beta_1 & a_1 \alpha_2+b_1 \beta_2 & a_1 \alpha_3+b_1 \beta_3 \\ a_2 \alpha_1+b_2 \beta_1 & a_2 \alpha_2+b_2 \beta_2 & a_2 \alpha_3+b_2 \beta_3 \\ a_3 \alpha_1+b_3 \beta_1 & a_3 \alpha_2+b_3 \beta_2 & a_3 \alpha_3+b_3 \beta_3 \end{vmatrix} =0 \end{aligned} $

Ex. 25 . Express $ \begin{vmatrix} (a-x)^{2} & (a-y)^{2} & (a-z)^{2} \\ (b-x)^{2} & (b-y)^{2} & (b-z)^{2} \\ (c-x)^{2} & (c-y)^{2} & (c-z)^{2}\end{vmatrix} $ as a product of two determinants and hence factorize it.

Sol. $ \begin{vmatrix} (a-x)^{2} & (a-y)^{2} & (a-z)^{2} \\ (b-x)^{2} & (b-y)^{2} & (b-z)^{2} \\ (c-x)^{2} & (c-y)^{2} & (c-z)^{2}\end{vmatrix} = \begin{vmatrix} a^{2}-2 a x+x^{2} & a^{2}-2 a y+y^{2} & a^{2}-2 a z+z^{2} \\ b^{2}-2 b x+x^{2} & b^{2}-2 b y+y^{2} & b^{2}-2 b z+z^{2} \\ c^{2}-2 c x+x^{2} & c^{2}-2 c y+y^{2} & c^{2}-2 c z+z^{2}\end{vmatrix} $

$= \begin{vmatrix} a^{2} & -2 a & 1 \\ b^{2} & -2 b & 1 \\ c^{2} & -2 c & 1\end{vmatrix} \begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix} $.

(Row by row multiplication)

$=-2 \begin{vmatrix} a^{2} & a & 1 \\ b^{2} & b & 1 \\ c^{2} & c & 1\end{vmatrix} \begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix} =2 \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{vmatrix} \begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix} $

(Interchanging $C_1$ and $C_3$ of first determinant)

Now $ \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{vmatrix} = \begin{vmatrix} 0 & a-b & a^{2}-b^{2} \\ 0 & b-c & b^{2}-c^{2} \\ 1 & c & c^{2}\end{vmatrix} $

(Applying $R_1 \longrightarrow R_1-R_2 ; R_2 \longrightarrow R_2-R_3$ )

$ \begin{aligned} &= \begin{vmatrix} 0 & (a-b) & (a-b)(a+b) \\ 0 & (b-c) & (b-c)(b+c) \\ 1 & c & c^{2} \end{vmatrix} \\ &=(a-b)(b-c) \begin{vmatrix} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^{2} \end{vmatrix} \quad \text{ [Taking }(a-b) \text{ common from } R_1,(b-c) \text{ common from } R_2 \text{ ] } \\ &=(a-b)(b-c) .1 \cdot \begin{vmatrix} 1 & a+b \\ 1 & b+c \end{vmatrix} \\ &=(a-b)(b-c)(b+c-a-b)=(a-b)(b-c)(c-a) \\ & \text{ Similarly, we have } \begin{vmatrix} 1 & y & y^{2} \\ 1 & z & z^{2} \end{vmatrix} =(x-y)(y-z)(z-x) \\ & \therefore 2 \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} \begin{vmatrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{vmatrix} =2(a-b)(b-c)(c-a)(x-y)(y-z)(z-x) . \end{aligned} $

PRACTICE SHEET

1. If $ a_{ij} =\big[ \frac{i}{j}\big]$, where $[x]$ stands for the greatest integer function, then a matrix $A_{2 \times 2}=[a_{i j}]$ will be

(a) $ \begin{bmatrix} 1 & 1 \\ 2 & 1\end{bmatrix} $

(b) $ \begin{bmatrix} 2 & 1 \\ 3 & 2\end{bmatrix} $

(d) $ \begin{bmatrix} 1 & -1 \\ 2 & 1\end{bmatrix} $

2. If $A= \begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix} $, then $A^{n}$ is equal to

(a) $ \begin{bmatrix} n & n k \\ 0 & n\end{bmatrix} $

(b) $ \begin{bmatrix} 1 & n k \\ 0 & 1\end{bmatrix} $

(d) $ \begin{bmatrix} 1 & k^{n} \\ 0 & 1\end{bmatrix} $

3. If $A= \begin{bmatrix} x & 1 \\ 1 & 0\end{bmatrix} $ and $A^{2}$ is the unit matrix, then the value of $x^{3}+x-2$ is equal to

(a) -8

(b) -2

(d) 6

4. If $A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix} $ and $I$ is the unit matrix of order 3 , then $A^{2}+2 A^{4}+4 A^{6}$ is equal to:

(a) $7 I$

(b) $8 A^{7}$

(d) $7 A^{8}$

5. Matrix $A$ is such that $A^{2}=2 A-I$, where $I$ is the identity matrix, then for $n \geq 2, A^{n}$ is equal to

(a) $n A-(n-1) I$

(b) $n A-I$

(d) $2^{n-1} A-(n-1) I$

6. If $ \begin{bmatrix} 1 & x & 1\end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 1 \\ 0 & 3 & 2\end{bmatrix} \begin{bmatrix} x \\ 1 \\ -2\end{bmatrix} =0$, then $x$ is equal to

(a) $\frac{1}{4}$

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

(d) $\frac{5}{4}$

(Odisha JEE 2008)

7. Let $M$ be a $3 \times 3$ matrix satisfying $M \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3\end{bmatrix} , M \begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1\end{bmatrix} $ and $M \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12\end{bmatrix} $. Then the sum of the diagonal entries of $M$ is

(a) 1

(b) 3

(d) 9

(IIT 2011)

8.. If $A=diag(\begin{matrix} 1 & -4 & 8\end{matrix} ), B=diag(\begin{matrix} -2 & 3 & 5\end{matrix} )$ and $C=diag(\begin{matrix} -3 & 7 & 10\end{matrix} )$, find $B+2 C-A$

(a) diag $(\begin{matrix} -4 & 1 & 12\end{matrix} )$

(b) diag $(\begin{matrix} -9 & 21 & 17\end{matrix} )$

(d) $diag(\begin{matrix} -4 & -9 & -7\end{matrix} )$

9. If $A= \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix} $, then $A^{n}=$

(a) $2^{n+1} \cdot A$

(b) $2^{n-1} \cdot A$

(d) $2^{n-2} \cdot A$

10. If the matrix $ \begin{bmatrix} 4 & -2 \\ k & -4\end{bmatrix} $ is nilpotent of order 2 , then $k$ equals

(a) 2

(b) 8

(d) 0

11. If $A= \begin{bmatrix} 3 & 1 \\ -1 & 2\end{bmatrix} $, then

(a) $A^{2}-5 A-7 I_2=0$

(b) $A^{2}+5 A-7 I_2=0$

(d) $A^{2}-5 A+7 I_2=0$

12. If $A= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix} $, then $A^{T} . A$ is a

(a) Null matrix

(b) Identity matrix

(d) None of these

13. If $A= \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{bmatrix} $ is a matrix satisfying $A A^{T}=9 I_3$, then the values of $\boldsymbol{a}$ and $\boldsymbol{b}$ are respectively

(a) $-2,-1$

(b) $-1,2$

(d) $2,-1$

(Kerala PET 2011)

14. If the orders of the matrices $A, B$ and $C$ are $5 \times 4,5 \times 6$ and $7 \times 4$ respectively, then the order of $(A^{T} \times B)^{T} \times C^{T}$ is of order

(a) $4 \times 6$

(b) $6 \times 5$

(d) $4 \times 4$

15. If $A$ is $3 \times 4$ matrix and $B$ is a matrix such that $A^{T} B$ and $B^{T} A$ are both defined, then $B$ is of the order

(a) $3 \times 3$

(b) $3 \times 4$

(d) $4 \times 4$

16. If matrix $A$ is symmetric as well as skew-symmetric, then $A$ is a

(a) Unit matrix

(b) Null matrix

(d) Diagonal matrix

17. Let $A$ and $B$ be symmetric matrices of the same order. Then, (a) $A+B$ is a symmetric matrix

(b) $A B-B A$ is a skew-symmetric matrix

(d) All of these

18. If matrix $A= \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b\end{bmatrix} $, where $a, b$ and $c$ are real positive numbers, $a b c=1$ and $A^{T}$. $A=I$, then find the value of $a^{3}+b^{3}+c^{3}$.

(a) 0

(b) 1

(d) 4

19. The matrix $A= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} $ is which of the following:

(a) Nilpotent

(b) Orthogonal

(d) Involuntary

20. If $A$ and $B$ are square matrices of the same order such that $A B=A$ and $B A=B$, then $A$ and $B$ are both

(a) Singular

(b) Idempotent

(d) Non-singular

21. If $A= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} $, then det. $A$ is equal to

(a) 0

(b) 1

(d) 2

22. Find $x$ if $ \begin{bmatrix} x & 0 & 1 \\ 2 & -1 & 4 \\ 1 & 2 & 0\end{bmatrix} $ is a singular matrix?

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

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

(d) $\frac{1}{8}$

23. For the matrix $A= \begin{bmatrix} 1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10\end{bmatrix} , A .(adj . A)$ is equal to

(a) $|A| I_3$

(b) $I_3$

(d) None of these

24. Let $A= \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix} $. Then the only correct statement about the matrix $A$ is

(a) $A$ is a zero matrix

(b) $A^{2}=I$

(d) $A^{-1}$ does not exist.

25. If $A= \begin{bmatrix} 2 & 3 \\ 5 & -2\end{bmatrix} $, then $A^{-1}$ equals

(a) $A$

(b) $\frac{1}{11} A$

(d) $A^{T}$

26. If $A= \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1\end{bmatrix} $, then $(A^{T})^{-1}$ equals

(a) $ \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{bmatrix} $

(b) $ \begin{bmatrix} -9 & 8 & 2 \\ -8 & 7 & -2 \\ -5 & 4 & 1\end{bmatrix} $

(d) $ \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1\end{bmatrix} $

27. The value of determinant $ \begin{bmatrix} \log _{y} x & 1 \\ 1 & \log _{x} y\end{bmatrix} $ is equal to

(a) -1

(b) 0

(d) 3

28. If $A= \begin{bmatrix} 0 & 3 \\ 0 & 0\end{bmatrix} $ and $f(x)=1+x+x^{2}+\ldots+x^{20}$, then $f(A)=$

(a) $ \begin{bmatrix} 1 & 3 \\ 0 & 0\end{bmatrix} $

(b) $ \begin{bmatrix} 0 & 3 \\ 1 & 3\end{bmatrix} $

(d) $ \begin{bmatrix} 1 & 3 \\ 0 & 1\end{bmatrix} $

29. If $ \begin{vmatrix} 2 a & 1 \\ b c+a b & c\end{vmatrix} =0$, then $a, b, c$ are in

(a) A.P.

(b) G.P.

(d) None of these

30. If $A= \begin{bmatrix} -1 & 1 \\ 1 & -1\end{bmatrix} $, then $I+A+A^{2}+\ldots . .+\infty=\ldots .$.

(a) $ \begin{bmatrix} 2 / 3 & -1 / 3 \\ -1 / 3 & 2 / 3\end{bmatrix} $

(b) $ \begin{bmatrix} 2 / 3 & 1 / 3 \\ 1 / 3 & 2 / 3\end{bmatrix} $

(d) $ \begin{bmatrix} 2 / 3 & 1 / 3 \\ -1 / 3 & 2 / 3\end{bmatrix} $

31. If $A=\frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4\end{bmatrix} $ then $A^{-1}$ is equal to

(a) $A^{2}$

(b) $I$

(d) 0

32. The value of the determinant $ \begin{vmatrix} 0 & \tan \theta & 1 \\ 1 & -\sec \theta & 0 \\ \sec \theta & \tan \theta & 1\end{vmatrix} $ is

(a) $\tan ^{2} \theta$

(b) $\cos ^{2} \theta$

(d) 1

33. $ \begin{vmatrix} 9 & 9 & 12 \\ 1 & -3 & -4 \\ 1 & 9 & 12\end{vmatrix} $ is equal to

(a) -121

(b) 136

(d) 10

34. The determinant $ \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{vmatrix} $ is equal to

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

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

(d) $(a+b)(b-c)(c+a)$

35. If $a+b+c=0$, then the determinant

$ \begin{vmatrix} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{vmatrix} $ is equal to

(a) 0

(b) $a b c$

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

36. The value of the determinant $ \begin{vmatrix} \log e & \log e^{2} & \log e^{3} \\ \log e^{2} & \log e^{3} & \log e^{4} \\ \log e^{3} & \log e^{4} & \log e^{5}\end{vmatrix} $ is

(a) 0

(b) 1

(d) $5 \log e$

(EAMCET 2006)

37. The value of the determinant $ \begin{vmatrix} 1 & a b & c(a+b) \\ 1 & b c & a(b+c) \\ 1 & c a & b(c+a)\end{vmatrix} $ is equal to

(a) 0

(b) $a b c$

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

38. One root of the equation $ \begin{vmatrix} x+a & b & c \\ b & x+c & a \\ c & a & x+b\end{vmatrix} =0$ is

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

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

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

39. The value of the determinant $ \begin{vmatrix} 1+\log a & \log b & \log c \\ \log a & 1+\log b & \log c \\ \log a & \log b & 1+\log c\end{vmatrix} $ is

(a) $\log (a b c)$

(b) $1-\log (a b c)$

(d) $1+\log (a b c)$

40. If $A$ is a $2 \times 2$ matrix and $|A|=2$, then the matrix represented by $A(adj . A)$ is equal to

(a) $ \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $

(b) $ \begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix} $

(J&K CET 2011)

41. If $l, m, n$ are the $p$ th, $q$ th and $r$ th terms of a $G P$, then $ \begin{vmatrix} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1\end{vmatrix} $ is equal to

(a) 0

(b) $l+m+n$

(d) $lmn$

(EAMCET 2009)

42. If $A$ is an invertible matrix which satisfies the relation $A^{2}+A-I=0$, then $A^{-1}$ equals

(a) $A^{2}$

(b) $I+A$

(d) $I-A^{2}$

(MPPET 2009)

43. For non-singular square matrices $A, B$ and $C$ of same order, $(A B^{-1} C)^{-1}$ is equal to

(a) $C^{-1} B A$

(b) $C^{-1} B A^{-1}$

(d) $C B A^{-1}$

44. If $A= \begin{bmatrix} 2 x & 0 \\ x & x\end{bmatrix} $ and $A^{-1}= \begin{bmatrix} 1 & 0 \\ -1 & 2\end{bmatrix} $, then $x$ equals

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

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

(d) 2

(UPSEE 2008)

45. If the system of equations $x+a y=0, a z+y=0$ and $a x+z=0$ has infinite solutions, then the value of $a$ is

(a) -1

(b) 0

(d) no real values

(IIT 2003)

46. If the system of linear equations

$x+2 a y+a z=0, x+3 b y+b z=0, x+4 c y+c z=0$ has a non-zero solution, then $a, b, c$.

(a) are in A.P.

(b) are in G.P.

(d) satisfy $a+2 b+3 c=0$.

(AIEEE 2003)

47. If $3 x-2 y=5$ and $6 x-4 y=9$, then the system of equations has

(a) Unique solution

(b) No solution

(d) None of these

48. If $x+5 y=3,2 x+10 y=6$, then the system of equations has

(a) Unique solution

(b) No solution

(d) None of these

49. The system of equations $5 x+3 y+z=16,2 x+y+3 z=19$ and $x+2 y+4 z=25$ has

(a) No solution

(b) Unique solution

(d) None of these.

50. Let $P=[a_{i j}]$ be a $3 \times 3$ matrix and let $Q=[b_{i j}]$, where $b_{i j}=$ $2^{i+j} a_{i j}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is 2 , then the determinant of matrix $Q$ is

(a) $2^{10}$

(b) $2^{11}$

(d) $2^{13}$

(IIT JEE 2012)

ANSWERS

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

HINTS AND SOLUTIONS

1. Given, $a _{i j}=[\frac{i}{j}]$ where $[x]$ stands for greatest integer function.

$ \begin{aligned} & \therefore \quad A _{2 \times 2}= \begin{bmatrix} a _{11} & a _{12} \\ a _{21} & a _{22} \end{bmatrix} \\ & a _{11}=[\frac{1}{1}]=1, a _{12}=[\frac{1}{2}]=[0.5]=0 \\ & a _{21}=[\frac{2}{1}]=2, a _{22}=[\frac{2}{2}]=1 \\ & \therefore \quad A _{2 \times 2}= \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \text{. } \end{aligned} $

$ \begin{aligned} A= \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \\ & \therefore \quad A^{2}= \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 1+0 & k+k \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 k \\ 0 & 1 \end{bmatrix} \\ & A^{3}=A^{2} \cdot A= \begin{bmatrix} 1 & 2 k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 1+0 & k+2 k \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 k \\ 0 & 1 \end{bmatrix} \end{aligned} $

$\therefore$ On generalisation, $A^{n}= \begin{bmatrix} 1 & n k \\ 0 & 1\end{bmatrix} $.

3. Given, $A^{2}=I$

$\Rightarrow \begin{bmatrix} x & 1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} x & 1 \\ 1 & 0\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $ $\Rightarrow \begin{bmatrix} x^{2}+1 & x+0 \\ x+0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $

( $\because$ Two matrices are equal if their corresponding elements are equal)

$\Rightarrow x^{2}+1=1$ and $x=0 \Rightarrow x^{2}=0 \Rightarrow x=0$.

$\therefore x^{3}+x-2=\mathbf{- 2}$.

4. Given, $A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{bmatrix} $

$ \begin{aligned} \therefore \quad A^{2}=A . A & = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \\ & = \begin{bmatrix} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ a+0-a & 0+b-b & 0+0+1 \end{bmatrix} \end{aligned} $

$= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ =I .

$ A^{4}=A^{2} \cdot A^{2}=I \cdot I=I $

$A^{6}=A^{2} \cdot A^{2} \cdot A^{2}=I \cdot I \cdot I=I$

$A^{8}=A^{2} \cdot A^{2} \cdot A^{2} \cdot A^{2}=I $

$\Rightarrow A^{2}+2 A^{4}+4 A^{6}=I+2 I+4 I=7 I=\mathbf{A}^{\mathbf{8}}$

5. $A^{2}=2 A-I$

$ \begin{aligned} & \therefore \quad A^{3}=A \cdot A^{2}=A \cdot(2 A-I) \\ & =2 A^{2}-A I=2 A^{2}-A \\ & =2(2 A-I)-A=3 A-2 I \\ & A^{4}=A . A^{3}=A(3 A-2 I) \\ & =3 A^{2}-2 A I=3(2 A-I)-2 A \\ & =6 A-3 I-2 A=4 A-3 I \end{aligned} $

~~ 6. $ \begin{bmatrix} 1 & x & 1\end{bmatrix} _{1 \times 3} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 1 \\ 0 & 3 & 2\end{bmatrix} _{3 \times 3} \begin{bmatrix} x \\ 1 \\ -2\end{bmatrix} _{3 \times 1}=0$

Using associative law and multiplying the first two matrices, we have

$ \begin{bmatrix} 1+0+0 & 2+5 x+3 & 3+x+2 \end{bmatrix} _{1 \times 3} \begin{bmatrix} x \\ 1 \\ -2 \end{bmatrix} _{3 \times 1}=0 $

$\Rightarrow \begin{bmatrix} 1 & 5+5 x & 5+x\end{bmatrix} _{1 \times 3} \begin{bmatrix} x \\ 1 \\ -2\end{bmatrix} _{3 \times 1}=0$

$\Rightarrow[x+5+5 x-10-2 x] _{1 \times 1}=0$

$\Rightarrow[4 x-5]=[0]$

$\Rightarrow 4 x-5=0 \Rightarrow x=\frac{\mathbf{5}}{\mathbf{4}}$.

~~ 7. Let $M= \begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33}\end{bmatrix} $ be the required $3 \times 3$ matrix.

Then, according to the first condition,

$\begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33}\end{bmatrix} $ $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ $=\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$

$ \Rightarrow \begin{bmatrix} a_{12} \\ a_{22} \\ a_{32} \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \Rightarrow \begin{rcases} \begin{matrix} a _{12}=-1 \\ a _{22}=2 \\ a _{32}=3 \end{matrix} \end{rcases}\quad …(i) $

According to second condition,

$ \begin{aligned} & { \begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33} \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} } \\ \Rightarrow & { \begin{bmatrix} a _{11}-a _{12} \\ a _{21}-a _{22} \\ a _{31}-a _{32} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \Rightarrow \begin{matrix} a _{11}-a _{12}=1 \\ a _{21}-a _{22}=1 \\ a _{31}-a _{32}=-1 \end{matrix} } \end{aligned} $

$\therefore \quad$ Using $(i)$, we have

$ \begin{aligned} a _{11}-(-1) & =1 \Rightarrow a _{11}=0 \\ a _{21}-2 & =1 \Rightarrow a _{21}=3 \\ a _{31}-3 & =-1 \Rightarrow a _{31}=2 \end{aligned} $

According to the third condition,

$ \begin{aligned} & { \begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} } \\ \Rightarrow & { \begin{bmatrix} a _{11}+a _{12}+a _{13} \\ a _{21}+a _{22}+a _{23} \\ a _{31}+a _{32}+a _{33} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} } \\ \Rightarrow & a _{11}+a _{12}+a _{13}=0, a _{21}+a _{22}+a _{23}=0, \\ & a _{31}+a _{32}+a _{33}=12 \\ \Rightarrow & 0-1+a _{13}=0,3+2+a _{23}=0, \\ \Rightarrow & 2+3+a _{33}=12 \\ \Rightarrow & a _{13}=1, a _{23}=-5, a _{33}=7 \end{aligned} $

$\therefore$ Sum of diagonal elements of

$ M=a _{11}+a _{22}+a _{33}=0+2+7=\mathbf{9} \text{. } $

~~ 8. Given, $A=$ diag. $ \begin{bmatrix} 1 & -4 & 8\end{bmatrix} , B=diag \begin{bmatrix} -2 & 3 & 5\end{bmatrix} $ and $C=diag \begin{bmatrix} -3 & 7 & 10\end{bmatrix} $. Then,

$ \begin{aligned} B+2 C-A & = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} +2 \begin{bmatrix} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} \\ & = \begin{bmatrix} -2-6-1 & 0 & 0 \\ 0 & 3+14+4 & 0 \\ 0 & 0 & 5+20-8 \end{bmatrix} \\ & = \begin{bmatrix} -9 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 17 \end{bmatrix} =diag \begin{bmatrix} -9 & 21 & 17 \end{bmatrix} . \end{aligned} $

~~ 9. $A= \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix} $

$ \begin{aligned} \Rightarrow \quad A^{2} & =A \cdot A= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \\ & = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} =2 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \\ & =2^{2-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} =2^{2-1} \cdot A \end{aligned} $

$ \begin{aligned} A^{3} & =A \cdot A^{2}= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 2+2 & 2+2 \\ 2+2 & 2+2 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} \\ & =4 \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} =2^{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \\ & =2^{3-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} =2^{3-1} \cdot A \end{aligned} $

$\therefore \quad A^{n}=2^{n-1} A$.

~~ 10. Given, a matrix $A$ is nilpotent of order $2 \Rightarrow A^{2}=0$.

$ \begin{aligned} & \Rightarrow \begin{bmatrix} 4 & -2 \\ k & -4 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ k & -4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 16-2 k & -8+8 \\ 4 k-4 k & -2 k+16 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ & \Rightarrow 16-2 k=0 \Rightarrow k=\mathbf{8} . \end{aligned} $

~~ 11. Given, $A= \begin{bmatrix} 3 & 1 \\ -1 & 2\end{bmatrix} $, then

$ \begin{aligned} A^{2} & = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \\ A^{2}-5 A & = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} -5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\ & = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5+5 & 3-10 \end{bmatrix} \end{aligned} $

$ = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} =-7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =-7 I $

$\therefore \quad A^{2}-5 A+7 I=0$.

~~ 12. $A= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix} $

$\Rightarrow \quad A^{T}= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix} $

$\Rightarrow A^{T} \cdot A= \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix} $

$= \begin{bmatrix} \cos ^{2} \alpha+\sin ^{2} \alpha & \cos \alpha \sin \alpha-\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha-\cos \alpha \sin \alpha & \sin ^{2} \alpha+\cos ^{2} \alpha\end{bmatrix} $

$= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} =I$.

~~ 13. Given, $A A^{T}=9 I_3$

$ \begin{aligned} & \Rightarrow \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} =9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 1+4+4 & 2+2-4 & a+4+2 b \\ 2+2-4 & 4+1+4 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & a^{2}+4+b^{2} \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} 9 & 0 & a+4+2 b \\ 0 & 9 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & a^{2}+4+b^{2} \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \end{aligned} $

$\therefore$ Equating the corresponding elements of both the matrices,

$ \begin{aligned} a+4+2 b & =0 \\ 2 a+2-2 b & =0 \\ a^{2}+4+b^{2} & =9 \\ \Rightarrow \quad a+2 b & =-4 \\ 2 a-2 b & =-2 \end{aligned} $

On adding $(i)$ and (ii)

$ 3 a=-6 \Rightarrow a=-2 $

Substituting $a=-2$ in (i), we get

$ -2+2 b=-4 \Rightarrow 2 b=-2 \Rightarrow b=-1 $

$\therefore a=-2, b=-1$.

~~ 14. $A$ is of order $5 \times 4 \Rightarrow A^{T}$ is of order $4 \times 5$

$B$ is of order $5 \times 6 \Rightarrow B^{T}$ is of order $6 \times 5$

$C$ is of order $7 \times 4 \Rightarrow C^{T}$ is of order $4 \times 7$

$\therefore(A^{T} \times B)$ is of order $4 \times 5 \quad 5 \times 64 \times 6$

$\therefore \quad(A^{T} \times B)^{T}$ is of order $6 \times 4$

$\Rightarrow(A^{T} \times B)^{T} \times C^{T}$ is of order $6 \times 4 \quad 4 \times 7=6 \times 7$.

~~ 15. $A$ is of order $3 \times 4 \Rightarrow A^{T}$ is of order $4 \times 3$.

Let the order of $B$ be $p \times q$.

Since, $A^{T} B$ is defined. So,

No. of rows of $B=$ No. of columns of $A^{T} \Rightarrow p=3$ Also, $B A^{T}$ is defined. So,

No. of columns of $B=$ No. of rows of $A^{T} \Rightarrow q=4$

$\therefore \quad B$ is of order $\mathbf{3} \times \mathbf{4}$.

~~ 16. $A$ is a symmetric matrix $\Rightarrow A^{T}=A$

$A$ is a skew-symmetric matrix $\Rightarrow A^{T}=-A$

$\therefore$ From $(i)$ and (ii),

$ A=-A \Rightarrow A+A=0 \Rightarrow 2 A=0 \Rightarrow A=0 $

$\Rightarrow A$ is null matrix.

~~ 17. Given, $A$ and $B$ are symmetric matrices of same order.

$ \begin{aligned} \therefore \quad A^{T} & =A \text{ and } B^{T}=B . \\ \therefore \quad(A+B)^{T} & =A^{T}+B^{T} \\ & =A+B \Rightarrow(A+B) \text{ is a symmetric matrix } \\ (A B-B A)^{T} & =(A B)^{T}-(B A)^{T} \\ & =B^{T} A^{T}-A^{T} B^{T}=B A-A B=-(A B-B A) \end{aligned} $

$\Rightarrow A B-B A$ is a skew-symmetric matrix

$ \begin{aligned} (A B+B A)^{T} & =(A B)^{T}+(B A)^{T} \\ & =B^{T} A^{T}+A^{T} B^{T}=B A+A B=A B+B A \end{aligned} $

$\Rightarrow(A B+B A)$ is a symmetric matrix.

$\therefore \quad$ All the given options hold.

~~ 18. $A^{T}$. $A=I$

$ \begin{aligned} \Rightarrow & { \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} } \\ \Rightarrow & { \begin{bmatrix} a^{2}+b^{2}+c^{2} & a b+b c+c a & a c+a b+c b \\ b a+c b+a c & b^{2}+c^{2}+a^{2} & c b+c a+a b \\ c a+a b+b c & c b+a c+b a & c^{2}+a^{2}+b^{2} \end{bmatrix} } \\ & = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ \Rightarrow & { \begin{bmatrix} a^{2}+b^{2}+c^{2} & a b+b c+c a & a b+b c+a c \\ a b+b c+a c & a^{2}+b^{2}+c^{2} & a b+b c+a c \\ a b+b c+a c & a b+b c+a c & a^{2}+b^{2}+c^{2} \end{bmatrix} } \\ \Rightarrow & a^{2}+b^{2}+c^{2}=1, a b+b c+c a=0 \end{aligned} $

(On equating corresponding elements of equal matrices) Now, we know that

$ \begin{aligned} & (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\ & =1+0=1 \\ & \Rightarrow a+b+c=1 \\ & \text{ Also, } a^{3}+b^{3}+c^{3}-3 a b c \\ & \quad=(a+b+c){a^{2}+b^{2}+c^{2}-(a b+b c+c a)} \\ & \quad=1{1-0}=1 \\ & \therefore \quad a^{3}+b^{3}+c^{3}=1+3 a b c=1+3 \times 1=\mathbf{4} \end{aligned} $

$(\because$ Given, $a b c=1)$ ~~ 19. $A= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} $

$ A^{T}= \begin{bmatrix} \cos \theta & +\sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} $

$A \cdot A^{T}= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix} $

$ = \begin{bmatrix} \cos ^{2} \theta+\sin \theta & 0 \\ 0 & \sin ^{2} \theta+\cos ^{2} \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =I $

$\therefore A$ is an orthogonal matrix.

~~ 20. $A B=A$

$\Rightarrow \quad(A B) A=A \cdot A$ (Multiplying by $A$ on both the sides)

$\Rightarrow \quad A(B A)=A^{2}$

(Associative law)

$\Rightarrow \quad A B=A^{2}$

$(\because B A=B)$

$\Rightarrow \quad A=A^{2}$

$(\because A B=A)$

$\Rightarrow A$ is idempotent.

Similarly,

$ \begin{aligned} B A=B & \Rightarrow(B A) B=B \cdot B \Rightarrow B(A B)=B^{2} \\ & \Rightarrow B A=B^{2} \Rightarrow B=B^{2} \Rightarrow B \text{ is idempotent. } \end{aligned} $

~~ 21 Given, $A= \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix} $

Then, det. $A= \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{vmatrix} $

$ =\cos \theta \cdot \cos \theta-(-\sin \theta) \cdot \sin \theta $

$ =\cos ^{2} \theta+\sin ^{2} \theta=1 $

$ \begin{bmatrix} \because A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \\ \Rightarrow|A|= \begin{vmatrix} a & b \\ c & d \end{vmatrix} =a d-b c \end{bmatrix} $

~~ 22 Let, $A= \begin{bmatrix} x & 0 & 1 \\ 2 & -1 & 4 \\ 1 & 2 & 0\end{bmatrix} $

Then, det. $A= \begin{vmatrix} x & 0 & 1 \\ 2 & -1 & 4 \\ 1 & 2 & 0\end{vmatrix} $

Now expanding along the first row, we have

$ \begin{aligned} det . A & =x \begin{vmatrix} -1 & 4 \\ 2 & 0 \end{vmatrix} -0 \begin{vmatrix} 2 & 4 \\ 1 & 0 \end{vmatrix} +1 \begin{vmatrix} 2 & -1 \\ 1 & 2 \end{vmatrix} \\ & =x(-1 \times 0-4 \times 2)+1(2 \times 2-(-1) \times 1) \\ & =x \times(-8)+5=-8 x+5 \end{aligned} $

For matrix $A$ to be a singular matrix det. $A=0$

$\therefore \quad-8 x+5=0 \Rightarrow x=\frac{\mathbf{5}}{\mathbf{8}}$.

~~ 23 We know that $A^{-1}=\frac{adj . A}{|A|}$ where $|A|=det$. $A$

$\Rightarrow \quad A^{-1} \cdot|A|=adj . A$

$ \begin{aligned} \Rightarrow \quad A \cdot(adj \cdot A) & =A \cdot(A^{-1} \cdot|A|) \\ & =A A^{-1} \cdot|A| \\ & =I|A|=|A| I ; \end{aligned} $

where $I$ is the identity matrix $(A A^{-1}=I)$

So to find $A$. adj . $A$. we need to find det. $A$.

$ \begin{matrix} \therefore & A= \begin{bmatrix} 1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix} \\ \therefore & det A= \begin{vmatrix} 1 & -1 & 1 \\ 2 & 3 & 0 \\ 18 & 2 & 10 \end{vmatrix} \end{matrix} $

Expanding along the first row, we have

$ \begin{aligned} & det . A=|A|=1 \begin{vmatrix} 3 & 0 \\ 2 & 10 \end{vmatrix} -(-1) \begin{vmatrix} 2 & 0 \\ 18 & 10 \end{vmatrix} +1 \begin{vmatrix} 2 & 3 \\ 18 & 2 \end{vmatrix} \\ &=1 \times 30+1 \times 20+1 \times(4-54) \\ &=30+20-50=0 \\ & \therefore \quad A \cdot adj A=|A| I=0 \times I=0 \\ & \Rightarrow \quad A \cdot adj A \text{ is a null matrix. } \end{aligned} $

~~ 24 Let us examine each statement separately.

$A$ is not a zero matrix
$A^{2}=A \cdot A= \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix} $

$ = \begin{bmatrix} 0+0+1 & 0 & 0 \\ 0 & 0+1+0 & 0 \\ 0 & 0 & 0+0+1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =I $

$\quad(-1) I=(-1) \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix} \neq A$
For $A^{-1}$ to exist, $det A=|A| \neq 0$ (Expanding $|A|$ along the first row $R_1$ ), we have

$\therefore \quad|A|=-1 \begin{vmatrix} 0 & -1 \\ -1 & 0\end{vmatrix} =-1 \times-1=1 \neq 0$

$\Rightarrow A^{-1}$ exists.

$\therefore$ The only correct statement is $(b)$.

~~ 25 $A= \begin{bmatrix} 2 & 3 \\ 5 & -2\end{bmatrix} $

$ |A|=(2 \times-2)-(3 \times 5)=-4-15=-19 \neq 0 $

As $|A| \neq 0$, therefore $A$ is non-singular and hence $A^{-1}$ exists.

$ \begin{aligned} & A^{-1}=\frac{adj A}{|A|} \\ & adj A= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} \\ \therefore \quad A _{11}=(-1)^{1+1}|-2|=-2, & A _{12}=(-1)^{1+2}|5|=-5 \\ A _{21}=(-1)^{2+1}|3|=-3, & A _{22}=(-1)^{2+2}|2|=2 \end{aligned} $

$ \therefore \quad A^{-1}=-\frac{1}{19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} =\frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} =\frac{1}{19} A . $

Note: Do not confuse here | |, i.e., determinant sign with absolute value sign.

~~ 26 $A= \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ 2 & 2 & 1\end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} 1 & 0 & 2 \\ -2 & -1 & 2 \\ 3 & 4 & 1\end{bmatrix} $

$\therefore$ Let

$ \begin{aligned} B & =det . A^{T}= \begin{vmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{vmatrix} \\ & =1 \begin{vmatrix} -1 & 2 \\ 4 & 1 \end{vmatrix} -0 \begin{vmatrix} -2 & 2 \\ 3 & 1 \end{vmatrix} -2 \begin{vmatrix} -2 & -1 \\ 3 & 4 \end{vmatrix} \\ & =1(-1-8)-2(-8+3) \\ & =-9-(2 \times-5)=-9+10=1 \neq 0 . \end{aligned} $

$|B| \neq 0 \Rightarrow B$ is non-singular $\Rightarrow B^{-1}$, i.e, $(A^{T})^{-1}$ exists

$\therefore \quad B^{-1}=\frac{adj B}{|B|}$

$adj B= \begin{bmatrix} B _{11} & B _{12} & B _{13} \\ B _{21} & B _{22} & B _{23} \\ B _{31} & B _{32} & B _{33}\end{bmatrix} ^{T}= \begin{bmatrix} B _{11} & B _{21} & B _{31} \\ B _{12} & B _{22} & B _{32} \\ B _{13} & B _{23} & B _{33}\end{bmatrix} $

$B _{11}=(-1)^{1+1} \begin{vmatrix} -1 & 2 \\ 4 & 1\end{vmatrix} =(-1-8)=-9$

$B _{12}=(-1)^{1+2} \begin{vmatrix} -2 & 2 \\ 3 & 1\end{vmatrix} =-(-2-6)=8$

$B _{13}=(-1)^{1+3} \begin{vmatrix} -2 & -1 \\ 3 & 4\end{vmatrix} =-8+3=-5$

$B _{21}=(-1)^{2+1} \begin{vmatrix} 0 & -2 \\ 4 & 1\end{vmatrix} =-(0+8)=-8$

$B _{22}=(-1)^{2+2} \begin{vmatrix} 1 & -2 \\ 3 & 1\end{vmatrix} =1+6=7$

$B _{23}=(-1)^{2+3} \begin{vmatrix} 1 & 0 \\ 3 & 4\end{vmatrix} =-(4-0)=-4$

$B _{31}=(-1)^{3+1} \begin{vmatrix} 0 & -2 \\ -1 & 2\end{vmatrix} =(0-2)=-2$

$B _{32}=(-1)^{3+2} \begin{vmatrix} 1 & -2 \\ -2 & 2\end{vmatrix} =-(2-4)=+2$

$B _{33}=(-1)^{3+3} \begin{vmatrix} 1 & 0 \\ -2 & -1\end{vmatrix} =-1$

$\therefore \quad(A^{T})^{-1}=B^{-1}=\frac{1}{\mathbf{1}} \begin{bmatrix} -\mathbf{9} & -\mathbf{8} & -\mathbf{2} \\ \mathbf{8} & \mathbf{7} & \mathbf{2} \\ -\mathbf{5} & -4 & -1\end{bmatrix} $.

~~ 27 $ \begin{vmatrix} \log _{y} x & 1 \\ 1 & \log _{x} y\end{vmatrix} =\log _{y} x \times \log _{x} y-1 \times 1$

$ =\frac{\log x}{\log y} \cdot \frac{\log y}{\log x}-1=1-1=\mathbf{0} . $

~~ 28 $A= \begin{bmatrix} 0 & 3 \\ 0 & 0\end{bmatrix} $

$\Rightarrow \quad A^{2}=A \cdot A= \begin{bmatrix} 0 & 3 \\ 0 & 0\end{bmatrix} \begin{bmatrix} 0 & 3 \\ 0 & 0\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix} =O$

$\Rightarrow \quad A^{3}=A^{2} \cdot A=O \cdot A=O$

$\Rightarrow \quad A^{2}=A^{3}=A^{4}=A^{5}=$ $=O$

$\therefore f(A)=I+A+A^{2}+\ldots \ldots .+A^{20}$

$ = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} +O+\ldots \ldots . .+O= \begin{bmatrix} \mathbf{1} & \mathbf{3} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} . $

~~ 29 $ \begin{vmatrix} 2 a & 1 \\ b c+a b & c\end{vmatrix} =0$

$ \begin{aligned} & \Rightarrow 2 a c-(b c+a b)=0 \\ & \Rightarrow 2 a c=a b+b c \Rightarrow 2 a c=b(a+c) \\ & \Rightarrow b=\frac{2 a c}{a+c} \Rightarrow \frac{1}{b}=\frac{a+c}{2 a c}=\frac{1}{2}[\frac{1}{c}+\frac{1}{a}] \\ & \Rightarrow a, b, c \text{ are in H.P. } \end{aligned} $

~~ 30 $I+A+A^{2}+\ldots \ldots . .+\infty$ is the sum of an infinite G.P. with first term $(a)=I$ and common ration $(r)=A$.

As we know that, sum of an infinite G.P. $=\frac{a}{1-r}$

$\therefore I+A+A^{2}+\ldots \ldots .+\infty=\frac{I}{I-A}=I .(I-A)^{-1}=(I-A)^{-1}$

Now, $\quad I-A= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 1 & -1\end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -1 & 2\end{bmatrix} $

$ \therefore \quad(I-A)^{-1}=\frac{adj .(I-A)}{det .(I-A)} $

det. $(I-A)=4-1=3 \neq 0$.

$ \begin{aligned} & \text{ adj. }(I-A)= \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} ^{T}= \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \\ \therefore \quad & (I-A)^{-1}=\frac{1}{3} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 / 3 & -1 / 3 \\ -1 / 3 & 2 / 3 \end{bmatrix} . \end{aligned} $

~~ 31 If $A^{-1}=A^{T}$ then,

$ \begin{aligned} A . A^{-1} & =I \Rightarrow A A^{T}=I . \text{ Where } I=\text{ unit matrix }= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ A \cdot A^{T} & =\frac{1}{9} \begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix} \cdot \frac{1}{9} \begin{bmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{bmatrix} \\ & =\frac{1}{81} \begin{bmatrix} 64+1+16 & -32+4+28 & -8-8+16 \\ -32+4+28 & 16+16+49 & 4-32+28 \\ -8-8+16 & 4-32+28 & 1+64+16 \end{bmatrix} \\ & =\frac{1}{81} \begin{bmatrix} 81 & 0 & 0 \\ 0 & 81 & 0 \\ 0 & 0 & 81 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} =I . \\ \therefore \boldsymbol{A}^{T} & =\boldsymbol{A}^{-1} . \end{aligned} $

~~ 32 Let $\Delta= \begin{vmatrix} 0 & \tan \theta & 1 \\ 1 & -\sec \theta & 0 \\ \sec \theta & \tan \theta & 1\end{vmatrix} $

Expanding along Row $1(R_1)$

Then, $\quad \Delta=0 \begin{vmatrix} -\sec \theta & 0 \\ \tan \theta & 1\end{vmatrix} -\tan \theta \begin{vmatrix} 1 & 0 \\ \sec \theta & 1\end{vmatrix} $

$ +1 \begin{vmatrix} 1 & -\sec \theta \\ \sec \theta & \tan \theta \end{vmatrix} $

$ =-\tan \theta \times 1+(\tan \theta+\sec ^{2} \theta)=\sec ^{2} \boldsymbol{\theta} . $

~~ 33 Let $\Delta= \begin{vmatrix} 9 & 9 & 12 \\ 1 & -3 & -4 \\ 1 & 9 & 12\end{vmatrix} $.

Taking out 3 common from $C_2$ and 4 common from $C_3$, we have

~~ 34 Let $\Delta= \begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{vmatrix} $

Operating $R_1 \to R_1-R_2$ and $R_2 \to R_2-R_3$, we have

$ \Delta= \begin{vmatrix} 0 & a-b & a^{2}-b^{2} \\ 0 & b-c & b^{2}-c^{2} \\ 1 & c & c^{2} \end{vmatrix} =(a-b)(b-c) \begin{vmatrix} 0 & 1 & (a+b) \\ 0 & 1 & (b+c) \\ 1 & c & c^{2} \end{vmatrix} $

[Taking out $(a-b)$ common from $R_1$ and $(b-c)$ common from $R_2$ ]

$ =(a-b)(b-c) \cdot 1 \begin{vmatrix} 1 & (a+b) \\ 1 & (b+c) \end{vmatrix} \quad(\text{ Expanding along } C_1) $

$=(a-b)(b-c)(b+c-a-b)$

$=(\boldsymbol{a}-\boldsymbol{b})(\boldsymbol{b}-\boldsymbol{c})(\boldsymbol{c}-\boldsymbol{a})$.

~~ 35 Let $\Delta= \begin{vmatrix} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{vmatrix} $

Operating $C_1 \to C_1-C_3$ and $C_2 \to C_2-C_3$, we have

$ \begin{aligned} \Delta & = \begin{vmatrix} -a-b-c & 0 & 2 a \\ 0 & -b-c-a & 2 b \\ c+a+b & c+a+b & c-a-b \end{vmatrix} \\ & = \begin{vmatrix} -(a+b+c) & 0 & 2 a \\ 0 & -(a+b+c) & 2 b \\ a+b+c & a+b+c & c-a-b \end{vmatrix} \\ & = \begin{vmatrix} 0 & 0 & 2 a \\ 0 & 0 & 2 b \\ 0 & 0 & c-a-b \end{vmatrix} =0 \quad(\because a+b+c=0 \text{ is given }) \end{aligned} $

~~ 36 Let $\Delta= \begin{vmatrix} \log e & \log e^{2} & \log e^{3} \\ \log e^{2} & \log e^{3} & \log e^{4} \\ \log e^{3} & \log e^{4} & \log e^{5}\end{vmatrix} $

$ = \begin{vmatrix} \log e & 2 \log e & 3 \log e \\ 2 \log e & 3 \log e & 4 \log e \\ 3 \log e & 4 \log e & 5 \log e \end{vmatrix} = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} $

$(\because \log e=1)$

Operating $C_2 \to C_2-C_1, C_3 \to C_3-C_1$, we have

$\Delta= \begin{vmatrix} 1 & 1 & 2 \\ 2 & 1 & 2 \\ 3 & 1 & 2\end{vmatrix} =2 \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & 1 \\ 3 & 1 & 1\end{vmatrix} $

(Taking out 2 common from $C_3$ )

$=2 \times 0=0$

( $\because$ Two columns are identical)

~~ 37 Let $\Delta= \begin{vmatrix} 1 & a b & c(a+b) \\ 1 & b c & a(b+c) \\ 1 & c a & b(c+a)\end{vmatrix} $

Operating $R_1 \to R_1-R_2$ and $R_2 \to R_2-R_3$, we have

$ \begin{aligned} \Delta & = \begin{vmatrix} 0 & a b-b c & c a+c b-a b-a c \\ 0 & b c-c a & a b+a c-b c-b a \\ 1 & c a & b(c+a) \end{vmatrix} \\ & = \begin{vmatrix} 0 & b(a-c) & -b(a-c) \\ 0 & c(b-a) & -c(b-a) \\ 1 & c a & b(c+a) \end{vmatrix} \end{aligned} $

Taking out $b(a-c)$ common from $R_1$ and $c(b-a)$ common from $R_2$, we have

$ \Delta=b(a-c) . c(b-a) \begin{vmatrix} 0 & 1 & -1 \\ 0 & 1 & -1 \\ 1 & c a & b(c+a) \end{vmatrix} =0 $

$(\because$ Two rows are identical)

~~ 38 Let $\Delta= \begin{vmatrix} x+a & b & c \\ c & x+b & a \\ a & b & x+c\end{vmatrix} $

Operating $C_1 \to C_1+C_2+C_3$

$ \begin{aligned} \Delta & = \begin{vmatrix} x+a+b+c & b & c \\ x+a+b+c & x+b & a \\ x+a+b+c & b & x+c \end{vmatrix} \\ & =(x+a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & x+b & a \\ 1 & b & x+c \end{vmatrix} \end{aligned} $

Operating $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_2$, we have

$ \Delta=(x+a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & x & a-c \\ 0 & -x & x+c-a \end{vmatrix} $

$ \begin{aligned} & =(x+a+b+c)[x(x+c-a)-x(c-a)] \\ & =(x+a+b+c) x^{2} \quad(\text{ On expanding by } C_1) \\ \therefore & \begin{vmatrix} x+a & b & c \\ c & x+b & a \\ a & b & x+c \end{vmatrix} =0 \Rightarrow x^{2}(x+a+b+c)=0 \\ \Rightarrow & x=\mathbf{0} \text{ or } x=-(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}) . \end{aligned} $

~~ 39 Let $\Delta= \begin{vmatrix} 1+\log a & \log b & \log c \\ \log a & 1+\log b & \log c \\ \log a & \log b & 1+\log c\end{vmatrix} $

Operating $C_1 \to C_1+C_2+C_3$, we have

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

Taking out $1+\log a b c$ common from $C_1$

$ \Delta=(1+\log (a b c)) \begin{vmatrix} 1 & \log b & \log c \\ 1 & 1+\log b & \log c \\ 1 & \log b & 1+\log c \end{vmatrix} $

Operating $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$, we have

$ \begin{aligned} \Delta & =(1+\log (a b c)) \begin{vmatrix} 1 & \log b & \log c \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ & =1+\log (a b c) \cdot 1 \cdot \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} \\ & =1+\log (a b c) \end{aligned} $

~~ 40 $A(adj . A)=|A| I$ where $I$ is the identity matrix, so

$ A(adj . A)=2 \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = \begin{vmatrix} \mathbf{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{2} \end{vmatrix} $

~~ 41 Let the first term of the $G P$ be $a$ and common ratio $k$. As,

$ T_n(n \text{th term of a } G P)=a k^{n-1} \text{, here } $

$l=a k^{p-1}, \quad m=a k^{q-1}, \quad n=a k^{r-1}$

Let, $\quad \Delta= \begin{vmatrix} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1\end{vmatrix} = \begin{vmatrix} \log a k^{p-1} & p & 1 \\ \log a k^{q-1} & q & 1 \\ \log a k^{r-1} & r & 1\end{vmatrix} $

$ \begin{aligned} & = \begin{vmatrix} \log a+(p-1) \log k & p & 1 \\ \log a+(q-1) \log k & q & 1 \\ \log a+(r-1) \log k & r & 1 \end{vmatrix} \\ & = \begin{vmatrix} \log a & p & 1 \\ \log a & q & 1 \\ \log a & r & 1 \end{vmatrix} + \begin{vmatrix} (p-1) \log k & p & 1 \\ (q-1) \log k & q & 1 \\ (r-1) \log k & r & 1 \end{vmatrix} \end{aligned} $

Taking out $\log a$ common from $C_1$ of first determinant and $\log k$ common from $C_1$ of second determinant we have

$ \begin{aligned} & \Delta=\log a \begin{vmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{vmatrix} +\log k \begin{vmatrix} (p-1) & p & 1 \\ (q-1) & q & 1 \\ (r-1) & r & 1 \end{vmatrix} \\ & =\underset{(\begin{matrix} \text{ Two columns } \\ \text{ are identical } \end{matrix} )}{\log a \times 0}+\log k \begin{vmatrix} p-1-p+1 & p & 1 \\ q-1-q+1 & q & 1 \\ r-1-r+1 & r & 1 \end{vmatrix} \end{aligned} $

Using $C_1 \to C_1-C_2+C_3$ in second determinant.

$ =\log k \begin{vmatrix} 0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{vmatrix} =\mathbf{0} . $

~~ 42 Given, $A^{2}+A-I=O$

Premultiplying by $A^{-1}$ on both the sides, we have

$ \begin{aligned} & A^{-1} A^{2}+A^{-1} A-A^{-1} I=A^{-1} O \\ & \Rightarrow(A^{-1} A) A+I-A^{-1}=O \quad(\because A^{-1} A=I, A^{-1} O=O) \\ & \Rightarrow I A+I-A^{-1}=O \\ & \Rightarrow A+I-A^{-1}=O \\ & (\because I A=A) \\ & \Rightarrow A+I=O+A^{-1} \\ & \Rightarrow A+I=A^{-1} \\ & (\because O+A^{-1}=A^{-1}) \end{aligned} $

~~ 43 $(A B^{-1} C)^{-1}$

$ \begin{matrix} =[(A B^{-1}) C]^{-1}=C^{-1} \cdot(A B^{-1})^{-1} & (\because(X Y)^{-1}=Y^{-1} X^{-1}) \\ =C^{-1} \cdot(B^{-1})^{-1} A^{-1} & \\ =\boldsymbol{C}^{-1} \boldsymbol{B} \boldsymbol{A}^{-1} . & (\because(X^{-1})^{-1}=X) \end{matrix} $

~~ 44 We know that $(A^{-1})^{-1}=A$

$ |A^{-1}|=2 \neq 0 \text{, Hence }(A^{-1})^{-1} \text{ exists. } $

Now, we find the cofactors of the elements of the matrix $A^{-1}$.

Let $B=A^{-1}= \begin{vmatrix} 1 & 0 \\ -1 & 2\end{vmatrix} $. Then,

$ \begin{aligned} & B _{11}=2, B _{12}=-(-1)=1, B _{21}=-(0)=0, B _{22}=1 \\ \therefore & adj B=adj(A^{-1})= \begin{vmatrix} B _{11} & B _{21} \\ B _{12} & B _{22} \end{vmatrix} = \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} \\ \therefore & (A^{-1})^{-1}=\frac{1}{2} \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 1 / 2 & 1 / 2 \end{vmatrix} \\ \therefore & A=(A^{-1})^{-1} \\ \Rightarrow & \begin{vmatrix} 2 x & 0 \\ x & x \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 1 / 2 & 1 / 2 \end{vmatrix} \Rightarrow \boldsymbol{x}=\frac{\mathbf{1}}{\mathbf{2}} . \end{aligned} $

~~ 45 The given system of equations can be written as:

$ \begin{matrix} x+a y+0 . z=0 \\ 0 . x+y+a \cdot z=0 \\ a x+0 . y+z=0 \end{matrix} $

These in matrix form can be written as: $A X=O$, where,

$ A= \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } O= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $

Now, the given system of equations will have an infinite number of solutions, if $|A|=0$.

$ \begin{gathered} |A|= \begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} =1 \begin{vmatrix} 1 & a \\ 0 & 1 \end{vmatrix} -a \begin{vmatrix} 0 & a \\ a & 1 \end{vmatrix} +0 \begin{vmatrix} 0 & 1 \\ a & 0 \end{vmatrix} \\ \quad(\text{ Expanding along } R_1) \\ =1 \times 1-a(0-a^{2})+0=1+a^{3} \\ \therefore \quad|A|=0 \Rightarrow 1+a^{3}=0 \Rightarrow a^{3}=-1 \Rightarrow a=-1 . \end{gathered} $

~~ 46 The given system of linear equations are:

$ \begin{aligned} & x+2 a y+a z=0 \\ & x+3 b y+b z=0 \\ & x+4 c y+c z=0 \end{aligned} $

These can be written in the matrix form as $A X=O$, i.e.,

$ \begin{gathered} { \begin{bmatrix} 1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} } \\ \text{ Here } A= \begin{bmatrix} 1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c \end{bmatrix} , X \begin{bmatrix} x \\ y \\ z \end{bmatrix} . \end{gathered} $

The system will have a non-zero solution if $|A|=0$

$ \begin{gathered} |A|= \begin{vmatrix} 1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c \end{vmatrix} =1 \begin{vmatrix} 3 b & b \\ 4 c & c \end{vmatrix} -1 \begin{vmatrix} 2 a & a \\ 4 c & c \end{vmatrix} +1 \begin{vmatrix} 2 a & a \\ 3 b & b \end{vmatrix} \\ =3 b c-4 c b-(2 a c-4 a c)+2 a b-3 b a \\ =-b c+2 a c-a b \\ \therefore \quad|A|=0 \Rightarrow-b c+2 a c-a b=0 \Rightarrow 2 a c=a b+b c \\ \Rightarrow \frac{2}{b}=\frac{1}{c}+\frac{1}{a} \Rightarrow a, b, c \text{ are in H.P. } \end{gathered} $

~~ 47 The system of equations is:

$ \begin{aligned} & 3 x-2 y=5 \\ & 6 x-4 y=9 \end{aligned} $

Writing the system of equations in matrix form, we have

$ \begin{bmatrix} 3 & -2 \\ 6 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 9 \end{bmatrix} $

A $\quad X=B$

where $A= \begin{bmatrix} 3 & -2 \\ 6 & -4\end{bmatrix} , X= \begin{bmatrix} x \\ y\end{bmatrix} , B= \begin{bmatrix} 5 \\ 9\end{bmatrix} $

To check the consistency of system of equations, find $|A|$

$ |A|=-12-(-12)=0 $

$\Rightarrow$ Either the system of equations has infinitely many solutions or no solution.

Now we find $(adj . A) B$

$ \begin{aligned} \text{ adj. } A & = \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} \\ A _{11} & =-4, A _{12}=-6, A _{21}=2, A _{22}=3 \\ \therefore \quad(adj A) B & = \begin{bmatrix} -4 & 2 \\ -6 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ 9 \end{bmatrix} = \begin{bmatrix} -20+18 \\ -30+27 \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \end{bmatrix} \neq 0 \end{aligned} $

$\Rightarrow$ No solution (inconsistent).

~~ 48 The system of equations is:

$ \begin{aligned} x+5 y & =3 \\ 2 x+10 y & =6 \end{aligned} $

Writing in matrix form, we have

$ \begin{aligned} A X & =B \Rightarrow \begin{bmatrix} 1 & 5 \\ 2 & 10 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 6 \end{bmatrix} \\ A & = \begin{bmatrix} 1 & 5 \\ 2 & 10 \end{bmatrix} , X= \begin{bmatrix} x \\ y \end{bmatrix} , B= \begin{bmatrix} 3 \\ 6 \end{bmatrix} \\ |A| & =10-10=0 \end{aligned} $

$\Rightarrow$ Either the system of equations have no solution or infinitely many solutions.

$ \begin{aligned} adj A & = \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} 10 & -5 \\ -2 & 1 \end{bmatrix} \\ (adj . A) B & = \begin{bmatrix} 10 & -5 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 6 \end{bmatrix} = \begin{bmatrix} 30-30 \\ -6+6 \end{bmatrix} =0 \end{aligned} $

$\therefore$ The system of equations has infinitely many solutions.

~~ 49 The system of equations is:

$ \begin{aligned} & 5 x+3 y+z=16 \\ & 2 x+y+3 z=19 \\ & x+2 y+4 z=25 \end{aligned} $

Here, $D= \begin{vmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{vmatrix} $

$ \begin{aligned} & =5 \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} -3 \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} +1 \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} \\ & =5(4-6)-3(8-3)+1(4-1) \\ & =5 \times-2-3 \times 5+1 \times 3=-10-15+3=-22 \neq 0 \end{aligned} $

$\because \quad D \neq 0$, the system has a unique solution.

Note: $x=\frac{D x}{D}, y=\frac{D y}{D}, z=\frac{D z}{D}$, where

$ D x= \begin{vmatrix} 16 & 3 & 1 \\ 19 & 1 & 3 \\ 25 & 2 & 4 \end{vmatrix} , D y \begin{vmatrix} 5 & 16 & 1 \\ 2 & 19 & 3 \\ 1 & 25 & 4 \end{vmatrix} , D z= \begin{vmatrix} 5 & 3 & 16 \\ 2 & 1 & 19 \\ 1 & 2 & 25 \end{vmatrix} $

~~ 50 Here,

$ \begin{aligned} P=[a _{i j}] _{3 \times 3} & = \begin{bmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33} \end{bmatrix} \\ Q=[b _{i j}] _{3 \times 3} & = \begin{bmatrix} b _{11} & b _{12} & b _{13} \\ b _{21} & b _{22} & b _{23} \\ b _{31} & b _{32} & b _{33} \end{bmatrix} \text{ where } b _{i j}=2^{i+j} a _{i j} \\ \therefore \quad|Q| & = \begin{vmatrix} 2^{2} a _{11} & 2^{3} a _{12} & 2^{4} a _{13} \\ 2^{3} a _{21} & 2^{4} a _{22} & 2^{5} a _{23} \\ 2^{4} a _{31} & 2^{5} a _{32} & 2^{6} a _{33} \end{vmatrix} \\ & =2^{2} \cdot 2^{3} \cdot 2^{4} \begin{vmatrix} a _{11} & 2 a _{12} & 4 a _{13} \\ a _{21} & 2 a _{22} & 4 a _{23} \\ a _{31} & 2 a _{32} & 4 a _{33} \end{vmatrix} \\ & =2^{9} \times 2 \times 4 \begin{vmatrix} a _{11} & a _{12} & a _{13} \\ a _{21} & a _{22} & a _{23} \\ a _{31} & a _{32} & a _{33} \end{vmatrix} \\ & =2^{12} .|P|=2^{12} .2=\mathbf{2}^{\mathbf{1 3}} . \end{aligned} $

SELF ASSESSMENT SHEET

~~ 1. If $A= \begin{bmatrix} 4 & 3 \\ 2 & 5\end{bmatrix} $, find $x$ and $y$ such that $A^{2}-x A+y I=0$

(a) $x=9, y=-14$

(b) $x=14, y=9$

(d) $x=-9, y=14$ ~~ 2. If $A$ and $B$ are two square matrices of the same order such that $A B=B$ and $B A=A$, then $A^{2}+B^{2}$ is always equal to

(a) $I$

(b) $A+B$

(d) $2 B A$ ~~ 3. If $ \begin{bmatrix} a & -1 & 5 \\ b & 3 & -4\end{bmatrix} \begin{bmatrix} 4 & 3 \\ 1 & 0 \\ 7 & -6\end{bmatrix} = \begin{bmatrix} 34 & -30 \\ -1 & 42\end{bmatrix} $, then $(a, b)$ is equal to

(a) $(1,3)$

(b) $(-2,4)$

(d) $(2,-3)$ ~~ 4. If $A= \begin{bmatrix} 0 & -3 & -4 / 3 \\ 3 & 0 & -1 / 4 \\ 4 / 3 & 1 / 4 & 0\end{bmatrix} $, then $det(A+A^{T})$ is equal to

(a) 0

(b) 1

(d) 3 ~~ 5. If every element of a determinant of order 3 of value $\Delta$ is multiplied by 5 , then the value of the new determinant is

(a) $\Delta$

(b) $5 \Delta$

(d) $125 \Delta$ ~~ 6. If $M= \begin{vmatrix} a & l & p \\ b & m & q \\ c & n & r\end{vmatrix} $ and $N= \begin{vmatrix} p & q & r \\ a & b & c \\ l & m & n\end{vmatrix} $, then

(a) $M^{\prime}=N$

(b) $M=-N$

(d) $M=N^{3}$ ~~ 7. If $ \begin{vmatrix} -a^{2} & a b & a c \\ a b & -b^{2} & b c \\ a c & b c & -c^{2}\end{vmatrix} =k a^{2} b^{2} c^{2}$, then $k$ is equal to

(a) -4

(b) 2

(d) 8 ~~ 8. The matrix $ \begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{bmatrix} $ is invertible if

(a) $\lambda \neq-20$

(b) $\lambda \neq-19$

(d) $\lambda \neq-17$ ~~ 9. The element in the first row and third column of the inverse of the matrix $ \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix} $ is

(a) -2

(b) 0

(d) 7 ~~ 10. The system of linear equations $2 x+2 y=5,5 x+k y=9$ has a unique solution if

(a) $k \neq 5$

(b) $k=0$

(d) $k \neq 2$ ~~ 11. The system of equations $3 x+4 y=2$ and $6 x+8 y=4$ has

(a) No solution

(b) a unique solution

(d) infinitely many solutions. ~~ 12. The system of equations $x+3 y=5,2 x+6 y=8$ has

(a) No solution

(b) a unique solution

(d) None of these

ANSWERS

~~ 1. (c) ~~ 2. (b) ~~ 3. (c) ~~ 4. $(a)$ ~~ 5. $(d)$ ~~ 6. (a) ~~ 7. (c) ~~ 8. (d) ~~ 9. $(d)$ ~~ 10. (a) ~~ 11. $(d)$ ~~ 12. (a)

HINTS AND SOLUTIONS

$ \text{ 1. } \begin{aligned} A= \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \Rightarrow A^{2} & =A \cdot A= \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} \\ & = \begin{bmatrix} 16+6 & 12+15 \\ 8+10 & 6+25 \end{bmatrix} = \begin{bmatrix} 22 & 27 \\ 18 & 31 \end{bmatrix} \end{aligned} $

Given $A^{2}-x A+y I=0 \Rightarrow A^{2}-x A=-y I$

$\Rightarrow \begin{bmatrix} 22 & 27 \\ 18 & 31\end{bmatrix} -x \begin{bmatrix} 4 & 3 \\ 2 & 5\end{bmatrix} =-y \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $

$\Rightarrow \begin{bmatrix} 22-4 x & 27-3 x \\ 18-2 x & 31-5 x\end{bmatrix} = \begin{bmatrix} -y & 0 \\ 0 & -y\end{bmatrix} $

$\Rightarrow 22-4 x=-y, 27-3 x=0$

Now $27-3 x=0 \Rightarrow 3 x=27 \Rightarrow \boldsymbol{x}=\mathbf{9}$

$\therefore 22-4 x=-y \Rightarrow 22-36=-y$

$ \Rightarrow-y=-14 \Rightarrow y=14 $

~~ 2. We know that if $A B=B$ and $B A=A$, then $A$ and $B$ are idempotent matrices, i.e, $A^{2}=A$ and $B^{2}=B$.

$ A^{2}+B^{2}=A+B $

Alternatively,

$ \begin{aligned} A^{2}+B^{2} & =A \cdot A+B \cdot B=A \cdot(B A)+B \cdot(A B) \\ & =(A B) A+(B A) B \end{aligned} $

(Matrix multiplication is associative)

$ \begin{matrix} =B A+A B & (\because A B=B, B A=A) \\ =\boldsymbol{A}+\boldsymbol{B} & (\because B A=A, A B=B) \end{matrix} $

~~ 3. $ \begin{bmatrix} a & -1 & 5 \\ b & 3 & -4\end{bmatrix} _{2 \times 3} \begin{bmatrix} 4 & 3 \\ 1 & 0 \\ 7 & -6\end{bmatrix} _{3 \times 2}= \begin{bmatrix} 34 & -30 \\ -1 & 42\end{bmatrix} _{2 \times 2}$

$LHS= \begin{bmatrix} 4 a-1+35 & 3 a+0-30 \\ 4 b+3-28 & 3 b+0+24\end{bmatrix} = \begin{bmatrix} 4 a+34 & 3 a-30 \\ 4 b-25 & 3 b+24\end{bmatrix} $

RHS $= \begin{bmatrix} 34 & -30 \\ -1 & 42\end{bmatrix} $.

Equating corresponding elements we have

$4 a+34=34 \Rightarrow a=0 \quad 4 b-25=-1 \Rightarrow 4 b=24 \Rightarrow b=6$ $3 a-30=-30 \Rightarrow a=0$ and $3 b+24=42 \Rightarrow 3 b=18 \Rightarrow b=6$

$\therefore a=0, b=6$.

~~ 4. $A+A^{T}= \begin{bmatrix} 0 & -3 & -4 / 3 \\ 3 & 0 & -1 / 4 \\ 4 / 3 & 1 / 4 & 0\end{bmatrix} + \begin{bmatrix} 0 & 3 & 4 / 3 \\ -3 & 0 & 1 / 4 \\ -4 / 3 & -1 / 4 & 0\end{bmatrix} $

$ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $

$\therefore \quad det(A+A^{T})=0$.

~~ 5. Let $\Delta= \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix} $

Since each element of $\Delta$ is multiplied by 5 , the new determinant

$ \begin{aligned} & \Delta^{\prime}= \begin{vmatrix} 5 a_1 & 5 a_2 & 5 a_3 \\ 5 b_1 & 5 b_2 & 5 b_3 \\ 5 c_1 & 5 c_2 & 5 c_3 \end{vmatrix} \\ & =(5 \times 5 \times 5) \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \end{aligned} $

(Taking out 5, common from each $=125 \Delta$.

of $R_1, R_2, R_3$ or $C_1, C_2, C_3$ )

~~ 6. $M= \begin{vmatrix} a & l & p \\ b & m & q \\ c & n & r\end{vmatrix} \Rightarrow M^{\prime}= \begin{vmatrix} a & b & c \\ l & m & n \\ p & q & r\end{vmatrix} $

(Interchanging $R_1$ and $R_3$ )

$=- \begin{vmatrix} p & q & r \\ l & m & n \\ a & b & c\end{vmatrix} = \begin{vmatrix} p & q & r \\ a & b & c \\ l & m & n\end{vmatrix} (.$ Interchanging $R_2$ and $.R_3)$

$=N$.

~~ 7. Let $\Delta= \begin{vmatrix} -a^{2} & a b & a c \\ a b & -b^{2} & b c \\ a c & b c & -c^{2}\end{vmatrix} =a b c \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c\end{vmatrix} $

(Taking out $a, b, c$ common from $R_1, R_2, R_3$ respectively)

$ =a^{2} b^{2} c^{2} \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} $

(Taking out $a, b, c$ common from $C_1, C_2, C_3$ respectively)

$ \begin{aligned} & =a^{2} b^{2} c^{2} \begin{vmatrix} 0 & 1 & 1 \\ 0 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \quad(\text{ Operating } C_1 \to C_1+C_2) \\ & =a^{2} b^{2} c^{2} \times 2 \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} =a^{2} b^{2} c^{2} \times 2(1+1) \end{aligned} $

$ =4 a^{2} b^{2} c^{2} $

$ \therefore \quad k=4 $

(Expanding along $C_1$ )

~~ 8. Let $A= \begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{bmatrix} $, then det. $A=|A|= \begin{vmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{vmatrix} $

The matrix $A$ is invertible if $|A| \neq 0$

$ \begin{aligned} |A| & =\lambda \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} -(-1) \begin{vmatrix} -3 & 1 \\ -1 & 2 \end{vmatrix} +4 \begin{vmatrix} -3 & 0 \\ -1 & 1 \end{vmatrix} \\ & =\lambda(0-1)+1(-6+1)+4(-3-0) \\ & =-\lambda-5-12=-\lambda-17 \\ \therefore & |A| \neq 0 \Rightarrow-\lambda-17 \neq 0 \Rightarrow \lambda \neq-\mathbf{1 7 .} \end{aligned} $

~~ 9. Let $A= \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{bmatrix} $, then $|A|= \begin{vmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{vmatrix} $

$=1 \begin{vmatrix} 1 & 2 \\ 0 & 1\end{vmatrix} =1 \neq 0 \Rightarrow A^{-1}$ exists.

Now we find the cofactor matrix of $A$. So,

$A _{11}= \begin{vmatrix} 1 & 2 \\ 0 & 1\end{vmatrix} =1, A _{12}=- \begin{vmatrix} 0 & 2 \\ 0 & 1\end{vmatrix} =0, A _{13}= \begin{vmatrix} 0 & 1 \\ 0 & 0\end{vmatrix} =0$

$A _{21}=- \begin{vmatrix} 2 & -3 \\ 0 & 1\end{vmatrix} =-2, A _{22}= \begin{vmatrix} 1 & -3 \\ 0 & 1\end{vmatrix} =1, A _{23}=- \begin{vmatrix} 1 & 2 \\ 0 & 0\end{vmatrix} =0$

$A _{31}= \begin{vmatrix} 2 & -3 \\ 1 & 2\end{vmatrix} =7, A _{32}=- \begin{vmatrix} 1 & -3 \\ 0 & 2\end{vmatrix} =-2, A _{33}= \begin{vmatrix} 1 & 2 \\ 0 & 1\end{vmatrix} =1$

$\therefore \quad adj A= \begin{bmatrix} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33}\end{bmatrix} ^{T}$

$= \begin{bmatrix} A _{11} & A _{21} & A _{31} \\ A _{12} & A _{22} & A _{32} \\ A _{13} & A _{23} & A _{33}\end{bmatrix} = \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{bmatrix} $

$\therefore \quad A^{-1}=\frac{1}{|A|}$ adj $A= \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1\end{bmatrix} $

$\therefore$ The element in the first row and third column of $A^{-1}$ is 7 .

~~ 10. The given system of equations can be written in the matrix form as:

Now there will be a unique solution for $A X=B$ if $|A| \neq 0$ i.e, $2 k-10 \neq 0$

$\Rightarrow k \neq 5$

~~ 11. The given system of equations can be written in the matrix form as:

$ \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} $

Now, $|A|=24-24=0$

Hence the given system of equations either have infinite solutions or no solution.

Now we find $(adj A) B$.

Cofactors of A are:

$ \begin{matrix} A _{11}=8, A _{12}=-6, A _{21}=-4, A _{22}=3 \\ \therefore \quad \text{ adj } A= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ -6 & 3 \end{bmatrix} \end{matrix} $

$(adj A) B= \begin{bmatrix} 8 & -4 \\ -6 & 3\end{bmatrix} \begin{bmatrix} 2 \\ 4\end{bmatrix} = \begin{bmatrix} 16-16 \\ -12+12\end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix} =0$

Hence $(adj A) B=0 \Rightarrow$ Infinitely many solutions.

~~ 12..

$\begin{matrix} x+3 y=5 \\ 2 x+6 y=8\end{matrix}\biggl ] $ $\to $ $\begin{bmatrix}1 & 3 \\ 2 & 6\end{bmatrix}$ $\begin{bmatrix} x \\ y\end{bmatrix}$ = $\begin{bmatrix} 5 \\ 8\end{bmatrix}$ $\quad$ (in matrix form)

$ |A|=6-6=0 $

$\Rightarrow$ Either the system of equations has no solution or infinite solutions.

$ \begin{aligned} & adj A= \begin{bmatrix} A _{11} & A _{21} \\ A _{12} & A _{22} \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \\ & \therefore \quad(adj A) B= \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq 0 \end{aligned} $

Hence, the given system of equations is inconsistent with no solution.

Binomial Theorem

KEY FACTS

1. Expansion of a Binomial

If $\boldsymbol{x}$ and $\boldsymbol{a}$ are real numbers, then for all $n \in N$,

$ \begin{aligned} \quad(x+a)^{n} & ={ }^{n} C_0 x^{n} a^{0}+{ }^{n} C_1 x^{n-1} a^{1}+{ }^{n} C_2 x^{n-2} a^{2}+\ldots \ldots+{ }^{n} C_r x^{n-r} a^{r}+\ldots \ldots+{ }^{n} C _{n-1} x^{1} a^{n-1}+{ }^{n} C_n x^{0} a^{n} \\ \Rightarrow \quad(x+a)^{n} & =\sum _{r=0}^{n}{ }^{n} C_r x^{n-r} a^{r} . \end{aligned} $

2. Properties of Binomial Expansion

(a) The total number of terms in the expansion of $(x+a)^{n}$ is $\boldsymbol{n}+\mathbf{1}$.

(b) The sum of the indices of $\boldsymbol{x}$ and $\boldsymbol{a}$ in each term is $\boldsymbol{n}$.

Thus, the $(r+1)$ th term $={ }^{n} C_r x^{n-r} a^{r}$ and sum of indices $=(n-r+r)=\boldsymbol{n}$.

(c) The coefficients of terms equidistant from the beginning and end are equal. These coefficients are known as binomial coefficients.

$ { }^{n} C_r={ }^{n} C _{n-r} \forall r=0,1,2, \ldots \ldots, n \Rightarrow{ }^{n} C_0={ }^{n} C_n,{ }^{n} C_1={ }^{n} C _{n-1},{ }^{n} C_2={ }^{n} C _{n-2} \text{ and so on. } $

(d) Replacing $\boldsymbol{a}$ by $-\boldsymbol{a}$ in the expansion of $(x+a)^{n}$, we have

$ \begin{aligned} \quad(x-a)^{n} & ={ }^{n} C_0 x^{n} a^{0}-{ }^{n} C_1 x^{n-1} a^{1}+{ }^{n} C_2 x^{n-2} a^{2}-{ }^{n} C_3 x^{n-3} a^{3}+\ldots \ldots .+(-1)^{r}{ }^{n} C_r x^{n-r} a^{r} \ldots \ldots+(-1)^{r} C_n x^{0} a^{n} . \\ \Rightarrow \quad(x-a)^{n} & =\sum _{r=0}^{n}(-\mathbf{1})^{r}{ }^{n} C_r \cdot \boldsymbol{x}^{n-r} \boldsymbol{a}^{r} \end{aligned} $

(e) $(x+a)^{n}+(x-a)^{n}=2({ }^{n} C_0 x^{n} a^{0}+{ }^{n} C_2 x^{n-2} a^{2}+\ldots \ldots \ldots.)$

$(x+a)^{n}-(x-a)^{n}=2({ }^{n} C_1 x^{n-1} a^{1}+{ }^{n} C_3 x^{n-3} a^{3}+\ldots \ldots \ldots ..)$

( $f$ ) The $(r+1)$ th term called the general term in the expansion of $(x+a)^{n}$ is given by $T _{r+1}={ }^{n} C_r x^{n-r} a^{r}$. In the expansion of $(x-a)^{n}, T _{r+1}=(-1)^{r}{ }^{n} C_r x^{n-r} a^{r}$.

(g) Middle Term: In the expansion of $(x+a)^{n}$, if

$\boldsymbol{n}$ is even natural number, then middle term $=(\frac{n}{2}+1)$ th term
$\boldsymbol{n}$ is odd natural number, then $(\frac{n+1}{2})$ th term and $(\frac{n+3}{2})$ th term are the two middle terms.

(h) Term from the end: In the expansion of $(x+a)^{n}$, the $r$ th term from the end is

$=[(n+1)-(r-1)]$ th term from the beginning

$=(\boldsymbol{n}-\boldsymbol{r}+\mathbf{2})$ th term from the beginning.

(i) The sum of the binomial coefficients $=2^{n}$, i.e.,

$ C_0+C_1+C_2+\ldots \ldots+C_n=2^{n} $

(j) $C_1+C_3+C_5+\ldots \ldots=C_0+C_2+C_4+\ldots \ldots=2^{n-1}$

General Term and Expansion of Binomial Theorem

SOLVED EXAMPLES

Ex. 1 . Expand $(2 x+\frac{y}{2})^{5}$.

Sol. $\quad(2 x+\frac{y}{2})^{5}={ }^{5} C_0(2 x)^{5}+{ }^{5} C_1(2 x)^{4}(\frac{y}{2})+{ }^{5} C_2(2 x)^{3}(\frac{y}{2})^{2}+{ }^{5} C_3(2 x)^{2}(\frac{y}{2})^{3}+{ }^{5} C_4(2 x)(\frac{y}{2})^{4}+{ }^{5} C_5(\frac{y}{2})^{5}$

$ \begin{aligned} & =32 x^{5}+5 \times 16 x^{4} \times \frac{y}{2}+\frac{5 \times 4}{2 \times 1} \times 8 x^{3} \times \frac{y^{2}}{4}+\frac{5 \times 4}{2 \times 1} \times 4 x^{2} \times \frac{y^{3}}{8}+5 \times 2 x \times \frac{y^{4}}{16}+\frac{y^{5}}{32} \\ & =\mathbf{3 2} \boldsymbol{x}^{\mathbf{5}}+\mathbf{4 0} \boldsymbol{x}^{\mathbf{4}} \boldsymbol{y}+\mathbf{2 0} \boldsymbol{x}^{\mathbf{3}} \boldsymbol{y}^{\mathbf{2}}+\mathbf{5} \boldsymbol{x}^{\mathbf{2}} \boldsymbol{y}^{\mathbf{3}}+\frac{\mathbf{5}}{\mathbf{8}} \boldsymbol{x} \boldsymbol{y}^{\mathbf{4}}+\frac{\boldsymbol{y}^{\mathbf{5}}}{\mathbf{3 2}} . \end{aligned} $

Ex. 2 . Expand $(\frac{2}{x}-\frac{x}{2})^{6}, x \neq 0$.

Sol.

$ \begin{aligned} (\frac{2}{x}-\frac{x}{2})^{6} & ={ }^{6} C_0(\frac{2}{x})^{6}+{ }^{6} C_1(\frac{2}{x})^{5}(-\frac{x}{2})+{ }^{6} C_2(\frac{2}{x})^{4}(-\frac{x}{2})^{2}+{ }^{6} C_3(\frac{2}{x})^{3}(-\frac{x}{2})^{3}+{ }^{6} C_4(\frac{2}{x})^{2}(-\frac{x}{2})^{4} \\ & +{ }^{6} C_5(\frac{2}{x})(-\frac{x}{2})^{5}+{ }^{6} C_6(-\frac{x}{2})^{6} \\ & \frac{64}{x^{6}}+6 \times \frac{32}{x^{5}} \times(-\frac{x}{2})+\frac{6 \times 5}{2 \times 1} \times \frac{16}{x^{4}} \times \frac{x^{2}}{4}+\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{8}{x^{3}} \times(-\frac{x^{3}}{8})+\frac{6 \times 5}{2 \times 1} \times \frac{4}{x^{2}} \times \frac{x^{4}}{16} \\ & =\frac{\mathbf{6 4}}{\boldsymbol{x}^{6}}-\frac{\mathbf{9 6}}{\boldsymbol{x}^{4}}+\frac{\mathbf{6 0}}{\boldsymbol{x}^{\mathbf{2}}}-\mathbf{2 0}+\frac{\mathbf{1 5}}{\mathbf{4}} \boldsymbol{x}^{\mathbf{2}}-\frac{\mathbf{3}}{\mathbf{8}} \boldsymbol{x}^{4}+\frac{\boldsymbol{x}^{6}}{\mathbf{6 4}} . \end{aligned} $

Ex. 3 . Find the coefficient of $x^{5}$ in the expansion of $(1+2 x)^{6}(1-x)^{7}$.

Sol. $\quad(1+2 x)^{6}(1-x)^{7}=[1+{ }^{6} C_1(2 x)+{ }^{6} C_2(2 x)^{2}+{ }^{6} C_3(2 x)^{3}+{ }^{6} C_4(2 x)^{4}+{ }^{6} C_5(2 x)^{5}+(2 x)^{6}] \times[1-{ }^{7} C_1 x+{ }^{7} C_2 x^{2}.$

$ .-{ }^{7} C_3 x^{3}+{ }^{7} C_4 x^{4}-{ }^{7} C_5 x^{5}+{ }^{7} C_6 x^{6}-x^{7}] $

$ =[1+12 x+60 x^{2}+160 x^{3}+240 x^{4}+192 x^{5}+64 x^{6}] \times[1-7 x+21 x^{2}-35 x^{3}+35 x^{4}-21 x^{5}+7 x^{6}-x^{7}] $

The terms containing $x^{5}$ in the product are obtained an multiplying constant term by the term containing $x^{5}$, term containing $x$ by the term containing $x^{4}$ and so on.

These products are $(1 \times(-21 x^{5}))+(12 x \times 35 x^{4})+(60 x^{2} \times(-35 x^{3}))+(160 x^{3} \times 21 x^{2})+(240 x^{4} \times(-7 x))+(192 x^{5} \times 1)$

$ =-21 x^{5}+420 x^{5}-2100 x^{5}+3360 x^{5}-1680 x^{5}+192 x^{5} $

$\therefore$ Coefficient of $x^{5}=(-21+420-2100+3360-1680+192)=\mathbf{1 7 1}$.

Ex. 4 . Find the 6th term in the expansion of $(2 x^{2}-\frac{1}{3 x^{2}})^{10}$.

(Karnataka CET 2007)

Sol. The general term $((n+1)$ th term $)$ in the expansion of $(2 x^{2}-\frac{1}{3 x^{2}})^{10}$ is

$ T _{r+1}=(-1)^{r}{ }^{10} C_r(2 x^{2})^{10-r}(\frac{1}{3 x^{2}})^{r}=(-1)^{r}{ }^{10} C_r 2^{10-r} x^{20-2 r}(3)^{-r} x^{-2 r}=(-1)^{r}{ }^{10} C_r 2^{10-r}(3)^{-r} x^{20-4 r} $

$ \therefore \quad T_6=T _{5+1}=(-1)^{510} C_5 2^{10-5}(3)^{-5} x^{20-20}=(-1) \times \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{2^{5}}{3^{5}} \times x^{0}=(-1) \times 252 \times \frac{32}{243}=-\frac{\mathbf{8 9 6}}{\mathbf{2 4 3}} . $

Ex. 5 . Find the coefficient of $x^{20}$ in the expansion of $(1+3 x+3 x^{2}+x^{3})^{20}$

(DCE 2007)

Sol. $(1+3 x+3 x^{3}+x^{3})^{20}=((1+x^{3}))^{20}=(1+x)^{60}$

$\because \quad(1+x)^{n}={ }^{n} C_0+{ }^{n} C_1 x+{ }^{n} C_2 x^{2}+\ldots \ldots \ldots+{ }^{n} C_n x^{n}$,

The coefficient of $x^{20}$ in the expansion of $(1+x)^{60}={ }^{60} C _{20}$ or ${ }^{60} C _{40} \quad(\because{ }^{n} C_r={ }^{n} C _{n-r})$

Ex. 6 . If the coefficients of $(m+1)$ th term and $(m+3)$ th term in the expansion of $(1+x)^{20}$ are equal, then find the value of $m$ ?

(UPSEE 2007)

Sol.

$ \text{ General term } \quad T _{r+1}={ }^{20} C_r x^{r} \therefore \Rightarrow T _{m+1}={ }^{20} C_m x^{m} $

$ T _{m+3}={ }^{20} C _{m+2} x^{m+2} $

Given, coeff. of $(m+1)$ th term $=$ coeff. of $(m+3)$ th term

$ \begin{matrix} \Rightarrow & { }^{20} C_m={ }^{20} C _{m+2} \Rightarrow{ }^{20} C _{20-m}={ }^{20} C _{m+2} \\ \Rightarrow & 20-m=m+2 \Rightarrow 2 m=18 \Rightarrow m=\mathbf{9} . \end{matrix} \quad(\because{ }^{n} C_r={ }^{n} C _{n-r}) $

Ex. 7 . If second, third and fourth terms in the expansion of $(x+a)^{n}$ are 240,720 and 1080 respectively, then find the value of $n$.

Sol. $\quad(x+a)^{n}={ }^{n} C_0 x^{n}+{ }^{n} C_1 x^{n-1} a+{ }^{n} C_2 x^{n-2} a^{2}+{ }^{n} C_3 x^{n-3} a^{3}+$ $+{ }^{n} C_n a^{n}$

$\therefore \quad T_2={ }^{n} C_1 x^{n-1} a=240$

$T_3={ }^{n} C_2 x^{n-2} a^{2}=720$

$ T_4={ }^{n} C_3 x^{n-3} a^{3}=1080 $

On dividing (ii) by (i), we get $\frac{{ }^{n} C_2 x^{n-2} a^{2}}{{ }^{n} C_1 x^{n-1} a}=\frac{720}{240} \Rightarrow \frac{\frac{n(n-1)}{2 \times 1} \times a}{n \times x}=3 \Rightarrow \frac{(n-1) a}{2 x}=3$

On dividing (iii) by (ii) we get $\frac{{ }^{n} C_3 x^{n-3} a^{3}}{{ }^{n} C_2 x^{n-2} a^{2}}=\frac{1080}{720} \Rightarrow \frac{\frac{n(n-1)(n-2)}{3 \times 2 \times 1} \times a}{\frac{n(n-1)}{2} \times x}=\frac{3}{2} \Rightarrow \frac{(n-2) a}{(3 x)}=\frac{3}{2}$

Now dividing ( $v$ ) by (iv) we get $\frac{(\frac{n-2}{3}) \frac{a}{x}}{(\frac{n-1}{2}) \frac{a}{x}}=\frac{3 / 2}{3} \Rightarrow \frac{2(n-2)}{3(n-1)}=\frac{1}{2} \Rightarrow 4 n-8=3 n-3 \Rightarrow n=\mathbf{5}$.

Ex. 8 . What is the coefficient of $x^{4}$ in the expansion of $(1+x)^{11}+(1+x)^{12}+\ldots . .+(1+x)^{20}$ ?

Sol. $(1+x)^{11}+(1+x)^{12}+\ldots . .+(1+x)^{20}$

$ \begin{aligned} & =\frac{(1+x)^{11}{(1+x)^{10}-1}}{(1+x)-1} \quad \begin{bmatrix} \because \text{ This is the sum of a G.P. with } 10 \text{ terms whose } \\ \text{ first term }=(1+x)^{11} \text{ and common ratio }=(1+x) \end{bmatrix} \\ & =\frac{1}{x}[(1+x)^{21}-(1+x)^{11}] . \end{aligned} $

$\therefore$ Coeff. of $x^{4}$ in the expansion of $\frac{1}{x}[(1+x)^{21}-(1+x)^{11}]$

$ =\text{ Coeff. of } x^{5} \text{ in }[(1+x)^{21}-(1+x)^{11}]={ }^{\mathbf{2 1}} \mathbf{C} _5-{ }^{11} \mathbf{C} _{\mathbf{5}} \text{. } $

Ex. 9 . If the coefficients of the second, third and fourth terms in the expansion of $(1+x)^{2 n}$ are in A.P., show that $2 n^{2}-9 n+7=0$.

(AMU, IIT)

Sol.

$ \begin{aligned} (1+x)^{2 n} & ={ }^{2 n} C_0+{ }^{2 n} C_1 x+{ }^{2 n} C_2 x^{2}+\ldots . .+{ }^{2 n} C_n x^{n}+\ldots . .+{ }^{2 n} C _{2 n} x^{2 n} \\ T_2 & ={ }^{2 n} C_1 x, T_3={ }^{2 n} C_2 x^{2}, T_4={ }^{2 n} C_3 x^{3} \end{aligned} $

Given, coefficients of $T_2, T_3$ and $T_4$ are in A.P.

$ \begin{aligned} & \Rightarrow \quad 2 \cdot{ }^{2 n} C_2={ }^{2 n} C_1+{ }^{2 n} C_3 \Rightarrow 2 \cdot \frac{(2 n) !}{(2 n-2) ! 2 !}=\frac{(2 n) !}{(2 n-1) !}+\frac{(2 n) !}{(2 n-3) ! 3 !} \\ & \Rightarrow \quad(2 n)(2 n-1)=2 n+\frac{2 n(2 n-1)(2 n-2)}{6} \Rightarrow 6(2 n-1)=6+(2 n-1)(2 n-2) \\ & \Rightarrow \quad 12 n-6=6+4 n^{2}-6 n+2 \Rightarrow 4 n^{2}-18 n+14=0 \Rightarrow 2 n^{2}-9 n+7=0 . \end{aligned} $

PRACTICE SHEET-1
(General Term and Expansion of Binomial Theorem)

~~ 1. The ninth term in the expansion $(3 x-\frac{1}{2 x})^{8}$ is

(a) $\frac{1}{512 x^{9}}$

(b) $-\frac{1}{512 x^{9}}$

(d) $\frac{1}{256 x^{8}}$

(KCET 2007)

~~ 2. If the fourth term in the expansion of $(a x+\frac{1}{x})^{n}$ is $\frac{5}{2}$, then

(a) $a=\frac{1}{2}, n=6$

(b) $a=\frac{1}{3}, n=5$

(d) $a=\frac{1}{4}, n=1$

(J&K CET 2003, AMU 2013)

~~ 3. The two consecutive terms in the expansion of $(3+2 x)^{74}$, whose coefficients are equal are

(a) 11,12

(b) 7,8

(d) None of these

(Manipal Engg. 2009)

~~ 4. The coefficient of $x^{-10}$ in $(x^{2}-\frac{1}{x^{3}})^{10}$ is

(a) -252

(b) 210

(d) -120

(WB JEE 2009)

~~ 5. The coefficient of $x^{4}$ in the expansion of $(\frac{x}{2}-\frac{3}{x^{2}})^{10}$ is

(a) $\frac{405}{256}$

(b) $\frac{450}{263}$

(d) $\frac{540}{269}$

(RPET 2001)

~~ 6. If in the expansion of $(a-2 b)^{n}$, the sum of the $5^{\text{th }}$ and $6^{\text{th }}$ term is zero, then the value of $\frac{a}{b}$ is

(a) $\frac{n-4}{5}$

(b) $\frac{2(n-4)}{5}$

(d) $\frac{5}{2(n-4)}$

(BCECE 2009) ~~ 7. If $x^{-7}$ occurs in the $r$ th term of $(a x-\frac{b}{x^{2}})^{11}$, then the value of $r$ is

(a) 6

(b) 7

(d) 9

(MP PET 2011)

~~ 8. In the expansion of $(x-\frac{1}{x})^{6}$, the coefficient of $x^{0}$ is

(a) 20

(b) -20

(d) -30

(UPSEE 2009)

~~ 9. If the coefficients of $5^{\text{th }}, 6^{\text{th }}$ and $7^{\text{th }}$ terms in the expansion of $(1+x)^{n}$ are in A.P., then $\boldsymbol{n}$ equals

(a) 7

(b) 5

(d) 10

(UPSEAT 2000)

~~ 10. If $r$ th and $(r+1)$ th terms in the expansion of $(p+q)^{n}$ are equal, then $\frac{(n+1) q}{r(p+q)}$ is

(a) 0

(b) 1

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

(KCET 2011)

~~ 11. The first three terms in the expansion of $(1+a x)^{n}(n \neq 0)$ are $1,6 x$ and $16 x^{2}$. Then the value of $\boldsymbol{a}$ and $\boldsymbol{n}$ are respectively

(a) 2 and 9

(b) 3 and 2

(d) $3 / 2$ and 6 ~~ 12. In the expansion of $(2-3 x^{3})^{20}$, if the ratio of the 10 th term to the 11 th term is $45 / 22$, then $\boldsymbol{x}$ is equal to

(a) $-\frac{2}{3}$

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

(d) $-\sqrt[3]{\frac{3}{2}}$

(Odisha JEE 2012)

~~ 13. If the second term in the expansion of $[\sqrt[13]{a}+\frac{a}{\sqrt{a^{-1}}}]^{n}$ is $14 a^{5 / 2}$, then the value of $\frac{{ }^{n} C_3}{{ }^{n} C_2}$ is

(a) 4

(b) 3

(d) 6

(DCE 2006)

~~ 14. Let $t_n$ denote the $n$th term in a binomial expansion. If $\frac{t_6}{t_5}$ in the expansion of $(a+b)^{n+4}$ and $\frac{t_5}{t_4}$ in the expansion of $(a+b)^{n}$ are equal, then $n$ equals

(a) 9

(b) 11

(d) 15

(Kerala PET 2013)

~~ 15. If $T_r$ denotes the $r$ th term in the expansion of $(x+\frac{1}{x})^{23}$, then

(a) $T _{12}=x^{2} T _{13}$

(b) $x^{2}-T _{13}=T _{12}$

(d) $T _{12}+T _{13}=25$ ~~ 16. Let the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ be $\boldsymbol{p}$ and the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n-1}$ be $\boldsymbol{g}$, then

(a) $p=2 g$

(b) $2 p=3 g$

(d) $3 p=2 g$

(WBJEE 2011)

~~ 17. If $0 \leq r \leq n$, then the coefficient of $x^{r}$ in the expansion of $P=1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots . .+(1+x)^{n}$ is

(a) ${ }^{n} C_r$

(b) ${ }^{n} C _{r+1}$

(d) ${ }^{n+1} C _{r+1}$

(AMU 2007)

~~ 18. The coefficient of $\frac{1}{x}$ in the expansion of $(\frac{1}{x}+1)^{n}(1+x)^{n}$

(a) ${ }^{2 n} C_n$

(b) ${ }^{2 n} C _{n-1}$

(d) ${ }^{n} C _{n-1}$

(Odisha JEE 2004)

~~ 19. If the magnitude of the coefficient of $x^{7}$ in the expansion of $(a x^{2}+\frac{1}{b x})^{8}$, where $a, b$ are positive numbers is equal in magnitude of the coefficient of $x^{-7}$ in the expansion of $(a x-\frac{1}{b x^{2}})^{8}$, then $\boldsymbol{a}$ and $\boldsymbol{b}$ are connected by the relation:

(a) $a b=1$

(b) $a b=2$

(d) $a b^{2}=2$

(WBJEE 2008)

~~ 20. In the expansion of $(1+x+x^{2}+x^{3})^{6}$, the coefficient of $x^{14}$ is

(a) 130

(b) 120

(d) 125

(Kerala CEE 2007)

ANSWERS

~~ 1. (d) ~~ 2. (a) ~~ 3. (c) ~~ 4. (b) ~~ 5. (a) ~~ 6. (b) ~~ 7. (b) ~~ 8. (b) ~~ 9. (a) ~~ 10. $(b)$ ~~ 11. (c) ~~ 12. (a) ~~ 13. (a) ~~ 14. $(d)$ ~~ 15. (a) ~~ 16. $(a)$ ~~ 17. $(d)$ ~~ 18. (b) ~~ 19. (a) ~~ 20. (b)

HINTS AND SOLUTIONS

~~ 1. Given, $(3 x-\frac{1}{2 x})^{8}$

General term $=T _{r+1}={ }^{8} C_r(3 x)^{8-r}(-\frac{1}{2 x})^{r}$

$\therefore$ Ninth term $=T_9={ }^{8} C_8(3 x)^{8-8}(-\frac{1}{2 x})^{8}$

$($ Here $r+1=9 \Rightarrow r=8$ )

$ =\frac{1}{256 x^{8}} \text{. } $

~~ 2. Given, $(a x+\frac{1}{x})^{n}$

General term $=T _{r+1}={ }^{n} C_r(a x)^{n-r}(\frac{1}{x})^{r}$

$\therefore \quad T_4={ }^{n} C_3(a x)^{n-3}(\frac{1}{x})^{3}$

$ ={ }^{n} C_3 a^{n-3} \cdot x^{n-3} \cdot x^{-3}={ }^{n} C_3 a^{n-3} x^{n-6} $

Given, $\quad T_4=\frac{5}{2} \Rightarrow{ }^{n} C_3 a^{n-3} x^{n-6}=\frac{5}{2}$

Clearly, the fourth term does not contain $x$, so power of $x=0$.

$ \begin{matrix} \therefore & n-6=0 \Rightarrow n=6 \\ \Rightarrow & { }^{6} C_3 a^{6-3}=\frac{5}{2} \end{matrix} $

(From (i))

$ \begin{aligned} & \Rightarrow \quad \frac{6}{3} a^{3}=\frac{5}{2} \Rightarrow \frac{6 \times 5 \times 4}{3 \times 2} a^{3}=\frac{5}{2} \\ & \Rightarrow \quad a^{3}=\frac{5}{5 \times 4 \times 2}=\frac{1}{8} \Rightarrow a=\frac{\mathbf{1}}{\mathbf{2}} \\ & \therefore \quad a=\frac{1}{2}, n=6 . \end{aligned} $

~~ 3. General term of $(3+2 x)^{74}$ is $T _{r+1}={ }^{74} C_r(3)^{74-r}(2 x)^{r}$ $={ }^{74} C_r(3)^{74-r} 2^{r} \cdot x^{r}$

Let the two consecutive terms be $(r+1)$ th term and $(r+2)$ th term.

Then, $T _{r+2}={ }^{74} C _{r+1}(3)^{74-(r+1)} 2^{r+1} x^{r+1}$

Given, Coefficient of $T _{r+1}=$ Coefficient of $T _{r+2}$

$\Rightarrow{ }^{74} C_r(3)^{74-r} \cdot 2^{r}={ }^{74} C _{r+1}(3)^{73-r} \cdot 2^{r+1}$

$\Rightarrow \frac{{ }^{74} C _{r+1}}{{ }^{74} C_r}=\frac{3^{74-r}}{3^{73-r}} \cdot \frac{2^{r}}{2^{r+1}} \Rightarrow \frac{\frac{\mid 74}{|74-(r+1)| r+1}}{\frac{\mid 74}{|74-r| r}}=\frac{3}{2}$

$\Rightarrow \frac{74-r}{r+1}=\frac{3}{2} \Rightarrow 148-2 r=3 r+3$

$ \Rightarrow \quad 5 r=145 \Rightarrow r=29 . $

$\therefore \quad$ The two consecutive terms are 30th and 31st.

~~ 4. Given, $(x^{2}-\frac{1}{x^{3}})^{10}$

General term $=T _{r+1}={ }^{10} C_r(x^{2})^{10-r}(-\frac{1}{x^{3}})^{r}$

$ \begin{aligned} & ={ }^{10} C_r x^{20-2 r}(-1)^{r}(x^{-3 r}) \\ & ={ }^{10} C_r x^{20-5 r}(-1)^{r} . \end{aligned} $

Since the term contains $x^{-10}, \therefore 20-5 r=-10$

$\Rightarrow \quad 5 r=30 \Rightarrow r=6$

$\therefore$ Coefficient of $x^{-10}={ }^{10} C_6(-1)^{6}=\frac{\underline{10}}{\underline{6} \underline{4}}$

$ =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2}=\mathbf{2 1 0} \text{. } $

~~ 5. Given, $(\frac{x}{2}-\frac{3}{x^{2}})^{10}$

General term $=T _{r+1}={ }^{10} C_r(\frac{x}{2})^{10-r}(-\frac{3}{x^{2}})^{r}$

$ \begin{aligned} & ={ }^{10} C_r(\frac{1}{2})^{10-r}(-3)^{r} x^{10-r} \cdot x^{-2 r} \\ & ={ }^{10} C_r(\frac{1}{2})^{10-r}(-3)^{r} x^{10-3 r} \end{aligned} $

Since the term contains $x^{4}, 10-3 r=4 \Rightarrow 3 r=6 \Rightarrow r=2$

$\therefore$ Coefficient of $x^{4}={ }^{10} C_2(\frac{1}{2})^{10-2}(-3)^{2}$

$ \begin{aligned} & =\frac{\lfloor 10}{\underline{8} \underline{2}} \times(\frac{1}{2})^{8} \cdot(-3)^{2} \\ & =\frac{10 \times 9}{2} \times \frac{1}{256} \times 9=\frac{\mathbf{4 0 5}}{\mathbf{2 5 6}} . \end{aligned} $

~~ 6. Given, $(a-2 b)^{n}$.

General term $=T _{r+1}={ }^{n} C_r(a)^{n-r}(-2 b)^{r}$

$ ={ }^{n} C_r(a)^{n-r}(-2)^{r} b^{r} $

$ \begin{matrix} T_5={ }^{n} C_4 a^{n-4}(-2)^{4} b^{4} & (\because r+1=5 \Rightarrow r=4) \\ T_6={ }^{n} C_5 a^{n-5}(-2)^{5} b^{5} & (\because r+1=6 \Rightarrow r=5) \end{matrix} $

Given, $\quad T_5+T_6=0$

$\Rightarrow{ }^{n} C_4 a^{n-4} .16 b^{4}+{ }^{n} C_5 a^{n-5}(-32) b^{5}=0$

$\Rightarrow 16{ }^{n} C_4 a^{n-4} b^{4}=32 \cdot{ }^{n} C_5 a^{n-5} b^{5}$

$\Rightarrow \frac{{ }^{n} C_5 a^{n-5} b^{5}}{{ }^{n} C_4 a^{n-4} b^{4}}=\frac{16}{32}=\frac{1}{2} \Rightarrow \frac{b}{a}=\frac{1}{2} \times \frac{{ }^{n} C_4}{{ }^{n} C_5}$

$ =\frac{1}{2} \times \frac{\frac{\underline{\underline{4}}}{\frac{\lfloor n}{\underline{n}-4}}}{\frac{5 ! n-5}{\square}} \times \frac{5}{n-4}=\frac{5}{2(n-4)} $

$\Rightarrow \quad \frac{a}{b}=\frac{\mathbf{2}(\boldsymbol{n}-\mathbf{4})}{\mathbf{5}}$.

~~ 7. Given, $(a x-\frac{b}{x^{2}})^{11}$

General Term $=T _{r+1}={ }^{11} C_r(a x)^{11-r}(-\frac{b}{x^{2}})^{r}$

$ \begin{aligned} & ={ }^{11} C_r a^{11-r} \cdot x^{11-r}(-b)^{r} \cdot x^{-2 r} \\ & ={ }^{11} C_r a^{11-r}(-b)^{r} x^{11-3 r} \end{aligned} $

$\therefore \quad r$ th term $=T_r={ }^{11} C _{r-1} a^{11-(r-1)}(-b)^{r-1} x^{11-3(r-1)}$

$ ={ }^{11} C _{r-1} a^{12-r} x^{14-3 r}(-b)^{r-1} $

Since $x^{-7}$ occurs in the $r$ th term

$ 14-3 r=-7 \Rightarrow 3 r=21 \Rightarrow r=7 . $

~~ 8. Let $(r+1)$ th term be the coefficient of $x^{0}$ in the expansion of $(x-\frac{1}{x})^{6}$.

$ \begin{aligned} \therefore \quad T _{r+1} & ={ }^{6} C_r x^{6-r}(-\frac{1}{x})^{r}=(-1)^{r}{ }^{6} C_r x^{6-r} x^{-r} \\ & =(-1)^{r}{ }^{6} C_r x^{6-2 r} \end{aligned} $

As the power of $x$ is 0 , this means that the given term is a constant term.

$ \begin{aligned} & \therefore \quad 6-2 r=0 \Rightarrow r=3 \\ & \therefore \quad T_4=(-1)^{3}{ }^{6} C_3=-\mathbf{2 0 .} \end{aligned} $

~~ 9. $(1+x)^{n}={ }^{n} C_0+{ }^{n} C_1 x+{ }^{n} C_2 x^{2}+\ldots . .+{ }^{n} C _{n-1} x^{n-1}+{ }^{n} C_n x^{n}$ Given, coefficients of 5th, 6th and 7th terms are in A.P.

$ \begin{aligned} & \Rightarrow{ }^{n} C_4,{ }^{n} C_5,{ }^{n} C_6 \text{ are in A.P } \Rightarrow 2 \cdot{ }^{n} C_5={ }^{n} C_4+{ }^{n} C_6 \\ & \Rightarrow 2 \cdot \frac{n !}{5 !(n-5) !}=\frac{n !}{4 !(n-4) !}+\frac{n !}{6 !(n-6) !} \\ & \Rightarrow 2 \times \frac{1}{5 \times 4 ! \times(n-5) \times(n-6) !} \\ & =\frac{1}{4 ! \times(n-4) \times(n-5) \times(n-6) !}+\frac{1}{6 \times 5 \times 4 !(n-6) !} \\ & \Rightarrow \frac{2}{5(n-5)}=\frac{1}{(n-4)(n-5)}+\frac{1}{30} \\ & \Rightarrow 12(n-4)=30+(n-4)(n-5) \\ & \Rightarrow n^{2}-9 n+50=12 n-48 \Rightarrow n^{2}-21 n+98=0 \\ & \Rightarrow(n-14)(n-7)=0 \Rightarrow n=14,7 . \end{aligned} $

~~ 10. Given, $(p+q)^{n}$

$\therefore$ General term $=T _{r+1}={ }^{n} C_r p^{n-r} q^{r}$

$\therefore \quad r$ th term $=T_r={ }^{n} C _{r-1}(p)^{n-(r-1)} q^{r-1}$

$ ={ }^{n} C _{r-1} p^{n-r+1} q^{r-1} $

$(r+1)$ th term $=T _{r+1}={ }^{n} C_r p^{n-r} q^{r}$

Given, $T_r=T _{r+1}$

$ \begin{aligned} & \Rightarrow{ }^{n} C _{r-1} p^{n-r+1} q^{r-1}={ }^{n} C_r p^{n-r} q^{r} \\ & \Rightarrow \frac{\mid n}{|n-r+1| r-1} p^{n-r+1} q^{r-1}=\frac{\mid n}{n-r \mid r} p^{n-r} q^{r} \\ & \Rightarrow \frac{1}{(n-r+1)|n-r| r-1} \times \underline{n-r} \cdot r \cdot \underline{r-1} \end{aligned} $

$ =\frac{p^{n-r} q^{r}}{p^{n-r+1} q^{r-1}} $

$\Rightarrow \frac{r}{n-r+1}=\frac{q}{p} \Rightarrow p r=q n-q r+q$

$\Rightarrow \quad(p+q) r=(n+1) q \Rightarrow \frac{(\boldsymbol{n}+\mathbf{1}) \boldsymbol{q}}{r(\boldsymbol{p}+\boldsymbol{q})}=\mathbf{1}$.

~~ 11. $(1+a x)^{n}={ }^{n} C_0+{ }^{n} C_1(a x)+{ }^{n} C_2(a x)^{2}+{ }^{n} C_3(a x)^{3}+\ldots$

Given, $T_2=6 x$ and $T_3=16 x^{2}$

$\Rightarrow{ }^{n} C_1 a x=6 x$ and ${ }^{n} C_2(a x)^{2}=16 x^{2}$

$\Rightarrow n a x=6 x$ and $\frac{n(n-1)}{2} a^{2} x^{2}=16 x^{2}$

$\Rightarrow n a=6 \Rightarrow n^{2} a^{2}=36$ and $n(n-1) a^{2}=32$

$\therefore \frac{n(n-1) a^{2}}{n^{2} a^{2}}=\frac{32}{36} \Rightarrow \frac{n-1}{n}=\frac{8}{9}$

$\Rightarrow 9 n-9=8 n \Rightarrow n=9$

$\because \quad n a=6 \quad \therefore \quad a=\frac{6}{n}=\frac{6}{9}=\frac{\mathbf{2}}{\mathbf{3}}$.

~~ 12. General Term of $(2-3 x^{3})^{20}$ is $T _{r+1}={ }^{20} C_r(2)^{20-r}(-3 x^{3})^{r}$

$\therefore \quad T _{10}={ }^{20} C_9 2^{11}(-3 x^{3})^{9}={ }^{20} C_9(-1)^{9} \cdot 2^{11} \cdot 3^{9} \cdot x^{27}$

$T _{11}={ }^{20} C _{10} 2^{10}(-3 x^{3})^{10}={ }^{20} C _{10} 2^{10} \cdot 3^{10} \cdot x^{30}$

Given, $\frac{T _{10}}{T _{11}}=\frac{{ }^{20} C_9 \cdot(-1)^{9} 2^{11} \cdot 3^{9} \cdot x^{27}}{{ }^{20} C _{10} \cdot 2^{10} \cdot 3^{10} \cdot x^{30}}=\frac{45}{22}$

$\Rightarrow-\frac{10}{11} \times \frac{2}{3 x^{3}}=\frac{45}{22} \Rightarrow x^{3}=-\frac{8}{27} \Rightarrow x=-\frac{2}{\mathbf{3}}$.

~~ 13. Given, $(\sqrt[13]{a}+\frac{a}{\sqrt{a^{-1}}})^{n}$

$ =(a^{1 / 13}+\frac{a^{1}}{a^{-1 / 2}})^{n}=(a^{1 / 13}+a^{3 / 2})^{n} $

$\Rightarrow T_2=14 a^{5 / 2} \Rightarrow{ }^{n} C_1(a^{\frac{1}{13}})^{n-1}(a^{\frac{3}{2}})=14 a^{\frac{5}{2}}$

$\Rightarrow n a^{\frac{n-1}{13}+\frac{3}{2}}=14 a^{\frac{5}{2}} \Rightarrow n a^{\frac{2 n-2+39}{26}}=14 a^{\frac{5}{2}}$

$\Rightarrow n a^{\frac{2 n-37}{26}}=14 a^{\frac{5}{2}} \Rightarrow n=14$

$\therefore \quad \frac{{ }^{n} C_3}{{ }^{n} C_2}=\frac{{ }^{14} C_3}{{ }^{14} C_2}=\frac{\frac{14 !}{11 ! 3 !}}{\frac{14 !}{12 ! 2 !}}=\mathbf{4}$.

~~ 14. General term is expansion of $(a+b)^{n+4}$ is $T _{r+1}$

$ \therefore \quad \frac{t_6}{t_5}=\frac{{ }^{n+4} C_5 a^{n+4-5} b^{5}}{{ }^{n+4} C_4 a^{n+4-4} b^{4}}=\frac{{ }^{n+4} C_5}{{ }^{n+4} C_4} \cdot \frac{b}{a} $

General term in the expansion of $(a+b)^{n}$ is $T _{r+1}={ }^{n} C_r a^{n-r} b^{r}$

$\therefore \frac{t_5}{t_4}=\frac{{ }^{n} C_4 a^{n-4} b^{4}}{{ }^{n} C_3 a^{n-3} b^{3}}=\frac{{ }^{n} C_4}{{ }^{n} C_3} \cdot \frac{b}{a}$

According to the given condition, from (i) and (ii),

$ \begin{aligned} & \frac{{ }^{n+4} C_5}{{ }^{n+4} C_4} \cdot \frac{b}{a}=\frac{{ }^{n} C_4}{{ }^{n} C_3} \cdot \frac{b}{a} \\ \Rightarrow & \frac{(n+4) !}{\frac{(n-1) ! 5 !}{n ! 4) !}}=\frac{\frac{n !}{(n-4) ! 4 !}}{\frac{n !}{(n-3) ! 3 !}} \Rightarrow \frac{n}{5}=\frac{n-3}{4} \Rightarrow n=\mathbf{1 5 .} \end{aligned} $

~~ 15. The general term in the expansion of $(x+\frac{1}{x})^{23}$ is

$ T _{r+1}={ }^{23} C_r x^{23-r}(\frac{1}{x})^{r}={ }^{23} C_r x^{23-2 r} $

$ \text{ Now, } \quad T _{12}={ }^{23} C _{11} x^{23-22}={ }^{23} C _{11} x $

$ (\because r+1=12 \Rightarrow r=11) $

$ \begin{aligned} T _{13} & ={ }^{23} C _{12}(\frac{1}{x})^{23-24}={ }^{23} C _{12}(\frac{1}{x}) \\ & (\because r+1=13 \Rightarrow r=12) \\ = & { }^{23} C _{11}(\frac{1}{x}) \quad \ldots(iii) \end{aligned} $

$(\because{ }^{23} C _{11}={ }^{23} C _{23-11}={ }^{23} C _{12})$

$ \therefore \quad \frac{T _{12}}{T _{13}}=\frac{x}{\frac{1}{x}}=x^{2} \Rightarrow \boldsymbol{T} _{12}=\boldsymbol{x}^{2} \cdot \boldsymbol{T} _{\mathbf{1 3}} . $

~~ 16. $(1+x)^{2 n}={ }^{2 n} C_0+{ }^{2 n} C_1 x+{ }^{2 n} C_2 x^{2}+{ }^{2 n} C_3 x^{3}+\ldots . .+{ }^{2 n} C_n x^{n}$

$ +\ldots . .+{ }^{2 n} C _{2 n} x^{2 n} $

$(1+x)^{2 n-1}={ }^{2 n-1} C_0+{ }^{2 n-1} C_1 x+{ }^{2 n-1} C_2 x^{2}+{ }^{2 n-1} C_3 x^{3}+\ldots .$.

$ +{ }^{2 n-1} C_n x^{n}+\ldots . .+{ }^{2 n-1} C _{2 n-1} x^{2 n-1} $

Given, $p=$ coeff. of $x^{n}$ in expansion of $(1+x)^{2 n}={ }^{2 n} C_n$

$g=$ coeff. of $x^{n}$ in expansion of $(1+x)^{2 n-1}={ }^{2 n-1} C_n$

Now $p={ }^{2 n} C_n=\frac{\lfloor 2 n}{\lfloor\underline{n} \underline{n}}=\frac{2 n \times\lfloor 2 n-1}{n\lfloor(n-1)\lfloor\underline{n}}=2[\frac{\lfloor 2 n-1}{\lfloor n-1\lfloor n}]$…

$ g={ }^{2 n-1} C_n=\frac{\mid 2 n-1}{\lfloor n-1-n ! n}=\frac{2 n-1}{\lfloor n-1\lfloor n} $

$\therefore \quad$ From (i) and (ii) $\boldsymbol{p}=\mathbf{2 g}$.

~~ 17. Given, $P=1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots . .+(1+x)^{n}$ $P$ is the sum of a G.P of $(n+1)$ terms with first term $=1$, common ratio $=(1+x)$.

$ \begin{aligned} \therefore P & =\frac{.1(1+x)^{n+1}-1}{(1+x)-1}[\because S_n=\frac{a(r^{n}-1)}{r-1} \text{ when } n>1] \\ & .=\frac{1}{x}\lbrace(1+x)^{n+1}-1 \rbrace \\ & =\frac{1}{x}\lbrace(1+{ }^{n+1} C_1 x+{ }^{n+1} C_2 x^{2}+\ldots .+{ }^{n+1} C_r x^{r}.. \\ & ..\quad+{ }^{n+1} C _{r+1} x^{r+1}+\ldots . x^{n+1})-1\rbrace \\ & =\lbrace{ }^{n+1} C_1+{ }^{n+1} C_2 x+\ldots .+{ }^{n+1} C_r x^{r-1}+{ }^{n+1} C _{r+1} x^{r}. \end{aligned} $

$.+\ldots . x^{n}\rbrace$

Thus, the coefficient of $x^{r}$ in this expansion is ${ }^{n+1} C _{r+1}$ or ${ }^{n+1} C _{n-r^{*}}$

~~ 18. $(\frac{1}{x}+1)^{n}(1+x)^{n}=(\frac{1+x}{x})^{n}(1+x)^{n}=\frac{1}{x^{n}}(1+x)^{2 n}$

$ =\frac{1}{x^{n}}(1+{ }^{2 n} C_1 x+{ }^{2 n} C_2 x^{2}+\ldots . .+{ }^{2 n} C _{n-1} x^{n-1}. $

$ .+\ldots \ldots+{ }^{2 n} C _{2 n} x^{2 n}) $

$=(\frac{1}{x^{n}}+{ }^{2 n} C_1 \frac{1}{x^{n-1}}+{ }^{2 n} C_2 \frac{1}{x^{n-2}}+\ldots . .+{ }^{2 n} C _{n-1} \frac{1}{x}.$

The coefficient of $\frac{1}{x}$ is ${ }^{2 n} \boldsymbol{C} _{n-1}$.

$ .+\ldots . .+{ }^{2 n} C _{2 n} \frac{1}{x^{n}}) $

~~ 19. Let the term containing $x^{7}$ in the expansion of $(a x^{2}+\frac{1}{b x})^{8}$ is $T _{r+1}$.

Then, $T _{r+1}={ }^{8} C_r(a x^{2})^{8-r}(\frac{1}{b x})^{r}$

$ ={ }^{8} C_r \frac{a^{8-r}}{b^{r}} \cdot x^{16-2 r} \cdot x^{-r}={ }^{8} C_r \frac{a^{8-r}}{b^{r}} x^{16-3 r} $

Since the term contains $x^{7}, 16-3 r=7 \Rightarrow 3 r=9 \Rightarrow r=3$.

$\therefore$ Coefficient of $x^{7}$ in the expansion of $(a x^{2}+\frac{1}{b x})^{8}$

$ ={ }^{8} C_3 \frac{a^{8-3}}{b^{3}}={ }^{8} C_3 \frac{a^{5}}{b^{3}} $

Now, let the term containing $x^{-7}$ in expansion of $(a x-\frac{1}{b x^{2}})^{8}$ is $T _{R+1}$

Then, $T _{R+1}={ }^{8} C_R(a x)^{8-R}(-\frac{1}{b x^{2}})^{R}$

$=(-1)^{R}{ }^{8} C_R \frac{a^{8-R}}{b^{R}} x^{8-R} \cdot x^{-2 R}=(-1)^{R}{ }^{8} C_R \frac{a^{8-R}}{b^{R}} x^{8-3 R}$ Since this term contains $x^{-7}$

$\therefore 8-3 R=-7 \Rightarrow 3 R=15 \Rightarrow R=5$

$\therefore$ Coefficient of $x^{-7}$ in the expansion of $(a x-\frac{1}{b x^{2}})^{8}$

$ =(-1)^{5}{ }^{8} C_5 \frac{a^{8-5}}{b^{5}}=-{ }^{8} C_5 \frac{a^{3}}{b^{5}} $

Given, $\quad|{ }^{8} C_3 \frac{a^{5}}{b^{3}}|=|{ }^{8} C_5 \frac{a^{3}}{b^{5}}|$

$ \begin{matrix} \Rightarrow & \frac{a^{5}}{b^{3}}=\frac{a^{3}}{b^{5}} \quad(\because{ }^{8} C_3={ }^{8} C _{8-3}={ }^{8} C_5) \\ \Rightarrow & a^{2} b^{2}=1 \quad \Rightarrow \quad \boldsymbol{a} \boldsymbol{b}=\mathbf{1} . \end{matrix} $

~~ 20. $(1+x+x^{2}+x^{3})^{6}=[(1+x)+x^{2}(1+x)]^{6}$

$ \begin{aligned} & =[(1+x)(1+x^{2})]^{6}=(1+x)^{6}(1+x^{2})^{6} \\ & =({ }^{6} C_0+{ }^{6} C_1 x+{ }^{6} C_2 x^{2}+{ }^{6} C_3 x^{3}+{ }^{6} C_4 x^{4}+{ }^{6} C_5 x^{5}+{ }^{6} C_6 x^{6}) \\ & ({ }^{6} C_0+{ }^{6} C_1 x^{2}+{ }^{6} C_2 x^{4}+{ }^{6} C_3 x^{6}+{ }^{6} C_4 x^{8}+{ }^{6} C_5 x^{10}+{ }^{6} C_6 x^{12}) \\ & \therefore \quad \text{ Coefficient of } x^{14}={ }^{6} C_6 \cdot{ }^{6} C_4+{ }^{6} C_4 \cdot{ }^{6} C_5+{ }^{6} C_2 \cdot{ }^{6} C_6 \\ & =\frac{6 \times 5}{2}+\frac{6 \times 5}{2} \times 6+\frac{6 \times 5}{4}=30+90=\mathbf{1 2 0} . \end{aligned} $

(Middle Term, Term independent of $x$, Greatest term, $p^{\text{th }}$ term from the end)

SOLVED EXAMPLES

Ex. 1. . Write the middle term in the expansion of $(x-\frac{1}{2 x})^{10}$.

Sol. The general term in the expansion of $(x-\frac{1}{2 x})^{6}$ is

$ T _{r+1}=(-1)^{r}{ }^{6} C_r x^{6-r}(\frac{1}{2 x})^{r}=(-1)^{r} \cdot{ }^{6} C_r x^{6-r} \cdot 2^{-r} x^{-r}=(-1)^{r} \cdot{ }^{6} C_r x^{6-2 r} \cdot 2^{-r} $

Now the power of binomial expansion being 6 , (even), the middle term is $(\frac{6}{2}+1)$ th term $=4$ th term.

$\therefore \quad T_4=T _{3+1}=(-1)^{3}{ }^{6} C_3 x^{6-6} 2^{-3}=(-1) \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times x^{0} \times \frac{1}{8}=-\frac{\mathbf{5}}{\mathbf{2}}$.

Ex. 2 . Find the coefficient of the term independent of $x$ in the expansion of $[\sqrt{\frac{x}{3}}+\frac{\sqrt{3}}{x}]^{10}$.

Sol. Given, $[\sqrt{\frac{x}{3}}+\frac{\sqrt{3}}{x^{2}}]^{10}$,

General term $=T _{r+1}={ }^{10} C_r(\sqrt{\frac{x}{3}})^{10-r}(\frac{\sqrt{3}}{x^{2}})^{r}={ }^{10} C_r(\frac{x}{3})^{\frac{10-r}{2}}(\sqrt{3})^{r} x^{-2 r}={ }^{10} C_r(\frac{1}{3})^{\frac{10-r}{2}}(\sqrt{3})^{r} x^{\frac{10-r}{2}-2 r}$

For the term independent of $x$,

$ \frac{10-r}{2}-2 r=0 \Rightarrow \frac{10-r}{2}=2 r \Rightarrow 10-r=4 r \Rightarrow 5 r=10 \Rightarrow r=2 $

$\therefore \quad T _{2+1}$, i.e., $T_3$ is the term independent of $x$ and

$ T_3={ }^{10} C_2(\frac{1}{3})^{\frac{10-2}{2}}(\sqrt{3})^{2}=\frac{10 \times 9}{2} \times(\frac{1}{3})^{4} \times 3=45 \times \frac{1}{27}=\frac{5}{\mathbf{3}} . $

Ex. 3. . Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is the sum of the coefficients of the two middle terms in the expansion of $(1+x)^{2 n-1}$.

Sol. The expansion of $(1+x)^{2 n}$ contains $(2 n+1)$ terms. The middle term here is $(\frac{2 n}{2}+1)$ th term, i.e., $(n+1)$ th term.

$\therefore \quad T _{n+1}={ }^{2 n} C_n x^{n}=\frac{(2 n) !}{n ! n !} x^{n}=\frac{(2 n) !}{(n !)^{2}} x^{n}$

In the expansion of $(1+x)^{2 n-1},(2 n-1)$ being odd, there are two middle term $(\frac{2 n-1+1}{2})$ th term and $(\frac{2 n-1+3}{2})$ th term, i.e., $n$th term and $(n+1)$ th term.

$\therefore$ In case of $(1+x)^{2 n-1}$

$ t_n={ }^{2 n-1} C _{n-1} x^{n-1}, t _{n+1}={ }^{2 n-1} C_n x^{n} $

$\therefore$ Sum of coefficients of $t_n$ and $t _{n+1}={ }^{2 n-1} C _{n-1}+{ }^{2 n-1} C_n={ }^{2 n-1+1} C_n={ }^{2 n} C_n \quad(\because{ }^{n} C_r+{ }^{n} C _{r+1}={ }^{n+1} C _{r+1})$

$ =\frac{2 n !}{(n !)^{2}}=\text{ coefficient of middle term of }(1+x)^{2 n} $

Ex. 4. . Find the 5 th term from the end in the expansion of $(\frac{2}{x}-\frac{x^{3}}{5})^{9}$.

Sol. Using $r$ th term from the end $=(m-r+2)$ th term, we have 5 th term from the end in the expansion of $(\frac{2}{x}-\frac{x^{3}}{5})^{9}$ $=(9-5+2)$ th from the end $=6$ th term from the beginning in the given expansion

$ \begin{aligned} & =T_6=T _{5+1}={ }^{9} C_5(\frac{2}{x})^{9-5}(-\frac{x^{3}}{5})^{5}=\frac{-9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times(\frac{2}{x})^{4} \times(\frac{x^{3}}{5})^{5} \quad T_1 T_2 T_3 T_4 T_5 T_6 T_7 T_8 T_9 T _{10} \\ & =-126 \times \frac{16}{x^{4}} \times \frac{x^{15}}{3125}=\frac{-\mathbf{2 0 1 6}}{\mathbf{3 1 2 5}} \boldsymbol{x}^{\mathbf{1 1}} \end{aligned} $

6th term from the beginning

PRACTICE SHEET-2 (Middle Term, Term independent of $x$, Greatest term, $p^{\text{th }}$ term from the end)

~~ 1. The middle term in the expansion of $(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}})^{10}$ is

(a) 252

(b) 260

(d) 450

(MP PET 2013)

~~ 2. The middle term in the expansion of $(x^{2}+\frac{1}{x^{2}}+2)^{n}$ is

(a) $\frac{n}{[(n / 2) !]^{2}}$

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

(d) $\frac{(2 n) !}{(n !)^{2}}$

(Manipal Engg. 2012)

~~ 3. In the expansion of $(1-3 x+3 x^{2}-x^{3})^{2 n}$, the middle term is

(a) $(n+1)$ th term

(b) $(2 n+1)$ th term

(d) None of these

(MP PET 2011)

~~ 4. The coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is (a) $\frac{2^{n}(2 n-1)(2 n-3) \ldots .3 .1}{n(n-1)(n-2) \ldots .3 .2 .1}$

(b) $n(n+1)(n+2) \ldots . .2 n$

(d) $\frac{2^{n}(n)(n+1)(n+2) \ldots . .(2 n)}{n(n-1)(n-2) \ldots . .3 .2 .1}$

(Odisha JEE 2011)

~~ 5. The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1+\alpha x)^{4}$ and of $(1-\alpha x)^{6}$ is the same if $\alpha$ equals

(a) $-\frac{3}{10}$

(b) $-\frac{5}{3}$

(d) $\frac{10}{3}$

(AIEEE 2004)

~~ 6. The numerically greatest term in the expansion of $(3+2 x)^{44}$, when $x=\frac{1}{5}$ is

(a) 4th term

(b) 5 th term

(d) 7 th term

(AMU 2009)

~~ 7. The greatest term in the expansion of $(1+3 x)^{54}$, where $x=\frac{1}{3}$ is

(a) $T _{28}$

(b) $T _{25}$

(d) $T _{24}$ (DCE 2007) ~~ 8. The term idependent of $x$ in the expansion of $(\frac{2 \sqrt{x}}{5}-\frac{1}{2 x \sqrt{x}})^{11}$ is

(a) 5 th term

(b) 6th term

(J&K CET 2007)

~~ 9. The term independent of $x$ in $(\sqrt{x}-\frac{2}{x})^{18}$ is

(a) ${ }^{18} C _{12} 2^{8}$

(b) ${ }^{18} C_6 2^{12}$

(d) ${ }^{18} C _{12} 2^{6}$

(MP PET 2009)

~~ 10. The term idependent of $x$ in the expansion of $(1+x)^{n}$ $[1+(\frac{1}{x})]^{n}$ is

(a) $C_1+C_2+C_3+\ldots . .+C_n$

(b) $C_1+2 C_2+3 C_3+\ldots . .+{ }^{n} C_n$

(d) $C_0{ }^{2}+2 C_1{ }^{2}+3 C_2{ }^{2}+\ldots . .+(n+1) C_n{ }^{2}$

(Kerala PET 2000)

~~ 11. If in the expansion of $(\frac{3 \sqrt{x}}{7}-\frac{5}{2 x \sqrt{x}})^{13 n}$ contains a term independent of $x$, then $n$ should be a multiple of

(a) 3

(b) 4

(d) 6

(Kerala PET 2008)

~~ 12. The $13^{\text{th }}$ term in the expansion of $(x^{2}+\frac{2}{x})^{n}$ is independent of $x$, then the sum of the divisors of $\boldsymbol{n}$ is

(a) 36

(b) 37

(d) 39

(Karnataka CET 2012)

~~ 13. The term independent of $x$ in the expansion $(x-\frac{1}{x})^{4}(x+\frac{1}{x})^{3}$ is

(a) -3

(b) 0

(d) 1

(Gujarat CET 2007)

~~ 14. What is the ratio of coefficient of $x^{15}$ to the term independent of $x$ in $(x^{2}+\frac{2}{x})^{15}$ ?

(a) $\frac{1}{64}$

(b) $\frac{1}{32}$

(d) $\frac{1}{4}$

(NDA/NA 2011)

~~ 15. The term independent of $x$ in the expansion of $(x+\frac{1}{x}+2)^{11}$ is

(a) $\frac{11 !}{6 ! 6 !}$

(b) $\frac{11 !}{5 ! 6 !}$

(d) ${ }^{22} C _{11}$ ~~ 16. 5 th term from the end in the expansion of $(\frac{x^{3}}{2}-\frac{2}{x^{2}})^{12}$ is

(a) $-7920 x^{-4}(b)$

(b) $7920 x^{4}$

(d) $-7920 x^{4}$ ~~ 17. In the expansion of $(3 x-\frac{2}{x^{2}})^{15}$, if the $p$ th term from the end does not depend on the value of $x$, then the value of $\boldsymbol{p}$ is

(a) 9

(b) 10

(d) 12

(Rajasthan PET 2009)

~~ 18. Find the sixth term of the expansion of $(y^{1 / 2}+x^{1 / 3})^{n}$, if the binomial coefficient of the third term from the end is 45 .

(a) $240 y^{3 / 2} x^{2 / 3}$

(b) $252 y^{5 / 2} x^{5 / 3}$

(d) $240 y^{5 / 2} x^{5 / 3}$

ANSWERS

~~ 1. (a) ~~ 2. $($ d) ~~ 3. (c) ~~ 4. (a) ~~ 5. (a) ~~ 6. (c) ~~ 7. (a) ~~ 8. (c) ~~ 9. (c) ~~ 10. (b) ~~ 11. (d) ~~ 12. (d) ~~ 13. (c) ~~ 14. (b)

HINTS AND SOLUTIONS

~~ 1. In the expansion of $(\frac{2 x^{2}}{3}+\frac{3}{2 x^{2}})^{10}$, the middle term is the $(\frac{10}{2}+1)$ th term, i.e., 6th term

Now $T_6=T _{5+1}={ }^{10} C_5(\frac{2 x^{2}}{3})^{10-5}(\frac{3}{2 x^{2}})^{5}$ $(\because T _{r+1}={ }^{n} C_r(a)^{n-r} b^{r})$.

$=\frac{10 !}{5 ! 5 !} \frac{2^{5} x^{10}}{3^{5}} \cdot \frac{3^{5}}{2^{5} \cdot x^{10}}=\frac{10 !}{5 ! 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\mathbf{2 5 2}$. ~~ 2. $(x^{2}+\frac{1}{x^{2}}+2)^{n}=((x+\frac{1}{x})^{2})^{n}=(x+\frac{1}{x})^{2 n}$

$\therefore \quad$ The middle term in the expansion of $(x+\frac{1}{x})^{2 n}$ $=(\frac{2 n}{2}+1)$ th term $=(n+1)$ th term.

Now $T _{n+1}={ }^{2 n} C_n(x)^{2 n-n}(\frac{1}{x})^{n}=\frac{(2 n !)}{n ! n !} x^{n} \cdot \frac{1}{x^{n}}=\frac{(\mathbf{2 n !})}{(\boldsymbol{n} !)^{2}}$.

~~ 3. $(1-3 x+3 x^{2}-x^{3})^{2 n}=((1-x)^{3})^{2 n}=(1-x)^{6 n}$

$\therefore \quad$ Middle term $=(\frac{6 n}{2}+1)$ th term $=(3 n+1)$ th term.

~~ 4. The middle term in the expansion of $(1+x)^{2 n}$ is $(\frac{2 n}{2}+1)$ th term, i.e., $(n+1)$ th tern.

$\therefore$ Coefficient of $(n+1)$ th term in the expansion of $(1+x)^{2 n}={ }^{2 n} C_n$

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

$=\frac{{2 n(2 n-2)(2 n-4) \ldots . .4 .2 .}{(2 n-1)(2 n-3) \ldots .3 .1}}{{n(n-1)(n-2) \ldots 3.2 .1}^{2}}$

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

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

~~ 5. Middle term in the expansion of $(1+\alpha x)^{4}$ is the $(\frac{4}{2}+1)$ th term, i.e., 3rd term.

$\therefore \quad t_3=t _{2+1}={ }^{4} C_2(\alpha x)^{2}$

Middle term in the expansion of $(1-\alpha x)^{6}$ is the $(\frac{6}{2}+1)$ th term, i.e., 4 th term.

$\therefore \quad T_4=T _{3+1}={ }^{6} C_3(-1)^{3}(\alpha x)^{3}$

$\therefore$ Coefficient of $t_3=$ Coefficient of $T_4$

$\Rightarrow{ }^{4} C_2 \alpha^{2}={ }^{6} C_3(-1)^{3} \alpha^{3}$

$\Rightarrow \frac{4 \times 3}{2} \alpha^{2}=(-1) \times \frac{6 \times 5 \times 4}{3 \times 2} \alpha^{3} \Rightarrow 6 \alpha^{2}=-20 \alpha^{3}$

$\Rightarrow \alpha=-\frac{6}{20}=-\frac{\mathbf{3}}{\mathbf{1 0}}$.

~~ 6. The general term in the expansion of $(3+2 x)^{44}$ is

$ \begin{aligned} T _{r+1} & ={ }^{44} C_r(3)^{44-r}(2 x)^{r} \\ \therefore \quad T_r & ={ }^{44} C _{r-1}(3)^{44-(r-1)}(2 x)^{r-1}={ }^{44} C _{r-1} 3^{45-r}(2 x)^{r-1} \end{aligned} $

For $T _{r+1}$ to be the greatest term, $(.$ where $.x=\frac{1}{5})$

$ \begin{aligned} & \frac{T _{r+1}}{T_r}>1 \Rightarrow \frac{{ }^{44} C_r 3^{44-r}(2 x)^{r}}{{ }^{44} C _{r-1} 3^{45-r}(2 x)^{r-1}}>1 \\ \Rightarrow & \frac{\frac{(44) !}{(44-r) ! r !}}{(44-r+1) !(r-1) !} \times \frac{2 x}{3}>1 \Rightarrow \frac{(45-r)}{r} \times \frac{2 x}{3}>1 \\ \Rightarrow & \frac{(45-r)}{r} \times \frac{2}{3} \times \frac{1}{5}>1 \Rightarrow 90-2 r>15 r \\ \Rightarrow & 90>17 r \text{ or } 17 r<90 \Rightarrow r<\frac{90}{17}=5 \frac{5}{7} \\ \Rightarrow & \frac{T _{r+1}}{T_r}>1 \text{ for all value of } r \leq 5 \\ \Rightarrow & T_6 \text{, i.e., the 6th term is numerically the greatest term. } \end{aligned} $

~~ 7. Similar to Q. 6 . ~~ 8. The general term in the expansion of $(\frac{2 \sqrt{x}}{5}-\frac{1}{2 x \sqrt{x}})^{11}$ is

$ \begin{aligned} T _{r+1} & ={ }^{11} C_r(\frac{2 \sqrt{x}}{5})^{11-r}(-\frac{1}{2 x \sqrt{x}})^{r} \\ & ={ }^{11} C_r(-1)^{r} 2^{11-r}(x^{1 / 2})^{11-r}(5^{-1})^{11-r} \cdot 2^{-r} \cdot(x^{-3 / 2})^{r} \\ & ={ }^{11} C_r(-1)^{r} 2^{11-2 r} 5^{r-11} x^{\frac{11}{2}-\frac{r}{2}-\frac{3 r}{2}} \\ & ={ }^{11} C_r(-1)^{r} 2^{11-2 r} 5^{r-11} x^{\frac{11}{2}-2 r} \end{aligned} $

This term will be independent of $x$, if $\frac{11}{2}-2 r=0 \Rightarrow 2 r=\frac{11}{2}$ $\Rightarrow r=\frac{11}{4}$. Since $r=\frac{11}{4}$ is not an integral value, there is no term in the expansion of $(\frac{2 \sqrt{x}}{5}-\frac{1}{2 x \sqrt{x}})^{11}$ which is independent of $x$.

~~ 9. The general term in the expansion of $[\sqrt{x}-\frac{2}{x}]^{18}$ is

$ \begin{aligned} T _{r+1} & ={ }^{18} C_r(\sqrt{x})^{18-r}(-\frac{2}{x})^{r}={ }^{18} C_r x^{\frac{18-r}{2}}(-1)^{r} 2^{r} x^{-r} \\ & ={ }^{18} C_r x^{\frac{18-3 r}{2}}(-1)^{r} 2^{r} \end{aligned} $

This term will be independent of $x$ if $\frac{18-3 r}{2}=0 \Rightarrow 18-3 r$ $=0 \Rightarrow r=6$.

$\therefore$ Reqd. term $=T _{6+1}={ }^{18} C_6(-1)^{6} 2^{6}={ }^{18} C_6 2^{6}={ }^{18} C _{12} 2^{6}$

~~ 10. $(1+x)^{n}(1+\frac{1}{x})^{n}$ $(\because{ }^{18} C_6={ }^{18} C _{12})$

$\therefore \quad$ The term independent of $x$ in this expansion is

$[{ }^{n} C_0 \cdot{ }^{n} C_0+{ }^{n} C_1 \cdot{ }^{n} C_1+{ }^{n} C_2 \cdot{ }^{n} C_2+\ldots \ldots .+{ }^{n} C_n \cdot{ }^{n} C_n]$

$=({ }^{n} C_0)^{2}+({ }^{n} C_1)^{2}+({ }^{n} C_2)^{2}+\ldots . .+({ }^{n} C_n)^{2}$

$=C_0{ }^{2}+C_1{ }^{2}+C_2{ }^{2}+\ldots . .+C_n{ }^{2}$.

~~ 11. The general term in the expansion of $(\frac{3 \sqrt{x}}{7}-\frac{5}{2 x \sqrt{x}})^{13 n}$ is

$ \begin{aligned} T _{r+1} & ={ }^{13 n} C_r(\frac{3 \sqrt{x}}{7})^{13 n-r}(\frac{-5}{2 x \sqrt{x}})^{r} \\ & ={ }^{13 n} C_r(-1)^{r}(3)^{13 n-r}(7)^{r-13 n}(5)^{r}(2)^{-r} x^{\frac{13 n-r}{2}} \cdot x^{-\frac{-3 r}{2}} \\ & ={ }^{13 n} C_r(-1)^{r}(3)^{13 n-r}(7)^{r-13 n}(5)^{r}(2)^{-r} x^{\frac{13 n}{2}-2 r} \end{aligned} $

This term is independent of $x$ if $\frac{13 n}{2}-2 r=0$

$\Rightarrow \frac{13 n}{2}=2 r \Rightarrow n=4(\frac{r}{13})$

$\Rightarrow n$ should be a multiple of 4 .

~~ 12. 13th term in the expansion of $(x^{2}+\frac{2}{x})^{n}$ is given by

$ \begin{aligned} T _{13} & ={ }^{n} C _{12}(x^{2})^{n-12}(\frac{2}{x})^{12}={ }^{n} C _{12} x^{2 n-24} \frac{2^{12}}{x^{12}} \\ & ={ }^{n} C _{12} x^{2 n-36} \cdot 2^{12} \end{aligned} $

If the 13th term is independent of $x$, then $2 n-36=0$

$\Rightarrow n=18$.

The divisors of $n=18$ are $1,2,3,6,9,18$ and their sum $=1+2+3+6+9+18=39$.

~~ 13. $(x-\frac{1}{x})^{4}(x+\frac{1}{x})^{3}$

$ \begin{aligned} =({ }^{4} C_0 x^{4}-{ }^{4} C_1. & .x^{2}+{ }^{4} C_2-{ }^{4} C_3 \frac{1}{x^{2}}+{ }^{4} C_4 \frac{1}{x^{4}}) \\ & \times({ }^{3} C_0 x^{3}+{ }^{3} C_1 x+{ }^{3} C_2 \frac{1}{x}+{ }^{3} C_3 \frac{1}{x^{3}}) \end{aligned} $

As can be seen from the given product, there is no term free of $x$ on RHS, therefore the term independent of $x$ is $\mathbf{0}$.

~~ 14. Given, binomial expression $=(x^{2}+\frac{2}{x})^{15}$

General term $=T _{r+1}={ }^{15} C_r(x^{2})^{15-r}(\frac{2}{x})^{r}$

$={ }^{15} C_r x^{30-2 r} 2^{r} \cdot x^{-r}={ }^{15} C_r x^{30-3 r} \cdot 2^{r}$

For coefficient of $x^{15}$, put $30-3 r=15 \Rightarrow 3 r=15 \Rightarrow r=5$

$ \therefore \quad T _{5+1}=T_6={ }^{15} C_5 x^{15} \cdot 2^{5} $

For term independent of $x$, put $30-3 r=0 \Rightarrow 3 r=30$ $\Rightarrow r=10$

$\therefore \quad T _{10+1}=T _{11}={ }^{15} C _{10} 2^{10}$

$\therefore$ Required ratio $=\frac{T_6}{T _{11}}=\frac{{ }^{15} C_5 2^{5}}{{ }^{15} C _{10} 2^{10}}=\frac{{ }^{15} C _{10}}{{ }^{15} C _{10}} \times \frac{1}{2^{5}}=\frac{1}{32}$

$ (\because{ }^{n} C_r={ }^{n} C _{n-r}) $

~~ 15. $(x+\frac{1}{x}+2)^{11}=((\sqrt{x}+\frac{1}{\sqrt{x}})^{2})^{11}=(\sqrt{x}+\frac{1}{\sqrt{x}})^{22}$

General term in the expansion of $(\sqrt{x}+\frac{1}{\sqrt{x}})^{22}$ is

$ \begin{aligned} T _{r+1} & ={ }^{22} C_r(\sqrt{x})^{22-r}(\frac{1}{\sqrt{x}})^{r}={ }^{22} C_r(x^{1 / 2})^{22-2 r} \\ & ={ }^{22} C_r x^{11-r} \end{aligned} $

This term is independent of $x$, if $11-r=0 \Rightarrow r=11$

$\therefore$ Term independent of $x$ is $T _{11+1}=T _{12}={ }^{22} C _{11} x^{22-22}={ }^{\mathbf{2 2}} \boldsymbol{C} _{\mathbf{1 1}}$.

~~ 16. 5 th term from the end

$=(12-5+2)$ th term from the beginning

$=9$ th term from the beginning in the expansion of

$ (\frac{x^{3}}{2}-\frac{2}{x^{2}})^{12} $

$ \begin{aligned} & ={ }^{12} C_8(\frac{x^{3}}{2})^{12-8}(-\frac{2}{x^{2}})^{8}={ }^{12} C_8 x^{-4} \cdot 2^{4} \\ & =\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \times 2^{4} \times x^{-4}=\mathbf{7 9 2 0} \boldsymbol{x}^{-4} . \end{aligned} $

~~ 17. The general term in the expansion $(3 x-\frac{2}{x^{2}})^{15}$ is

$ \begin{aligned} T _{r+1} & ={ }^{15} C_r(3 x)^{15-r}(-\frac{2}{x^{2}})^{r} \\ & ={ }^{15} C_r(3)^{15-r}(x)^{15-r}(-1)^{r}(2)^{r}(x)^{-2 r} \\ & ={ }^{15} C_r(-1)^{r}(2)^{r}(3)^{15-r}(x)^{15-3 r} \end{aligned} $

This term is independent of $x$ if $15-3 r=0 \Rightarrow r=5$

$\therefore T _{5+1}=T_6=6$ th term from the beginning is independent of $x$.

$\Rightarrow(15-6+2)$ th term is independent of $x$ from the end

$\Rightarrow 11$ th term from the end is independent of $x$.

$\Rightarrow p=11$.

~~ 18. 3 rd term from the end in the expansion of $(y^{1 / 2}+x^{1 / 3})^{n}$

$=(n-3+2)$ th term from the beginning in the given expansion

$=(n-1)$ th term from the beginning in the expansion of

$ (y^{1 / 2}+x^{1 / 3})^{n} $

$=T _{n-1}={ }^{n} C _{n-2}(y^{1 / 2})^{n-(n-2)}(x^{1 / 3})^{n-2}=\frac{n(n-1)}{2} \times y \times x^{\frac{n-2}{3}}$

Given, Binomial Coeff. of 3 rd term, i.e., $T _{2+1}$ from the end

$=45$, i.e., ${ }^{n} C_2=45 \Rightarrow \frac{n(n-1)}{2}=45$

$\Rightarrow n(n-1)=90 \Rightarrow n=10$

$\therefore \quad T_6$ in the expansion of $(y^{1 / 2}+x^{1 / 3})^{10}$

$={ }^{10} C_5(y^{1 / 2})^{10-5}(x^{1 / 3})^{5}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times y^{5 / 2} \times x^{5 / 3}$

$=252 y^{5 / 2} x^{5 / 3}$.

KEY FACTS (Contd.)

1. Binomial theorem for negative or fractional index

Let $\boldsymbol{n}$ be a negative integer or a fraction (+ve or $-ve)$ and $x$ be a real number such that $|x|<1$, then

$ (1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots+\frac{n(n-1)(n-2) \ldots(n-r+1)}{r !} x^{r}+\ldots \infty $

Note 1. The expansion is valid only if $|x|<1$.

~~ 2. The first term of the binomial is unity, i.e., it is of the form $(1+x)$ where $x \in R$ and $|x|<1$. Also the expansion of $(1+x)$ to negative or fractional index contains infinite terms.

$ \begin{aligned} \therefore \quad(x+a)^{n} & =a^{n}(1+\frac{x}{a})^{n} \quad(n \text{ being a negative integer or fraction }(+ve \text{ or }-ve)) \\ & =a^{n}(1+n \cdot \frac{x}{a}+\frac{n(n-1)}{2 !}(\frac{x}{a})^{2}+\ldots .)=a^{n}+n a^{n-1} x+\frac{n(n-1) a^{n-2}}{2 !} x^{2}+\ldots . . \end{aligned} $

This expression is valid only when $|\frac{x}{a}|<1$ or $|x|<|a|$.

~~ 2. General term in the expansion of $(1+x)^{n}$, when $\boldsymbol{n}$ is an integer or a fractional rational number.

$ T _{r+1}=\frac{n(n-1)(n-2) \ldots . .(n-r+1)}{r !} \cdot x^{r} $

3. Some special expansions and their general terms.

a. $(1-x)^{-1}=1+x+x^{2}+\ldots . .+x^{r}+\ldots .$. b. $(1+x)^{-1}=1-x+x^{2}-\ldots . .+(-1)^{r} x^{r}+\ldots .$. c. $(1-x)^{-2}=1+2 x+3 x^{2}+\ldots . .+(r+1) x^{r}+\ldots .$. d. $(1+x)^{-2}=1-2 x+3 x^{2}-\ldots . .+(-1)^{r}(r+1) x^{r}+\ldots .$. e. $(1-x)^{-3}=1+3 x+6 x^{2}+\ldots . .+\frac{(r+1)(r+2)}{2} x^{r}+\ldots$. f. $(1+x)^{-3}=1-3 x+6 x^{2}-\ldots . .+(-1)^{r} \frac{(r+1)(r+2)}{2} x^{r}+\ldots .$.

SOLVED EXAMPLES

Ex. 1 . Write down and simplify the first four terms in the binomial expansion of $(1-2 x)^{2 / 3}$.

Sol.

$ \begin{aligned} (1-2 x)^{2 / 3} & =1+\frac{2}{3}(-2 x)+\frac{\frac{2}{3}(\frac{2}{3}-1)}{2.1}(-2 x)^{2}+\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{3.2 .1}(-2 x)^{3}+\ldots . . \\ & =1-\frac{4}{3} x-\frac{4}{9} x^{2}-\frac{32}{81} x^{3}+\ldots . . \end{aligned} $

Ex. 2 . Find the coefficient of $x^{4}$ in the expansion of $(\frac{1-x}{1+x})^{2}$.

Sol. $\quad(\frac{1-x}{1+x})^{2}=(1-x)^{2}(1+x)^{-2}=(1-2 x+x^{2})(1-2 x+3 x^{2}-4 x^{3}+5 x^{4} \ldots ..)$

The coefficient of $x^{4}$ in this expansion or product is

$ (1 \times 5)+(-2) \times(-4)+1 \times 3=5+8+3=\mathbf{1 6} \text{. } $

$ (1-2 x+x^{2})(1-2 x+3 x^{2}-4 x^{3}+5 x^{4}) $

Ex. 3 . Find the coefficient of $x$ in the expansion of $(1+x+x^{2}+x^{3})^{-3}$.

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

$ ={1-3 x+6 x^{2}-10 x^{3}+\ldots . .}{1-3 x^{2}+6 x^{4}-10 x^{6}+\ldots . .} $

$\therefore \quad$ The coefficient of $x$ here $=-\mathbf{3}$.

Ex. 4 . Find the square root of $1+2 x+3 x^{2}+4 x^{3}+\ldots .$. ?

(RPET 2007)

Sol. $1+2 x+3 x^{2}+4 x^{3}+\ldots . .=(1-x)^{-2}$

$\therefore \quad$ Reqd. square root $=[(1-x)^{-2}]^{\frac{1}{2}}=(1-x)^{-1}$

Ex. 5 . Find the coefficient of $x^{n}$ in the expansion of $(1-4 x)^{-1 / 2}$.

Sol. $(1-4 x)^{-1 / 2}=1+(-\frac{1}{2})(-4 x)+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)}{1.2}(-4 x)^{2}+\frac{(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{1.2 .3}(-4 x)^{3}+\ldots$.

$ \begin{aligned} & +\frac{(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2) \ldots \ldots(-\frac{1}{2}-r+1)}{1.2 \cdot 3 \ldots \ldots r} \cdot(-4 x)^{r}+\ldots \ldots \\ & =1+\frac{1}{2}(4 x)+\frac{(\frac{1}{2})(\frac{3}{2})}{2 !}(4 x)^{2}+\frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})}{3 !}(4 x)^{3}+\ldots .+\frac{(\frac{1}{2})(\frac{3}{2})(\frac{5}{2}) \ldots(\frac{2 r-1}{2})}{r !}(4 x)^{r}+\ldots \ldots \\ \therefore \quad T _{r+1} & =\text{ General term }=\frac{(\frac{1}{2})(\frac{3}{2}) \ldots(\frac{2 r-1}{2})}{r !}(4 x)^{r}=\frac{1.3 \cdot 5 \ldots(2 r-1)}{2^{r} \cdot r !} \cdot 2^{2 r} \cdot x^{r}=\frac{1.3 \ldots(2 r-1)}{r !} 2^{r} \cdot x^{r} \\ & =\frac{1.2 .3 \ldots . .(2 r-1) \cdot 2 r}{2.4 .6 \ldots \ldots .2 r} \cdot \frac{2^{r} \cdot x^{r}}{r !}=\frac{(2 r) !}{2^{r}(1.2 .3 \ldots \ldots r)} \cdot \frac{2^{r} \cdot x^{r}}{r !}=\frac{(2 r) !}{r ! \cdot r !} \cdot x^{r} \end{aligned} $

$ \therefore \quad \text{ coefficient of } x^{r}=\frac{(2 r) !}{r ! r !} \Rightarrow \text{ coefficient of } x^{n}=\frac{(2 n) !}{(n !)^{2}} \text{. } $

PRACTICE SHEET-3 (Binomial Theorem for Negative or Fractional Index)

~~ 1. The coefficient of $x^{4}$ in the expansion of $(1+x^{-2})$, where $|x|<1$

(a) -5

(b) -3

(d) 5

(J&K CET 2010)

~~ 2. If $|x|<1$, then the coefficient of $x^{6}$ in the expansion of $(1+x+x^{2})^{-3}$ is

(a) 3

(b) 6

(d) 12

(Kerala PET 2009)

~~ 3. Coefficient of $\boldsymbol{x}$ in the expansion of $(\frac{1+x}{1-x})^{n}$ is

(a) $2 n$

(b) $4 n$

(d) $\frac{n(n+1)}{2}$

(AMU 2003)

~~ 4. If $|x|<1$, then the coefficient of $x^{n}$ in $(1+2 x+3 x^{2}+4 x^{3}.$ $+\ldots.)^{1 / 2}$ is

(a) 1

(b) $-n$

(d) $n+1$

(WB JEE 2009)

~~ 5. The coefficient of $x^{4}$ in the expansion of $\frac{(1-3 x)^{2}}{(1-2 x)}$ is

(a) 1

(b) 2

(d) 4

(EAMCET 2001)

~~ 6. For $|x|<1$, the constant term in the expansion of $\frac{1}{(x-1)^{2}(x-2)}$ is

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

(b) 0

(d) 2 ~~ 7. The coefficient of $x^{n}$ in the expansion of $(1-2 x)^{-1 / 2}$ is

(a) $\frac{(2 n) !}{(n !)^{2}}$

(b) $\frac{(2 n) !}{2^{n} n !}$

(c) $\frac{(2 n) !}{2^{n}(n !)^{2}}(d) \frac{(2 n) !}{n !}$ ~~ 8. If $|x|<\frac{1}{2}$, then the coefficient of $x^{r}$ in the expansion of $\frac{1+2 x}{(1-2 x)^{2}}$ is

(a) $r 2^{r}$

(b) $r \cdot 2^{2 r+1}$

(d) $(2 r+1) \cdot 2^{r}$

(EAMCET 2005)

ANSWERS

~~ 1. $($ d) ~~ 2. (a) ~~ 3. (a) ~~ 4. (a) ~~ 5. (d) ~~ 6. (a) ~~ 7. (c) ~~ 8. (d)

HINTS AND SOLUTIONS

~~ 1. $(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+5 x^{4}$

$\Rightarrow$ Coefficient of $x^{4}$ in the expansion of $(1+x)^{-2}=\mathbf{5}$.

~~ 2. $(1+x+x^{2})^{-3}=(\frac{1-x^{3}}{1-x})^{-3}=(\frac{1-x}{1-x^{3}})^{3}=(1-x)^{3}(1-x^{3})^{-3}$ $=({ }^{3} C_0-{ }^{3} C_1 x+{ }^{3} C_2 x^{2}-{ }^{3} C_3 x^{3})$

$ [1+(-3) \cdot(-x^{3})+\frac{(-3)(-3-1)}{1.2}(-x^{3})^{2}. $

$ .+\frac{(-3)(-3-1)(-3-2)}{1.2 .3}(-x^{3})^{3}+\ldots] $

$=(1-3 x+3 x^{2}-x^{3})(1+3 x^{3}+6 x^{6}+10 x^{9}+\ldots.)$

$\Rightarrow$ Coefficient of $x^{6}$ in the given expansion

$ =(1 \times 6+(-1) \times 3)=6-3=3 \text{. } $

~~ 3. $(\frac{1+x}{1-x})^{n}=(1+x)^{n}(1-x)^{-n}$

$ \begin{aligned} & =(1+n x+\frac{n(n-1)}{2 !} x^{2}+\ldots) \\ & \quad(1+(-n)(-x)+\frac{-n(-n-1)}{1.2 .}(-x)^{2}+\ldots) \end{aligned} $

$\therefore$ Coefficient of $x$ in the given expansion $=n+n=\mathbf{2 n}$.

~~ 4. $(1+2 x+3 x^{2}+4 x^{3}+\ldots .)^{1 / 2}=[(1-x)^{-2}]^{1 / 2}$

$=(1-x)^{-1}=1+x+x^{2}+x^{3}+\ldots$.

$\Rightarrow$ coefficient of each term $=1$

$\Rightarrow$ coefficient of $x^{n}=1$.

~~ 5. $=(1-3 x)^{2}(1-2 x)^{-1}$

$ \begin{aligned} & =(1-6 x+9 x^{2})(1+(-1))(-2 x)+\frac{(-1)(-1-1)}{1.2}(-2 x)^{2} \\ & +\frac{(-1)(-1-1)(-1-2)}{1.2 .3} \cdot(-2 x)^{3} \\ & \quad+\frac{(-1)(-1-1)(-1-2)(-1-3)}{1.2 .3 .4}(-2 x)^{4}+\ldots . \\ & =(1-6 x+9 x^{2})(1+2 x+4 x^{2}+8 x^{3}+16 x^{4}+\ldots .) \\ & \Rightarrow \text{ coefficient of } x^{4} \text{ in the given product } \end{aligned} $

$=(1 \times 16)+(-6 \times 8)+(9 \times 4)=-48+36+16=4$.

~~ 6. $\frac{1}{(x-1)^{2}(x-2)}=\frac{1}{-2(1-x)^{2}(1-x / 2)}$

$=-\frac{1}{2}[(1-x)^{-2}(1-\frac{x}{2})^{-1}]$

[Note: The steps of taking 2 out from $(x-2)$ ]

$=-\frac{1}{2}[(1+2 x+3 x^{2}+\ldots ..)(1+\frac{x}{2}+\ldots ..)]$ $\therefore \quad$ Coefficient of constant term $=-\frac{1}{2}$.

We have done it because expansion of $(x+a)^{n}$ when $n$ is a -ve integer or fraction is valid only when $|a|<1$.

$\therefore$ We write $(x+a)^{n}=a^{n}(1+\frac{x}{a})^{n}$, where $|\frac{x}{a}|<1$.

~~ 7. General term in the expansion of $(1-2 x)^{-1 / 2}$

$ \begin{aligned} T _{r+1} & =\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2) \cdots(-\frac{1}{2}-r+1)}{r !}(-2 x)^{r} \\ & =\frac{(-1)^{r}(\frac{1}{2})(\frac{3}{2})(\frac{5}{2}) \cdots \cdot(\frac{2 r-1}{2})}{r !}(-1)^{r} 2^{r} \cdot x^{r} \\ & =(-1)^{2 r} \frac{1 \cdot 3 \cdot 5 \cdot 7 \ldots \ldots(2 r-1)}{r ! 2^{r}} \cdot 2^{r} \cdot x^{r} \\ & =\frac{1.3 .5 \cdot 7 \ldots \ldots(2 r-1)}{r !} x^{r} \end{aligned} $

For this term to contain $x^{n}, r=n$

$ \begin{aligned} \therefore \quad T _{n+1} & =\frac{1 \cdot 3 \cdot 5 \cdot 7 \ldots \ldots(2 n-1)}{n !} x^{n} \\ & =\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots \ldots(2 n-2)(2 n-1)(2 n)}{2 \cdot 4 \cdot 6 \ldots \ldots(2 n-2)(2 n) n !} x^{n} \\ & =\frac{(2 n) !}{2^{n}(1 \cdot 2 \cdot 3 \ldots \ldots(n-1) \cdot n) n !} x^{n}=\frac{(2 n) !}{2^{n} \cdot n ! n !} x^{n} \end{aligned} $

$\therefore \quad$ Required coefficient $=\frac{2 n !}{2^{n}(n !)^{2}}$.

~~ 8. $\frac{1+2 x}{(1-2 x)^{2}}=(1+2 x)(1-2 x)^{-2}$

$ =(1+2 x)(1+-2(-2 x))+\frac{-2(-2-1)}{2 !}(-2 x)^{2} $

$+\ldots . . \frac{-2(-2-1)(-2-2) \ldots . .(-2-\overline{r-1}+1)}{(r-1) !}(-2 x)^{r-1}+$

$\frac{-2(-2-1) \ldots \ldots(-2-r+1)}{r !}(-2 x)^{r}+\ldots .\rbrace $

$\therefore \quad$ Coefficient of $x^{r}$ in the given expansion $=(r+1) 2^{r}+2 \cdot r \cdot 2^{r-1}=(r+1) 2^{r}+r \cdot 2^{r}=(\mathbf{2 r}+\mathbf{1}) \cdot \mathbf{2}^{r}$.

Properties of Binomial Coefficients

SOLVED EXAMPLES

Ex. 1 . If $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$, then find the values of

(i) $C_1+C_2+C_3+\ldots . .+C_n$

(ii) $C_1-C_2+C_3-\ldots . .+(-1)^{n} C_n$

(iii) $\frac{C_1}{C_o}+\frac{2 C_2}{C_1}+\frac{3 C_3}{C_2}+\ldots \ldots+\frac{n C_n}{C _{n-1}}$

Sol. (i) $\quad(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$

Sum of binomial coefficients $=2^{n}$

$\therefore C_0+C_1+C_2+\ldots . .+C_n=2^{n} \Rightarrow C_1+C_2+\ldots . .+C_n=2^{n}-C_0=2^{n}-1$

(ii) $\quad(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$

Put $x=-1$, in the above equation, we have

$ \begin{aligned} & 0=C_0-C_1+C_2-C_3+\ldots . .+(-1)^{n} C_n \Rightarrow C_0-C_1+C_2-C_3+\ldots . .+(-1)^{n} C_n=0 \\ \Rightarrow & C_1-C_2+C_3+\ldots . .-(-1)^{n} C_n=C_0=\mathbf{1} \end{aligned} $

(iii)

$ \begin{aligned} & \frac{C_1}{C_0}+\frac{2 C_2}{C_1}+\frac{3 C_3}{C_2}+\ldots . .+\frac{n C_n}{C _{n-1}}=\frac{n}{1}+\frac{2 n(n-1)}{2.1 . n}+\frac{3 n(n-1)(n-2) \times 2}{3.2 .1 . n(n-1)}+\ldots . .+\frac{n \times 1}{n} \\ = & n+(n-1)+(n-2)+\ldots . .+1=1+2+3+\ldots . .+(n-2)+(n-1)+n \\ = & \frac{\boldsymbol{n}(\boldsymbol{n}+\mathbf{1})}{\boldsymbol{2}}(\text{ Sum of an A.P.) } \end{aligned} $

Ex. 2 . If $(1+x-3 x^{2})^{10}=1+a_1 x+a_2 x^{2}+\ldots . .+a _{20} x^{20}$, then find $a_2+a_4+a_6+\ldots . .+a _{20}$.

(Kerala PET 2007)

Sol.

$ (1+x-3 x^{2})^{10}=1+a_1 x+a_2 x^{2}+\ldots . .+a _{20} x^{20} $

Putting $x=1$ in $(i)$, we get:

$ 1+a_1+a_2+\ldots . .+a _{20}=(-1)^{10}=1 $

Putting $x=-1$ in (i), we get

$ 1-a_1+a_2-\ldots . .+a _{20}=(-3)^{10}=3^{10} $

Adding (ii) and (iii), we get

$ \begin{gathered} 2(1+a_2+a_4+\ldots .+a _{20})=3^{10}+1 \\ \Rightarrow 2(a_2+a_4+\ldots . .+a _{20})=3^{10}-1 \Rightarrow a_2+a_4+\ldots . .+a _{20}=\frac{\mathbf{3}^{\mathbf{1 0}}-\mathbf{1}}{\mathbf{2}} . \end{gathered} $

Ex. 3 . Given that $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$ and $C_0+C_1+C_2+C_3+\ldots . .+C_n=1024$, find the value of $n$ and hence find the term in the expansion of $(x^{2}+\frac{1}{x})^{n}$ which contains $x^{11}$.

Sol. Sum of coefficients in the expansion of $(1+x)^{n}$

$ =C_0+C_1+C_2+\ldots . .+C_n=2^{n} $

Given,

$ 2^{n}=1024 \Rightarrow 2^{n}=2^{10} \Rightarrow n=10 . $

Now, general term $T _{r+1}$ in the expansion of $(x^{2}+\frac{1}{x})^{10}$

$ ={ }^{10} C_r(x^{2})^{10-r}(\frac{1}{x})^{r}={ }^{10} C_r x^{20-3 r} $

$\therefore \quad$ For term containing $x^{11}, 20-3 r=11 \Rightarrow 3 r=9 \Rightarrow r=3$

$\therefore \quad T_4=T _{3+1}={ }^{10} C_3 x^{20-9}=\frac{10 \times 9 \times 8}{3 \times 2} \times x^{11}=\mathbf{1 2 0} \boldsymbol{x}^{\mathbf{1 1}}$.

Ex. 4 . If $C_0, C_1, C_2, \ldots ., C_n$ be the coefficients in the expansion of $(1+x)^{n}$, then find the value of $C_0{ }^{2}+C_1{ }^{2}+C_2{ }^{2}$ $+\ldots . .+C_n^{2}$.

Sol. $\quad(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+C_3 x^{3}+\ldots . .+C_n x^{n}$

Also $\quad(1+x)^{n}=C_n x^{n}+C _{n-1} x^{n-1}+\ldots . .+C_2 x^{2}+C_1 x+C_0$

Multiplying both the sides of eqn. (i) and (ii), we have

$ (1+x)^{2 n}=(C_0+C_1 x+C_2 x^{2}+\ldots . .+C _{n-1} x^{n-1}+C_n x^{n}) \times(C_n x^{n}+C _{n-1} x^{n-1}+\ldots . .+C_2 x^{2}+C_1 x+C_0) $

Now equating coefficients of $x^{n}$ on both the sides, we have

$ \begin{aligned} &{ }^{2 n} C_n=C_0 C_n+C_1 C _{n-1}+C_2 C _{n-2}+\ldots . .+C _{n-1} C_1+C_n C_0 \\ & \Rightarrow \quad{ }^{2 n} C_n=C_0{ }^{2}+C_1{ }^{2}+C_2{ }^{2}+\ldots \ldots+(C _{n-1})^{2}+C_n{ }^{2} \\ & \Rightarrow C_0{ }^{2}+C_1{ }^{2}+C_2{ }^{2}+\ldots . .+C_n{ }^{2}={ }^{2 n} C_n=\frac{\underline{2 n}}{(\lfloor n)^{2}.}=\frac{(\mathbf{2 n}) !}{(\mathbf{n} !)^{2}} . \end{aligned} $

Ex. 5 . If $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$, then find the value of $C_0+\frac{1}{2} C_1+\frac{1}{3} C_2+\ldots . .+\frac{1}{(n+1)} C_n$.

Sol.

$ \begin{aligned} & C_0+\frac{1}{2} C_1+\frac{1}{3} C_2+\ldots . .+\frac{1}{n+1} C_n={ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots .+\frac{1}{n+1}{ }^{n} C_n \\ &= 1+\frac{n}{2}+\frac{n(n-1)}{3.2}+\ldots .+\frac{1}{n+1}=\frac{1}{n+1}[(n+1)+\frac{(n+1) n}{2}+\frac{(n+1) n(n-1)}{3.2}+\ldots . .+1] \\ &= \frac{1}{n+1}[{ }^{n+1} C_1+{ }^{n+1} C_2+\ldots . .+{ }^{n+1} C _{n+1}]=\frac{1}{n+1}[{ }^{n+1} C_0+{ }^{n+1} C_1+{ }^{n+1} C_2+\ldots .+{ }^{n+1} C _{n+1}-{ }^{n+1} C_0] \\ &= \frac{1}{n+1}(\mathbf{2}^{n+1}-\mathbf{1}) . \\ & \quad(\because \text{ Sum of coefficients of }(1+x)^{n}=2^{n}) \end{aligned} $

PRACTICE SHEET-4
(Properties of Binomial Coefficients)

~~ 1. The value of ${ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7$ is

(a) -1

(b) 0

(d) None of these

(Odisha JEE 2009)

~~ 2. The value of $({ }^{7} C_0+{ }^{7} C_1)+({ }^{7} C_1+{ }^{7} C_2)+\ldots .+({ }^{7} C_6+{ }^{7} C_7)$ is

(a) $2^{8}-1$

(b) $2^{8}+1$

(d) $2^{8}$

(Kerala CEE 2008)

~~ 3. If $n=5$, then

$({ }^{n} C_0)^{2}+({ }^{n} C_1)^{2}+({ }^{n} C_2)^{2}+\ldots .+({ }^{n} C_5)^{2}$ is equal to

(a) 250

(b) 254

(d) 245

~~ 4. The value of ${ }^{50} C_4+\sum _{r=1}^{6}{ }^{56-r} C_3$ is

(a) ${ }^{56} C_4$

(b) ${ }^{56} C_3$

(d) ${ }^{55} C_4$ 5. $\frac{{ }^{8} C_0}{6}-{ }^{8} C_1+{ }^{8} C_2 \cdot 6-{ }^{8} C_3 \cdot 6{ }^{2}+\ldots . .+{ }^{8} C_8 \cdot 6^{7}$ is equal to

(a) 0

(b) $6^{7}$

(d) $\frac{5^{8}}{6}$

(J&K CET 2009)

~~ 6. The sum of the last eight coefficients in the expansion of $(1+x)^{15}$ is

(DCE 2007)

(a) $2^{16}$

(b) $2^{15}$

(d) None of these ~~ 7. ${ }^{20} C_4+2 \cdot{ }^{20} C_3+{ }^{20} C_2-{ }^{22} C _{18}$ is equal to

(a) 0

(b) 1242

(d) 6345

(RPET 2005)

~~ 8. The sum of the series

${ }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-{ }^{20} C_3+\ldots \ldots .+{ }^{20} C _{10}$ is

(a) ${ }^{20} C _{10}$

(b) $\frac{1}{2}{ }^{20} C _{10}$

(d) ${ }^{20} C _{10}$

(AIEEE 2007)

~~ 9. In the expansion of $(1+x)^{30}$, the sum of the coefficients of the odd powers of $x$ is

(a) $2^{30}$

(b) $2^{31}$

(d) $2^{29}$ ~~ 10. If $(1+x+x^{2})^{n}=a_0+a_1 x+a_2 x^{2}+\ldots \ldots+a _{2 n} x^{2 n}$, then $a_0-a_1+a_2-a_3+\ldots . .+a _{2 n}$ is equal to

(a) 0

(b) 1

(d) ${ }^{2 n} C_n$ ~~ 11. If $(1-x+x^{2})^{n}=a_0+a_1 x+a_2 x^{2}+\ldots \ldots+a _{2 n} x^{2 n}$, then $a_0+a_2+a_4+\ldots . .+a _{2 n}$ equals

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

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

(d) $\frac{3^{n}+1}{2}$ ~~ 12. If $n$ is an odd positive integer and $(1+x+x^{2}+x^{3})^{n}=$ $\sum _{r=0}^{3 n} a_r x^{r}$, then $a_0-a_1+a_2-a_3+\ldots .-a _{3 n}$ equals

(a) -1

(b) 1

(d) 0

(Karnataka CET 2011)

~~ 13. If $(1+x-2 x^{2})^{6}=1+a_1 x+a_2 x^{2}+\ldots . .+a _{12} x^{12}$, then the value of $a_1+a_3+\ldots . .+a _{11}$ is

(a) 32

(b) -32

(d) -64 ~~ 14. Let $(1+x)^{n}=1+a_1 x+a_2 x^{2}+\ldots . .+a_n x^{n}$. If $a_1, a_2$ and $a_3$ are in A.P., then the value of $\boldsymbol{n}$ is

(a) 4

(b) 5

(d) 7

(BITSAT 2010)

~~ 15. If $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots .+C_n x^{n}$, then the sum $C_1+2 \cdot C_2+3 \cdot C_3+\ldots .+n \cdot C_n$ equals

(a) $2^{n-1}$

(b) $(n-1) 2^{n-1}$

(d) $(n-1) 2^{n}$ ~~ 16. If $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots .+C_n x^{n}$, then

$C_0+3 \cdot C_1+5 \cdot C_2+\ldots .+(2 n+1) \cdot C_n$ equals

(a) $2^{2 n-1}$

(b) $(n+1) 2^{n}$

(d) $(n-1) 2^{n-1}$ ~~ 17. $(1+\frac{C_1}{C_0})(1+\frac{C_2}{C_1})(1+\frac{C_3}{C_2}) \ldots . .(1+\frac{C_n}{C _{n-1}})$ is equal to

(a) $\frac{n+1}{n !}$

(b) $\frac{(n+1)^{n}}{(n-1) !}$

(d) $\frac{(n+1)^{n}}{n !}$

(Kerala CEE 2013)

~~ 18. If $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$, then

$C_0 C_1+C_1 C_2+\ldots .+C _{n-1} C_n$ is equal to

(a) $\frac{(2 n) !}{(n-1) !(n+1) !}$

(b) $\frac{(2 n-1) !}{(n-1) !(n+1) !}$

(d) None of these

(Odisha JEE 2008)

ANSWERS

~~ 1. (b) ~~ 2. (c) ~~ 3. (c) ~~ 4. (a) ~~ 5. (d) ~~ 6. (c) ~~ 7. $($ a) ~~ 8. (b) ~~ 9. (d) ~~ 10. (b) ~~ 11. $($ d) ~~ 12. (d) ~~ 13. (b) ~~ 14. $($ d $)$ ~~ 15. (c) ~~ 16. (b) ~~ 17. (d) ~~ 18. $($ a)

HINTS AND SOLUTIONS

~~ 1. ${ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7$

$={ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_9-{ }^{15} C_8$

$=0$

$(\because{ }^{n} C_r={ }^{n} C _{n-r})$

~~ 2. $({ }^{7} C_0+{ }^{7} C_1)+({ }^{7} C_1+{ }^{7} C_2)+\ldots . .+({ }^{7} C_6+{ }^{7} C_7)$

$={ }^{8} C_1+{ }^{8} C_2+{ }^{8} C_3+\ldots . .+{ }^{8} C_7 \quad(\because{ }^{n} C_r+{ }^{n} C _{r+1}={ }^{n+1} C _{r+1})$

$={ }^{8} C_1+{ }^{8} C_2+{ }^{8} C_3+\ldots . .+{ }^{8} C_7+({ }^{8} C_0+{ }^{8} C_8)-({ }^{8} C_0+{ }^{8} C_8)$

$={ }^{8} C_0+{ }^{8} C_1+{ }^{8} C_2+{ }^{8} C_3+\ldots . .+{ }^{8} C_7+{ }^{8} C_8-(2)$

$(\because{ }^{n} C_0={ }^{n} C_n=1)$

$=\mathbf{2}^{\mathbf{8}} \mathbf{- 2}$. $\quad(\because{ }^{n} C_0+{ }^{n} C_2+\ldots . .+{ }^{n} C_n=2^{n})$

~~ 3. $({ }^{n} C_0)^{2}+({ }^{n} C_1)^{2}+({ }^{n} C_2)^{2}+\ldots .+({ }^{n} C_5)^{2}$

$=({ }^{5} C_0)^{2}+({ }^{5} C_1)^{2}+({ }^{5} C_2)^{2}+\ldots .+({ }^{5} C_5)^{2}$

$=1^{2}+5^{2}+10^{2}+10^{2}+5^{2}+1^{2}$

$=1+25+100+100+25+1=\mathbf{2 5 2}$.

~~ 4. ${ }^{50} C_4+\sum _{r=1}^{6}{ }^{56-r} C_3$

$={ }^{50} C_4+{ }^{55} C_3+{ }^{54} C_3+{ }^{53} C_3+{ }^{52} C_3+{ }^{51} C_3+{ }^{50} C_3$

$={ }^{50} C_3+{ }^{50} C_4+{ }^{51} C_3+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{51} C_4+{ }^{51} C_3+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$(\because{ }^{n} C_r+{ }^{n} C _{r+1}={ }^{n+1} C _{r+1})$

$={ }^{52} C_4+{ }^{52} C_3+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{53} C_4+{ }^{53} C_3+{ }^{54} C_3+{ }^{55} C_3={ }^{54} C_4+{ }^{54} C_3+{ }^{55} C_3$

$={ }^{55} C_4+{ }^{55} C_3={ }^{56} \boldsymbol{C}_4$.

~~ 5. $\frac{{ }^{8} C_0}{6}-{ }^{8} C_1+{ }^{8} C_2 \cdot 6-{ }^{8} C_3 \cdot 6{ }^{2}+\ldots . .+{ }^{8} C_8 \cdot 6^{7}$

$ \begin{matrix} =\frac{1}{6}[{ }^{8} C_0-6 \cdot{ }^{8} C_1+6{ }^{2} \cdot{ }^{8} C_2-6^{3} \cdot{ }^{8} C_3+\ldots .+6^{8} \cdot{ }^{8} C_8] \\ =\frac{1}{6}[(1-6)^{8}]=\frac{5^{8}}{6} \quad[\because(1-x)^{n}={ }^{n} C_0-x \cdot{ }^{n} C_1+x^{2} \cdot{ }^{n} C_2. \\ .-x^{3} \cdot{ }^{n} C_3+\ldots . .+x^{n} \cdot{ }^{n} C_n] \end{matrix} $

~~ 6. $(1+x)^{15}={ }^{15} C_0+{ }^{15} C_1 \cdot x+{ }^{15} C_2 \cdot x^{2}+{ }^{15} C_3 \cdot x^{3}+\ldots . .+{ }^{15} C_n \cdot x^{n}$

As, sum of binomial coefficients $=2^{n}$

$ \begin{aligned} \therefore & { }^{15} C_0+{ }^{15} C_1+{ }^{15} C_2+\ldots . .+{ }^{15} C _{14}+{ }^{15} C _{15}=2^{15} \\ \Rightarrow & { }^{15} C _{15}+{ }^{15} C _{14}+{ }^{15} C _{13}+\ldots . .+{ }^{15} C_8+{ }^{15} C_8+{ }^{15} C _{14}+{ }^{15} C _{15} \\ & =2^{15} \quad(\because{ }^{n} C_r={ }^{n} C _{n-r}) \\ \Rightarrow & 2({ }^{15} C_8+{ }^{15} C_9+\ldots . .+{ }^{15} C _{14}+{ }^{15} C _{15})=2^{15} \\ \Rightarrow & ({ }^{15} C_8+{ }^{15} C_9+\ldots . .+{ }^{15} C _{15})=2^{14} \\ \Rightarrow & \text{ Required sum = Sum of last eight coefficients }=\mathbf{2}^{\mathbf{1 4}} . \end{aligned} $

~~ 7. ${ }^{20} C_4+2 \cdot{ }^{20} C_3+{ }^{20} C_2-{ }^{22} C _{18}$

$ \begin{aligned} & ={ }^{20} C_4+{ }^{20} C_3+{ }^{20} C_3+{ }^{20} C_2-{ }^{22} C _{18} \\ & ={ }^{21} C_4+{ }^{21} C_3-{ }^{22} C _{18} \quad(\because{ }^{n} C_r+{ }^{n} C _{r+1}={ }^{n+1} C _{r+1}) \\ & ={ }^{22} C_4-{ }^{22} C _{18}={ }^{22} C _{18}-{ }^{22} C _{18}=\mathbf{0} \\ & \quad(\because{ }^{n} C_r={ }^{n} C _{n-r} \Rightarrow{ }^{22} C_4={ }^{22} C _{22-4}={ }^{22} C _{18}) \end{aligned} $

~~ 8. $(1+x)^{20}={ }^{20} C_0+{ }^{20} C_1 x+{ }^{20} C_2 x^{2}+{ }^{20} C_3 x^{3}+\ldots$.

$ +{ }^{20} C _{19} x^{19}+{ }^{20} C _{20} x^{20} $

On putting $x=-1$, we get

$ \begin{aligned} 0= & { }^{20} C_0-{ }^{20} C_1+{ }^{20} C_2-{ }^{20} C_3+\ldots . .-{ }^{20} C _{19}+{ }^{20} C _{20} \\ \Rightarrow \quad 0= & { }^{20} C_0-{ }^{20} C_1+\ldots . .-{ }^{20} C_9+{ }^{20} C _{10}-{ }^{20} C_9+{ }^{20} C_8 \ldots . \\ & +{ }^{20} C_0 \\ & (\because{ }^{n} C_r={ }^{n} C _{n-r}) \end{aligned} $

$\Rightarrow 0=2({ }^{20} C_0-{ }^{20} C_1+\ldots . .-{ }^{20} C_9)+{ }^{20} C _{10}$

$\Rightarrow 2({ }^{20} C_0-{ }^{20} C_1+\ldots . .-{ }^{20} C_9)=-{ }^{20} C _{10}$

$\Rightarrow 2({ }^{20} C_0-{ }^{20} C_1+\ldots . .-{ }^{20} C_9+{ }^{20} C _{10})={ }^{20} C _{10}$.

(Adding $2 .{ }^{20} C _{10}$ on both the sides)

$\Rightarrow{ }^{20} C_0-{ }^{20} C_1+\ldots . .-{ }^{20} C_9+{ }^{20} C _{10}=\frac{1}{2}{ }^{20} C _{10}$.

~~ 9. Sum of the coefficients of odd powers of $x$ in $(1+x)^{30}$

$ =C_1+C_3+C_5+\ldots \ldots+C _{29}=2^{30-1}=\mathbf{2}^{\mathbf{2 9}} $

~~ 10. Given, $(1+x+x^{2})^{n}=a_0+a_1 x+a_2 x^{2}+\ldots . .+a _{2 n} x^{2}$

Putting $x=-1$ in the above equation, we have

$(1-1+1)^{n}=a_0-a_1+a_2-a_3+\ldots . .+a _{2 n}$

$\Rightarrow a_0-a_1+a_2-a_3+\ldots . .+a _{2 n}=1^{n}=\mathbf{1}$.

~~ 11. Given, $(1-x+x^{2})^{n}=a_0+a_1 x+a_2 x^{2}+\ldots . .+a _{2 n} x^{2 n}$

Putting $x=1$ in the given equation $(i)$, we have

$ 1=a_0+a_1+a_2+a_3+\ldots . .+a _{2 n} $

Putting $x=-1$ in the above given equation (i), we have

$(1-(-1)+1)^{n}=a_0-a_1+a_2-a_3+\ldots . .+a _{2 n}$

$\Rightarrow 3^{n}=a_0-a_1+a_2-a_3+\ldots . .+a _{2 n}$

Adding eqns (ii) and (iii), we have

$ \begin{aligned} & 1+3^{n}=2(a_0+a_2+a_4+\ldots .+a _{2 n}) \\ \Rightarrow & a_0+a_2+a_4+\ldots .+a _{2 n}=\frac{3^{n}+1}{2} \end{aligned} $

~~ 12. Given, $(1+x+x^{2}+x^{3})^{n}=\sum _{r=0}^{3 n} a_r x^{r}$

$\Rightarrow(1+x+x^{2}+x^{3})^{n}=a_0+a_1 x+a_2 x^{2}+\ldots . .+a _{3 n} x^{3 n}$

Putting $x=-1$ in the above given equation, we have

$ 0=a_0-a_1+a_2-a_3+\ldots \ldots-a _{3 n} $

~~ 13. Given, $(1+x-2 x^{2})^{6}=1+a_1 x+a_2 x^{2}+\ldots . .+a _{12 n} x^{12}$

Putting $x=1$, in $(i)$ we have

$ \begin{aligned} & (1+1-2 \times 1)^{6}=1+a_1+a_2+\ldots . .+a _{12} \\ \Rightarrow & 1+a_1+a_2+\ldots . .+a _{12}=0 \end{aligned} $

Putting $x=-1$, in $(i)$, we have

$ \begin{aligned} & (1+(-1)-2(-1)^{2})^{6}=1-a_1+a_2-\ldots . .+a _{12} \\ \Rightarrow & 1-a_1+a_2-\ldots . .+a _{12}=(-2)^{6}=64 \end{aligned} $

Subtracting eqn. (ii) from eqn. (i), we have

$ \begin{aligned} & 2(a_1+a_3+a_5+\ldots . a _{11})=-64 \\ \Rightarrow & a_1+a_3+a_5+\ldots . . a _{11}=-32 . \end{aligned} $

~~ 14. $(1+x)^{n}={ }^{n} C_0+{ }^{n} C_1 x+{ }^{n} C_2 x^{2}+{ }^{n} C_3 x^{3}+\ldots .+{ }^{n} C_n x^{n}$

$ =1+a_1 x+a_2 x^{2}+a_3 x^{3}+\ldots \ldots+a_n x^{n} $

Given, $a_1, a_2, a_3$ are in A.P.

$\Rightarrow{ }^{n} C_1,{ }^{n} C_2,{ }^{n} C_3$, are in A.P $\Rightarrow 2 .{ }^{n} C_2={ }^{n} C_1+{ }^{n} C_3$

$ \begin{aligned} & \Rightarrow 2 \times \frac{n !}{2 !(n-2) !}=\frac{n !}{(n-1) ! 1 !}+\frac{n !}{(n-3) ! 3 !} \\ & \Rightarrow \frac{2 \times n(n-1)}{2}=n+\frac{n(n-1)(n-2)}{3 \times 2} \\ & \Rightarrow 6(n^{2}-n)=6 n+n(n^{2}-3 n+2) \\ & \Rightarrow 6 n^{2}-6 n=6 n+n^{3}-3 n^{2}+2 n \Rightarrow n^{3}-9 n^{2}+14 n=0 \\ & \Rightarrow n(n^{2}-9 n+14)=0 \Rightarrow n(n-2)(n-7)=0 \\ & \Rightarrow n=2 \text{ or } 7 . \end{aligned} $

Rejecting $n=2$ as there are only three terms in the expansion of $(1+x)^{2}$, we have $n=7$.

~~ 15. $(1+x)^{n}={ }^{n} C_0+{ }^{n} C_1 x+{ }^{n} C_2 x^{2}+\ldots .+{ }^{n} C_n x^{n}$

$ =C_0+C_1 x+C_2 x^{2}+\ldots \ldots+C_n x^{n} $

$\therefore C_1+2 . C_2+3 . C_3+\ldots . .+n . C_n$

$=n+2 \times \frac{n(n-1)}{2 !}+3 \times \frac{n(n-1)(n-2)}{3 !}+\ldots . .+n$

$=n[1+(n-1)+\frac{(n-1)(n-2)}{2 !}+\ldots . .+1]$

$=n[{ }^{n-1} C_0+{ }^{n-1} C_1+{ }^{n-1} C_2+\ldots . .+{ }^{n-1} C _{n-1}]=\boldsymbol{n} \cdot \mathbf{2}^{\boldsymbol{n}-1}$.

~~ 16. $C_0+3 C_1+5 C_2+\ldots . .(2 n+1) C_n$

$=(C_0+C_1+C_2+\ldots . .+C_n)+(2 C_1+4 C_2+\ldots . .+2 n C_n)$

$=2^{n}+2(C_1+2 C_2+3 . C_3+\ldots . .+n C_n)$

$=2^{n}+2 \times n \cdot 2^{n-1}$

(Proved in Q.15)

$=2^{n}+n .2^{n}=\mathbf{2}^{n}(\boldsymbol{n}+\mathbf{1})$.

~~ 17. $(1+\frac{C_1}{C_0})(1+\frac{C_2}{C_1})(1+\frac{C_3}{C_2}) \ldots .(1+\frac{C_n}{C _{n-1}})$

$=(1+\frac{n}{1})(1+\frac{n(n-1)}{2 n})(1+\frac{2 n(n-1)(n-2)}{6 n(n-1)}) \cdots$

$(1+\frac{1}{n})$

$=(\frac{n+1}{1})(\frac{n+1}{2})(\frac{n+1}{3}) \cdots(\frac{n+1}{n})=\frac{(\boldsymbol{n}+\mathbf{1})^{n}}{n !}$.

~~ 18. Given, $(1+x)^{n}=C_0+C_1 x+C_2 x^{2}+\ldots . .+C_n x^{n}$

Also, $(x+1)^{n}=C_0 x^{n}+C_1 x^{n-1}+C_2 x^{n-2}+\ldots$.

$ +C _{n-1} x+C_n $

Multiplying $(i)$ by (ii) we get

$ \begin{aligned} (1+x)^{2 n}= & (C_0+C_1 x+C_2 x^{2}+\ldots . .+C _{n-1} x^{n-1}+C_n x^{n}) \\ & \times(C_0 x^{n}+C_1 x^{n-1}+C_2 x^{n-2}+\ldots \ldots+C _{n-1} x+C_n) \end{aligned} $

Now equating the coefficient of $x^{n-1}$ on both the sides, we get

$C_0 C_1+C_1 C_2+\ldots . .+C _{n-1} C_n={ }^{2 n} C _{n-1}=\frac{(\mathbf{2 n}) !}{(\boldsymbol{n}-\mathbf{1}) !(\boldsymbol{n}+\mathbf{1}) !}$.

SELF ASSESSMENT SHEET

~~ 1. If the $r$ th term in the expansion of $(\frac{x}{3}-\frac{2}{x^{2}})^{10}$ contains $x^{4}$, then $r$ is equal to

(a) 3

(b) 0

(d) 5 ~~ 2. If the coefficient of $r$ th and $(r+1)$ th terms in the expansion of $(3+7 x)^{29}$ are equal, then $r$ equals

(a) 15

(b) 21

(d) None of these ~~ 3. The sum of coefficients of the expansion $(\frac{1}{x}+2 x)^{n}$ is 6561 .

The coefficient of term independent of $x$ is

(a) $16 .{ }^{8} C_4$

(b) ${ }^{8} C_4$

(d) None of these

(BCECE 2009)

~~ 4. The value of $(\frac{{ }^{50} C_0}{1}+\frac{{ }^{50} C_2}{3}+\frac{{ }^{50} C_4}{5}+\ldots .+\frac{{ }^{50} C _{50}}{51})$ is

(a) $\frac{2^{50}}{51}$

(b) $\frac{2^{50}-1}{51}$

(d) $\frac{2^{50}-1}{50}$

(Kerala PET 2007) ~~ 5. ${ }^{47} C_4+\sum _{r=1}^{5}{ }^{52-r} C_3$ is equal to

(a) ${ }^{47} C_6$

(b) ${ }^{52} C_5$

(d) None of these

~~ 6. The coefficent of $x^{5}$ in $(1+2 x+3 x^{2}+\ldots)^{3 / 2}$ is

(a) 19

(b) 20

(d) 22 ~~ 7. If $|x|<\mid$, then the coefficient of $x^{n}$ in the expansion of $(1+x+x^{2}+x^{3}+\ldots . .)^{2}$ is

(a) $n-1$

(b) $n$

(d) $n+2$ ~~ 8. What is the coefficient of $x^{5}$ in the expansion of $(1-2 x+3 x^{2}-4 x^{3}+\ldots . \infty)^{-5}$.

(a) $\frac{10 !}{(5 !)^{2}}$

(b) $5^{-5}$

(d) $\frac{10 !}{6 ! 4 !}$

ANSWERS

~~ 1. (a) ~~ 2. (b) ~~ 3. (a) ~~ 4. (a) ~~ 5. (c) ~~ 6. $(c)$ ~~ 7. $(c)$ ~~ 8. (a)

HINTS AND SOLUTIONS

~~ 1. General Term $=T _{r+1}={ }^{n} C_r(x)^{n-r}(a)^{r}$ in the expansion of $(x+a)^{n}$.

$\therefore \quad T_r$ in the expansion of $(\frac{x}{3}-\frac{2}{x^{2}})^{10}$

$ \begin{aligned} & ={ }^{10} C _{r-1}(\frac{x}{3})^{10-(r-1)}(-\frac{2}{x^{2}})^{r-1} \\ & ={ }^{10} C _{r-1}(x)^{13-3 r}(3)^{-11+r}(-1)^{r} \cdot(2)^{r-1} \end{aligned} $

For $x^{4}, 13-3 r=4 \Rightarrow 3 r=9 \Rightarrow r=3$.

~~ 2. In the expansion of $(3+7 x)^{29}$

$ T _{r+1}={ }^{29} C_r(3)^{29-r}(7 x)^{r}={ }^{29} C_r 3^{29-r} \cdot 7^{r} \cdot x^{r} $

$\therefore$ Coefficient of $(r+1)$ th term $={ }^{29} C_r 3^{29-r} 7^{r}$

$\therefore$ Coefficient of $r$ th term $={ }^{29} C _{r-1} 3^{29-(r-1)} 7^{r-1}$

$ ={ }^{29} C _{r-1} 3^{30-r} 7^{r-1} $

Given, $\quad{ }^{29} C_r \times 3^{29-r} \times 7^{r}={ }^{29} C _{r-1} \times 3^{30-r} \times 7^{r-1}$

$\Rightarrow \frac{{ }^{29} C_r}{{ }^{29} C _{r-1}}=\frac{3}{7} \Rightarrow \frac{30-r}{r}=\frac{3}{7}$

$\Rightarrow 210-7 r=3 r \quad \Rightarrow \quad 210=10 r \quad \Rightarrow \quad r=21$.

~~ 3. Sum of the coefficients of the expansion $(\frac{1}{x}+2 x)^{n}=6561$

Putting $x=1$,

$ (1+2)^{n}=6561 \Rightarrow 3^{n}=3^{8} \Rightarrow n=8 $

$\therefore \quad T _{r+1}$ in the expansion of $(\frac{1}{x}+2 x)^{8}$

$ ={ }^{8} C_r(\frac{1}{x})^{8-r}(2 x)^{r}={ }^{8} C_r 2^{r} x^{2 r-8} $

Since this term is independent of $x, 2 r-8=0 \Rightarrow r=4$.

$\therefore$ Reqd. term $={ }^{8} C_4 \cdot 2^{4}=\mathbf{1 6} \cdot{ }^{8} \boldsymbol{C} _{\mathbf{4}}$. ~~ 4. $\frac{{ }^{50} C_0}{1}+\frac{{ }^{50} C_2}{3}+\frac{{ }^{50} C_4}{5}+\ldots . .+\frac{{ }^{50} C _{50}}{51}$

$.=\frac{1}{1}+\frac{50 \times 49}{3 \times 2 !}+\frac{50 \times 49 \times 48 \times 47}{5 \times 4 !}+\ldots . .+\frac{1}{51})$

$=\frac{1}{51}[51+\frac{51 \times 50 \times 49}{3 !}+\frac{51 \times 50 \times 49 \times 48 \times 47}{5 !}+\ldots ..$.

$=\frac{1}{51}[{ }^{51} C_1+{ }^{51} C_3+{ }^{51} C_5+\ldots ..]=\frac{1}{51} \times 2^{51-1}=\frac{2^{50}}{51}$

~~ 5. ${ }^{47} C_4+\sum _{r=1}^{5}{ }^{52-r} C_3$

$(\because.$ Sum of odd coefficient $.=2^{n-1})$

$={ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+{ }^{48} C_3+{ }^{47} C_3+{ }^{47} C_4$

$={ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+{ }^{48} C_3+{ }^{48} C_4$

$(\because{ }^{n} C_r+{ }^{n} C _{r+1}={ }^{n+1} C _{r+1})$

$={ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+{ }^{49} C_4={ }^{51} C_3+{ }^{50} C_3+{ }^{50} C_4$

$={ }^{51} C_3+{ }^{51} C_4={ }^{52} \boldsymbol{C} _{\mathbf{4}}$.

~~ 6. $(1+2 x+3 x^{2}+\ldots . .)^{3 / 2}=((1-x)^{-2})^{3 / 2}=(1-x)^{-3}$

$=1+3 x+\frac{3.4}{2 !} x^{2}+\frac{3.4 .5}{3 !} x^{3}+\frac{3.4 \cdot 5 \cdot 6}{4 !} x^{4}+\frac{3 \cdot 4 \cdot 5 \cdot 6.7}{5 !} x^{5}+\ldots$.

$\therefore \quad$ Coefficient of $x^{5}=\frac{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{5 !}=\mathbf{2 1}$.

~~ 7. $.(1+x+x^{2}+x^{3}+\ldots .)^{2}={(1-x)^{-1})}^{2}=(1-x)^{-2}$

$=1+2 x+3 x^{2}+4 x^{3}+\ldots . .+(n+1) x^{n}+\ldots .$.

$\therefore$ Coefficient of $x^{n}$ in this expansion $=(\boldsymbol{n}+\mathbf{1})$.

~~ 8. $(1-2 x+3 x^{2}-4 x^{3}+\ldots . . \infty)^{-5}={(1+x)^{-2}}^{-5}=(1+x)^{10}$

$\therefore \quad$ Coefficient of $x^{5}={ }^{10} C_5=\frac{10 !}{5 ! 5 !}=\frac{\mathbf{1 0}}{(\mathbf{5} !)^{\mathbf{2}}}$.

$ (\because(1+x)^{n}=1+{ }^{n} C_1 x+{ }^{n} C_2 x^{2}+{ }^{n} C_3 x^{2}+\ldots .) $

Mathematical Induction

KEY FACTS

~~ 1. A statement $T(n)$ is true for all $n \in N$, where $N$ is the set of natural numbers, provided:

(i) $T(1)$ is true (ii) $T(k)$ is true $\Rightarrow T(k+1)$ is true.

~~ 2. The proof of a proposition $T(n)$ by the method of mathematical induction consists of the following steps:

(a) Step I: (Basic Step): Actual verification of the proposition $[T(1)],[T(2)]$, etc., for particular positive integral values of $n$ say $n=1,2, \ldots$.

(b) Step II: (Induction Step): Assuming the proposition to be true for some positive integral value $k$ of $n$ i.e., $T(k)$ and then proving that it is true for the value $(k+1)$ which is the next higher integer, i.e., proving $T(k+1)$ true whenever $T(k)$ holds.

SOLVED EXAMPLES

Type I: Summation of Series

Ex. 1 . Using the method of induction, show that $1+2+3+\ldots+n=\frac{1}{2} n(n+1)$, for all $n \in N$.

Sol. Let $T(n)=1+2+3+\ldots+n=\frac{1}{2} n(n+1)$

Basic Step: For $n=1$,

LHS $=T(1)=1, \quad$ RHS $=\frac{1}{2} \times 1 \times 2=1 \Rightarrow$ LHS $=$ RHS $\Rightarrow T(1)$ is true.

Induction Step: Assume that $T(k)$ is true, i.e.,

$1+2+3+\ldots+k=\frac{1}{2} k(k+1)$

To obtain $T(k+1)$, we add $(k+1)$ th term $=(k+1)$ to both the sides, i.e.,

$1+2+3+\ldots+k+(k+1)=\frac{1}{2} k(k+1)+(k+1)$

$\Rightarrow 1+2+3+\ldots+k+(k+1)=(k+1)(\frac{k}{2}+1)$

$\Rightarrow 1+2+3+\ldots+k+(k+1)=\frac{1}{2}(k+1)(k+2)$

$\Rightarrow$ Thus the statement $T(n)$ is true for $n=k+1$ under the assumption that $T(k)$ is true. Therefore, by the principle of mathematical induction, the statement is true for every + ve integer $n$.

Ex. 2 . Prove that for all +ve integral values of $n, 1+3+5+\ldots .+(2 n-1)=n^{2}$.

Sol. Let $T(n)$ be the statement: $1+3+5+\ldots+(2 n-1)=n^{2}$

Basic Step: For $n=1$, LHS $=1$, RHS $=1^{2} \Rightarrow$ LHS $=$ RHS $\Rightarrow T(1)$ is true

Induction Step: Assume that $T(k)$ is true, i.e.,

$1+3+5+\ldots+(2 k-1)=k^{2}$

To obtain $T(k+1)$, add the $(k+1)$ th term $=2(k+1)-1=2 k+2-1=2 k+1$ to both the sides. Then,

$1+3+5+\ldots+(2 k-1)+(2 k+1)=k^{2}+2 k+1 \Rightarrow 1+3+5+\ldots$ to $(k+1)$ terms $=(k+1)^{2}$

Thus the statement is true for $n=k+1$ under the assumption that statement is true for $n=k$

Therefore, the statement $1+3+5+\ldots$ to $n$ terms $=n^{2}$ for every positive integer $n$.

Ex. 3 . Prove that for every natural number $n$.

$1^{3}+2^{3}+3^{3}+\ldots+n^{3}=[\frac{n(n+1)}{2}]^{2}$

Sol. Let $T(n)$ be the statement,

$1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}$

Basic Step: For $n=1,1^{3}=\frac{1^{2}(1+1)^{2}}{4}=1 \Rightarrow T(1)$ is true.

Induction Step: Let $T(k)$ hold for a natural number $k$, that is

$1^{3}+2^{3}+3^{3}+\ldots+k^{3}=\frac{k^{2}(k+1)^{2}}{4}$

Now, to obtain $T(k+1)$, add the $(k+1)$ th term $=(k+1)^{3}$ to both the sides of $T(k)$, i.e.,

$ \begin{aligned} 1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3} & =\frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}=\frac{(k+1)^{2}}{4}[k^{2}+4(k+1)]=\frac{(k+1)^{2}(k+2)^{2}}{4} \\ & =[\frac{(k+1)(k+2)}{2}]^{2} \end{aligned} $

Hence $T(k+1)$ is true, whenever $T(k)$ is true.

Ex. 4 . Use the principal of mathematical induction to prove the following statement true for all $n \in N$.

$ x+4 x+7 x+\ldots+(3 n-2) x=\frac{1}{2} n(3 n-1) x $

Sol. Let $T(n)$ be the statement:

$x+4 x+7 x+\ldots+(3 n-2) x=\frac{1}{2} n(3 n-1) x$

Basic Step: For $n=1$,

$x=\frac{1}{2} \times 1 \times(3 \times 1-1) \times x \Rightarrow x=x \Rightarrow T(1)$ is true.

Induction Step: Assume $T(k)$ holds for a natural number $k$, i.e.,

$x+4 x+7 x+\ldots+(3 k-2) x=\frac{1}{2} k(3 k-1) x$

Now to show that $T(k+1)$ holds, add the $(k+1)$ th term $=[3(k+1)-2] x=(3 k+1) x$ to both the sides of $T(k)$, i.e., $x+4 x+7 x+\ldots+(3 k-2) x+(3 k+1) x=\frac{1}{2} k(3 k-1) x+(3 k+1) x$

$ \begin{aligned} & =\frac{1}{2}[k(3 k-1) x+2(3 k+1) x]=\frac{1}{2}[(3 k^{2}+5 k+2) x]=\frac{1}{2}(k+1)(3 k+2) x \\ & =\frac{1}{2}(k+1)[3(k+1)-1] x \\ \Rightarrow & T(k+1) \text{ is true, whenever } T(k) \text{ is true. } \end{aligned} $

Ex. 5 . Prove by the method of mathematical induction that $a+(a+d)+(a+2 d)+\ldots+(a+(n-1) d)$ $=\frac{n}{2}{2 a+(n-1) d}$ for all $n \in N$, where $a, d \in R$.

Sol. Let $T(n)$ be the statement

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

Basic Step: For $n=1$, $LHS=a, RHS=\frac{1}{2}[2 a]=a$

$\Rightarrow$ LHS $=$ RHS $\Rightarrow T(1)$ is true.

Induction Step: Let $T(k)$ hold true, i.e.,

$a+(a+d)+(a+2 d)+\ldots+(a+(k-1) d)=\frac{k}{2}[2 a+(k-1) d]$

Now to show that $T(k+1)$ holds true, we add the $(k+1)$ th term, i.e., $a+{(k+1)-1} d=a+k d$ to both the sides of $T(k)$, i.e.,

$ \begin{aligned} a+(a+d) & +(a+2 d)+\ldots+(a+(k-1) d)+(a+k d) \\ = & \frac{k}{2}[2 a+(k-1) d]+(a+k d)=a k+\frac{k(k-1) d}{2}+a+k d \\ = & a(k+1)+\frac{1}{2}{k(k-1) d+2 k d}=(k+1) a+\frac{1}{2}{k^{2} d+k d} \\ = & (k+1) a+\frac{1}{2} k(k+1) d=\frac{(k+1)}{2}[2 a+{(k+1)-1} d] . \end{aligned} $

Thus, $T(k+1)$ is true, whenever $T(k)$ is true.

Ex. 6 . Using the method of mathematical induction, show that for all $n \in N, a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a(1-r^{n})}{(1-r)}$, $r \neq 1$.

Sol. Let $T(n)$ be the statement:

$a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a(1-r^{n})}{(1-r)}, r \neq 1$

Basic Step: For $n=1, \Rightarrow$ LHS $=a$, RHS $=\frac{a(1-r^{1})}{1-r}=a$.

$ \Rightarrow \text{ LHS }=\text{ RHS } \Rightarrow T(1) \text{ is true. } $

Induction Step: Let the statement hold true for $n=k$, i.e., let $T(k)$ be true, i.e., $a+a r+a r^{2}+\ldots+a r^{k-1}=\frac{a(1-r^{k})}{1-r}$ Then to show $T(k+1)$ holds, add the $(k+1)$ th term $=a r^{(k+1)-1}=a r^{k}$ to both the sides of $T(k)$, i.e., $a+a r+a r^{2}+\ldots+a r^{k-1}+a r^{k}=\frac{a(1-r^{k})}{(1-r)}+a r^{k}$

$ =\frac{a-a r^{k}+a r^{k}(1-r)}{(1-r)}=\frac{a-a r^{k}+a r^{k}-a r^{k+1}}{(1-r)}=\frac{a-a r^{k+1}}{(1-r)}=\frac{a(1-r^{k+1})}{(1-r)} $

Thus, $T(k+1)$ is true, whenever $T(k)$ holds true.

Ex. 7 . Prove that for all +ve integral values of $n, \frac{1}{1.2}+\frac{1}{2.3}+\ldots . .+\frac{1}{n(n+1)}=\frac{n}{n+1}$.

Sol. Let $T(n)$ be the statement: $\frac{1}{1.2}+\frac{1}{2.3}+\ldots . .+\frac{1}{n(n+1)}=\frac{n}{n+1}$

Basic Step: For $n=1, \frac{1}{1 \cdot 2}=\frac{1}{1+1} \Rightarrow T(1)$ is ture.

Induction Step: Assume $T(k)$ is true, i.e., $\frac{1}{1.2}+\frac{1}{2.3}+\ldots . .+\frac{1}{k(k+1)}=\frac{k}{k+1}$

To obtain $T(k+1)$, add the $(k+1)$ th term, i.e., $\frac{1}{(k+1)(k+2)}$ to both sides of $T(k)$. Then,

$\frac{1}{1.2}+\frac{1}{2.3}+\ldots . .+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$

$=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{k^{2}+2 k+1}{(k+1)(k+2)}=\frac{(k+1)^{2}}{(k+1)(k+2)}=\frac{(\boldsymbol{k}+\mathbf{1})}{(\boldsymbol{k}+\mathbf{2})}$

Thus $T(k+1)$ is true with the assumption that $T(k)$ is true. Hence the statement $T(n)$ holds for all positive integral values of $n$.

Ex. 8 . Prove by the principle of induction that

~~ 1. 4. $7+2.5 .8+3.6 .9+\ldots+n(n+3)(n+6)=\frac{n}{4}(n+1)(n+6)(n+7)$.

Sol. Let $T(n)$ denote the given statement

~~ 1. $4.7+2.5 .8+3.6 .9+\ldots+n(n+3)(n+6)=\frac{n}{4}(n+1)(n+6)(n+7)$

Basic Step: For $n=1$, LHS $=1.4 .7=28$

$ RHS=\frac{1}{4}(1+1)(1+6)(1+7)=28 $

$\Rightarrow$ LHS $=$ RHS $\Rightarrow T(1)$ is true.

Induction Step: Assume $T(k)$ is true for all $k \in N$, i.e.,

$ 1.4 .7+2.5 .8+3.6 .9+\ldots+k(k+3)(k+6)=\frac{k}{4}(k+1)(k+6)(k+7) $

Now we shall show that $T(k+1)$ is also true.

To obtain $T(k+1)$ add the $(k+1)$ th term, i.e., $(k+1)(k+1+3)(k+1+6)=(k+1)(k+4)(k+7)$ to both the sides of $T(k)$. Then,

$ \begin{aligned} & \text{ 1. 4. } 7+2.5 .8+3.6 .9+\ldots+k(k+3)(k+6)+(k+1)(k+4)(k+7) \\ & =\frac{k}{4}(k+1)(k+6)(k+7)+(k+1)(k+4)(k+7)=(k+1)(k+7)[\frac{k}{4}(k+6)+(k+4)] \\ & =\frac{(k+1)(k+7)}{4}[k^{2}+6 k+4 k+16]=\frac{(k+1)}{4}(k+7)(k^{2}+10 k+16) \\ & =\frac{(k+1)}{4}(k+7)(k+2)(k+8)=\frac{1}{4}(k+1)(k+2)(k+7)(k+8) \\ & \Rightarrow T(k+1) \text{ is true, assuming } T(k) \text{ is true. } \\ & \Rightarrow T(n) \text{ is true for all } n \in N . \end{aligned} $

Ex. 9 . Prove by the principle of mathematical induction that $\boldsymbol{n}<2^{n}$ for all $n \in N$.

Sol. Let the statement $T(n)=n<2^{n}$.

Basic Step: For $n=1,1<2^{1} \Rightarrow T(1)$ is true.

Induction Step: Let $T(k)$ be true $\Rightarrow k<2^{k}$ for all $k \in N$.

$ \begin{aligned} k<2^{k} & \Rightarrow 2 k<2.2^{k} \\ & \Rightarrow 2 k<2^{k+1} \Rightarrow(k+k)<2^{k+1} \\ & \Rightarrow(k+1) \leq(k+k)<2^{k+1} \\ & \Rightarrow(k+1)<2^{k+1} \\ & \Rightarrow T(k+1) \text{ is true, whenever } T(k) \text{ is true. } \end{aligned} $

$ \Rightarrow(k+1) \leq(k+k)<2^{k+1} \quad(\because k \in N \Rightarrow k \geq 1) $

$\therefore T(n)$ is true $\forall n \in N$.

Ex. 10 . Prove by the principle of mathematical induction that $1+2+3+\ldots+n<\frac{(2 n+1)^{2}}{8}$.

Sol. Let $T(n)$ be the statement

$ 1+2+3+\ldots+n<\frac{(2 n+1)^{2}}{8} $

Basic Step: For $n=1$, we have $\frac{(2 n+1)^{2}}{8} \frac{(2 \times 1+1)^{2}}{8}=\frac{9}{8}>1 \Rightarrow T(1)$ is true.

Induction Step: Let $T(k)$ be true. Then,

$ 1+2+\ldots+k<\frac{(2 k+1)^{2}}{8} $

Now we need to prove $T(k+1)$ to be true. To obtain $T(k+1)$ add the $(k+1)$ th term $=(k+1)$ to both the sides of $T(k)$. Then,

$ \begin{aligned} & 1+2+3+\ldots+k+(k+1)<\frac{(2 k+1)^{2}}{8}+(k+1) \\ \Rightarrow & 1+2+3+\ldots+k+(k+1)<\frac{4 k^{2}+12 k+9}{8}=\frac{(2 k+3)^{2}}{8} \\ \Rightarrow & 1+2+3+\ldots+k+(k+1)<\frac{[2(k+1)+1]^{2}}{8} \end{aligned} $

Hence $T(k+1)$ is true, whenever $T(k)$ is true $\Rightarrow T(n)$ is true for all $n \in N$.

Ex. 11 . Using the principle of mathematical induction prove that $1 \underline{1}+2 \underline{2}+3 \underline{3}+\ldots . .+n \underline{n}=\mid n+1-1$ for all $n \in N$.

Sol. Let the statement $T(n)$ be

$ 1|1+2 \underline{2}+3 \underline{3}+\ldots+n \underline{n}=| n+1-1 $

Basic Step: For $n=1$, LHS $=1\lfloor 1 \times 1=1$

$ \text{ RHS }=\lfloor 1+1-1=\lfloor 2-1=2-1=1 \Rightarrow \text{ LHS }=\text{ RHS } \Rightarrow T(1) \text{ is true. } $

Induction Step: Assume the statement $T(k)$ to be true for $n=k, k \in N$. Then,

$ 1 \underline{1}+2 \underline{2}+3 \underline{3}+\ldots+k \underline{k}=\underline{k}+1-1 $

Now we need to prove $T(k+1)$ to be true.

To obtain $T(k+1)$, add the $(k+1)$ th term $=(k+1) \mid k+1$ to both sides of $T(k)$, i.e.,

$\Rightarrow 1\lfloor\underline{1}+2 \underline{2}+3 \underline{3}+\ldots+k|\underline{k}+(k+1) \underline{k+1}=| k+1-1+(k+1) \mid k+1$

$ \begin{aligned} & =|k+1+(k+1)| k+1-1 \\ & =k+1(1+k+1)-1 \\ & =k+1(k+2)-1 \\ & =k+2-1=(k+1)+1-1 \end{aligned} $

$\Rightarrow$ The result is true for $n=k+1$

$\Rightarrow T(k+1)$ is true on the assumption that $T(k)$ is true.

$\Rightarrow T(n)$ holds for all $n \in N$.

Ex. 12 . If $A= \begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix} $, then prove that $A^{n}= \begin{bmatrix} 1+2 n & -4 n \\ n & 1-2 n\end{bmatrix} $.

(WBJEE 2008)

Sol. Let the statement $T(n)$ be: If $A= \begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix} $, then $A^{n}= \begin{bmatrix} 1+2 n & -4 n \\ n & 1-2 n\end{bmatrix} $

Basic Step: For $n=1, A= \begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix} $

$ A^{1}= \begin{bmatrix} 1+2 \times 1 & -4 \times 1 \\ 1 & 1-2 \times 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \Rightarrow T(1) \text{ is true } $

Induction Step: Assume $T(k)$ to be true, i.e.,

If $A= \begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix} $, then $A^{k}= \begin{bmatrix} 1+2 k & -4 k \\ k & 1-2 k\end{bmatrix} $

Now we need to show $T(k+1)$ is true. $\quad A^{k}= \begin{bmatrix} 1+2 k & -4 k \\ k & 1-2 k\end{bmatrix} $

$ \begin{aligned} \therefore A^{k+1}=A^{k} \cdot A & = \begin{bmatrix} 1+2 k & -4 k \\ k & 1-2 k \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k+2 k-1 \end{bmatrix} = \begin{bmatrix} 3+2 k & -4-4 k \\ k+1 & -1-2 k \end{bmatrix} \\ & = \begin{bmatrix} 1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1) \end{bmatrix} . \Rightarrow T(k+1) \text{ is true, whenever } T(k) \text{ is true. } \end{aligned} $

PRACTICE SHEET-1

~~ 1. Let $S(k)=1+3+5+\ldots .+(2 k-1)=3+k^{2}$. Then, which of the following is true?

(a) $S(1)$ is correct

(b) $S(k) \Rightarrow S(k+1)$

(d) Principle of mathematical induction can be used to prove the above formula.

(AIEEE 2004)

~~ 2. If ’ $n$ ’ be any positive integer, then $n(n+1)(2 n+1)$ is

(a) an odd integer

(b) an integral multiple of 6

(d) None of these

(EAMCET 2005)

~~ 3. For all natural number $n, 2+4+6+\ldots . .+2 n$ equals

(a) $2(n+1)$

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

(d) $(n+2)(n+4)$ ~~ 4. For all $n \in N$, the sum of the series $1^{2}+2^{2}+3^{2}+\ldots .+n^{2}$ is equal to

(a) $\frac{n}{3}(n+1)(2 n+1)$

(b) $\frac{n}{6}(n+1)(n+3)$

(d) $\frac{n}{6}(n+1)(2 n+1)$ ~~ 5. For all $n \in N$, the sum $1.3+3.5+5.7+\ldots .+(2 n-1)(2 n+1)$ equals

(a) $\frac{n(2 n^{2}+3 n+1)}{6}$

(b) $\frac{n(4 n^{2}+6 n-1)}{3}$

(d) $\frac{1}{3} n^{2}(4 n^{2}+5)$

ANSWERS

~~ 1. (b) ~~ 2. (b) ~~ 3. $(c)$ ~~ 4. (d) ~~ 5. (b)

HINTS AND SOLUTIONS

~~ 1. $S(k)=1+3+5+\ldots .+(2 k-1)=3+k^{2}$

Putting $k=1$ on both the sides, we get

LHS $=1$, RHS $=3+1=4 \Rightarrow$ LHS $\neq$ RHS $\Rightarrow S(1)$ is not true.

Assume $S(k)=1+3+5+\ldots+(2 k-1)=3+k^{2}$ is true. Then,

To find $S(k+1)$, add the $(k+1)$ th term $=(2(k+1)-1)=$ $2 k+1$ on both the sides of $S(k)$.

$\therefore S(k+1)=1+3+5+\ldots .+(2 k-1)+(2 k+1)=3+k^{2}+$ $2 k+1$

$\Rightarrow 1+3+5+\ldots .+(2 k-1)+(2 k+1)=3+(k+1)^{2}$

$\Rightarrow S(k+1)$ is also true.

$\therefore S(k) \Rightarrow S(k+1)$ is true.

~~ 2. When $n=1, n(n+1)(2 n+1)=(1)(2)(3)=6$, which is an integral multiple of 6. It is neither an odd integer nor a perfect square. Using the principle of mathematical induction, we shall now show that the expression $n(n+1)$ $(2 n+1)$ is an integral multiple of $6 \forall n \in N$. Assume $T(n)=n(n+1)(2 n+1)=6 x$ where $x \in N$.

Basic Step: $T(1)$ is true as shown above.

Induction Step: Let $T(k)$ be true for all $k \in N$.

$\Rightarrow \quad k(k+1)(2 k+1)=6 x$, where $x \in N$

For $T(k+1)$, we replace $k$ by $(k+1)$ in the given expression, i.e.,

$ \begin{aligned} T(k+1) & =(k+1)(k+2)(2(k+1)+1) \\ & =(k+1)(k+2)((2 k+1)+2) \\ & =(k+1)(k+2)(2 k+1)+2(k+1)(k+2) \\ & =k(k+1)(2 k+1)+2(k+1)(2 k+1) \\ & =k(k+1)(2 k+1)+2(k+1)[(2 k+1)+(k+2)] \\ & =6 x+2(k+1)(3 k+3)=6 x+6(k+1)^{2} \\ & =6(x+(k+1)^{2}) \\ & =6 \times \text{ a positive integer } \end{aligned} $

$\therefore T(k)$ is true $\Rightarrow T(k+1)$ is true.

$\therefore T(n)$ is true for all $n \in N$.

~~ 3. Let $S_n=2+4+6+\ldots .+2 n$

When $n=1, S_n=2$

Now from the options given, when $n=1$,

$2(n+1)=4, \frac{1}{2} n(n+2)=\frac{3}{2}, n(n+1)=2,(n+2)(n+4)=15$

$\therefore \quad S_n \neq 2(n+1), S_n \neq \frac{1}{2} n(n+2), S_n \neq(n+2)(n+4)$ for $n=1$

$ S_n=n(n+1) \text{ for } n=1 $

$\therefore \quad$ We need to prove $2+4+6+\ldots .+2 n=n(n+1) \forall n \in N$.

Let $T(n)=2+4+6+\ldots .+2 n=n(n+1)$

Basic Step: For $n=1$, LHS = $2 \times 1=2$, RHS = $1 \times(1+1)=2$ $\Rightarrow$ LHS $=$ RHS $\Rightarrow T(1)$ is true. Induction Step: Assume $T(k)$ is true, i.e.,

$2+4+6+\ldots .+2 k=k(k+1)$

To obtain $T(k+1)$, we add the $(k+1)$ th term, i.e, $2(k+1)$ to both the sides of $T(k)$, i.e.,

$2+4+6+\ldots+2 k+2(k+1)$

$=k(k+1)+2(k+1)$

$=(k+1)(k+2)=(k+1)((k+1)+1)$

Thus the statement $T(n)$ is true for $n=k+1$, whenever it is true for $n=k$.

Therefore by the principle of mathematical induction it is true for all $\boldsymbol{n} \in \boldsymbol{N}$.

~~ 4. Let $S_n=1^{2}+2^{2}+3^{2}+\ldots . .+n^{2}$.

When $n=1, S_n=1^{2}=1$.

Now from the options given, when $\boldsymbol{n}=\mathbf{1}$,

$\frac{n}{3}(n+1)(2 n+1)=\frac{1}{3} \times 2 \times 3=2$

$\frac{n}{6}(n+1)(n+3)=\frac{1}{6} \times 2 \times 4=\frac{4}{3}$

$\frac{n}{3}(2 n-1)(n+2)=\frac{1}{3} \times 1 \times 3=1$

$\frac{n}{6}(n+1)(2 n+1)=\frac{1}{6} \times 2 \times 3=1$

$\therefore S_n \neq \frac{n}{3}(n+1)(2 n+1), S_n \neq \frac{n}{6}(n+1)(n+3)$ for $n=1$

$S_n=\frac{n}{3}(2 n-1)(n+2)$ and $S_n=\frac{n}{6}(n+1)(2 n+1)$ for $n=1$

When $n=2, S_n=1^{2}+2^{2}=5$

$\frac{n}{3}(2 n-1)(n+2)=\frac{2}{3} \times 3 \times 4=8 \neq S_n$

$\frac{n}{6}(n+1)(2 n+1)=\frac{2}{6} \times 3 \times 5=5$

$\therefore S_n=\frac{n}{6}(n+1)(2 n+1)$ when $n=2$.

Clearly $S_n=\frac{n}{6}(n+1)(2 n+1)$ for $n=1$, and $n=2$.

$\therefore$ We shall show by mathematical induction that

$1^{2}+2^{2}+3^{2}+\ldots .+n^{2}=\frac{n}{6}(n+1)(2 n+1)$ for all $n \in N$.

Let $T(n)=1^{2}+2^{2}+3^{2}+\ldots .+n^{2}=\frac{n}{6}(n+1)(2 n+1)$

Basic Step: For $n=1$,

$ \begin{aligned} & \text{ LHS }=1^{2}=1 \\ & \text{ RHS }=\frac{1}{6} \times 2 \times 3=1 \end{aligned} $

$\Rightarrow$ LHS $=$ RHS $\Rightarrow T(1)$ is true

Induction Step: Let $T(k)$ be true for all $k \in N$.

$\Rightarrow 1^{2}+2^{2}+3^{2}+\ldots .+k^{2}=\frac{k}{6}(k+1)(2 k+1)$

To obtain $T(k+1)$, add the $k$ th term $=(k+1)^{2}$ to both the sides of $T(k)$ i.e.,

$ \begin{aligned} 1^{2}+ & 2^{2}+3^{2}+\ldots+k^{2}+(k+1)^{2} \\ & =\frac{k}{6}(k+1)(2 k+1)+(k+1)^{2} \\ & =\frac{1}{6}(k+1)[k(2 k+1)+6(k+1)] \\ & =\frac{1}{6}(k+1)[2 k^{2}+k+6 k+6] \\ & =\frac{1}{6}(k+1)(2 k^{2}+7 k+6) \\ & =\frac{1}{6}(k+1)(k+2)(2 k+3) \\ & =\frac{1}{6}(k+1)(k+2)[2(k+1)+1] \end{aligned} $

$\Rightarrow T(k+1)$ is true, whenever $T(k)$ is true, $k \in N$

$\Rightarrow T(n)$ is true for all $n \in N$.

~~ 5. Let $S_n=1.3+3.5+5.7+\ldots .+(2 n-1)(2 n+1)$

When $n=1, S_n=1.3=3$

From the given options, when $n=1$,

$ \begin{aligned} & \frac{n(2 n^{2}+3 n+1)}{6}=\frac{1 \times(2+3+1)}{6}=1 \neq S_n \\ & \frac{n(4 n^{2}+6 n-1)}{3}=\frac{1 \times(4+6-1)}{3}=\frac{9}{3}=3=S_n \\ & \frac{n(n^{2}+4)}{6}=\frac{1 \times 5}{6}=\frac{5}{6} \neq S_n \\ & \frac{n^{2}(4 n^{2}+5)}{3}=\frac{1 \times(4+5)}{3}=\frac{9}{3}=3=S_n \end{aligned} $

When $\boldsymbol{n}=\mathbf{2}, S_n=1.3+3.5=3+15=18$

$ \begin{aligned} & \frac{n(4 n^{2}+6 n-1)}{3}=\frac{2(4 \times 4+12-1)}{3}=18=S_n \\ & \frac{n^{2}(4 n^{2}+5)}{3}=\frac{4 \times(4 \times 4+5)}{3}=\frac{4 \times 21}{3}=28 \neq S_n . \\ & \therefore S_n=\frac{\boldsymbol{n}}{\mathbf{3}}(4 n^{2}+\mathbf{6 n}-\mathbf{1}) \text{ for } \boldsymbol{n}=\mathbf{1} \text{ and } \boldsymbol{n}=\mathbf{2} \end{aligned} $

Now we shall show that $S_n=\frac{n}{3}(4 n^{2}+6 n-1)$ for all $n \in N$.

Using the principle of mathematical induction.

Let $T(n)=1.3+3.5+5.7+\ldots .+(2 n-1)(2 n+1)$

$=\frac{n}{3}(4 n^{2}+6 n-1)$

Basic Step: For $n=1$, LHS $=1.3=3$

$ RHS=\frac{1 \times(4+6-1)}{3}=\frac{9}{3}=\mathbf{3} . $

$\therefore$ LHS $=$ RHS $\Rightarrow T(1)$ is true.

Induction Step: Assume $T(n)$ to be true for $n=k, k \in N$

$\Rightarrow 1.3+3.5+5.7+\ldots .+(2 k-1)(2 k+1)=\frac{k}{3}(4 k^{2}+6 k-1)$

To obtain $T(k+1)$,

add the $(k+1)$ th term $=[(2(k+1)-1)(2(k+1)+1)]$

$=(2 k+1)(2 k+3)$ to both the sides of $T(k)$, i.e.,

$1.3+3.5+5.7+\ldots .+(2 k-1)(2 k+1)+(2(k+1)-1)$

$(2(k+1)+1)$

$=\frac{1}{3} k(4 k^{2}+6 k-1)+(2 k+1)(2 k+3)$

$=\frac{1}{3} k(4 k^{2}+6 k-1)+(4 k^{2}+8 k+3)$

$=\frac{1}{3}{4 k^{3}+6 k^{2}-k+12 k^{2}+24 k+9}$

$=\frac{1}{3}{4 k^{3}+18 k^{2}+23 k+9}$

$=\frac{1}{3}(k+1)(4 k^{2}+14 k+9)$

$=\frac{1}{3}(k+1){4(k+1)^{2}+6(k+1)-1}$

$\therefore T(k+1)$ is true, whenever $T(k)$ is true.

Hence, by the principle of mathematical induction, $T(n)$ is true for all $n \in N$.

$ \begin{gathered} \Rightarrow \quad 1.3+3.5+5.7+\ldots .+(2 n-1)(2 n+1) \\ \quad=\frac{1}{3} n(4 n^{2}+6 n-1) \text{ is true } \forall n \in N . \end{gathered} $

Type II: Proving Divisibility

Ex. 1 . Prove that $x^{n}-y^{n}$ is divisible by $x-y$, when $\boldsymbol{n}$ is a + ve integer.

Sol. Let $T(n)$ be the statement: $x^{n}-y^{n}$ is divisible by $x-y$.

Basic Step: For $n=1, x^{1}-y^{1}=x-y$ is divisible by $(x-y) \Rightarrow T(1)$ is true

Induction Step: Assume that $T(k)$ is true, i.e., for $k \in N$

$ x^{k}-y^{k} \text{ is divisible by }(x-y) $

Now, we prove $T(k+1)$ is true.

$ \begin{aligned} x^{k+1}-y^{k+1} & =x^{k} \cdot x-y^{k} \cdot y=x^{k} \cdot x-x^{k} \cdot y+x^{k} \cdot y-y^{k} \cdot y(\text{ Adding and subtracting } x^{k} \cdot y) \\ & =x^{k}(x-y)+y(x^{k}-y^{k}) \end{aligned} $

Since $x^{k}(x-y)$ is divisible by $(x-y)$ and $(x^{k}-y^{k})$ is divisible by $(x-y)$ (By induction step, i.e., assuming $T(k)$ is true), therefore, $x^{k+1}-y^{k+1}=x^{k}(x-y)+y(x^{k}-y^{k})$ is divisible by $(x-y)$

$\Rightarrow T(k+1)$ is true, whenever $T(k)$ is true.

$\Rightarrow T(n)$ holds for all positive integral values of $n$.

Ex. 2 . Prove that $3^{2 n+2}-8 n-9$ is divisible by 64 for any positive integer $\boldsymbol{n}$.

Sol. Let $T(n)$ be the statement: $3^{2 n+2}-8 n-9$ is divisible by 64 .

Basic Step: For $n=1,3^{2 \times 1+2}-8 \times 1-9=81-17=64$ which is divisible by 64 .

$ \Rightarrow T(1) \text{ holds. } $

Induction Step: Let $T(k), k \in N$ hold, i.e.,

$3^{2 k+2}-8 k-9$ is divisible by 64 .

Then, $T(k+1)=3^{2(k+1)+2}-8(k+1)-9=3^{2} .3^{2 k+2}-8 k-17$

$ =9(3^{2 k+2}-8 k-9)+64 k+64=9 . T(k)+64(k+1) $

$\Rightarrow T(k+1)$ is divisible by 64 , whenever $T(k)$ is divisible by 64 .

$\Rightarrow T(n)$ is true for every natural number $n$.

Ex. 3 . Use the principle of mathematical induction to prove, for all $n \in N, 10^{2 n-1}+1$ is divisible by 11 .

Sol. Let the given statement $T(n)=10^{2 n-1}+1$ be a multiple of 11

$ \Rightarrow \quad 10^{2 n-1}+1=M(11) $

Basic Step: For $n=1,10^{2 \times 1-1}+1=10+1=11$ which is divisible by 11 .

Induction Step: Assume that $T(k)=10^{2 k-1}+1$ is divisible by 11 .

$ \Rightarrow \quad 10^{2 k-1}+1=M(11) \forall n \in N $

Then, we now show that $T(k+1)$ is true.

$ \begin{aligned} T(k+1) & =10^{2(k+1)-1}+1=10^{2 k-1+2}+1=10^{2} \cdot 10^{2 k-1}+1 \\ & =100(M(11)-1)+1 \\ & =100 \cdot M(11)-100+1=100 \cdot M(11)-99 \end{aligned} $

$\Rightarrow T(k+1)$ is divisible by 11 , when $T(k)$ is divisible by 11 .

$\Rightarrow T(n)$ holds true for all $n \in N$.

Ex. 4 . If $\boldsymbol{n}$ is an integer, $n \geq 1$, then show that $3^{2^{n}}-1$ is divisible by $2^{n+2}$.

Sol. Let $T(n)$ be the statement: $3^{2^{n}}-1$ is divisible by $2^{n+2}$

Basic Step: For $n=1$,

$ 3^{2^{1}}-1=8 \text{ and } 2^{n+2}=8 \Rightarrow T(1) \text{ is true } $

Induction Step: Assume $T(k)$ to be true, i.e.,

$ \begin{aligned} T(k) & =3^{2^{k}}-1 \text{ is divisible by } 2^{k+2} \\ & =3^{2^{k}}-1=m \cdot 2^{k+2} \text{ when } m \in N \\ & =3^{2^{k}}=m \cdot 2^{k+2}+1 \end{aligned} $

Now we need to prove that $T(k+1)$ holds true.

$ \begin{aligned} \therefore 3^{2^{k+1}}-1=3^{2^{k} \cdot 2}-1 & =(m \cdot 2^{k+2}+1)^{2}-1 \quad(using(i)) \\ & =m^{2}(2^{k+2})^{2}+2 m \cdot 2^{k+2}+1-1=2^{k+2}(m^{2} \cdot 2^{k+2}+2 m) \end{aligned} $

$\Rightarrow T(k+1)=3^{2^{k+1}}-1$ is divisible by $2^{k+2}$, whenever $T(k)$ holds.

Thus $3^{2^{n}}-1$ is divisible by $2^{n+2}$ for all integers $n \geq 1$.

PRACTICE SHEET-2

~~ 1. For all $n \in N,(2^{3 n}-1)$ will be divisible by

(a) 25

(b) 8

(d) 3

(WBJEE 2010)

~~ 2. If $n$ is a positive integer, then $n^{3}+2 n$ is divisible by

(a) 2

(b) 6

(d) 3

(Karnataka CET 2009)

~~ 3. For each $n \in N, 49^{n}+16 n-1$ is divisible by

(a) 3

(b) 29

(d) 64

(BCECE 2009)

~~ 4. If $n$ is a positive integer, then $5^{2 n+2}-24 n-25$ is divisible by

(a) 574

(b) 576

(d) 575

(Kerala CEE 2009)

~~ 5. For all integers $n \geq 1$, which of the following is divisible by 9 ?

(a) $8^{n}+1$

(b) $10^{n}+1$

(d) $3^{2 n}+3 n+1$

(EAMCET 2006)

~~ 6. For all $n \in N, 2^{3 n}-7 n-1$ is divisible by

(a) 64

(b) 36

(d) 25

(AIEEE 2006)

~~ 7. $10^{n}+3(4^{n+2})+5$ is divisible by (for all $n \in N$ )

(a) 5

(b) 7

(d) 13

(Kerala PET 2005)

~~ 8. For all natural numbers $n$, the expression $2.7^{n}+3.5^{n}-5$ is divisible by

(a) 16

(b) 24

(d) 21

(IIT 1985)

ANSWERS

~~ 1. (c) ~~ 2. $(d)$ ~~ 3. $(d)$ ~~ 4. (b) ~~ 5. (c) ~~ 6. (c) ~~ 7. $(c)$ ~~ 8. (b)

HINTS AND SOLUTIONS

~~ 1. For $n=1,2^{3 n}-1=2^{3}-1=8-1=7$, which is divisible by 7 , and not divisible by any other alternative given.

$\therefore$ We shall prove $2^{3 n}-1$ divisible by 7 for all $n \in N$.

Let $T(n)=2^{3 n}-1$ is divisible by 7 .

Basic Step: For $n=1, T(1)=2^{3}-1=8-1=7$ is divisible by 7 is true.

Induction Step: Assume $T(k)$ to be true, i.e.,

$ \begin{aligned} & T(k)=2^{3 k}-1 \text{ is divisible by } 7 \\ & \Rightarrow 2^{3 k}-1=7 m, m \in N \\ & \Rightarrow 2^{3 k}=7 m+1 \end{aligned} $

Now $2^{3(k+1)}-1=2^{3 k+3}-1=2^{3} \cdot 2^{3 k}-1=8.2^{3 k}-1$

$ =8 .(7 m+1)-1=56 m+7=7(8 m+1) $

$\Rightarrow 2^{3(k+1)}-1$ is divisible by 7

$\therefore T(k+1)$ is true whenever $T(k)$ is true.

$\Rightarrow 2^{3 n}-1$ is divisible by 7 for all $n \in N$.

~~ 2. For $n=1, n^{3}+2 n=1+2=3$ which is divisible by 3 and none of the other given alternatives.

$\therefore$ We shall prove $n^{3}+2 n$ divisible by 3 for all $n \in N$.

Let $T(n)=n^{3}+2 n$ is divisible by 3 .

Basic Step: For $n=1, T(1)=n^{3}+2 n=1+2=3$ is divisible by 3 is true.

Induction Step: Assume $T(k)$ to be true, i.e.,

$ \begin{aligned} T(k) & =k^{3}+2 k \text{ is divisible by } 3 \\ & =k^{3}+2 k=3 m, \text{ where } m \in N . \end{aligned} $

Now we need to prove that $T(k+1)$ holds true, i.e., $(k+1)^{3}+2(k+1)$ is divisible by 3 .

$(k+1)^{3}+2(k+1)=k^{3}+3 k^{2}+3 k+1+2 k+2$

$ \begin{aligned} & =(k^{3}+2 k)+(3 k^{2}+3 k+3) \\ & =3 m+3(k^{2}+k+1) \end{aligned} $

(From $(i)$ )

$\Rightarrow T(k+1)=(k+1)^{3}+2(k+1)$ is divisible by 3 , whenever $T(k)=k^{3}+2 k$ is divisible by 3 .

$\Rightarrow n^{3}+2 n$ is divisible by $3 \forall n \in N$.

~~ 3. For $n=1,49^{1}+16 \times 1-1=49+15=64$

$\therefore$ For $n=1,49^{n}+16 n-1$ is divisible by 64 and not by any of the other given alternatives.

$\therefore$ We shall prove using mathematical induction, that $49^{n}+16 n-1$ is divisible by $64 \forall n \in N$.

Let $T(n)$ be the statement: $49^{n}+16 n-1$ is divisible by 64 Basic Step: For $n=1, T(1)$ is divisible by 64 as proved above.

Induction Step: Assume $T(k)$ to be true i.e.,

$ \begin{aligned} T(k)= & 49^{k}+16 k-1 \text{ is divisible by } 64, \text{ i.e., } \\ & 49^{k}+16 k-1=64 m, m \in N . \end{aligned} $

$\therefore \quad T(k+1)=49^{k+1}+16(k+1)-1$

$ =49.49^{k}+16 k+16-1 $

$ \begin{aligned} & =49.49^{k}+16 k+15 \\ & =49(49^{k}+16 k-1)-48(16 k)+64 \\ & =49(64 m)-12(64 k)+64 \\ & =64(49 m-12 k+1) \end{aligned} $

$\Rightarrow 49^{k+1}+16(k+1)-1$ is divisible by 64 .

$\Rightarrow T(k+1)$ is true whenever $T(k)$ is true.

$\Rightarrow 49^{n}+16 n-1$ is divisible by $64 \forall n \in N$.

~~ 4. For $n=1,5^{2 n+2}-24 n-25=5^{4}-24-25=625-49$ $=576$ which is divisible by 576 and none of the other given alternative. $\therefore$ To prove: $5^{2 n+2}-24 n-25$ is divisible by 576 using mathematical induction.

Let $T(n)$ be the statement: $5^{2 n+2}-24 n-25$ is divisible by $576 \forall n \in N$.

Basic Step: For $n=1, T(1)=5^{4}-24-25=576$ which is divisible by 576 .

$ \Rightarrow T(1) \text{ is true. } $

Induction Step: Assume $T(k)$ where $n=k, k \in N$ to be true i.e.,

$T(k)=5^{2 k+2}-24 k-25$ is divisible by 576 is true, i.e.,

$5^{2 k+2}-24 k-25=576 m, m \in N$

$\therefore \quad T(k+1)=5^{2(k+1)+2}-24(k+1)-25$

$=5^{2 k+2} .25-24 k-24-25$

$=5^{2 k+2} .25-24 k-49$

$=25(5^{2 k+2}-24 k-25)+24 .(24 k)+576$

$=25 .(576 m)+576 k+576 \quad($ From $(i))$

$=576(25 m+k+1)$

$\Rightarrow 2^{2(k+1)+2}-24(k+1)-25$ is divisible by 576

$\Rightarrow T(k+1)$ is true, whenever $T(k)$ is true.

$\Rightarrow 5^{2 n+2}-24 k-25$ is divisible by $576 \forall n \in N$.

~~ 5. For $n=1$,

$8^{n}+1=8^{1}+1=9$ divisible by 9

$10^{n}+1=10^{1}+1=11$ not divisible by 9

$4^{n}-3 n-1=4-3-1=0$ divisible by 9

$3^{2 n}+3 n+1=13$ not divisible by 9

For $n=2$

$8^{n}+1=8^{2}+1=65$ not divisible by 9

$4^{n}-3 n-1=4^{2}-3 \times 2-1=16-6-1=9$ divisible by 9

$\therefore$ We need to prove $4^{n}-3 n-1$ to be divisible by $9 \forall n \in N$. using mathematical induction.

Let $T(n): 4^{n}-3 n-1$ is divisible by 9 ,

Basic Step: $T(1)=0$ which is divisible by $9 \Rightarrow T(1)$ is true.

Induction Step: Assume $T(k)$ to be true, i.e.,

$4^{k}-3 k-1$ is divisible by $9 k \in N$

$\Rightarrow 4^{k}-3 k-1=9 m, m \in N$

$\therefore 4^{k+1}-3(k+1)-1=4.4^{k}-3 k-3-1=4.4^{k}-3 k-4$

$=4(4^{k}-3 k-1)+9 k=4.9 m+9 k=9(4 m+k)$

$\Rightarrow 4^{k+1}-3(k+1)-1$ is divisible by 9

$\Rightarrow T(k+1)$ is true whenever $T(k)$ is true, $k \in N$

$\Rightarrow 4^{n}+3 n-1$ is divisible by $9 \forall n \in N$.

~~ 6. For $n=1,2^{3 n}-7 n-1=2^{3}-7-1=8-8=0$ which is divisible by all the given alternatives.

For $n=2,2^{3 n}-7 n-1=2^{6}-7 \times 2-1=64-14-1=49$, which is divisible by only 49 out of the given alternatives.

$\therefore$ We need to prove $2^{3 n}-7 n-1$ is divisible by $49 \forall n \in N$. Let $T(n)$ be the statement: $2^{3 n}-7 n-1$ is divisible by 49

Basic Step: For $n=1,2^{3 n}-7 n-1=0$, divisible by 49

$ \Rightarrow T(1) \text{ is true. } $

Induction Step: Assume $T(k)$ is true $\forall k \in N$, i.e., $2^{3 k}-7 k-1$ is divisible by 49 , i.e.,

$ 2^{3 k}-7 k-1=49 m, m \in N $

Now $2^{3(k+1)}-7(k+1)-1=2^{3 k} \cdot 2^{3}-7 k-7-1$

$=8.2^{3 k}-7 k-8=8(.2^{3 k}-7 k-1)+49 k$

$=8.49 m+49 k=49(8 m+k)$

$\Rightarrow 2^{3(k+1)}-7(k+1)-1$ is divisible by 49

$\Rightarrow T(k+1)$ is true whenever $T(k)$ is true

$\Rightarrow \mathbf{2}^{3 n}-\mathbf{7 n}-\mathbf{1}$ is divisible by $49 \forall n \in N$.

~~ 7. For $n=1,10^{n}+3(4^{n+2})+5=10+3 \times 4^{3}+5$

$ =10+192+5=207 $

which is divisible by only 9 and none of the other given alternatives.

$\therefore \quad$ We need to prove $10^{n}+3(4^{n+2})+5$ is divisible by $9 \forall n \in N$.

Let $T(n)$ be the statement $10^{n}+3(4^{n+2})+5$ is divisible by 9 .

Basic Step: For $n=1, T(1)$ holds true as prove above.

Induction Step: Assume $T(k)$ to be true, $k \in N$ i.e.,

$10^{k}+3(4^{k+2})+5$ is divisible by 9 , i.e.,

$10^{k}+3(4^{k+2})+5=9 m, m \in N$

Now, $10^{k+1}+3(4^{k+1+2})+5$

$=10^{k+1}+3(4^{k+3})+5$

$=10.10^{k}+12.4^{k+2}+5$

$=4(10^{k}+3(4^{k+2})+5)+6 \cdot 10^{k}-15$

$=4 .(9 m)+6(10^{k}-1)-9$

$=4 .(9 m)+6 \cdot(9 x)-9 \quad(\because 10^{k}-1.$ is always

$=9(4 m+6 x-1)$

divisible by 9 )

$\Rightarrow 10^{k+1}+3(4^{(k+1)+2})+5$ is divisible by 9 .

$\Rightarrow T(k+1)$ is true whenever $T(k)$ is true, $\forall k \in N$

$\Rightarrow 10^{n}-3(4^{n+2})+5$ is divisible by $9 \forall k \in N$.

~~ 8. When $n=1,2.7^{n}+3.5^{n}-5=2.7+3.5-5=24$ which is divisible by 24 and none of the other given alternatives.

$\therefore$ We need to prove $2.7^{n}+3.5^{n}-5$ is divisible

by $24 V$

$n \in N$.

Let $T(n)$ be the statement $2.7^{n}+3.5^{n}-5$ is divisible by 24 . $T(1)$ holds true as shown above.

Assume $T(k)$ to be true, i.e., $2.7^{k}+3.5^{k}-5$ is divisible by 24 , i.e.,

$ 2.7^{k}+3.5^{k}-5=24 m, m \in N $

Now

$ \begin{aligned} 2.7^{k+1}+3.5^{k+1}-5 & =2.7 .7^{k}+3.5 .5^{k}-5 \\ & =(2.7^{k}+3.5^{k}-5)+12(7^{k})+12(5^{k}) \\ & =24 m+12(7^{k}+5^{k}) \end{aligned} $

$ \begin{bmatrix} \text{ Now } 7^{k} \text{ and } 5^{k}, k \in N \text{ being both odd, their sum is even. } \\ \text{ Let } 7^{k}+5^{k}=2 x, x \in N\end{bmatrix} $

$ =24 m+12(2 x) ; m, x \in N=24(m+x) $

$\Rightarrow 2.7^{k+1}+3.5^{k+1}-5$ is divisible by 24

$\Rightarrow T(k+1)$ is true whenever $T(k)$ is true, $k \in N$.

$\Rightarrow 2.7^{n}+3.5^{n}-5$ is divisible by 24 for all $n \in N$.

6 Plane Geometry: Circle

KEY FACTS

A. Definitions

~~ 1. - The paths (locus) traced out by a moving point, at a fixed distance from a fixed point is called a circle.

The path so traced out is called the circumference (abbreviation $\odot c e$ ), the fixed point is called the centre and the fixed distance is called the radius.

In the given Fig., $O \to$ centre; $O C \to$ radius; $A C B D \to$ circumference; $A B \to$ diameter $(2 \times$ radius $)$

~~ 2. - A diameter divides a circle into two equal parts, each part being a semi-circle i.e., $A P B$ and $A Q B$ are semi-circles.

The part of a circle enclosed by any two radii of a circle is called sector, i.e., $A O B$
A part of the circumference is called an arc, i.e., $\overgroup{AB}$.
A quadrant is one-fourth of a circle, where the two bounding radii are at

rt. $\angle s$ to each other.

Any two points on a circle divide the circle into two parts. The smaller part is called the minor arc and the larger part is called the major arc.

~~ 3. - A line segment whose end points lie on the circle is called a chord. $A B, P Q, R S$ are all chords.

The chord passing through the centre of the circle is the longest chord and is the diameter of the circle, e.g. $B S$ is the diameter.
A chord divides a circle into two regions called segments of the circle. The larger part, containing the centre i.e., $A P B$ in the given figure is called the major segment and the smaller part not containing the centre, i.e., $A Q B$ is called the minor segment.

~~ 4. - A line intersecting a circle in two distinct points is called a secant. Secant $A B$ intersects the given circle in points $A$ and $B$.

A line which intersects the circle in exactly one point is called a tangent. The point of intersection, $T$, is called the point of contact or the point of tangency.

~~ 5. - Circles having the same centre are called concentric circles.

Circles with equal radii are called congruent circles.
Points lying on the same circle are called concyclic points. $A, L, B$

and $N$ are concyclic points.

~~ 6. Central angle: An angle formed at the centre of the circle is called the central angle. $\angle A O B$ is the central angle.

When two chords have a common end point, then the angle included between these two chords at the common point is called the inscribed angle. $\angle P Q R$ is inscribed by the arc $P S R$.

~~ 7. - A quadrilateral whose all four vertices lies on a circle is called a cyclic quadrilateral.

A circle which passes through all the three vertices of a triangle is called a circumcircle. The circumcentre is always equidistant from the vertices of the triangle.

$O A=O B=O C$

A circle which touches all the three sides of a triangle, i.e., all the three sides of the triangle are tangents to the circle is called an incircle. Incentre is always equidistant from the sides of a triangle.

$O P=O Q=O R$.

THEOREMS

I. CHORD PROPERTIES

Theorem 1. A straight line, drawn from the centre of a circle perpendicular to the chord bisects the chord.

If $O D \perp A B$, then $A B=2 A D=2 B D$.

Theorem 2. The line joining the centre of the circle to the mid-point of the chord is perpendicular to the chord.

Given, $A D=D B$, then $O D \perp A B$.

Theorem 3. The perpendicular bisectors of two chords of a circle intersect at its centre. Theorem 4. The perpendicular bisectors of a chord of a circle always passes through the centre.

Theorem 5. One and only one circle can be drawn through three points not lying in the same straight line.

Theorem 6. Equal chords of a circle are (or of congruent circles) equidistant from the centre.

Theorem 4. The perpendicular bisectors of a chord of a circle always passes through the

$A B=P Q \Rightarrow O D=O R$

Theorem 7. Chords which are equidistant from the centre in a circle (or congruent circles) are equal.

$O E=O F \Rightarrow A B=P Q$.

Theorem 8. The angular bisector of the angle between two equal chords of a circle passes through the centre.

Theorem 9. If two circles intersect, then the line joining their centres is the perpendicular bisector of the common chord. $A B$ is the perpendicular bisector of $P Q$.

Theorem 10. If any two chords of a circle, the one which is greater is nearer to the circle.

$A B>C D \Rightarrow O P<O Q$

Conversely, of any two chords of a circle, the nearer to the center is greater.

$O P<O Q \Rightarrow A B>C D$.

SOLVED EXAMPLES

We now take up some examples to illustrate the properties and results discussed so far.

Ex. 1 . The distance between two points $A$ and $B$ is $3 cm$. $A$ circle of radius $1.7 cm$ is drawn to pass through these points. Find the distance of $A B$ from the centre of the circle.

Sol. Let $O$ be the centre of the circle of radius $1.7 cm$ which is drawn to pass through $A$ and $B$. From $O$ draw $O D \perp A B$. Then $O D$ is the required distance.

$\therefore \quad A D=D B=1.5 cm \quad$ (perp. from centre bisects chord)

$\therefore$ In rt. $\angle d \triangle O D B$.

$ \begin{aligned} & O D^{2}=O B^{2}-D B^{2} \\ & =(1.7)^{2}-(1.5)^{2}=2.89-2.25=0.64 \\ & \therefore \quad O D=\sqrt{0.64}=0.8 cm . \end{aligned} $

(Pythagoras Th.)

Ex. 2 . $A B$ and $C D$ are two parallel chords of a circle such that $A B=16 cm$ and $C D=30 cm$. If the chords are on the opposite sides of the centre and the distance between them is $23 cm$, find the radius of the circle.

Sol. Let $O$ be the centre of the circle and radius $r cm$. Draw $O M \perp A B$ and $O N \perp C D$.

Then, $M O N$ is a straight line and

$ A M=\frac{1}{2} A B=8 cm \text{ and } C N=\frac{1}{2} C D=15 cm . $

Let

$ O M=x cm \text{. Then, } O N=(23-x) cm \text{. } $

Join $O A$ and $O C$. Then $O A=O C=r cm$.

In right $\triangle O M A, O A^{2}=A M^{2}+O M^{2} \Rightarrow r^{2}=8^{2}+x^{2}$

In right $\triangle O N C, O C^{2}=C N^{2}+O N^{2} \Rightarrow r^{2}=15^{2}+(23-x)^{2}$

From $(i)$ and $(i i)$, we have $8^{2}+x^{2}=15^{2}+(23-x)^{2}$

$\Rightarrow 64+x^{2}=225+529-46 x+x^{2} \Rightarrow 46 x=754-64 \Rightarrow 46 x=690 \Rightarrow x=\frac{690}{46}=15 cm$.

$\therefore$ From $(i), r^{2}=8^{2}+15^{2}=64+225=289 \Rightarrow r=\sqrt{289}=17$

Hence, the radius of the circle is $\mathbf{1 7} \mathbf{~ c m}$.

Ex. 3 . In a circle of radius $5 cm, A B$ and $A C$ are two chords such that $A B=A C=6 cm$. Find the length of the chord $B C$.

Sol. Since, the angular bisector of the angle between two equal chords of a circle passes through the centre therefore, $A O$ and so $A M$ is the bisector of $\angle B A C$ and also is perpendicular bisector of chord $B C$.

$\therefore \quad \angle A M B=90^{\circ}$ and $B M=M C$

Let $O M=x$. Then $A M=5-x$

In right $\triangle A M B, A B^{2}=A M^{2}+M B^{2}$ (Pythagoras Theorem)

$ \begin{matrix} \Rightarrow & 6^{2} & =(5-x)^{2}+B M^{2} \\ \Rightarrow & B M^{2} & =36-(5-x)^{2} \end{matrix} $

In right $\triangle O M B, \quad B O^{2}=B M^{2}+M O^{2} \Rightarrow 5^{2}=B M^{2}+x^{2} \Rightarrow B M^{2}=25-x^{2}$

$\therefore$ From $(i)$ and (ii), we have $36-(5-x)^{2}=25-x^{2}$

$\Rightarrow \quad 36-(25+x^{2}-10 x)=25-x^{2}$

$\Rightarrow \quad 11+10 x=25 \Rightarrow 10 x=25-11=14 \Rightarrow x=\frac{14}{10}=1.4 cm$

$\therefore$ From (ii), $B M^{2}=25-x^{2}=25-(1 \cdot 4)^{2}=25-1.96=23.04 \Rightarrow B M=\sqrt{23.04}=4.8 cm$

Hence, length of the chord $B C=2 B M=2 \times 4 \cdot 8=\mathbf{9 . 6} \mathbf{~ c m}$.

Ex. 4 . If a line $l$ intersects two concentric circles at points $A, B, C$ and $D$ as shown in the figure, prove that $A B=C D$.

Prove that two concentric circles intercept equal portions on any straight line that cuts them.

Sol. Let $O$ be the centre of the two concentric circles and $O M$ the perpendicular from $O$ to the line $l . A D$ is the chord of the larger circle and $B C$, the chord of the smaller circle.

Since, perpendicular from the centre to a chord bisects the chord, therefore,

(i) and (ii) gives, $A M-B M=M D-M C \Rightarrow \boldsymbol{A B}=\boldsymbol{C D}$.

Ex. 5 . Prove that the line joining the mid-points of two equal chords of a circle makes equal angles with the chords.

Sol. Let $A B$ and $C D$ be the two equal chords of a circle with centre $O$. Let $L$ and $M$ be the mid-points of $A B$ and $C D$ respectively.

(a)

(b)

Then $\quad O L \perp A B$ and $O M \perp C D \quad$ (The line joining centre to the mid-point of a chord is perp. to the chord.) Also, $\quad A B=C D \therefore O L=O M \quad$ (Equal chords are equidistant from the centre) $\therefore$ In $\triangle O L M, \angle O L M=\angle O M L \quad$ (In a $\triangle$, angles opp. equal sides are equal)

$\therefore \quad \angle O L A=90^{\circ}$ and $\angle O M C=90^{\circ}$

$\Rightarrow \quad \angle A L M=90^{\circ}-\angle O L M$ and $\angle C M L=90^{\circ}-\angle O M L$

From (i) and (ii) it follows that $\angle \boldsymbol{A} \boldsymbol{L} \boldsymbol{M}=\angle \boldsymbol{C M L}$.

Ex. 6 . Two equal chords $A B$ and $C D$ of a circle with centre $O$, when produced meet at a point $P$ outside the circle. Prove that (i) $P B=P D$ and (ii) $P A=P C$.

Sol. Draw $O M \perp A B, O N \perp C D$ and join $O P$. Then

$ A M=B M=\frac{1}{2} A B $

and

$ C N=D N=\frac{1}{2} C D $

But

(The perpendicular from the centre of a circle bisects the chord.)

$ \Rightarrow \quad \frac{1}{2} A B=\frac{1}{2} C D \Rightarrow A M=B M=C N=D N $

Also, $\quad O M=O N$

(Equal chords are equidistant from the centre)

In right $\triangle s O M P$ and $O N P$, we have

$ \begin{matrix} O M & =O N, \angle O M P=\angle O N P(\text{ each } 90^{\circ}) \text{ and } O P \text{ is common } \\ \therefore & \triangle O M P & \cong \triangle O N P & (R H S) \\ \therefore & M P & =N P & (\text{ c.p.c.t. }) \end{matrix} $

$\therefore$ From $(i)$ and (ii), we get

and

$ \begin{aligned} & M P-B M=N P-D N \Rightarrow \boldsymbol{P B}=\boldsymbol{P} \boldsymbol{D} . \\ & M P+A M=N P+C N \Rightarrow \boldsymbol{P A}=\boldsymbol{P} . \end{aligned} $

II. THEOREMS ON ARCS AND ANGLES

Theorem 11. The angle subtended at the centre by an arc of a circle is double the angle which this are subtends at any point on the remaining part of the circle.

In all these cases,

$ \angle A O B=2 \angle A C B $

Note: in diagram (iii) Reflex $\angle A O B=2 \angle A C B$.

Theorem 12. Angle in a semicircle is a right angle.

(i)

(ii)

(iii)

$ \angle A C B=90^{\circ} $

Theorem 13. If an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a semi-circle.

Theorem 14. Angles in the same segment of a circle are equal.

$ \angle A C B=\angle A D B=\angle A E B . $

Theorem 15. If the straight line joining two points subtends equal angles at two other points on the same side of it, then the four points lie on a circle, i.e., are concyclic.

$\angle A P B=\angle A Q B \Rightarrow$ Points $A, P, Q, B$ lie on the same circle.

Theorem 16. The opposite angles of a cyclic quadrilateral are supplementary. $\angle A C D+\angle A B D=180^{\circ}, \angle C A B+\angle C D B=180^{\circ}$. Conversely, if a pair of opposite angles of a quadrilateral are supplementary, then the qudrilateral is cyclic.

Theorem 17. If the side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the interior opposite angle, e.g., $\angle C B X=\angle A D C$.

$\therefore$ If a parallelogram is inscribed in a circle, it is always a rectangle.

Theorem 18. Equal chords (or equal arcs) of a circle (or congruent circles) subtend equal angles at the centre.

$ A B=C D(\text{ or } \overline{A B}=\overline{C D}) \Rightarrow \angle A O B=\angle C O D . $

Ex. 7 . In a given circle $A B C D, O$ is the centre and $\angle B D C=42^{\circ}$. Calculate the $\angle A C B$.

Sol. $A O C$ is a diameter since $O$ is the centre of the circle.

$ \begin{matrix} \therefore & \angle A B C=90^{\circ} & (\angle \text{ in a semi-circle }) \\ & \angle B A C=\angle B D C=42^{\circ} & (\angle \text{ in the same segment }) \\ \therefore & \text{ In } \triangle A B C, & \\ & 42^{\circ}+90^{\circ}+\angle A C B=180^{\circ} & (\angle \text{ sum of } \triangle) \\ \therefore & \angle A C B=180^{\circ}-132^{\circ}=\mathbf{4 8}^{\circ} . & \end{matrix} $

Ex. 8 . (i) In Fig. (i), $O$ is the centre of the circle and the measure of $arc A B C$ is $110^{\circ}$. Using the above results, find $\angle A D C$ and $\angle A B C$.

Fig. (i)

Fig. (ii)

Fig. (iii)

(ii) In Fig. (ii), calculate the measure of $\angle A O C$.

(iii) In Fig. (iii), $A B C$ is a triangle in which $\angle B A C=30^{\circ}$. Show that $B C$ is equal to the radius of the circumcircle of $\triangle A B C$, whose centre is $O$.

Sol. (i) $Arc A C$ subtends $\angle A O C=110^{\circ}$ at the centre and $\angle A D C$ at the remaining part of the circumference,

$ \therefore \quad \angle A D C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 110^{\circ}=\mathbf{5 5}^{\circ} $

Now, $\quad$ reflex $\angle A O C=360^{\circ}-110^{\circ}=250^{\circ}$

Major arc $A D C$ subtends reflex $\angle A O C$ at the centre and $\angle A B C$ at the remaining part of the circumference

$\therefore \quad \angle A B C=\frac{1}{2}($ reflex $\angle A O C)=\frac{1}{2} \times 250^{\circ}=\mathbf{1 2 5}^{\circ}$

$\begin{matrix} \text{ Alternatively } & \angle A B C+\angle A D C=180^{\circ} \\ \Rightarrow & \angle A B C+55^{\circ}=180^{\circ}\end{matrix} \quad \Rightarrow \quad \begin{matrix} (\text{ opp. } \angle s \text{ of a cyclic quad. are supplementary) }\end{matrix} $

$ \text{ (ii) In } \triangle A O B, \quad O A=O B \quad \text{ (radii of the same circle) } $

$\therefore \quad \angle O B A=\angle O A B=30^{\circ}$

(angles opposite equal sides)

Similarly, in $\triangle B O C, \because O B=O C \therefore \angle O B C=\angle O C B=40^{\circ}$

$\therefore \quad \angle A B C=\angle O B A+\angle O B C=30^{\circ}+40^{\circ}=70^{\circ}$

Now, arc $A C$ subtends $\angle A O C$ at the centre and $\angle A B C$ at the remaining circumference.

$\therefore \quad \angle A O C=2 \angle A B C=2 \times 70^{\circ}=\mathbf{1 4 0}^{\circ}$

(iii) $Arc B C$ subtends $\angle B O C$ at the centre and $\angle B A C$ at the remaining circumference.

$\therefore \quad \angle B O C=2 \angle B A C=2 \times 30^{\circ}=60^{\circ}$ $\therefore \quad$ In $\triangle B O C, \angle O B C+\angle O C B=180^{\circ}-\angle B O C=180^{\circ}-60^{\circ}=120^{\circ}$

But $\quad \angle O B C=\angle O C B(\because O B=O C$, being radii $)$

$\therefore \quad \angle O B C=\angle O C B=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

$\therefore \quad \triangle O B C$ is equilateral (each $\angle=60^{\circ}$ ) $\Rightarrow B C=O B=O C$

$\Rightarrow \quad B C$ is equal to the radius of the circumcircle.

Ex. 9 . (i) In Fig. (i), $O$ is the centre of the circle. The angle subtended by the arc $B C D$ at the centre is $140^{\circ} . B C$ is produced to $P$. Determine $\angle B A D$ and $\angle D C B$, and $\angle D C P$.

(ii) In Fig. (ii), $C$ is a point on the minor $arc A B$ of the circle with centre $O$. Given $\angle A C B=x^{o}$ and $\angle A O B=y^{o}$, express $y$ in terms of $x$. Calculate $x$, if $A C B O$ is a parallelogram.

(iii) In Fig. (iii), $A B$ is a diameter of a circle with centre $O$ and radius $O D$ is perpendicular to $A B$. If $C$ is any point on $arc D B$, find $\angle B A D, \angle A C D$.

Fig. (i)

Fig. (ii)

Fig. (iii)

Sol. (i) $\angle B A D=\frac{1}{2} \angle B O D=\frac{1}{2} \times 140^{\circ}=\mathbf{7 0}^{\circ}$

(angle at the centre by arc $B C D=$ twice angle at the remaining circumference)

Now, arc $B A D$ makes reflex $\angle B O D=(360^{\circ}-140^{\circ})=220^{\circ}$ at the centre, and $\angle B C D$ at a point $C$ on the remaining circumference.

$ \begin{aligned} & \text{ nference. } \\ & \therefore \end{aligned} \quad \angle B C D=\frac{1}{2}(\text{ reflex } \angle B O D)=\frac{1}{2} \times 220^{\circ}=\mathbf{1 1 0}^{\circ} . $

Also, $\angle B C D+\angle D C P=180^{\circ}$ (Linear pair) $\Rightarrow \angle D C P=180^{\circ}-\angle B C D=180^{\circ}-110^{\circ}=\mathbf{7 0}^{\circ}$.

(ii) Major arc $A B$ subtends reflex $\angle A O B$ at the centre and $\angle A C B=x^{\circ}$ at a point $C$ on the remaining circumference

$\therefore \quad$ reflex $\angle A O B=2 \angle A C B$

$ \Rightarrow \quad 360^{\circ}-y=2 x \Rightarrow y=360-2 x $

If $A C B O$ is a parallelogram, then

$ x^{0} =y^{0} $

$\Rightarrow x =y \Rightarrow x=360-2 x $

$ 3 x =360 \Rightarrow \boldsymbol{x}=\mathbf{1 2 0}^{\circ}$.

(opp. $\angle s$ of a $|$ gm)

$ [From (i)]$

(iii) Arc $B D$ makes $\angle B O D$ at the centre and $\angle B A D$ at point $A$ on remaining circumference.

$\therefore \quad \angle B A D=\frac{1}{2} \angle B O D=\frac{1}{2} \times 90^{\circ}=\mathbf{4 5}^{\circ}$.

Also, arc $A D$ makes $\angle A O D$ at the centre and $\angle A C D$ at point $C$ on the remaining circumference.

$\therefore \quad \angle A C D=\frac{1}{2} \angle A O D=\frac{1}{2} \times 90^{\circ}=\mathbf{4 5}^{\circ}$.

Thus, $\quad \angle B A D=\angle A C D=45^{\circ}$.

Ex. 10 . (i) In Fig. (i), find the value of the angles $x$ and $y$.

Fig. (i)

Fig. (ii)

Fig. (iii) (ii) In Fig. (ii), $A B C D$ is a cyclic quadrilateral. $A E$ is drawn parallel to $C D$ and $B A$ is produced to $F$. If $\angle A B C=\mathbf{9 2}, \angle F A E=20^{\circ}$, find $\angle B C D$.

(iii) In Fig. (iii), $O$ is the centre of the circle. Arc $A B C$ subtends an angle of $130^{\circ}$ at the centre $O . A B$ is extended up to $P$. Find $\angle P B C$.

Sol. (i) Side $B C$ of cyclic quad. $A B C D$ is produced to $F$.

$ \begin{aligned} \therefore & & \angle D C F & =\angle B A D \\ & \Rightarrow & \boldsymbol{x} & =\mathbf{7 8}^{\mathbf{0}} \end{aligned} $

(ext. $\angle=$ int. opp. $\angle$ )

In cyclic quad. DCFE, $x+y=180^{\circ}$

$ \Rightarrow \quad 78^{\circ}+y=180^{\circ} \Rightarrow y=180^{\circ}-78^{\circ}=\mathbf{1 0 2}^{\mathbf{o}} $

(ii) In cyclic quad. $A B C D, \angle A D C+\angle A B C=180^{\circ}$

$ \Rightarrow \quad \angle A D C+92^{\circ}=180^{\circ} \Rightarrow \angle A D C=180^{\circ}-92^{\circ}=88^{\circ} $

Now, $A E | C D$ and $A D$ cuts them

$ \begin{aligned} & \therefore \quad \angle E A D=\angle A D C=88^{\circ} \\ & \therefore \quad \angle F A D=20^{\circ}+88^{\circ}=108^{\circ} \\ & \text{ So, } \quad \angle B C D=\angle F A D=\mathbf{1 0 8}^{\mathbf{0}} \text{. } \end{aligned} $

(alternate $\angle s$ )

(iii) Let $D$ be any point on the major $arc A C$.

Then,

$ \angle A D C=\frac{1}{2} \angle A O C=\frac{1}{2} \times 130^{\circ}=65^{\circ} $

(Angle subtended by an arc at the centre = twice the angle by the arc at the remaining circumference)

Now, $A B C D$ is a cyclic quadrilateral whose side $A B$ is produced to any point $P$.

$\therefore$ ext. $\angle P B C=$ int. opp. $\angle A D C \Rightarrow \angle P B C=\mathbf{6 5}^{\circ}$.

Ex. 11 . In the figure given, $A B$ is a diameter of the circle with centre $O$ and $C D | B A$. If $\angle C A B=x$, find the value of

(i) $\angle C O B$

(ii) $\angle D O C$

(iii) $\angle DAC$ (iv) $\angle A D C$.

Sol. (i) $\angle C O B=2 \angle C A B=\mathbf{2} \boldsymbol{x}^{\circ}$ (angle at the centre $=2 \times$ angle at the remaining part of the circumference)

$ \begin{aligned} \angle O C D & =\angle C O B=\mathbf{2} \boldsymbol{x}^{\circ} \\ O D & =O C \end{aligned} $

(alternate $\angle s, D C | A B$ )

$\Rightarrow \quad \angle O C D=\angle O D C$

$\Rightarrow \quad \angle O D C=2 x^{\circ}$

$\therefore \quad$ In $\triangle D O C, \angle D O C=180^{\circ}-(2 x^{\circ}+2 x^{\circ})=\mathbf{1 8 0}^{\circ}-\mathbf{4} x^{\circ} \quad(\angle$ sum prop. of a $\triangle)$

(iii)

$ \angle D A C=\frac{1}{2} \angle D O C=\frac{1}{2}(180-4 x)^{\circ} $

(angle made by arc DC at the centre $=$ Twice the angle at the remaining part of the circumference)

$ =(90-2 x)^{\circ} $

(iv) In $\triangle A D C, \quad \angle A C D=\angle C A B=x^{\circ}$

(alt $\angle s ; D C | A B)$

$\therefore \quad \angle A D C=180^{\circ}-(x^{\circ}+90^{\circ}-2 x^{\circ})=(\mathbf{9 0}+\boldsymbol{x})^{\circ}$.

( $\angle$ sum prop. of a $\triangle$ )

Ex. 12 . Two circles $A B C D, A B F E$ intersect at $A$ and $B$. $E A D$ and $F B C$ are straight lines. Prove that $E F$ is parallel to $D C$.

Sol. Given. Circles $A B C D, A B F E$ intersecting at $A$ and $B$, and $E A D$ and $F B C$ are st. lines.

To prove that $E F | D C$

Construction. Join $A B$

Proof. $\begin{matrix} & \angle D A B & =\angle E F B \\ & & \angle D A B+\angle B C D & =2 \text{ rt. } \angle s \\ & & \angle E F B+\angle B C D & =2 \text{ rt. } \angle s\end{matrix} $

(ext. $\angle$ of cyclic quad.)

(opp. $\angle$ s of cyclic quad.)

But $\angle s E F B$ and $B C D$ are int. $\angle s$ on the same side of the transversal $F C$.

$(\because \angle E F B=\angle D A B)$

$\therefore \quad E F | D C$.

Ex. 13 . Prove that the bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (provided they are not parallel) intersect at right angle.

Sol. Given. Cyclic quad. $A B C D$ whose opp. sides when produced intersect at $P$ and $Q . P M$ and $Q N$ are angular bisectors of $\angle s P$ and $Q$ respectively, and meet each other at $M$. Let $Q N$ intersect $C D$ at $L$.

To prove that $\angle Q M P=90^{\circ}$

Proof. $\angle 1=\angle L C Q+\angle C Q L$ (ext. $\angle$ of a $\triangle=$ sum of int opp $\angle s)$

But $\quad \angle L C Q=\angle B A D($ ext. $\angle$ of cyclic quad $A B C D)$

and $\quad \angle C Q L=\angle L Q D$

$\therefore \quad \angle 1=\angle B A D+\angle L Q D=\angle 7+\angle 3$

Now, $\quad \angle 2=\angle A Q N+\angle N A Q$

( $Q L$ is bisector of $\angle D Q C$, given)

$ =\angle 3+\angle 7 $

$\Rightarrow \quad \angle 1=\angle 2$

Now, in $\triangle s P L M$ and $P N M$, we have

$ \begin{aligned} & \angle 1=\angle 2 \\ & \angle 5=\angle 6 \end{aligned} $

Ex. 14 . Prove that any four vertices of a regular pentagon are concyclic.

Sol. Let $A B C D E$ be a regular pentagon.

Join $A C$ and $B D$

In $\triangle s A B C$ and $B C D$, we have

$ \begin{matrix} A B & =D C \\ & & & \\ \angle A B C & =\angle B C D \\ & B C \text{ is common } \\ \therefore & & \triangle A B C \cong \triangle B C D \\ \Rightarrow & \angle B A C & =\angle B D C \end{matrix} $

(sides of a regular pentagon) (angles of a regular pentagon)

$\Rightarrow A, B, C, D$ are cyclic (Since $\angle B A C$ and $\angle B D C$ are angles subtended by $B C$ on the same side of it.)

Hence, any four vertices of a regular pentagon are concyclic.

Ex. 15 . $D$ and $E$ are points on equal sides of $A B$ and $A C$ of an isosceles $\triangle A B C$ such that $A D=A E$. Prove that $B$, $C, E, D$ are concyclic.

Sol. Given. Isos. $\triangle A B C$ in which $A B=A C, D$ and $E$ are points on $A B$ and $A C$ respectively such that $A D=A E$.

To prove that points $B, C, E, D$ are concyclic

Proof. In $\triangle A B C, A B=A C \Rightarrow \angle B=\angle C$

In $\triangle A D E, A D=A E \Rightarrow \angle A D E=\angle A E D$

Now, In $\triangle A B C, \angle A+\angle B+\angle C=180^{\circ}$

In $\triangle A D E, \angle A+\angle A D E+\angle A E D=180^{\circ}$

$\therefore \quad \angle A+\angle B+\angle C=\angle A+\angle A D E+\angle A E D$

$\Rightarrow \quad \angle B+\angle C=\angle A D E+\angle A E D$

$\Rightarrow \quad 2 \angle B=2 \angle A E D \Rightarrow \angle A E D=\angle B \quad($ From (1) and (2))

Now, $\quad \angle A E D+\angle C E D=180^{\circ}$

$\Rightarrow \quad \angle B+\angle C E D=180^{\circ} \quad(\because \angle A E D=\angle B$ proved above $)$

$\Rightarrow$ quad. $B C E D$ is a cyclic quad, i.e., pts. $B, C, E, D$ are concylic.

(If sum of opp $\angle s$ of a quad is $180^{\circ}$ it is a cyclic quad.)

Ex. 16 . Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.

Sol. $P Q R S$ is a cyclic quad. and angles $\angle A, \angle B, \angle C$ and $\angle D$ are angles in the four exterior segments.

Join $A R$ and $A S$.

In cyclic quad. $A Q B R, \angle 1+\angle B=180^{\circ}$

In cyclic quad. $A R C S, \angle 2+\angle C=180^{\circ}$

In cyclic quad. $A P D S, \angle 3+\angle D=180^{\circ}$

$\therefore \angle 1+\angle 2+\angle 3+\angle B+\angle C+\angle D=180^{\circ}+180^{\circ}+180^{\circ}=6$ rt. $\angle s$

$\Rightarrow \angle A+\angle B+\angle C+\angle D=\mathbf{6}$ rt. $\angle$ s.

Ex. 17 . In the given figure two equal chords $A B$ and $C D$ of a circle with centre $O$, intersect each other at $E$. Prove that $A D=C B$.

Sol. We have

$ \begin{aligned} & \text{ chord } A B=\text{ chord } C D \text{ (given) } \\ \Rightarrow & \text{ minor arc } A B=\text{ minor arc } C D \\ \Rightarrow & \overgroup{A B}=\overgroup{C D} \\ \Rightarrow & \overgroup{A B}-\overgroup{B D}=\overgroup{C D}-\overgroup{B D} \\ \Rightarrow & \overgroup{A D}=\overgroup{C B} \Rightarrow A D=C B. \end{aligned} $

(In equal circles chords of the equal arcs are also equal.)

Ex. 18 . $A, B, C, D$ are four consecutive points on a circle such that $A B=C D$. Prove that $A C=B D$.

Sol.

$ \begin{aligned} & A B=C D \quad \text{ (given) } \\ & \Rightarrow \quad \overgroup{A B}=\overgroup{C D} \quad \text{ (In equal circles equal chords cut off equal arcs) } \\ & \Rightarrow \quad \overgroup{A B}+\overgroup{B C}=\overgroup{B C}+\overgroup{C D} \\ & \Rightarrow \quad arc A B C=arc B C D \\ & \Rightarrow \quad \text{ chord } A C=\text{ chord } B D \Rightarrow A C=B D \text{. Hence proved. } \end{aligned} $

(AEC is a st. line) $(\angle$ sum of $a \Delta)$ $\Rightarrow \quad Arc P R=Arc Q S \quad$ (In the same circle, arcs subtending equal angles at the centre are equal.)

Hence, proved.

Note: It is clear otherwise also.

$\because \quad P Q R=S R Q \therefore arc P R=arc Q S$.

Ex. 20 . In $\triangle A B C$, the perpendiculars from vertices $A$ and $B$ on their opposite sides meet (when produced) the circum-circle of $\triangle A B C$ at points $D$ and $E$ respectively. Prove that $arc C D=arc C E$.

Sol. $\quad \angle C A D=90^{\circ}-\angle C \quad \because A D \perp B C$

$\angle C B E=90^{\circ}-\angle C \quad \because B E \perp A C$

$\Rightarrow \quad \angle C A D=\angle C B E \Rightarrow \frac{1}{2} \angle C O D=\frac{1}{2} \angle C O E$

$\Rightarrow \quad \angle C O D=\angle C O E$

$\Rightarrow \quad arc C D=arc C E$.

Ex. 21 . In the given Fig., $\triangle A B C$ is equilateral, $P$ and $S$ are mid-points of arcs $A B$ and $A C$. Prove that $P Q=Q R=R S$.

Sol. Chord $A B=$ Chord $A C$ (sides of an equilateral $\triangle$ )

$\Rightarrow Arc A P B=Arc A S C$

Also, given, $Arc A P=Arc P B=\frac{1}{2} Arc A P B$

$Arc A S=Arc S C=\frac{1}{2} Arc A S C$

$\Rightarrow Arc A P=Arc P B=Arc A S=Arc S C$

Now $Arc A P B$ subtends $\angle A C B$ and $Arc A S C$ subtends angle $\angle A B C$ on the circumference.

$\because \quad Arc A P B=Arc A S C \Rightarrow \angle A C B=\angle A B C=2 x$ (say)

Now $Arc P B=Arc S C \Rightarrow \angle P A B=\angle C A S=\frac{1}{2} \times 2 x=x$.

$P S | B C, \angle A Q R=\angle A R Q=2 x$

$\therefore \quad \angle A P Q=\angle A Q R-\angle P A Q=2 x-x=x \quad($ ext. $\angle$ property of a $\triangle)$

Similarly $\angle A S R=x$.

Now, $\triangle A B C$ is equilateral $\Rightarrow 2 x=60^{\circ} \Rightarrow x=30^{\circ}$

$\Rightarrow \quad \angle A Q R=\angle A R Q=2 x=60^{\circ} \Rightarrow \triangle A Q R$ is equilateral.

Also, $\quad \angle P A Q=\angle A P Q \Rightarrow A Q=Q P \quad$ (In $\triangle A Q P$ )

$\angle R A S=\angle S R A \Rightarrow A R=R S \quad$ (In $\triangle A R S$ )

$\triangle A Q R$ being equilateral $A Q=A R=Q R$

$\therefore \quad$ From (i), (ii) and (iii),

$P Q=Q R=R S$.

III. THEOREMS ON TANGENTS AND SECANTS

Theorem 19. The tangent at any point of a circle is perpendicular to the radius through the point of contact. $O R \perp A B$.

Theorem 20. If two tangents are drawn to a circle from an external point then,

(i) the tangents are equal in length, i.e., $P A=P B$

(ii) the tangents subtend equal angles at the centre of the circle, i.e., $\angle P O A=\angle P O B$

(iii) the tangents are equally inclined to the line joining the point and the centre of the circle, i.e., $\angle A P O=\angle B P O$

(iv) the angle between the tangents is supplementary of the angle that they subtend at the centre, i.e., $\angle A O B+\angle A P B=180^{\circ}$.

Theorem 21. If two circles touch each other, the point of contact lies on the straight line joining the centres of the two circles.

Touching or tangent circles and common tangents

Definitions. Two circles are tangent if they are tangent to the same line at the same point. The two circles are also said to touch each other.

(i)

(ii)

(iii)

(iv)

(v)

If the circles are internally tangent, then there is just one line tangent to both of them Fig. ( $i$ ). If the circles are externally tangent, then there are three lines tangent to both circles. Fig. (ii)

If one circle is contained in the interior of another, then there is no line that is tangent to both circles. Fig. (iii)

If the circles intersect in two points. Then there are two lines tangent to both circles Fig. (iv).

If the two circles do not intersect, then there are four lines that are tangent to both circles. Fig. ( $v)$

If two circles are coplanar, and their centres are on the same side of their common tangent, then they are internally tangent, as in Fig. (ii).

If two circles are coplanar, and their centres are on opposite sides of their common tangent, then they are externally tangent as in Fig. (iii).

If a line is tangent to each of two circles it is called a common tangent to two circles. It is called an exterior (or direct) common tangent, if the circles lie on the same side of it, as in Fig. $(v)$ and it is called an interior (or transverse) common tangent, if the circles lie on opposite sides of it, as in Fig. (v).

Note : Two circles touch if the distance $(d)$ between their centres is equal to the sum of their radii (external contact) or equal to the difference of their radii (internal contact).

i.e., $\quad d=r_1+r_2$, if the circles touch externally;

$d=r_1-r_2$, if the circles touch internally.

SOLVED EXAMPLES

Ex. 22 . (a) In Fig. (i), the tangent to a circle of radius $1.5 cm$ from an external point $P$, is $2 cm$ long. Calculate the distance of $P$ from the nearest point of the circumference.

(b) In Fig. (ii) from an external point $P$, tangents $P A$ and $P B$ are drawn to circle $O$. $C D$ is tangent to the circle at $E$. If $A P=16 cm$, find the perimeter of $\triangle P C D$.

(c) In Fig. (iii), there are two concentric circles of radii $3 cm$ and $5 cm$ respectively. Find the length of the chord of the outer circle which touches the inner circle.

Fig. (i)

Fig. (ii)

Fig. (iii)

Sol. (a) $P B$ is the required distance

In right $\triangle O A P$,

( $\angle$ between tangent and radius through the pt. of contact.)

(a)

$ \angle O A P=90^{\circ} $

$ O P^{2}=O A^{2}+A P^{2} $

$ =(1 \cdot 5)^{2}+(2)^{2}=2 \cdot 25+4=6.25 cm^{2} $

$\Rightarrow$

$O P=\sqrt{6.25} cm=2.5 cm$

$\therefore$

$ P B=O P-O B=2.5 cm-1.5 cm=1 cm . $

$ \begin{matrix} C E=C A \text{ and } D E & =D B \quad(\text{ tangents to the circle from external points } C \text{ and } D) \ldots(i) \\ \text{ Perimeter of } \triangle P C D & =P C+C D+P D=P C+(C E+E D)+P D & \\ & =(P C+C A)+(D B+P D) & \\ & =P A+P B=16 cm+16 cm=\mathbf{3 2} \mathbf{~ c m .} \end{matrix} $

(Pythagoras)

(c) Let $O$ be the centre of the two concentric circles and let $A B$ be the chord of the outer circle which touches the inner circle at $M$.

Then,

Also,

$O M \perp A B$

(Tangent $\perp$ radius through the pt. of contact)

$ A M=M B \Rightarrow A B=2 A M $

$ \begin{aligned} O A & =5 cm \\ O M & =3 cm \end{aligned} $

Now, in right $\triangle O M A, A M^{2}=O A^{2}-O M^{2}$

$ \begin{matrix} \Rightarrow & A M^{2}=5^{2}-3^{2}=25-9=16 cm^{2} \Rightarrow A M=\sqrt{16}=4 cm \\ \therefore & A B=2 A M=2 \times 4=\mathbf{8} \mathbf{c m} . \end{matrix} $

$(\perp$ from centre bisects chord $)$ (radius of the outer circle) (radius of the inner circle) (Pythagoras)

Ex. 23 . In the adjoining Fig., $X Y$ is a diameter of the circle, $P Q$ is a tangent to the circle at $Y$. Given that $\angle A X B=50^{\circ}$ and $\angle A B X=70^{\circ}$, calculate $\angle B A Y$ and $\angle A P Y$.

Sol. In $\triangle A X B$,

$ \begin{aligned} & \angle X A B+\angle A X B+\angle A B X=180^{\circ} \\ \Rightarrow \quad & \angle X A B=180^{\circ}-(50^{\circ}+70^{\circ})=180^{\circ}-120^{\circ}=60^{\circ} . \end{aligned} $

$X Y$ being the diameter of the circle.

$ \begin{matrix} \Rightarrow & \angle X A Y=90^{\circ} \\ \therefore & \angle B A Y=\angle X A Y-\angle X A B=90^{\circ}-60^{\circ}=\mathbf{3 0 ^ { \circ } .} \end{matrix} $

$(\angle$ in a semi circle)

Now $\angle B X Y=\angle B A Y=30^{\circ}$ ( $\angle s$ in the same segment of the circle)

$\therefore \quad \angle A C X=\angle B X C+\angle C B X=30^{\circ}+70^{\circ}=100^{\circ}$ (ext. $\angle=$ sum of int. opp. $\angle s$ in $.\triangle B X C)$

Also, $\angle X Y P=90^{\circ}$

(radius through the point of contact is perpendicular to the tangent)

For $\triangle C P Y$,

$ \begin{aligned} & \angle A C X=\angle A P Y+\angle C Y P \\ \Rightarrow \quad & 100^{\circ}=\angle A P Y+90^{\circ} \\ \Rightarrow \quad & \angle A P Y=\mathbf{1 0}^{\circ} \end{aligned} $

(ext. $\angle=$ sum of int. opp. $\angle s$ )

Ex. 24 . Three circles have the centres at $A, B, C$ and each circle touches the other two externally. If $A B=5 cm$, $B C=7 cm, C A=6 cm$, find the radii of the three circles.

Sol. Let the radii of the circles with centres $A, B, C$ be $x cm, y cm$ and $z cm$ respectively. Then

$ \begin{aligned} A B & =x+y=5 \\ B C & =y+z=7 \\ C A & =z+x=6 \\ \text{ Adding, } \quad 2(x+y+z) & =18 \\ \therefore \quad x+y+z & =9 \end{aligned} $

Subtracting each equation in turn from (iv), we obtain

$ \begin{aligned} & z=9-5=4, \\ & x=9-7=2, y=9-6=\mathbf{3} . \end{aligned} $

Ex. 25 . In the given figure a circle is inscribed in quadrilateral $A B C D$. If $B C=38 cm, B Q=27 cm, D C=25 cm$ and $A D \perp D C$, find the radius of the circle.

Sol. Let the sides $A D, A B, B C$ and $C D$ touch the circle at point $P, Q, R$ and $S$ respectively. Since tangent to a circle is perpendicular to the radius through the point of contact.

$\therefore \quad O P \perp A D$ and $O S \perp D C$. Also $A D \perp D C$ (given)

$\therefore \quad O P D S$ is a square. $B R=B Q=27 cm$ (tangents from an external point to a circle are equal in length)

$\therefore \quad C R=B C-B R=(38-27) cm=11 cm$

Similarly, $C S=C R=11 cm$

$\therefore \quad D S=D C-C S=(25-11) cm=14 cm$

$\therefore$ Radius of circle $=O P=D S=\mathbf{1 4} \mathbf{~ c m}$. ( $\because O P D S$ is a square)

Ex. 26 . Two circles of radii $25 cm$ and $9 cm$ touch each other externally. Find the length of the direct common tangent.

Sol. Let the two circles with centres $A, B$ and of radii $25 cm$ and $9 cm$ touch each other externally at point $C$. Then, $A B=A C+C B=(25+9) cm=34 cm$.

Let $P Q$ be the direct common tangent. $\therefore B Q \perp P Q$ and $A P \perp P Q$.

Draw $B R \perp A P$. Then $B R P Q$ is a rectangle.

(Tangent $\perp$ radius at the point of contact)

In $\triangle A B R, A B^{2}=A R^{2}+B R^{2}$

(Pythagoras’ Theorem)

$\Rightarrow \quad(34)^{2}=(16)^{2}+B R^{2}$

$\Rightarrow \quad(B R)^{2}=1156-256=900$

$\Rightarrow \quad B R=\sqrt{900} cm=30 cm$

$\therefore \quad P Q=B R=\mathbf{3 0} \mathbf{~ c m}$.

Ex. 27 . The radii of two concentric circles are $13 cm$ and $8 cm$ respectively. $A B$ is a diameter of the bigger circle. $B D$ is a tangent to the smaller circle touching it at $D$. Find the length of $A D$.

Sol. Let the line $B D$ intersect the bigger circle at $C$. Join $A C$. Then, in the smaller circle.

$O D \perp B D$

$\Rightarrow O D \perp B C \Rightarrow B D=D C$

$\Rightarrow D$ is the mid-point of $B C$

Also, given $O$ is the mid-point of $A B$ ( $B C$ is the chord of the bigger circle and perpendicular from the centre of the circle to a chord bisects the chord)

$\therefore$ In $\triangle B A C, O$ is the mid-point of $A B$ and $D$ is the mid-point of $B C$.

$\therefore O D=\frac{1}{2} A C$ (segment joining the mid-points of any two sides of a triangle is half the third side) $\Rightarrow A C=2 O D \Rightarrow A C=2 \times 8=16 cm$

In right $\triangle O B D$

$ \begin{aligned} & O D^{2}+B D^{2}=O B^{2} \Rightarrow B D=\sqrt{O B^{2}-O D^{2}}=\sqrt{(13)^{2}-8^{2}}=\sqrt{169-64}=\sqrt{105} \\ \therefore & D C=B D=\sqrt{105} \end{aligned} $

Now $A D^{2}=A C^{2}+D C^{2}$

$\Rightarrow A D^{2}=16^{2}+(\sqrt{105})^{2}=256+105$

$\Rightarrow A D^{2}=361 \Rightarrow A D=\sqrt{361}=\mathbf{1 9} \mathbf{~ c m}$.

Ex. 28 . $P Q$ is a transverse common tangent to the circles with centres $A$ and $B$ touching them at $P$ and $Q$ respectively: Prove that $\frac{A P}{B Q}=\frac{A O}{B O}$ where $O$ is the point of intersection of the common tangent and the line joining the centres.

Sol. $P Q$ is the tangent to the circle with centre $A$ at point $P$ and to the circle with centre $B$ at point $Q$. $\therefore \angle A P Q=90^{\circ}$ Similarly $\angle B Q P=90^{\circ}$ $\therefore$ In $\triangle A P O$ and $B Q O$

$ \begin{matrix} \angle A P O=\angle B Q O & (\text{ each }=90^{\circ}) \\ \angle A O P=\angle B O Q & (\text{ vert. opp. } \angle s) \end{matrix} $

$ \begin{aligned} & \therefore \quad \triangle A P O \sim \triangle B Q O \quad \text{ (AA similarity) } \\ & \Rightarrow \frac{\boldsymbol{A P}}{\boldsymbol{B Q}}=\frac{\boldsymbol{A} \boldsymbol{O}}{\boldsymbol{B O}} . \end{aligned} $

(radius $\perp$ tangent at the point of contact)

Ex. 29 . In the given figure, two circles with centres $O$ and $O^{\prime}$ touch externally at a point $A$. A line through $A$ is drawn to intersect these circles in $B$ and $C$. Prove that the tangents at $B$ and $C$ are parallel.

Sol. The two circles with centres $O$ and $O^{\prime}$ touch externally at $A$. Line through $A$ intersects the circles at $B$ and $C$. Tangents $P B Q$ and $R C S$ are drawn. We have to prove $P B Q | R C S$. Join $O$ and $O^{\prime}$ to $A, B$ and $C$

$ \begin{aligned} O A & =O B \\ \Rightarrow \quad \angle O B A & =\angle O A B \\ \Rightarrow \quad \angle O B A & =\angle O^{\prime} A C \\ \Rightarrow \quad \angle O B A & =\angle O^{\prime} C A \end{aligned} $

(radii of the same circle) (angles opp. equal sides are equal) ( $\angle O A B=\angle O^{\prime} A C$, vert. opp. $\angle s$ ) $(\angle O^{\prime} A C=\angle O^{\prime} C A, \because O^{\prime} C=O^{\prime} A)$

Also, $O B Q=O^{\prime} C R$ (each $=90^{\circ}$, radius $\perp$ tangent at the point of contact $)$

$\Rightarrow \quad \angle O B A+\angle 1=\angle O^{\prime} C A+\angle 2$

$ \begin{matrix} \Rightarrow \quad \angle 1=\angle 2 & (\because \angle O B A=O^{\prime} C A) \\ \Rightarrow \quad P B Q | R C S & \text{ (alt. } \angle S \text{ are equal, BAC is the transversal) } \end{matrix} $

IV. ALTERNATE SEGMENT THEOREM

The alternate segment property: $T^{\prime} P T$ is a tangent to a circle at the point $P$ and $P A$ is a chord of contact. $\angle A P T$ and the segment $A X P$ lie on opposite sides of the chord of contact. Therefore, when dealing with $\angle A P T$, segment $A X P$ is called the alternate segment and any angle $A X P$ is called the angle in the alternate segment. (Abbreviation: $\angle s$ in alternate segments)

Theorem 22. If a st. line touches a circle and from the point of contact, a chord is drawn, the angles between the tangent and the chord are respectively equal to the angles in the alternate segments.

(Abbreviation: $\angle s$ in alternate segments)

e.g., $\angle R P B=\angle P S R, \angle A P R=\angle P Q R$

SOLVED EXAMPLES

Ex. 30 . In the given figure, line $P Q$ touches the circle at $A$. If $\angle P A C=80^{\circ}$, and $\angle Q A B=63^{\circ}$, calculate the angles of $\triangle A B C$.

Sol.

$ \begin{matrix} \angle A C B=\angle Q A B=\mathbf{6 3}^{\circ} \\ \angle A B C=\angle P A C=\mathbf{8 0}^{\circ} \end{matrix} $

( $\angle s$ in alternate segments)

Now, in $\triangle A B C, \angle C A B+\angle A B C+\angle A C B=180^{\circ}$

$(\angle$ sum of $a \Delta)$

$\Rightarrow \angle C A B+80^{\circ}+63^{\circ}=180^{\circ} \Rightarrow \angle C A B=180^{\circ}-143^{\circ}=37^{\circ}$

Ex. 31 . $P A$ and $P B$ are two tangents of a circle. $\angle A P B=50^{\circ}$ and chord $A C$ is drawn parallel to $P B$. Find by calculation the angles of $\triangle A B C$.

Sol.

$ P A=P B $

(Tangents from external point of a circle are equal)

$ \begin{aligned} & \Rightarrow \quad \angle P A B=\angle P B A=x \\ & \text{ In } \triangle P A B, \quad x+x+50^{\circ}=180^{\circ} \\ & \Rightarrow \quad 2 x=130^{\circ} \Rightarrow x=65^{\circ} \\ & \text{ Now, } \quad \angle C A P+\angle A P B=180^{\circ}(\text{ co-int } \angle s, A C | P B) \\ & \Rightarrow \quad \angle C A B+x+50^{\circ}=180^{\circ} \Rightarrow \angle C A B+65^{\circ}+50^{\circ}=180^{\circ} \\ & \Rightarrow \quad \angle C A B=180^{\circ}-115^{\circ}=\mathbf{6 5}^{\circ} \\ & \therefore \quad \angle C B A=180^{\circ}-(\angle A C B+\angle C A B)=180^{\circ}-(65^{\circ}+65^{\circ}) \\ & =180^{\circ}-130^{\circ}=\mathbf{5 0}^{\circ} \end{aligned} $

$ \therefore \quad \angle A C B=\angle P B A=x=65^{\circ} \quad(\angle s \text{ in alternate segments }) $

Hence, the angles of $\triangle A B C$ are $65^{\circ}, 65^{\circ}, 50^{\circ}$.

Ex. 32 . $P Q$ and $P R$ are two equal chords of a circle. Show that $S P T$, a tangent at $P$ is parallel to $Q R$.

$P$ is the mid-pt. of arc $Q P R$ of a circle. Show that the tangent at $P$ is parallel to the chord $Q R$.

Sol.

$ \begin{aligned} P Q & =P R \\ \angle P R Q & =\angle P Q R \\ \angle R P T & =\angle P Q R \\ \angle P R Q & =\angle R P T \end{aligned} $

But these are alternate angles $\therefore S P T | Q R$.

(Given)

( $\angle$ s opp. equal sides in a $\Delta$ )

( $\angle s$ in alternate segments)

Hence proved.

Ex. 33 . In the given figure, $S A T$ is the tangent to the circumcircle of $a \triangle A B C$ at the vertex $A$. A line parallel to $S A T$ intersects $A B$ and $A C$ at the points $D$ and $E$ respectively. Prove that $\triangle A B C \sim \triangle A E D$, and $A B \times A D=A E \times A C$.

Sol. In the given figure we have

$ \begin{aligned} & \angle S A D=\angle A D E \\ & \angle T A E=\angle A E D \end{aligned} $

Ex. 34 . Two lines $A B C$ and $A D E$ are intersected by two parallel lines in $B, D$ and $C, E$ respectively. Prove that the circumcircles of $\triangle A B D$ and $\triangle A C E$ touch each other at $A$.

Sol. Draw TA tangent to the circumcircle of $\triangle ACE$

Now,	$\angle T A C=\angle A E C$	( $\angle s$ in alternate segments)
	$\angle A E C=\angle A D B$	(Corr. $\angle s, B D \\| C E)$
$\therefore$	$\angle T A C=\angle A D B$
or

Since $\angle A D B$ is an angle in the alternate segment, therefore, $T A$ is a tangent to the circumcircle of $\triangle A D B$ also. Thus, $T A$ is a tangent to both the circles at the same point $A$, hence, the two circles touch each other at $A$.

Proved.

V. SEGMENTS OF A CHORD

Definition: If $A B$ is any chord of a circle, and if $X$ is any point either on $A B$ or $A B$ produced, then $A X$ and $B X$ are called the segments of the chord formed by the point of division $X$.

Theorem:

Theorem 23. If two chords of a circle intersect internally or externally, then the

product of the lengths of their segments are equal.

(i) When two chords $A B$ and $C D$ of a circle with centre $O$, intersect at a point $P$ inside the circle, then $\boldsymbol{A P} \cdot \boldsymbol{P B}=\boldsymbol{C P} . \boldsymbol{P D}$

(ii) Two chords $A B$ and $C D$ of a circle, when produced intersect at a point $P$ outside the circle, then $\boldsymbol{P A} . \boldsymbol{P B}=\boldsymbol{P C} . \boldsymbol{P D}$.

Theorem 24. Conversely, if two straight lines $A B$ and $C D$ cut each other either both internally or both externally at $P$ so that $P A . P B=P C . P D$, then the four points $A, B, C, D$ lie on a circle.

Theorem 25. If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

e.g., when, chord $A B$ and tangent $T P$ of a circle intersect at a point $P$ outside the circle, then $\boldsymbol{P A} \cdot \boldsymbol{P B}=\boldsymbol{P} \boldsymbol{T}^{\mathbf{2}}$.

Theorem 26. Conversely, iffrom any point $P$ on a line $A B$ produced another straight line is drawn and a point $T$ is taken on it such that $P A . P B=P T^{2}$, then the line PT is a tangent to the circle which passes through $A, B, T$.

SOLVED EXAMPLES

Ex. 35 . In the given figure, chords $A B$ and $C D$ intersect at point $P$. If $C P=6, C D=9$ and $A B=19$, what are the lengths of $A P$ and $P B$ ?

Sol.

$ \begin{aligned} & C P=6, & D P=C D-C P & =9-6=3 \\ \text{ Let } & A P & =x, \text{ then } & P B=A B-A P=19-x \\ \text{ Now } & A P \cdot P B & =C P \cdot D P & \\ \therefore & x(19-x) & =6 \times 3 \Rightarrow & \Rightarrow 19 x-x^{2}=18 \\ \Rightarrow & x^{2}-19 x+18 & =0 \Rightarrow & (x-1)(x-18)=0 \\ \Rightarrow & & x & =\mathbf{1}, \mathbf{1 8} \end{aligned} $

Hence, the lengths of $A P$ and $P B$ are $\mathbf{1}$ unit and $\mathbf{1 8}$ units.

Ex. 36 . In the given figure, $P$ is outside the circle and secants $P C A$ and $P D B$ intersect the circle at $C$ and $A, D$ and $B$ respectively. If $P A=24, C A=16$ and $D B=26$, find $P B$.

Sol. Let

$ P D=x $

$ P D \times P B=P C \times P A \Rightarrow x(x+26)=8 \times 24 $

$\Rightarrow \quad x^{2}+26 x-192=0$

$\Rightarrow \quad(x+32)(x-6)=0$

$\Rightarrow \quad x=6$, taking only positive value of $x$.

$\Rightarrow \quad P B=6+26=32$.

Ex. 37 . In the adjoining figure, the angle $A$ of the triangle $A B C$ is a right angle. The circle on $A C$ as diameter cuts $B C$ at $D$. If $B D=9$, and $D C=7$, calculate the length of $A B$.

Sol. Since $O A$ is a radius and $\angle B A C=90^{\circ}$, therefore $B A$ is a tangent to the circle, by the tangent-radius property.

Hence,

$B A^{2}=B D \times B C=9 \times 16=144$

$\Rightarrow \quad B A=\mathbf{1 2}$.

Ex. 38 . If the diagonals of the quadrilateral $A B C D$ cut at $O$, and if $O A=3 cm, O B=9 cm, A C=15 cm$, $B D=13 cm$, prove that $A B C D$ is cyclic.

Sol.

$ \begin{matrix} O C & =A C-O A=15-3=12 cm \\ & & O D & =B D-B O=13-9=4 cm \\ \therefore \quad & & A O \cdot O C & =3 \times 12=36 cm^{2} \\ \Rightarrow \quad & B O \cdot O D & =9 \times 4=36 cm^{2} \\ & & A O \cdot O C & =B O \cdot O D \end{matrix} $

Hence, $A B C D$ is a cyclic quadrilateral.

Ex. 39 . $A B C$ and $D B C$ are two right triangles with common hypotenuse $B C$ and with their sides $A C$ and $D B$ intersecting at $P$. Prove that $A P . P C=B P$. PD.

Sol. $\therefore \quad \angle B A C=\angle B D C($ each $=1 rt . \angle)$

$\therefore \quad$ Pts. $B, A, D, C$ are concyclic ( $\angle$ s on the same side of segment $B C$ are equal)

$\therefore \quad A P . P C=B P . P D$ (product of segments of intersecting chords of a circle)

Ex. 40 . In a trapezium $A B C D, A B | C D$ and $A D=B C$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, prove that $P A \times P C=P B \times P D$.

Sol. Draw $D E \perp A B$ and $C F \perp A B$

In $\triangle S E E A$ and $C F B$, we have

$ \begin{aligned} & A D=B C \quad \text{ (Given) } \\ & \angle D E A=\angle C F B \quad(\text{ each }=90^{\circ}) \\ & D E=C F \quad(\text{ Distance between two parallels }) \\ & \therefore \quad \triangle D E A \cong \triangle C F B \quad \text{ (R.H.S.) } \\ & \Rightarrow \quad \angle D A E=\angle C B F \\ & \text{ (c.p.c.t.) } \end{aligned} $

Thus $A C$ and $B D$ are two chords of the circle circumscribring the trapezium such that they intersect at $P$.

Hence, $\boldsymbol{P A} \times \boldsymbol{P} \boldsymbol{C}=\boldsymbol{P} \boldsymbol{B} \times \boldsymbol{P D}$.

Ex. 41 . The radius of the incircle of a triangle is $24 cm$. The segments into which one side is divided by the points of contact are $36 cm$ and $48 cm$. Find the lengths of the other two sides of the triangle.

Sol. Let the sides $Q R, P R$ and $Q P$ touch the incircle in points $A, B$ and $C$ respectively.

Suppose $Q R$ is divided by point $A$ into segments $Q A$ and $A R$ measuring $36 cm$ and $48 cm$ respectively.

$\because A Q$ and $Q C$ are tangents to the incircle from point touching it at points $A$ and $C$ respectively and lengths of tangents from the same external point are equal, $Q C=Q A=36 cm$

Similarly, $R B=R A=48 cm$.

Let $P C=P B=x cm$. Also let $Q R=a, P R=b, P Q=c$.

Then, $a=(36+48) cm, \quad b=(x+48) cm, \quad c=(x+36) cm$

$\therefore$ Semi-perimeter $(s) \quad=\frac{1}{2}(a+b+c)$

$ =\frac{(36+48+x+48+x+36) cm}{2}=(x+84) cm $

$(s-a)=x cm,(s-b)=36 cm,(s-c)=48 cm$

$\therefore \quad$ Area of $\triangle P Q R=\sqrt{s(s-a)(s-b)(s-c)}$

$ =\sqrt{(x+84) \cdot x \cdot 36 \cdot 48} \cdot cm^{2} $

$\therefore \quad$ In radius $r=\frac{\Delta}{s}=\frac{\text{ Area of } \triangle P Q R}{s}=\frac{\sqrt{(x+84) \cdot x \cdot 36 \cdot 48}}{x+84}=24 \sqrt{\frac{3 x}{x+84}}$

$\because r=24, \quad \therefore \quad 24=24 \sqrt{\frac{3 x}{x+84}} \Rightarrow \sqrt{\frac{3 x}{x+84}}=1$

$\Rightarrow \quad 3 x=x+84 \quad \Rightarrow \quad 2 x=84 \Rightarrow x=42$

$\therefore \quad b=(x+48) cm=\mathbf{9 0} \mathbf{~ c m}$

$\therefore \quad c=(x+36) cm=\mathbf{7 8} \mathbf{~ c m}$.

Ex. 42 . In a quadrilateral $A B C D$, a circle with centre at the mid-point of $A B$ touches the sides $B C, C D$ and $A D$. Show that $A B^{2}=4 A D B C$.

Sol. Let $O$ be the mid-point of $A B$. Let $X$ and $Y$ be respectively the points of contact of $A D$ and $B C$ with the circle. Then,

$O A=O B$

$(O$ is mid-point of $A B$ )

$O X=O Y$

(radii) $\Rightarrow \angle OAD=\angle OBC$

$\because A D$ and $D C$ are tangets,

$\angle A D C=2 \angle A D O$

(Tangents are equally inclined to the line joining the centre and pt. of contact of tangents)

Similarly,

$ \angle B C D=2 \angle B C O $

Now $\angle O A D+\angle A D C+\angle D C B+\angle O B C=180^{\circ}$

$ =2(\angle O A D+\angle A D O+\angle A O D) \quad(\text{ Area of } \triangle O A D) $

$\Rightarrow \angle O A D+2 \angle A D O+2 \angle B C O+\angle O A D=2(\angle O A D+\angle A D O+\angle A O D)$

$\Rightarrow \angle B C O=\angle A O D$

(using (i), (ii) and (iii))

Now $O A=O B, \angle O A D=\angle O B C, \angle B C O=\angle A O D \Rightarrow \triangle A O D \cong \triangle B C O$

$\Rightarrow \frac{A O}{B C}=\frac{A D}{B O} \Rightarrow A O$. BO $\Rightarrow A D$ BC

$\Rightarrow \frac{1}{2} A B \cdot \frac{1}{2} A B=A D \cdot B C$

$\Rightarrow A B^{2}=4 A D$. BC.

Ex. 43 . Two circles with radii $r$ and $R$ are externally tangent at a point $P$. Determine the length of the segment cut from the common tangent through $\boldsymbol{P}$ by the other common tangents.

Sol. Without loss of generality, we may assume that $r \leq R$. Let the circle with radius $r$ have centre $O_1$ and the circle with radius $R$ have centre $O_2$. Let $P$ be their point of tangency. Let the common external tangents meet the circles at $A, B, C$ and $D$, as in the diagram. Let the internal common tangent meet the external common tangents at $K$ and $L$.

Let $S$ be the point on $O_2 B$ such that $O_1 S \perp O_2 B$. Then $O_1 S=A B$ and $O_2 S=R-r$.

Also

$O_1 S=\sqrt{(O_1 O_2)^{2}-(O_2 S)^{2}}=\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \sqrt{R r}$.

Thus, $K P=\frac{1}{2} A B^{*}=\sqrt{R r}$. Similarly, since $C D=A B=2 \sqrt{R r}$, we have

(* Since $K P=K A=K B)$ $P L=\sqrt{R r}$, which implies that $K L=2 \sqrt{R r}$.

PRACTICE SHEET

Level-1

1. In the given figure, $O$ is the centre of the circle. $O A=3 cm, A C=3 cm$ and $O M \perp A C$. What is $\angle A B C$ equal to? (a) $60^{\circ}$ (b) $45^{\circ}$ (c) $30^{\circ}$ (d) None of these

(CDS 2011) 2. $A C$ is the diameter of the circumcircle of the cyclic quadrilateral $A B C D$. If $\angle B D C=42^{\circ}$, then what is $\angle A C B$ equal to? (a) $42^{\circ}$ (b) $45^{\circ}$ (c) $48^{\circ}$ (d) $58^{\circ}$

3. If $A, B, C$, are three consecutive points on the arc of a semicircle such that the angles subtended by the chords $A B$ and $A C$ at the centre $O$ are $60^{\circ}$ and $100^{\circ}$ respectively. Then $\angle B A C$ is equal to (a) $20^{\circ}$

(b) $50^{\circ}$

(d) $200^{\circ}$

4. In the given figure, $B T$ and $C T^{\prime}$ are two tangents at points $B$ and $C$ on the circle and $\angle B P C=80^{\circ}$. Then $\angle A$ is

(a) $80^{\circ}$ (b) $60^{\circ}$ (c) $50^{\circ}$ (d) $40^{\circ}$

5. In the given figure, $S T$ is the diameter of the circle with centre $O, P Q$ is the tangent at point $R$. If $\angle T R Q=40^{\circ}$, find $\angle R T S$. (a) $40^{\circ}$ (b) $50^{\circ}$ (c) $60^{\circ}$ (d) $30^{\circ}$

6. In the given figure, $T A S$ is a tangent $T \uparrow$ to the circle at the point $A$. If $\angle O B A$ $=32^{\circ}$, what is the value of $x$ ? (a) $64^{\circ}$ (b) $40^{\circ}$ (c) $58^{\circ}$ (d) $50^{\circ}$ 7. In a triangle $D E F, O$ is the centre of the incircle $A B C$. $\angle D E F=60^{\circ}, \angle D F E=75^{\circ}$. Find $\angle A O B$

(a) $75^{\circ}$

(b) $45^{\circ}$

(d) cannot be determined.

8. In the adjoining figure, $O$ is the centre of the circle and $A B$ is the diameter. Tangent $P Q$ touches the circle at $D$. $\angle B D Q=48^{\circ}$. Find the ratio of $\angle D B A: \angle D C B$. (a) $22 / 7$ (b) $7 / 22$ (c) $7 / 12$ (d) can’t be determined.

9. In the given diagram, $O$ is the centre of the circle and $C D$ is a tangent. $\angle C A B$ and $\angle A C D$ are supplementary to each other. $\angle O A C=30^{\circ}$. Find the value of $\angle O C B$ ? (a) $30^{\circ}$ (b) $20^{\circ}$ (c) $60^{\circ}$ (d) None of these 10. In the given figure, $O$ is the centre of the circle. $A C$ and $B D$ intersect at $P$. If $\angle A O B=100^{\circ}$ and $\angle D A P=30^{\circ}$, what is $\angle A P B$ ? (a) $77^{\circ}$ (b) $80^{\circ}$ (c) $85^{\circ}$ (d) $90^{\circ}$

(CDS 2010)

11. What is the distance (in $cm$ ) between two parallel chords of lengths $32 cm$ and $24 cm$ in a circle of radius $20 cm$ ? (a) 1 or 7 (b) 2 or 14 (c) 3 or 21 (d) 4 or 28

(CAT 2005)

12. In the given figure, $O$ is the centre of the circle and $A C$ the diameter. The line $F E G$ is tangent to the circle at point $E$. If $\angle G E C=52^{\circ}$, find the value of $\angle e+\angle c$. (a) $154^{\circ}$ (b) $156^{\circ}$ (c) $166^{\circ}$ (d) $180^{\circ}$ (CAT 2011)

13. $P B A$ and $P D C$ are two secants. $A D$ is the diameter of the circle with centre at O. $\angle A=40^{\circ}, \angle P=20^{\circ}$. Find the measure of $\angle D B C$.

(a) $30^{\circ}$ (b) $45^{\circ}$ (c) $50^{\circ}$ (d) $40^{\circ}$

14. In the given figure, $O$ is the centre of incircle for $\triangle P Q R$. Find $\angle Q O R$ if $\angle Q P R=40^{\circ}$. (a) $140^{\circ}$ (b) $110^{\circ}$ (c) $80^{\circ}$ (d) $120^{\circ}$

15. In the given figure, $P Q$ is the diameter of the circle whose centre is at $O$. If $\angle R O S=44^{\circ}$ and $O R$ is the bisector of $\angle P R Q$, then what is the value of $\angle R T S$ ? (a) $46^{\circ}$ (b) $64^{\circ}$ (c) $69^{\circ}$ (d) None of these

Level-2

16. $A C B$ is a tangent to a circle at $C . C D$ and $C E$ are chords such that $\angle A C E>\angle A C D$. If $\angle A C D=\angle B C E=50^{\circ}$, then

(a) $C D=C E$

(b) $E D$ is not parallel to $A B$

(d) $\triangle C D E$ is right angled triangle.

17. Two circles intersect each other at $O$ and $P . A B$ is a common tangent to the circles. Then the angles subtended by the line segment $A B$ at $O$ and $P$ are: (a) complementary (b) supplementary (c) equal (d) None of these 18. $O$ is the centre of the given circle. Then $\angle x+\angle y$ equals (a) $2 \angle z$ (b) $\frac{\angle z}{2}$ (c) $\angle z$ (d) None of these

19. In the given figure, $P T$ touches the circle whose centre is $O$, at $R$. Diameter $S Q$ when produced meets $P T$ at $P$. If $\angle S P R=x^{\circ}, \angle Q R P=$ $y^{\circ}$, then $x+2 y=$

(a) $100^{\circ}$ (b) $120^{\circ}$ (c) $80^{\circ}$ (d) $90^{\circ}$

20. In the given figure, $O$ is the centre of the circle. The line $U T V$ is a tangent to the circle at $T$. $\angle V T R$ $=52^{\circ}$ and $\triangle P T R$ is an isosceles triangle such that $T P=T R$. What is $\angle x+\angle y+\angle z$ equal to? (a) $175^{\circ}$ (b) $208^{\circ}$ (c) $218^{\circ}$ (d) $250^{\circ}$

21 $A, B, C, D$ are four distinct points on a circle whose centre is at $O$. If $\angle O B D-\angle C D B=\angle C B D-\angle O D B$, then what is $\angle A$ equal to? (a) $45^{\circ}$ (b) $60^{\circ}$ (c) $120^{\circ}$ (d) $135^{\circ}$

(CDS 2009)

22 $P Q$ is a common chord of two circles. $A P B$ is a secant line joining points $A$ and $B$ on the two circles. Two tangents $A C$ and $B C$ are drawn. If $\angle A C B=45^{\circ}$, then what is $\angle A Q B$ equal to? (a) $75^{\circ}$ (b) $90^{\circ}$ (c) $120^{\circ}$ (d) $135^{\circ}$

(CDS 2009)

23 In the given figure, $O$ is the centre of the circumcircle of $\triangle X Y Z$. Tangents at $X$ and $Y$ intersect at $T$. $\angle X T Y=80^{\circ}$, what is the value of $\angle Z X Y$ (a) $20^{\circ}$ (b) $40^{\circ}$ (c) $60^{\circ}$ (d) $80^{\circ}$

(CDS 2007)

24 In the figure given below (not drawn to scale) $A, B$ and $C$ are three points on a circle with centre $O$. The chord $B A$ is extended to a point $T$ such that $C T$ becomes a tangent to the circle at point C. If $\angle A T C=30^{\circ}$ and $\angle A C T=50^{\circ}$, then $\angle B O A$ is

(a) $100^{\circ}$ (b) $150^{\circ}$ (c) $80^{\circ}$

(d) not possible to determine

(CAT 2003)

25 Two circles touch internally at point $P$ and a chord $A B$ of the circle of longer radius intersects the other circle in $C$ and $D$. Which of the following holds good?

(a) $\angle C P A=\angle D P B$

(b) $2 \angle C P A=\angle C P D$

(d) $\angle B P Y=\angle C P D+\angle C P A$

26 If two equal circles of radius $5 cm$ have two common tangents $A B$ and $C D$ which touch the circle on $A, C$ and $B, D$ respectively and if $C D=24 cm$, find the length of $A B$.

27 Two circles $C(O, r)$ and $C(O^{\prime}, r^{\prime})$ intersect at two points $A$ and $B . O$ lies on $C$ $(O^{\prime}, r^{\prime})$. A tangents $C D$ is drawn to the circle $C(O^{\prime}, r^{\prime})$ at $A$. Then,

(a) $\angle O A C=\angle O A B$

(b) $\angle O A B=\angle A O^{\prime} O$

(d) None of these

28 $A B C$ is an equilateral triangle inscribed in a circle with $A B=5 cm$. Let the bisector of angle $A$ meet $B C$ in $X$ and the circle in Y. What is the value of $A X . A Y$ ? (a) $16 cm^{2}$ (b) $20 cm^{2}$ (c) $25 cm^{2}$ (d) $30 cm^{2}$

(CDS 2011)

29 In the given figure, $P T$ is a tangent to a circle of radius $6 cm$. If $P$ is at a distance of $10 cm$ from the centre $O$ and $P B=5 cm$, then what is the length of chord $B C$ ?

(a) $7.8 cm$ (b) $8 cm$ (c) $8.4 cm$ (d) $9 cm$

(CDS 2009)

30 In the given figure, $A P=3 cm$, $P B=5 cm, A Q=2 cm$ and $Q C=x$. What is the value of $x$ ? (a) $6 cm$ (b) $8 cm$ (c) $10 cm$ (d) $12 cm$

31 In a $\triangle A B C, A B=A C$. A circle through $B$ touches $A C$ at $D$ and intersects $A B$ at $P$. If $D$ is the mid-point of $A C$, which one of the following is correct?

(a) $A B=2 A P$

(b) $A B=3 A P$

(d) $2 A B=5 A P$

(CDS 2007)

32 In the given circle, $O$ is the centre of the circle and $A D, A E$ are the two tangents. $B C$ is also a tangent, then:

(a) $A C+A B=B C$

(b) $3 A E=A B+B C+A C$

(d) $2 A E=A B+B C+A C$

33 In the given figure, $A T$ and $B T$ are the two tangents at $A$ and $B$ respectively. $C D$ is also a tangent at $P$.

(a) $27 cm$ (b) $25 cm$ (c) $26 cm$ (d) $30 cm$

There are some more circles touching each other and the tangents $A T$ and $B T$ also. Which one of the following is true? (a) $P C+C T=P D+D T$ (b) $R G+G T=R H+H T$ (c) $P C+Q E=C E$ (d) All of these

34 Two circles with radii ’ $a$ ’ and ’ $b$ ’ respectively touch each other externally. Let ’ $c$ ’ be the radius of a circle that touches these two circles as well as a common tangent to the circles. Then, (a) $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{c}}$ (b) $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=\frac{-1}{\sqrt{c}}$ (c) $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{c}}$ (d) None of these 35 Triangle $P A B$ is formed by three tangents to circle $O$ and $\angle A P B=40^{\circ}$. Then $\angle A O B$ equals (a) $45^{\circ}$ (b) $60^{\circ}$ (c) $70^{\circ}$ (d) $55^{\circ}$

Level-3

36 In the given figure, $P A$ is tangent to semi-circle $S A R . P B$ is tangent to semi-circle $R B T ; S R T$ is a straight line, the lengths of the arcs are indicated in the figure Angle $A P B$ is measured

by (a) $\frac{1}{2}(a-b)$ (b) $a+b$ (c) $\frac{1}{2}(a+b)$ (d) $(a-b)$

37 Let $A B C D$ be a quadrilateral in which $A B$ is parallel to $C D$ and perpendicular to $A D, A B=3 C D$, the area of quadrilateral is 4 sq. unit. If a circle can be drawn touching all the sides of the quadrilateral, then the radius of the circle is (a) 2 units (b) $\sqrt{3}$ units (c) $\frac{\sqrt{3}}{2}$ units (d) $2 \sqrt{3}$ units

(RMO 2006)

38 Two fixed circles in a plane intersect in points $P$ and $Q$. A variable line through $P$ meets the circles again in $A$ and $B$. Prove that the angle $A Q B$ is of constant measure. 39 Let $A$ be one of the two points of intersection of two circles with centre $X, Y$ respectively. The tangents at $A$ to the two circles meet the circles again at $B, C$. Let the point $P$ be located so that $P X A Y$ is a parallelogram. Show that $P$ is also the circumcentre of $\triangle A B C$. 40 Let $A B C$ be a triangle and a circle $C^{\prime}$ be drawn lying inside the triangle, touching its incircle $C$ externally and also touching the two sides $A B$ and $A C$. Show that the ratio of the radii of the two circles $C$ ’ and $C$ is equal to $\tan ^{2}(\frac{\pi-2}{4})$. 41 Three circles touch each other externally and all the three touch a line. If two of them are equal and the third has radius $4 cm$. Find the radius of the equal circles. 42 $A B C$ is an equilateral triangle inscribed in a circle. $P$ is any point on the minor $arc B C$. Prove that $P A=P B+P C$. 43 Let $\triangle A B C$ be equilateral. On side $A B$ produced, we choose a point $P$ such that $A$ lies between $P$ and $B$. We now denote $\alpha$ as the lengths of sides of $\triangle A B C ; r_1$ as the radius of incircle of $\triangle P A C$ and $r_2$ as the exradius of $\triangle P B C$ with respect to side $B C$. Then prove that $r_1+r_2$ equals $\frac{\alpha \sqrt{3}}{2}$.

(Austrain Polish Mathematics Comptt.)

44 Let $A B C D$ be a cyclic quadrilateral and let $P$ and $Q$ be points on the sides $A B$ and $A D$ respectively, such that $A P=C D$ and $A Q=B C$. Let $M$ be the point of intersection of $A C$ and $P Q$. Then, show that $M$ is the midpoint of $P Q$.

(Australian Mathematical Olympiad)

45 Two disjoint circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ are given. A common exterior tangent touches circles $C_1$ and $C_2$ at $A$ and $B$ respectively and $O_1 O_2$ intersects circles $C_1$ and $C_2$ at points $C$ and $D$ respectively. Prove that:

(a) the points $A, B, C$ and $D$ are concyclic

(b) the straight lines $A C$ and $B D$ are perpendicular.

46 $A B C$ is a triangle with $\angle A>\angle C$ and $D$ is a point on $B C$ such that $\angle B A D=\angle A C B$. The perpendicular bisectors of $A D$ and $D C$ intersect in the point $E$. Prove that $\angle B A E=90^{\circ}$. 47 Points $D$ and $E$ are given on the sides $A B$ and $A C$ of $\triangle A B C$ in such a way that $D E | B C$ and tangent to the incircle of $\triangle A B C$. Prove that $D E \leq \frac{1}{8}(A B+B C+C A)$

(Italian Selection Test)

48 Two circles intersect each other in points $M$ and $N$. An arbitrary point $A$ of the first circle, which is not $M$ or $N$, is connected with $M$ and $N$, and the straight lines $A M$ and $A N$ intersect the second circle again in the points $B$ and $C$. Prove that the tangent to the first circle at $A$ is parallel to the straight line $B C$.

(Swiss Mathematical Test)

ANSWERS

1. (c)	2. $(c)$	3. (a)	4. (c)	5. (b)	6. (c)	7. (c)	8. $(b)$	9. $(a)$	10. $(b)$
11. $(d)$	12. (c)	13. (a)	14. $(b)$	15. $(d)$	16. (a)	17. $(b)$	18. (c)	19. $(d)$	20. $(c)$
21. (b)	22. $(d)$	23. $(d)$	24. (a)	25. $(a)$	26. $(c)$	27. $(a)$	28. $(c)$	29. (a)	30. $(c)$
31. (c)	32. $(d)$	33. $(d)$	34. $(c)$	35. $(c)$	36. $(b)$	37. $(c)$

HINTS AND SOLUTIONS

1. Given $O A=3 cm \Rightarrow O C=3 cm$ (radii of the circle)

Also $A C=3 cm \Rightarrow O A=O C=A C \Rightarrow \triangle A O C$ is equilateral $\Rightarrow \angle A O C=60^{\circ}$.

$\Rightarrow \angle A B C=\frac{1}{2} \times \angle A O C=30^{\circ} \quad$ (Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.)

2. $\angle C A B=\angle C D B=42^{\circ}$

( $\angle s$ in the same segment are equal) $\angle A B C=90^{\circ}$

$ (\angle \text{ in a semicircle }=90^{\circ}) $

$\therefore In \triangle A B C$

$ \angle A C B=180^{\circ}-(\angle C A B+\angle A B C) $

$ =180^{\circ}-(90^{\circ}+42^{\circ})=\mathbf{4 8}^{\circ} $

3. $\angle B O C=\angle A O C-\angle A O B$

$=100^{\circ}-60^{\circ}=40^{\circ}$

$\angle B A C=\frac{1}{2} \times \angle B O C=20^{\circ}$

(Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of circle.)

4. Join $B O$ and $O C$.

In quadrilateral $B O C P$,

$\angle O B P=\angle O C P=90^{\circ}$

(Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\angle B P C=80^{\circ}$

$\Rightarrow \angle B O C=360^{\circ}-(180^{\circ}+80^{\circ})=100^{\circ}$

$\Rightarrow \angle B A C=\theta=\frac{1}{2} \times \angle B O C=50^{\circ}$

5. $\angle T S R=\angle T R Q=40^{\circ}$ (Angles in alternate segment are equal) $S T$ being the diameter, $\angle S R T=90^{\circ}$ (Angle in a semi-circle)

$\therefore \angle R T S=180^{\circ}-(\angle T S R+\angle S R T)$

$ =180^{\circ}-(130^{\circ})=\mathbf{5 0}^{\circ} \text{. } $

6. $O A=O B \Rightarrow \angle O B A=\angle O A B=32^{\circ}$ (Isosceles $\triangle$ property) $\angle A O B=180^{\circ}-(\angle O B A+\angle O A B)=180^{\circ}-64^{\circ}=116^{\circ}$

$.\Rightarrow \angle A C B=\frac{1}{2} \times \angle A O B)=\frac{1}{2} \times 116^{\circ}=58^{\circ}$

Also, $\angle B A S=x=\angle A C B=58^{\circ}($ Angle in alternate segment are equal)

7. In $\triangle D E F, \angle EDF=180^{\circ}-(60^{\circ}+75^{\circ})$

$ =180^{\circ}-135^{\circ}=45^{\circ} $

$\angle O A D=\angle O B D=90^{\circ}$ (Tangents DE and DF are perpendicular to radii $O A$ and $O B$ respectively at $A$ and $B$ )

$\therefore$ In quad. $D A O B, \angle A O B=360^{\circ}-(90^{\circ}+90^{\circ}+45^{\circ})$

$ =360^{\circ}-225^{\circ}=135^{\circ} \text{. } $

8. $\angle D A B=\angle B D Q=48^{\circ}$ (Angles in alternate segment are equal) $\angle A D B=90^{\circ}$

(Angle in a semi-circle)

$ \begin{aligned} \therefore \angle A B D & =180^{\circ}-(\angle D A B+\angle A D B) \\ & =180^{\circ}-(48^{\circ}+90^{\circ})=42^{\circ} \end{aligned} $

$\therefore \angle D C B=180^{\circ}-\angle D A B=180^{\circ}-$

$48^{\circ}=132^{\circ}$

(opp. $\angle$ s of a cyclic quad are supp.)

$\therefore \frac{\angle D B A}{\angle D C B}=\frac{42}{132}=\frac{7}{\mathbf{2 2}}$.

9. $\angle O C D=90^{\circ}$

(Tangent $C D \perp$ Radius $O C$ ) $\angle O C A=\angle O A C=30^{\circ}$

$(O A=O C$, radii $)$

$\angle A C D=\angle O C D+\angle O C A=90^{\circ}+30^{\circ}$

$ =120^{\circ} $

$\angle B A C=180^{\circ}-120^{\circ}=60^{\circ}$

(Given $\angle A C D$ and $\angle B A C$ are supp.)

$\Rightarrow \angle B C D=\angle B A C=60^{\circ}$

(Angles in alternate segment are equal)

$\therefore \angle O C B=\angle O C D-\angle B C D=90^{\circ}-60^{\circ}=\mathbf{3 0}^{\circ}$.

10. $\angle A D B=\frac{1}{2} \times \angle A O B=50^{\circ}$

In $\triangle D P A, \angle A D P+\angle D A P+\angle D P A=180^{\circ}$

$\Rightarrow \angle D P A=180^{\circ}-(50^{\circ}+30^{\circ})=100^{\circ}$

Also, $D P B$ being a straight line,

$\angle A P B=180^{\circ}-\angle D P A=180^{\circ}-100^{\circ}=\mathbf{8 0}^{\circ}$.

11. Let $A B$ and $C D$ be chords of lengths $32 cm$ and $24 cm$ in a circle with centre $O$, on the same side of the centre.

Then $O A=O D=20 cm$

Let the perpendicular from the centre intersect the chords $A B$ and

$C D$ at $E$ and $F$ respectively. Then,

$E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. (Perpendicular from the centre of the circle to a chord bisects the chord.)

Now in rt. $\triangle O F D, O F=\sqrt{O D^{2}-F D^{2}}=\sqrt{20^{2}-12^{2}}$

$ \Rightarrow \quad O F=\sqrt{400-144}=\sqrt{256}=16 cm $

In rt. $\triangle O E B, O E=\sqrt{O B^{2}-E B^{2}}=\sqrt{20^{2}-16^{2}}$

$ =\sqrt{400-256}=\sqrt{144}=12 cm . $

$\therefore$ Required distance $=E F=O F-O E=(16-12) cm=4 cm$. So the option containing value 4 is correct. The other required distance is $28 cm$ when the chords lie on the opposite side of centre $O$.

12. $\angle O A E=\angle G E C=52^{\circ}$

(Angles in alternate segments are

$ \text{ equal) } $

$ \angle A E C=90^{\circ} $

(Angle in a semi circle)

$\therefore$ In $\triangle A C E, e=\angle A C E$

$=180^{\circ}-(90^{\circ}+52^{\circ})=38^{\circ}$

$A C D E$ is a cyclic quadrilateral

$\Rightarrow c=\angle E D C=180^{\circ}-\angle C A E=180^{\circ}-52^{\circ}=128^{\circ}$

(opp. $\angle$ s of a cyclic quad. are supp.)

$\therefore c+e=128^{\circ}+38^{\circ}=\mathbf{1 6 6}^{\circ}$.

13. In $\triangle A D P$, ext. $\angle A D C=$ Int. $\angle s(\angle A+\angle P)$

$ =40^{\circ}+20^{\circ}=60^{\circ} \text{. } $

$\angle A B C=\angle A D C=60^{\circ} \quad($ Angles in the same segment $)$

$\because A D$ is the diameter, $\angle A B D=90^{\circ}$

$\therefore \angle D B C=\angle A B D-\angle A B C=90^{\circ}-60^{\circ}=\mathbf{3 0}^{\circ}$. ~~ 14. $P O$ is joined.

Since the circle is the incircle for $\triangle A B C, P O, Q O, R O$ are the angle bisector of $\angle P, \angle Q$ and $\angle R$ respectively.

$ \begin{aligned} \angle C P O & =\angle A P O \\ & =\frac{1}{2} \times 40^{\circ}=20^{\circ} \end{aligned} $

Also, $O A \perp P R, O C \perp P Q, O B \perp Q R$

(Radii $\perp$ tangent at point of contact)

$\Rightarrow \angle O A P=90^{\circ}$

$\Rightarrow \angle A O P=y=180^{\circ}-(90^{\circ}+20^{\circ})=70^{\circ}$

Now, $y+y+x+x+z+z=360^{\circ}$

$ \begin{aligned} & \Rightarrow 2 y+2(x+z)=360^{\circ} \\ & \Rightarrow 2(x+z)=360^{\circ}-2 y=360^{\circ}-140^{\circ}=220^{\circ} \\ & \Rightarrow x+z=110^{\circ} \Rightarrow \angle Q O R=110^{\circ} . \end{aligned} $

~~ 15. Since $O R$ is the bisector of $\angle P R Q$

$\angle P R O=\angle O R Q=45^{\circ}$

$(\because \angle P R Q=90^{\circ}.$

$\angle$ in a semicircle)

Also, $O P=O R$ (radii)

$\therefore \angle O P R=\angle O R P=45^{\circ}$

In $\triangle O R S, O R=O S \Rightarrow \angle O R S=\angle O S R=\frac{180^{\circ}-44^{\circ}}{2}=68^{\circ}$

$\therefore \angle M R S=68^{\circ}-45^{\circ}=23^{\circ}$

$\Rightarrow \angle P R S=90^{\circ}+23^{\circ}=113^{\circ}$

$\angle P R S+\angle P Q S=180^{\circ}$

$\Rightarrow \angle P Q S=180^{\circ}-\angle P R S=180^{\circ}-113^{\circ}=67^{\circ}$

(opp. $\angle$ s of cyclic quad. PQSR)

In $\triangle P T Q, \angle P T Q=180^{\circ}-(\angle Q P T+\angle P Q T)$

$ =180^{\circ}-(45^{\circ}+67^{\circ})=68^{\circ} $

$\Rightarrow \angle R T S=\angle P T Q=\mathbf{6 8}^{\circ}$. ~~ 16. $\angle A C D=\angle C E D=50^{\circ}$

(Alternate Segment Theorem) $\angle B C E=\angle C D E=50^{\circ}$ $\Rightarrow \angle D=\angle E=50^{\circ} \Rightarrow C D=C E$ Also, $\angle A C D=\angle D=50^{\circ}$, but these are alternate interior angles $\Rightarrow E D | A B$ $\angle D C E=180^{\circ}-(50^{\circ}+50^{\circ})=80^{\circ}$

$\therefore \triangle C D E$ is an acute angled triangle.

~~ 17. $\angle O A B=\angle O P A$

$\angle O B A=\angle O P B$

$\lbrace \text {Angles in alternate segment are equal}. \rbrace$

$\therefore \angle O A B+\angle O B A$

$=\angle O P A+\angle O P B$

$=\angle A P B$.

In $\triangle A O B, \angle A O B$

$=180^{\circ}-(\angle O A B+\angle O B A)=180^{\circ}-\angle A P B$

$\Rightarrow \angle A O B+\angle A P B=180^{\circ}$.

i.e., $\angle S A O B$ and $A P B$ are supplementary

~~ 18. $\angle Q S R=\angle Q T R=\frac{1}{2} \times \angle Q O R=\frac{z}{2}$

(Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

$\angle P S M=\angle P T M=180^{\circ}-\frac{z}{2}$

(Straight line)

Also, $\angle S M T=\angle Q M R=y$ (vert. opp. $\angle s$ )

$\therefore$ In quad. PSMT, $\angle S M T+\angle P T M+\angle T P S+\angle P S M=360^{\circ}$

$ \begin{aligned} & \Rightarrow y+180^{\circ}-\frac{z}{2}+x+180^{\circ}-\frac{z}{2}=360^{\circ} \\ & \Rightarrow \boldsymbol{x}+\boldsymbol{y}=z . \end{aligned} $

~~ 19. $\angle Q S R=\angle Q R P=y^{\circ}$ (Angles in alternate segment are equal) Also, $\angle Q R S=90^{\circ} \quad($ Angle in a semi-circle $)$ $\angle P R S=\angle P R Q+\angle Q R S=y^{\circ}+90^{\circ}$

In $\triangle P R S, \angle S P R+\angle P R S+\angle P S R=180^{\circ}$

$\Rightarrow x^{\circ}+y^{\circ}+90^{\circ}+y^{\circ}=180^{\circ}$

$\Rightarrow x+2 y=\mathbf{9 0}^{\circ}$.

~~ 20. $x=\angle V T R=52^{\circ} \quad$ (Angles in alternate segment are equal) $x+z=180^{\circ}$

(PTMR is a cyclic quad.) $\Rightarrow z=180^{\circ}-x=180^{\circ}-52^{\circ}$ $=128^{\circ}$.

In $\triangle P T R, P T=T R \Rightarrow a=x$

$ =52^{\circ} $

$\Rightarrow \angle P T U=a=52^{\circ}$ (Angles in alternate segment are equal) $\angle Q T U=y+\angle P T U$ $\Rightarrow y+\angle P T U=90^{\circ}(\because \angle Q T U=90^{\circ}.$, rad. $O T \perp$ tangent $.U V)$

$\Rightarrow y=90^{\circ}-52^{\circ}=38^{\circ}$

$\therefore x+y+z=52^{\circ}+38^{\circ}+128^{\circ}=\mathbf{2 1 8}$.

~~ 21. Given, $\angle O B D-\angle C D B$

$ \begin{aligned} & =\angle C B D-\angle O D B \\ & \Rightarrow \angle O B D+\angle O D B=\angle C B D \\ & +\angle C D B \quad \ldots(\text{ i }) \\ & \because \quad O B=O D \text{ (radii) } \\ & \angle O B D=\angle O D B=\theta(\text{ say) } \\ & \text{ Let } \angle C B D=\theta_1, \angle C D B=\theta_2 \end{aligned} $

Then putting these value in eqn

(i), we have

$\theta+\theta=\theta_1+\theta_2 \Rightarrow 2 \theta=\theta_1+\theta_2$

Also, $\angle B O D=180^{\circ}-2 \theta$

$\Rightarrow$ Reflex $\angle B O D=360^{\circ}-(180^{\circ}-2 \theta)$

$\Rightarrow \angle B C D=\frac{1}{2} \times$ Reflex $\times \angle B O D$

$=\frac{1}{2} \times[360^{\circ}-(180^{\circ}-2 \theta)]($ Angle subtended at centre by an arc $=2 \times$ Angle subtended at any point on remaining part of the circle)

Also $\angle B C D=180^{\circ}-(\theta_1+\theta_2)$

$\therefore 180^{\circ}-(\theta_1+\theta_2)=\frac{360^{\circ}-(180^{\circ}-2 \theta)}{2}$

$\Rightarrow 180^{\circ}-2 \theta=90^{\circ}+\theta \quad(\because \theta_1+\theta_2=2 \theta)$

$\Rightarrow 3 \theta=90^{\circ} \Rightarrow \theta=30^{\circ}$

$\therefore \angle B O D=180^{\circ}-60^{\circ}=120^{\circ}$

$\Rightarrow \angle B A D=\frac{1}{2} \times \angle B O D=60^{\circ}$.

~~ 22. Since the tangents drawn on the two given circles, from the same external point are equal, $C A=C B$ $\Rightarrow \angle C A B=\angle C B A=x$ (say) In $\triangle C A B, \quad 45^{\circ}+x+x=180^{\circ}$

$ \Rightarrow 2 x=135^{\circ} \Rightarrow x=67 \frac{1^{\circ}}{2} $

$\angle A Q P=\angle B Q P=x=67 \frac{1^{\circ}}{2}($ Alternate Segment Theorem $)$

$\therefore \angle A Q B=\angle A Q P+\angle B Q P=67 \frac{1^{\circ}}{2}+67 \frac{1^{\circ}}{2}=\mathbf{1 3 5}^{\circ}$.

~~ 23. Given, $\angle X T Y=80^{\circ}$ $T X=T Y$ (Tangents from the same external point are equal)

$\Rightarrow \angle T X Y=\angle T Y X$

$=\frac{1}{2}(180^{\circ}-\angle X T Y)$

$=\frac{1}{2}(180^{\circ}-80^{\circ})=50^{\circ}$

$O X \perp X T$ (radii $\perp$ tangent at point of contact)

$\Rightarrow \angle O X T=90^{\circ} \Rightarrow \angle O X Y=\angle O X T-\angle T X Y=90^{\circ}-50^{\circ}$

Also, $O M \perp Z Y$

$\therefore$ In $\triangle X M Y, \angle X Y M=180^{\circ}-(\angle X M Y+\angle M X Y)$

$ =180^{\circ}-(90^{\circ}+40^{\circ})=50^{\circ} $

Also, by alternate segment theorem,

$ \angle X Z Y=\angle T X Y=50^{\circ} $

$\therefore$ In $\triangle X Z Y, \angle X=180^{\circ}-(\angle X Z Y+\angle X Y Z)$

$ =180^{\circ}-(50^{\circ}+50^{\circ})=\mathbf{8 0}^{\circ} \text{. } $

~~ 24. In $\triangle A T C, \angle C A T=180^{\circ}-(\angle A C T+\angle A T C)$

$ =180^{\circ}-(50^{\circ}+30^{\circ})=100^{\circ} . $

Also, $\angle C B A=\angle A C T=50^{\circ} \quad$ (Alternate Segment Theorem)

$\therefore$ ext $\angle C A T=$ int. opp. $\angle s(\angle C B A+\angle B C A)$

$\Rightarrow 100^{\circ}=50^{\circ}+\angle B C A \Rightarrow \angle B C A=50^{\circ}$

$\Rightarrow \angle B O A=2 \times \angle B C A=100^{\circ}$.

~~ 25. In the bigger circle, $\angle A P X=\angle A B P$

In the smaller circle, $\angle C P X=\angle P D C$

$\lbrace \text{Angles in alternate segment are equal}. \rbrace$

$ \begin{aligned} \Rightarrow \angle A P X+\angle C P A & =\angle C P X=\angle P D C \\ \Rightarrow \angle A B P+\angle C P A & =\angle P D C \quad(\because \angle A \\ \Rightarrow \angle A B P+\angle C P A & =\angle D B P+\angle D P B \\ & =\angle A B P+\angle D P B \\ \Rightarrow \quad \angle C P A & =\angle D P B . \end{aligned} $

$ \begin{aligned} & \Rightarrow \angle A B P+\angle C P A=\angle P D C \quad(\because \angle A P X=\angle A B P) \\ & \Rightarrow \angle A B P+\angle C P A=\angle D B P+\angle D P B(\text{ ext. } \angle \text{ theorem in } \end{aligned} $

$\triangle P D B)$

~~ 26. Hint

Refer to diagram: $\triangle O C P \cong \triangle O^{\prime} B P \Rightarrow O P=O^{\prime} P$

Let $C P=x$. Then, $P D=24-x$

$O P^{2}=5^{2}+x^{2}, O^{\prime} P^{2}=5^{2}+(24-x)^{2}$

$O P^{2}=O^{\prime} P^{2} \Rightarrow 25+x^{2}=25+(24-x)^{2} \Rightarrow x=12$

$\therefore O P^{2}=5^{2}+12^{2} \Rightarrow O P=13=O^{\prime} P$

$\therefore A B=O O^{\prime}=O P+P O^{\prime}=13+13=\mathbf{2 6} \mathbf{~ c m}$.

~~ 27. In $\triangle A O B, O A=O B \Rightarrow \angle O B A=\angle O A B$

(Isosceles $\Delta$ property)

Also, $\angle O A C=\angle O B A \quad$ (Alternate segment theorem)

$\Rightarrow \angle O A C=\angle O A B$.

~~ 28. $\because$ In an equilateral $\Delta$; angle bisector $A X$ bisects the base $B C$ at $X$.

$ \begin{aligned} \therefore B X & =C X=\frac{5}{2} cm \\ A X & =\sqrt{5^{2}-(5 / 2)^{2}} \\ & =\sqrt{25-\frac{25}{4}}=\sqrt{\frac{75}{4}}=\frac{5 \sqrt{3}}{2} \end{aligned} $

$A Y$ and $B C$ being the chords of the circle,

$A X . X Y=B X . X C$

$\Rightarrow \frac{5 \sqrt{3}}{2} \cdot X Y=\frac{5}{2} \cdot \frac{5}{2}$

$\Rightarrow X Y=\frac{5}{2 \sqrt{3}}$

$\therefore A X . A Y=\frac{5 \sqrt{3}}{2} \cdot(\frac{5 \sqrt{3}}{2}+\frac{5}{2 \sqrt{3}})$

$=\frac{75}{4}+\frac{25}{4}=\frac{100}{4}=\mathbf{2 5} \mathbf{~ c m}^{2}$.

~~ 29. Given $P O=10 cm$, radius $O T=6 cm, P B=5 cm$ In rt. $\triangle O T P,(\angle O T P=90^{\circ} P.$ radius $O T \perp$ tangent $P T)$

$ \begin{aligned} P T & =\sqrt{O P^{2}-O T^{2}} \\ & =\sqrt{100-36}=\sqrt{64}=8 \end{aligned} $

$\therefore$ By Tangent - Secant Theorem

$ \begin{aligned} & P T^{2}=P B \times P C \\ & \Rightarrow \quad 8^{2}=5 \times(B C+P B) \\ & \Rightarrow \quad 64=5(B C+5) \Rightarrow 5 B C=39 \\ & \Rightarrow \quad B C=\mathbf{7 . 8} cm . \end{aligned} $

~~ 30. If two chords $P B$ and $Q C$ intersect externally at a point $A$, then

$A B \times A P=A C \times A Q$

$\Rightarrow 8 \times 3=(2+x) \times 2$

$\Rightarrow 2+x=12 \Rightarrow x=10 cm$.

~~ 31. Using the tangent-secant theorem, we have

$A B \times A P=A D^{2}=(\frac{A C}{2})^{2} \quad(\because A D=D C)$

$\Rightarrow A B \times A P=\frac{1}{4} A C^{2}=\frac{1}{4} A B^{2}$

$(\because A B=A C)$

$\Rightarrow A B=\mathbf{4} A P$.

~~ 32. Since the lengths of the tangents from the same external point are equal, $C D=C P$ and $B P=B E$.

Also, $A E=A D$

Now $A D=A C+C D=A C+C P$

$A E=A B+B E=A B+B P$

$\therefore$ Adding eqns. (i) and (ii), we get

$A D+A E=A C+C P+A B+B P \quad(\because A D=A E)$

$\Rightarrow \quad 2 A E=A C+A B+B C$.

~~ 33. Since the lengths of the tangents from the same external point are equal, $A T=B T$

$A C=P C$ and $B D=D P$

$\therefore A T=B T \Rightarrow T C+C A=T D+D B$

$\Rightarrow T C+P C=T D+P D$

Hence option $(a)$ is true.

Similarly, we can prove the relations in option (b) and (c) for other circles also. ~~ 34. Let the centres of the three circles with radii $a, b, c$ be $A, B$ and $C$ respectively. Let the common tangent touch the three circles at points, $P, Q$ and $R$ respectively.

Since radius $\perp$ tangent at point of contact

$\angle A P R=\angle C R Q=\angle B Q R=90^{\circ}$

Draw a line $C M | P R$ meeting AP in $M$. Then, $\angle A M C=90^{\circ}$

$\therefore C M=P R$ and $M P=C R$ and $A M=A P-M P=a-c$ and $A C=a+c$

$(\therefore.$ Inrt $\triangle A M C, M C=\sqrt{A C^{2}-A M^{2}}=\sqrt{(a+c)^{2}-(a-c)^{2}}$ $=2 \sqrt{a c}$ )

Similarly we can show that $R Q=2 \sqrt{b c}$

$\Rightarrow P Q=P R+R Q=2 \sqrt{a c}+2 \sqrt{b c}$

Also, draw a line from $P | A B$ meeting $B Q$ in $N$. Then,

$P N=A B=a+b$,

$Q N=B Q-B N=b-a$

In rt. $\triangle P Q N, P Q=\sqrt{P N^{2}-Q N^{2}}$

$ =\sqrt{(a+b)^{2}-(a-b)^{2}}=4 a b $

$\Rightarrow P Q=2 \sqrt{a b}$

$\therefore$ From (i) and (ii)

$2 \sqrt{a c}+2 \sqrt{b c}=2 \sqrt{a b}$

$\Rightarrow \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}$.

~~ 35. $\angle P=40^{\circ}$

$\therefore \angle P A B+\angle P B A=180^{\circ}-40^{\circ}=140^{\circ}$ $\angle T A S=180^{\circ}-\angle P A B$ $\angle R B S=180^{\circ}-\angle P B A$

$\therefore \angle T A S+\angle R B S=360^{\circ}-(\angle P A B+\angle P B A)=360^{\circ}-140^{\circ}$ $=220^{\circ}$

Since $O A$ and $O B$ bisect angles $T A S$ and $R B S$ respectively.

$\angle O A S+\angle O B S=\frac{1}{2} \times 220^{\circ}=110^{\circ}$

$\therefore \angle A O B=180^{\circ}-110^{\circ}=\mathbf{7 0}^{\circ}$.

~~ 36. First, draw the line connecting $P$ and $R$ and denote its other inter-sections with the circles by $M$ and $N$; see accompanying figure. The arcs $M R$ and $N R$ contain the same number of degrees; so we may denote each arc by $x$. To verify this, note that we have two isosceles triangle with a base angle of one equal to a base angle of the other. $\therefore \angle N O R=\angle M O R$.

$\angle A P R=\frac{1}{2}{(c+a+c-x)-a}=\frac{1}{2}{2 c-x}$

$\angle B P R=\frac{1}{2}{b+d+d-(b-x)}=\frac{1}{2}{2 d+x}$

and the sum of angles $A P R$ and $B P R$ is

$ \angle B P A=c+d $

The desired angle is

$ \begin{aligned} 360^{\circ}-\angle B P A & =360^{\circ}-(c+d) \\ & =(180^{\circ}-c)+(180^{\circ}-d) \\ & =a+b . \end{aligned} $

~~ 37. Let the radius of the circle drawn inside the quadrilateral $A B C D$ be $r$. $\because A B | C D, \therefore A B C D$ is a trapezium.

Let $C D=x$, then

$A B=3 C D=3 x$

Draw a perpendicular $C M$ from $C$ on $A B$.

Area of a trapezium $=\frac{1}{2}$ (sum of parallel sides) $\times$ height

$\Rightarrow \frac{1}{2} \times(A B+C D) \times A D=4 \quad$ (Area $=4$ sq. units, given)

$\Rightarrow \frac{1}{2} \times(3 x+x) \times 2 r=4$

$\Rightarrow 4 x r=4 \Rightarrow x=\frac{1}{r}$

As all the sides i.e., $A B, B C, C D$ and $D A$ touch the incircle,

$D C+A B=D P+P C+A Q+Q B$

$=D R+C S+A R+S B \begin{bmatrix} \text{ Using, tangents from the } \\ \text{ same external point are equal. }\end{bmatrix} $

$=D R+A R+C S+S B$

$=A D+B C$

$\Rightarrow x+3 x=2 r+\sqrt{(2 r)^{2}+(2 x)^{2}}$

$(\because \angle M=90^{\circ}, C M=P Q=2 r, M B=A B-C D=2 x)$

$\Rightarrow 4 x=2 r+2 \sqrt{r^{2}+x^{2}}$

$\Rightarrow 2 x-r=\sqrt{r^{2}+x^{2}}$

On squaring both the sides, we have

$4 x^{2}-4 r x+r^{2}=r^{2}+x^{2}$ $\Rightarrow 3 x^{2}=4 x r$

$\Rightarrow \frac{3}{r^{2}}=\frac{4}{r} \times r$

$(\because x=\frac{1}{r})$

$\Rightarrow r^{2}=\frac{3}{4} \Rightarrow r=\frac{\sqrt{3}}{2}$ units.

~~ 38. Let the line $A^{\prime} B^{\prime}$ be another line through $P$ meeting the circles in $A^{\prime}$ and $B^{\prime}$.

Given, $A P B$ is a line through $P$ meeting the circle in $A$ and $B$

respectively.

$\angle P A Q=\angle P A^{\prime} Q$ ${Angles in the same segment are equal }$ $ \angle P B Q $ = $ \angle P B^{\prime} Q$

Now in $\triangle A Q B$,

$ \begin{aligned} \angle A Q B & =180^{\circ}-(\angle Q A B+\angle Q B A) \\ & =180^{\circ}-(\angle P A Q+\angle Q B P) \\ & =180^{\circ}-(\angle P A^{\prime} Q+\angle P B^{\prime} Q) \\ & =\angle A^{\prime} Q B^{\prime} \end{aligned} $

$\therefore \angle A Q B$ is the same for all lines $A P B$. Thus, $\angle A Q B$ is a constant angle.

~~ 39. To prove that $P$ is the circumcentre of $\triangle A B C$, we shall show that $P X$ and $P Y$ are the perpendicular bisector of $A B$ and $A C$ respectively.

Since $A B$ is tangent to circle II and $Y A$ is the

radius of circle II.

$ Y A \perp A B $

Also, $P X A Y$ is a parallelogram $\Rightarrow A Y | X P$

$\therefore A Y \perp A B$ and $A Y | X P \Rightarrow X P \perp A B$

Since $X$ is the centre of circle $I$ and $A B$ is a chord of circle $I$, and $X P \perp A B \Rightarrow X P$ bisects $A B \Rightarrow X P$ is the perpendicular bisector of $A B$

Similarly, we can show that $Y P$ is the perpendicular bisector of $A C$.

Since the perpendicular bisector of sides $A B$ and $A C$ of $\triangle A B C$ meet at $P, P$ is the circumcentre of $\triangle A B C$.

~~ 40. Let $C$ be the incentre, $r$ the inradius and $E$ the point of contact of the incircle with $A B$. Let $C^{\prime}$ be the centre of the circle touching $A B$, $A C$ and the incircle, $r^{\prime}$ the radius of this circle and $F$ its point of contact with $A B$. Since $A B$ and $A C$ both touch

this circle, its centre must also lie on $A C$.

From $C^{\prime}$ draw $C^{\prime} D \perp C E$. Then, in $\triangle C^{\prime} C D$

$C D=r-r^{\prime}$

$C C^{\prime}=r+r^{\prime}$

$\angle C D C^{\prime} \Rightarrow \pi / 2$ and $\angle D C^{\prime} C=\angle E A C=A / 2$

In $\triangle D C C^{\prime} \Rightarrow \sin A / 2=\frac{C D}{C C^{\prime}}=\frac{r-r^{\prime}}{r+r^{\prime}}$

$\Rightarrow \cos (\pi / 2-A / 2)=\frac{r-r^{\prime}}{r+r^{\prime}}$

$\Rightarrow \frac{r-r^{\prime}}{r+r^{\prime}}=\cos \theta \quad$ where $\theta=\frac{\pi-A}{2}$

$\Rightarrow \frac{r}{r^{\prime}}=\frac{1-\cos \theta}{1+\cos \theta}$ (on applying componendo and dividendo)

$\Rightarrow \frac{r}{r^{\prime}}=\frac{2 \sin ^{2} \theta / 2}{2 \cos ^{2} \theta / 2}=\tan ^{2} \theta / 2$

$\Rightarrow \frac{r}{r^{\prime}}=\tan ^{2} \frac{\pi-A}{4}$.

~~ 41. Consider the condition: what is the length of the common tangent when two circles of radii $r_1$ and $r_2$ touch externally?

Here $A B$ (the common tangent)

$=O^{\prime} C=\sqrt{O O^{\prime 2}-O C^{2}}$

$=\sqrt{(r_1+r_2)^{2}-(r_1-r_2)^{2}}$

$=\sqrt{4 r_1 r_2}=2 \sqrt{r_1 r_2}$

Therefore, according to the given figure, $P R$ is the length of the common tangent to circle of radii $r$ and 4 .

$\therefore P Q=2 \sqrt{4 r}=4 \sqrt{r}$

$Q R=2 \sqrt{4 r}=4 \sqrt{r}$

$\because P R=P Q+Q R$

$\therefore 2 r=4 \sqrt{r}+4 \sqrt{r} \Rightarrow r=4 \sqrt{r} \Rightarrow r^{2}=16 r \Rightarrow r=\mathbf{1 6} \mathbf{c m}$.

~~ 42. Given, $A B C$ is an equilateral triangle and $P$ is a point on the minor $arc B C$.

$\angle A B C=\angle B A C=\angle B C A=60^{\circ}$

Let $\angle B C P=x$

Produce $B P$ to $Q$ such that

$P Q=P C$. Join $C Q$.

$\angle C P Q$ is the external angle of the cyclic quadrilateral $A B P C$.

$\because \angle C P Q=\angle B A C=60^{\circ}$.

$\because P C=P Q$, and $\angle C P Q=60^{\circ}$, therefore $\triangle C P Q$ is equilateral.

Consider the triangles $A C P$ and $B C Q$.

$\angle A C P=60+x, \angle B C Q=60+x$ Now is $\triangle s A C P$ and $B C Q$

$\angle A C P=\angle B C Q=60+x($ Proved $)$

$\angle C A P=\angle C B P(\angle C B Q)$ (Angles in the same segment $P C$ )

$A C=B C($ sides of equilateral $\triangle A B C)$

$\therefore \triangle A C P \cong \triangle B C Q(A S A)$

$\Rightarrow A P=B Q \Rightarrow A P=B P+P Q \Rightarrow \boldsymbol{A P}=\boldsymbol{B P}+\boldsymbol{P C}$

$(\because P C=P Q)$

~~ 43. Let $O_1$ be the centre of the in-circle of $\triangle P A C$ and $O_2$ the centre of the circle which touches the triangle $P B C$ on side $B C$. Let the tangents from $P$ on these two circles touch them at points $T_1, T_1{ }^{\prime}$ and $T_2, T_2{ }^{\prime}$ respectively.

Looking at the figure, we see that $\angle T_1 O_1 R=60^{\circ}$ since each of $\angle s O_1 T_1 A$ and $O_1 R A$ being $=90^{\circ}$, it is the supplement of $\angle T_1 A R=120^{\circ}$ (as an exterior angle for $\triangle A B C)$. Hence, $\angle A O_1 R=30^{\circ}$. Similarly, we obtain $\angle B O_2 S=30^{\circ}$.

Since tangents drawn to a circle from an external point are equal, we have

$T_1 T_2=T_1 A+A B+B T_2=R A+A B+S B$

and

$ =r_1 \tan 30^{\circ}+\alpha+r_2 \tan 30^{\circ}=\frac{r_1+r_2}{\sqrt{3}}+\alpha, $

$ \begin{aligned} T_1^{\prime} T_2^{\prime} & =T_1^{\prime} C+C T_2^{\prime}=C R+C S=(\alpha-R A)+(\alpha-S B) \\ & =2 \alpha-\frac{r_1+r_2}{\sqrt{3}} . \end{aligned} $

Since common external tangents to two circles are equal, $T_1 T_2=T_1^{\prime} T_2^{\prime}$. Hence,

$ \frac{r_1+r_2}{\sqrt{3}}+\alpha=2 \alpha-\frac{r_1+r_2}{\sqrt{3}}, $

Hence we find that, $r_1+r_2=\frac{\alpha \sqrt{3}}{2}$.

~~ 44. Let $T$ be a point on $A D$ produced beyond $A$ such that $A T=B C$.

Since $A T=B C, A P=$ $C D$ and $\angle T A P=\angle T A B$ $=\angle B C D$, we get $\triangle A T P$ $\equiv \triangle C B D$, so that

$ \angle A T P=\angle C B D . $

Since $\angle C B D=\angle C A D$, we have

$ \angle A T P=\angle C A D . $

Thus, $T P | A C$; that is, $T P | A M$.

Hence, we get $P M: M Q=T A: A Q=B C: A Q=1: 1$.

Therefore, $P M=M Q$.

~~ 45. (a) Letting the base angles in isosceles triangles $A O_1 C$ and $BO_2 D$ be $x$ and $y$, respectively, the sum of the angles in quadrilateral $A B D C$ is

$(90^{\circ}-x)+(90^{\circ}-y)+(180^{\circ}-y)+(180^{\circ}-x)=360^{\circ}$, and we have

$ x+y=90^{\circ} \text{. } $

Hence, in $A B D C$, the angles at $A$ and $D$ add up to

$(90^{\circ}-x)+(180^{\circ}-y)=270^{\circ}-(x+y)=270^{\circ}-90^{\circ}=\mathbf{1 8 0}^{\circ}$ and thus, $A B D C$ is cyclic. This proves $(a)$.

(b) Let $A C$ and $B D$ when produced intersect at $E$. It follows from equation (1) that in triangle $C E D$ the angles at $C$ and $D$ add up to $90^{\circ}$. Thus, $C E D$ is a right-angled triangle with the right angle at $E$ and $A C$ and $B D$ are in fact perpendicular. ~~ 46. $\because$ The perpendicular bisector of $A D$ and $D C$ intersect in point $E$

$E$ is the circumcentre of $\Delta$ $A D C$.

Since $\angle D A B=\angle A C D$ we have that $A B$ is tangent to the circumcircle at $A$,

(Alternate Segment Theroem)

$\because$ radius $E A \perp$ tangent $A B$ at

point of contact $A$,

$\therefore \angle B A E=90^{\circ}$. ~~ 47. We set $B C=a, C A=b, A B=c$, and $2 s=a+b+c$. Let the incircle touch $B C, C A, A B$ at $P, Q, R$, respectively. Since $D E$ is parallel to $B C$, we have $\triangle A D E \sim \triangle A B C$. Thus,

$\frac{A D+D E+A E}{A B+B C+A C}=\frac{D E}{B C}=\frac{D E}{a}$.

Since $A D+D E+A E=A R+A Q=b+c-a$, we have

$ \frac{b+c-a}{a+b+c}=\frac{D E}{a} $

whence, $D E=\frac{a(b+c-a)}{a+b+c}$. Then

$\frac{1}{8}(A B+B C+C A)-D E$

$=\frac{a+b+c}{8}-\frac{a(b+c-a)}{a+b+c}=\frac{(a+b+c)^{2}-8 a(b+c-a)}{8(a+b+c)}$

$=\frac{(b+c)^{2}-6 a(b+c)+9 a^{2}}{8(a+b+c)}=\frac{(b+c-3 a)^{2}}{8(a+b+c)} \geq \mathbf{0}$.

Thus, $\frac{1}{\mathbf{8}}(A B+B C+C A) \geq D E$.

~~ 48. Let $A T$ be the tangent to the first circle at $A$. Then,

$\angle T A M=\angle A N M$

(Angles in alternate segment are equal)

$\Rightarrow \angle A N M=\angle M B C$

(ext. $\angle$ =int. opp. $\angle$ in a cyclic quad.)

we have $\angle T A B=\angle A B C \Rightarrow A T | B C$. (alt. $\angle s$ )

SELF ASSESSMENT SHEET

~~ 1. In the given figure, $O$ is the centre of the circle. $P Q$ is the tangent to the circle at $A$. If $\angle P A B=58^{\circ}$, then $\angle A Q B$ equals

(a) $32^{\circ}$

(b) $26^{\circ}$

(d) None of these

~~ 2. $A, B, C, D$ and $E$ are points on a circle. Point $C$ is due north of point $D$ and point $E$ is due west of point $D . \angle C A B=27^{\circ}$. The angle of elevation of point $B$ from point $E$ is $87^{\circ}$. The angle of elevation of point $B$ from point $D$ is

(a) $60^{\circ}$

(b) $33^{\circ}$

~~ 3. In the given figure, $C D$ is a direct common tangent to two circles intersecting each other at $A$ and $B$. Then, $\angle C A D+\angle C B D$ equals

(a) $120^{\circ}$

(b) $90^{\circ}$

(d) $180^{\circ}$

~~ 4. In the adjoining figure ’ $O$ ’ is the centre of the circle and $P Q, P R$ and $S T$ are the three tangents.

$\angle Q P R=50^{\circ}$, then $\angle S O T$ equals

(a) $35^{\circ}$

(b) $65^{\circ}$

(d) $50^{\circ}$

~~ 5. $A B C$ is an isosceles triangle. A circle is such that it passes through vertex $C$ and $A B$ acts as a tangent at $D$ for the same circle. $A C$ and $B C$ intersect

the circle at $E$ and $F$ respectively. $A C=B C=4 cm$ and $A B=6 cm$. Also $D$ is the mid-point of $A B$. What is the ratio of $E C:(A E+A D)$ ?

(a) $1: 2$

(b) $1: 3$

(d) None of these

~~ 6. In the given figure, $A D E C$ is a cyclic quadrilateral. $C E$ and $A D$ are extended to meet at $B . \angle C A D=60^{\circ}$ and $\angle C B A$ $=30^{\circ} . B D=6 cm$ and $C E=$

$5 \sqrt{3} cm$. What is the ratio $A C: A D$ ?

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

(b) $\frac{4}{5}$

(d) cannot be determind.

~~ 7. Two circles cut each other at $A$ and $B$. A straight line $C A D$ meets the circles at $C$ and $D$. If the tangents at $C$ and $D$ intersect at $E$, prove that $C, E, D, B$ lie on a circle. ~~ 8. $A B, B C, A D$ and $D F$ are four straight lines and their points of intersection $A, B, C, D, E$ and $F$ form four $\triangle s A D F, C D E$, $E B F$ and $A B C$. Show that the circumcircles of $4 \Delta s$ intersect at the same point.

ANSWERS

~~ 1. (b) ~~ 2. (c) ~~ 3. $(d)$ ~~ 4. (b) ~~ 5. (b) ~~ 6. (a)

HINTS AND SOLUTIONS

~~ 1. $\angle B A R=90^{\circ}$

(Angle in a semicircle) $\angle A R B=\angle P A B=58^{\circ}$

(Angles in alternate segments are equal) $\angle A B Q=180^{\circ}-(\angle B A R+$ $\angle A R B)$

( $\angle$ Sum property of $a \Delta$ )

$=180^{\circ}-(90^{\circ}+58^{\circ})$

$=180^{\circ}-148^{\circ}=32^{\circ}$

$\angle Q A R=\angle A B R=\angle A B Q=32^{\circ}$

(Angles in alternate segments are equal)

$ \begin{aligned} \angle A Q B & =180^{\circ}-(\angle A B Q+\angle B A Q) \\ & =180^{\circ}-(32^{\circ}+\angle B A R+\angle R A Q) \\ & =180^{\circ}-(32^{\circ}+90^{\circ}+32^{\circ}) \\ & =180^{\circ}-154^{\circ}=\mathbf{2 6}^{\circ} . \end{aligned} $

~~ 2. The required angle is $\angle B D E$. $C$ is north of $D$ and $E$ is west of $D$ $\Rightarrow \angle C D E=90^{\circ} \Rightarrow E C$ is the diameter of the circle

$(\because$ Angle in a semicircle is a rt. $\angle$ )

$\therefore \angle E B C=90^{\circ}$

(Angle in a semicircle) $\angle B E C=\angle B A C=27^{\circ}$ (Angle in

the same segment are equal)

$\therefore$ In $\triangle E B C, \angle E C B=180^{\circ}-(\angle E B C+\angle B E C)$

$=180^{\circ}-(90^{\circ}+27^{\circ})=63^{\circ} .($ Angle sum property of a $\triangle)$

$\Rightarrow \angle B D E=\angle B C E=63^{\circ}$

(Angles in the same segment are equal) ~~ 3. $\angle C A B=\angle B C D$ ) (Angles in alternate segments are $\angle D A B=\angle B D C$ equal)

$\therefore \angle C A D=\angle C A B+\angle D A B=\angle B C D+\angle B D C$

Now, $\angle C A D+\angle C B D=\angle B C D+\angle B D C+\angle C B D=180^{\circ}$

(Angle sum property in $\triangle B D C$ )

~~ 4. $\angle R O Q=180^{\circ}-50^{\circ}=130^{\circ}(\because \angle O Q P+\angle O R P+\angle O P R$

$ .+\angle R O Q=360^{\circ} \text{ and } \angle O Q P=\angle O R P=90^{\circ}) $

$R T=T M, Q S=S M \quad$ (Tangents to a circle from the same external point are equal)

Also, $O Q=O M=O R \quad$ (Radii of the given circle)

$\therefore \angle R O T=\angle T O M$ and $\angle M O S=\angle S O Q$.

$(\because$ Tangents from the an external point subtend equal angles at the centre)

$\Rightarrow \angle S O T=\angle S O M+\angle T O M=\frac{1}{2} \angle Q O M+\frac{1}{2} \angle R O M$

$\therefore \angle S O T=\frac{1}{2} \angle R O Q=\frac{1}{2} \times 130^{\circ}=\mathbf{6 5}^{\circ}$.

~~ 5. Here $A C$ and $B C$ are the secants of the circle and $A B$ is the tangent at $D$.

$\therefore A E \times A C=A D^{2} \Rightarrow A E \times 4=(3)^{2} \quad \Rightarrow A E=9 / 4$

$\therefore C E=A C-A E=4-\frac{9}{4}=\frac{7}{4}$

$\therefore C E:(A E+A D)=\frac{7}{4}:(\frac{9}{4}+3)=\frac{7}{4}: \frac{21}{4}=1: 3$. ~~ 6. $\angle C E D=120^{\circ}$ ( $\because$ CEDA is a cyclic quad.) $\Rightarrow \angle B E D=60^{\circ}$ $\therefore$ In $\triangle E D B, \angle E D B=90^{\circ}$

$\therefore \frac{B D}{B E}=\cos 30^{\circ} \Rightarrow \frac{6}{B E}=\frac{\sqrt{3}}{2} \Rightarrow B E=4 \sqrt{3} cm$. $B C=B E+C E=4 \sqrt{3}+5 \sqrt{3}=9 \sqrt{3} cm$

$\because A B$ and $C B$ are secants of the given circle,

$ \begin{aligned} & B D \times B A=B E \times E C \\ & \Rightarrow \quad 6 \times B A=4 \sqrt{3} \times 9 \sqrt{3} \\ & \Rightarrow \quad B A=18 cm . \end{aligned} $

$\therefore A D=A B-B D=12 cm$.

$\therefore A C: A D=9: 12=\mathbf{3}: 4$.

~~ 7. Join $A$ and $B, B$ and $C, B$ and $D$.

In $\triangle C D E$,

$\angle 1+\angle 2+\angle C E D=180^{\circ}$

$\because C E$ is a tangent to the circle $C B A$ at point $C$,

$\angle C B A$ is an angle in the alternate segment.

$\therefore \angle 1=\angle 3$

$\angle 2=\angle 4$

From (1), (2) and (3) we have $\angle 3+\angle 4+\angle CED=180^{\circ}$ $\Rightarrow \angle C B D+\angle C E D=180^{\circ}$

$\Rightarrow C, B, D, E$ are concyclic

~~ 8. Let us take the circumcircles of $\triangle D C E$ and $\triangle E B F$ meet at point $P$.

We have to now show that the circumcircles of $\triangle A D F$ and $\triangle A B C$ also pass through $P$,

i.e., $A D P F$ and $A B P C$ are cylic quadrilaterals.

$\angle D C P=\angle D E P$

(Angles in the same segment are equal)

Also, $\angle D E P=\angle F B P$

( $\because$ FBPE is a cyclic quadrilateral, ext $\angle=$ int. opp. $\angle$ )

$\therefore$ (i) and $($ ii $) \Rightarrow \angle D C P=\angle F B P=\angle A B P$. i.e.,

$ \begin{aligned} & \text{ ext } \angle=\text{ int. opp. } \angle \text{ of quad. } A B P C \\ & \Rightarrow A B P C \text{ is a cyclic quadrilateral. } \end{aligned} $

For cyclic quadrilateral $C D P E$, int opp. $\angle C D P=ext \angle P E B$ Also, $\angle P E B=\angle P F B \quad$ (Angles in the same segment)

$\therefore \angle C D P=\angle P F B \Rightarrow \angle A D P=\angle P F B$

$\Rightarrow$ int. opp. $\angle=$ ext $\angle$ in cyclic quad. $A D P F$.

$\therefore$ The circumcircles of $\triangle s A D F, C D E, E F B$ and $A B C$ intersect at point $P$.

7

Probability

(Further Continued from

Class IX)

KEY FACTS

1. Conditional Probability.

Suppose a red card is drawn from a pack of 52 cards, and is not put back, then the probability of drawing a red card in the first attempt is $\frac{26}{52}$ and in the second one it is $\frac{25}{51}$ as the red card is not replaced. Similarly in the above given case, if we draw a black card in the second attempt, then its probability $=\frac{26}{51}$ as number of black cards $=26$ but total number of remaining cards $=51$.

Hence the occurrence of the second event is fully dependent on the first event. Such events are called conditional events.

Definition: Let $A$ and $B$ be two events associated with a random experiment. Then, the probability of the occurrence of $A$ under the condition that $B$ has already occurred and $P(B) \neq 0$, is called the conditional probability of $\boldsymbol{A}$ given $\boldsymbol{B}$ and is written as $\boldsymbol{P}(\boldsymbol{A} / \boldsymbol{B})$

How to evaluate $P(A / B)$ or $P(B / A)$

If the event $A$ occurs when $B$ has already occurred, then $P(B) \neq 0$, then we may regard $B$ as a new (reduced) sample space for event $A$. In that case, the outcomes favourable to the occurrence of event $A$ are those outcomes which are favourable to $B$ as well as favourable to $A$, i.e, the outcomes favourable to $A \cap B$ and probability of occurrence of $A$ so obtained is the conditional probability of $A$ under the condition that $B$ has already occurred.

$ \therefore \quad P(A / B)=\frac{\text{ Number of outcomes favourable to both } A \text{ and } B}{\text{ Number of outcomes in sample space }(B, \text{ here })} $

$ =\frac{n(A \cap B)}{n(B)}=\frac{{\frac{n(A \cap B)}{n(S)}}}{{\frac{n(B)}{n(S)}}}=\frac{P(A \cap B)}{P(B)}, \text{ where } S \text{ is the sample space for the events } A \text{ and } B $

Similarly, $\quad P(B / A)=\frac{P(A \cap B)}{P(A)}, P(A) \neq 0$

where $P(B / A)$ is the conditional probability of occurrence of $B$, knowing that $A$ has already occurred.

Note: If $A$ and $B$ are mutually exclusive events, then,

$ \begin{matrix} P(A / B)=\frac{P(A \cap B)}{P(B)}=0 & \because P(A \cap B)=0 \\ P(B / A)=\frac{P(A \cap B)}{P(A)}=0 & \because P(A \cap B)=0 . \end{matrix} $

Ex. Two coins are tossed. What is the conditional probability of two tails given that at least one coin shows a tail.

Let $A$ : Getting two tails, $B$ : Getting at least one tail

Sample space $S={H H, H T, T H, T T}$

$\Rightarrow A={T T}, B={H T, T H, T T} . A \cap B={T T}$

$P(A)=\frac{1}{4}, P(B)=\frac{3}{4},(A \cap B)=\frac{1}{4}$

$\therefore$ Required probability $=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{3 / 4}=\frac{\mathbf{1}}{\mathbf{3}}$.

2. Multiplication Theorem of Probability.

If $A$ and $B$ are two events in a random experiment such that $P(A) \neq 0$ and $P(B) \neq 0$, then the probability of the simultaneous occurrence of the events $A$ and $B$ i.e., $P(A \cap B)$ is given by:

$ P(A \cap B)=P(A) \times P(B / A) \text{ or } P(A \cap B)=P(B) \times P(A / B) $

(This follows directly from the formula given for conditional probability in Key Fact No. 1)

Thus, the above given formulae hold true for dependent events.

Corollary 1: In case of independent events, occurrence of event $B$ does not depend on the occurrence of $A$, hence $P(B / A)=P(B)$.

$ \therefore \quad P(A \cap B)=P(A) \times P(B) $

Thus, we can say if $\boldsymbol{P}(\boldsymbol{A} \cap \boldsymbol{B})=\boldsymbol{P}(\boldsymbol{A}) \times \boldsymbol{P}(\boldsymbol{B})$, then the events $\boldsymbol{A}$ and $\boldsymbol{B}$ are independent.

Also, If $A$ and $B$ are two independent events associated with a random experiment having a sample space $S$, then

(a) $\bar{A}$ and $B$ are also independent events. So,

$ P(\bar{A} \cap B)=P(\bar{A}) \times P(B) $

(b) $A$ and $\bar{B}$ are also independent events, so,

$ P(A \cap \bar{B})=P(A) \times P(\bar{B}) $

$ P(\bar{A} \cap \bar{B})=P(\bar{A}) \times P(\bar{B}) $

Corollary 2: If $A_1, A_2, A_3, \ldots, A_n$ are $n$ independent events associated with a random experiment, then

$ P(A_1 \cap A_2 \cap A_3 \ldots \cap A_n)=P(A_1) \times P(A_2) \times P(A_3) \ldots \times P(A_n) $

Corollary 3: If $A_1, A_2, A_3, \ldots, A_n$ are $n$ independent events associated with a random experiment, then

$ P(A_1 \cup A_2 \cup A_3 \ldots \ldots \cup A_n)=1-P( \bar A_1) \times P( \bar A_2) \times P(\bar A_3) \times \ldots \times P(\bar A_n) $

Corollary 4: If the probability that an event will happen is $p$, the chance that it will happen in any succession of $\boldsymbol{r}$ trials is $p^{r}$. Also for the $\boldsymbol{r}$ repeated non-occurrence of the event we have the probability $=(\mathbf{1}-\boldsymbol{p})^{r}$.

3. Law of Total Probability

Let $E_1, E_2, \ldots ., E_n$ be $n$ mutually exclusive and exhaustive events associated with a random experiment. If $A$ is any event which occurs with $E_1$ or $E_2$ or $\ldots$ or $E_n$, then

$P(A)=P(E_1) \times P(A / E_1)+P(E_2) \times P(A / E_2)+\ldots+P(E_n) \times P(A / E_n)$.

Ex. There are two bags. One bag contains 4 white and 2 black balls. Second bag contains 5 white and 4 black balls. Two balls are transferred from first bag to second bag. Then one ball is taken from the second bag. Find the probability that it is white.

Sol. There are three mutually exclusive and exhaustive ways in which 2 balls can be transferred from first bag to second bag and then a white ball be drawn from the second bag. (i) Two white balls are transferred from first bag to second bag

(ii) Two black balls are transferred from first bag to second bag

(iii) One white and one black balls are transferred from first bag to second bag

Let the events be described as under:

$A: 2$ white balls drawn from 1st bag, transferred to 2 nd bag

$B: 2$ black balls drawn from 1 st bag, transferred to 2 nd bag

$C: 1$ white and 1 black ball drawn from 1st bag, transferred to 2 nd bag

$D: 1$ white ball drawn from second bag.

$ \therefore \quad \begin{aligned} & P(A)=\frac{{ }^{4} C_2}{{ }^{6} C_2}=\frac{4 \times 3}{6 \times 5}=\frac{6}{15} \\ & P(B)=\frac{{ }^{2} C_2}{{ }^{6} C_2}=\frac{1}{15} \\ & P(C)=\frac{{ }^{4} C_1 \times{ }^{2} C_1}{{ }^{6} C_2}=\frac{4 \times 2}{\frac{6 \times 5}{2}}=\frac{8}{15} \end{aligned} $

First Bag Second Bag

Now when $\boldsymbol{A}$ has occurred, we have 7 white and 4 black balls in 2 nd bag.

$\therefore \quad P($ Getting a white ball from 2nd bag $)=P(D / A)=\frac{7}{11}$

Similarly when $\boldsymbol{B}$ has occurred, we have 5 white and 6 black balls in 2nd bag

$\therefore \quad P($ Getting a white ball from 2nd bag $)=P(D / B)=\frac{5}{11}$

when $\boldsymbol{C}$ has occurred, we have 6 white and 5 black balls in 2nd bag,

$\therefore \quad P($ Getting a white ball from 2 nd bag $)=P(D / C)=\frac{6}{11}$

$\therefore \quad$ By law of total probability, $\boldsymbol{P}($ Ball drawn from 2nd bag is white)

$ \begin{aligned} P(D) & =P(A) \times P(D / A)+P(B) \times P(D / B)+P(C) \times P(D / C) \\ & =\frac{6}{15} \times \frac{7}{11}+\frac{1}{15} \times \frac{5}{11}+\frac{8}{15} \times \frac{6}{11} \\ & =\frac{42}{165}+\frac{5}{165}+\frac{48}{165}=\frac{95}{165}=\frac{\mathbf{1 9}}{\mathbf{3 3}} . \end{aligned} $

4. Some Useful Facts and Formulae

~~ 1. If $P_1, P_2, P_3, \ldots, P_n$ are the respective probabilities of the happening of certain $n$ independent events, then the probability of the failure of all these events is given by:

$ P=(1-P_1)(1-P_2) \ldots(1-P_n) $

~~ 2. Probability of the occurrence of at least one of the $\boldsymbol{n}$ independent events of a random experiment.

If $P_1, P_2, P_3, \ldots, P_n$ are the probabilities of the happening of ’ $n$ ’ independent events, then the (probability that at least one of the events must happen)

$ \begin{aligned} & =1-\text{ Probability of failure of all events } \\ & =1-(1-P_1)(1-P_2)(1-P_3) \ldots(1-P_n) \end{aligned} $

Ex. A problem in mathematics is given to 3 students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$. What is the probability that the problem is solved?

(AIEEE 2002, NDA 2002, SCRA 2002)

Sol. Let the respective events of solving the problem be denoted by $A, B, C$. Then

$ P(A)=\frac{1}{2}, P(B)=\frac{1}{3}, P(C)=\frac{1}{4} $

Clearly $A, B, C$ are independent events and the problem will be considered to have been solved if at least one student solves it.

$\therefore \quad$ Required probability $=P(A$ or $B$ or $C)=P(A \cup B \cup C)=1-P(\bar{A}) P(\bar{B}) P(\bar{C})$

$\bar{A}, \bar{B}, \bar{C}$ are the respective events of not solving the problem.

Also, $\quad P(\bar{A})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}, P(\bar{B})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}, P(\bar{C})=1-P(C)=1-\frac{1}{4}=\frac{3}{4}$.

$\therefore \quad$ Required probability $=1-P(\bar{A}) P(\bar{B}) P(\bar{C})=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}=1-\frac{1}{4}=\frac{\mathbf{3}}{\mathbf{4}}$.

~~ 5. Baye’s Theorem: Let $E_1, E_2, E_3, \ldots, E_n$ be $n$ mutually exclusive events associated with a random experiment. If $A$ is an event which occurs as a result of the events (cases) $E_1, E_2, E_3, \ldots, E_n$ then

$P(\frac{E i}{A})=$ Probability of occurrence of event $A$ as a result of a particular cause (event) $E_i$

$ =\frac{P(E_i) \cdot P(A)}{\sum _{i=1}^{n} P(E_i) \cdot P(A / E_i)} i=1,2, \ldots, n $

Ex. Three boxes contain 6 white, 4 blue; 5 white, 5 blue and 4 white, 6 blue balls respectively. One of the box is selected at random and a ball is drawn from it. If the ball drawn is blue, find the probability that it is from the second box.

Sol. Let $A, B, C, D$ be the events defined as:

$A$ : Selecting first box

$B$ : Selecting second box

$C:$ Selecting third box

$D$ : Event of drawing a blue ball.

Since there are three boxes and each box has an equally likely chance of selection, $P(A)=P(B)=P(C)=\frac{1}{3}$

If first box is chosen, i.e., $A$ has already occurred, then

Probability of drawing a blue ball from $A=\frac{4}{10} \quad \Rightarrow \quad P(D / A)=\frac{4}{10}$

If second box is chosen, i.e., $B$ has already occurred, then

Probability of drawing a blue ball from $B=\frac{5}{10} \quad \Rightarrow \quad P(D / B)=\frac{5}{10}$

Similarly $P(D / C)=\frac{6}{10}$

Now we are required to find the probability $(B / D)$, i.e., given that the ball drawn is blue, we need to find the probability that it is drawn from second box.

By Baye’s Theorem,

$ \begin{aligned} P(B / D) & =\frac{P(B) \times P(D / B)}{P(A) \times P(D / A)+P(B) \times P(D / B)+P(C) \times P(D / C)} \\ & =\frac{\frac{1}{3} \times \frac{5}{10}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{6}{10}}=\frac{\frac{1}{3} \times \frac{5}{10}}{\frac{1}{3} \times \frac{15}{10}}=\frac{\mathbf{1}}{\mathbf{3}} . \end{aligned} $

SOLVED EXAMPLES

Ex. 1 . Two coins are tossed. Find the conditional probability of getting two heads given that at least one coin shows a head.

Sol. Let $A$ : Getting two heads

$B:$ At least one coin showing a head.

$S={H H, H T, T H, H H}$

Then, $A={H H}, B={H T, T H, H H} \Rightarrow A \cap B={H H}$

$\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{1}{4}, P(B)=\frac{n(B)}{n(S)}=\frac{3}{4}, P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{1}{4}$

Now, Required probability $=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{3 / 4}=\frac{\mathbf{1}}{\mathbf{3}}$.

Ex. 2 . Two integers are selected at random from integers 1 through 11. If the sum is even, find the probability that both numbers are odd.

Sol. The integers from 1 through 11 are $1,2,3,4,5,6,7,8,9,10,11$. Out of these, there are 5 even and 6 odd integers.

Let $\boldsymbol{A}:$ Both numbers chosen are odd

$B$ : Sum of numbers is even at random

$S:$ Choosing 2 numbers from 11 numbers.

Then, $n(S)={ }^{11} C_2$

$n(A)={ }^{6} C_2(\because$ There are 6 odd integers $)$

As the sum of both chosen integers can be even if both are even or both are odd, so

$ \begin{matrix} n(B) & ={ }^{6} C_2+{ }^{5} C_2 \\ \text{ and } & n(A \cap B) & ={ }^{6} C_2 \\ \therefore & P(A) & =\frac{n(A)}{n(S)}=\frac{{ }^{6} C_2}{{ }^{11} C_2}=\frac{6 \times 5}{11 \times 10}=\frac{3}{11} \\ P(B) & =\frac{n(B)}{n(S)}=\frac{{ }^{6} C_2+{ }^{5} C_2}{{ }^{11} C_2}=\frac{6 \times 5+5 \times 4}{11 \times 10}=\frac{5}{11} \\ P(A \cap B) & =\frac{n(A \cap B)}{n(S)}=\frac{{ }^{6} C_2}{{ }^{11} C_2}=\frac{5}{11} \\ \therefore \quad & P(A / B) & =\frac{P(A \cap B)}{P(B)}=\frac{3 / 11}{5 / 11}=\frac{\mathbf{3}}{\mathbf{5}} . \end{matrix} $

Ex. 3 . A die is rolled. If the outcome is an odd number, what is the probability that it is a prime number ?

Sol. $S={1,2,3,4,5,6} \Rightarrow n(S)=6$

Let $A$ : Event of getting an odd number

$B$ : Event of getting a prime number

$A={1,3,5} \Rightarrow n(A)=3$

$B={2,3,5} \Rightarrow n(B)=3$

$A \cap B={3,5} \Rightarrow n(A \cap B)=2$

$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}, P(B)=\frac{n(B)}{n(S)}=\frac{3}{6}=\frac{1}{2}, P(A \cap B)=\frac{n(A \cap B)}{n(S)}=\frac{2}{6}=\frac{1}{3}$

$\therefore \quad P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{1 / 3}{1 / 2}=\frac{\mathbf{2}}{\mathbf{3}}$.

Ex. 4 . A man speaks truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

(DCE 2009)

Sol. Let $E_1, E_2$ and $A$ be the events defined as follows:

$E_1=$ Six occurs,

$E_2=$ Six does not occur

$A=$ man reports it is a six

Then, $\quad P(E_1)=\frac{1}{6}, P(E_2)=1-\frac{1}{6}=\frac{5}{6}$

$P(A / E_1)=$ Probability of man reporting it a six when six occurs $=$ Probability of speaking truth $=\frac{3}{4}$

$P(A / E_2)=$ Probability of man reporting a six when six does not occur

$=$ Probability of not speaking truth $=1-\frac{3}{4}=\frac{1}{4}$

$\therefore \quad P($ Throw is actually a six $)=\frac{P(E_1) \times P(A / E_1)}{P(E_1) \times P(A / E_1)+P(E_2) \times P(A / E_2)}$

(Baye’s Theoram)

$=\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{\frac{3}{24}}{\frac{8}{24}}=\frac{\mathbf{3}}{\mathbf{8}}$.

Ex. 5 . A person goes to office either by car, scooter, bus or train, the probabilities of which being $\frac{1}{7}, \frac{3}{7}, \frac{2}{7}$ and $\frac{1}{7}$ respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $\frac{2}{9}, \frac{1}{9}, \frac{4}{9}$ and $\frac{1}{9}$ respectively. Given that he reached office in time, then what is the probability that he travelled by a car.

(IIT 2005)

Sol. Let the events $E_1, E_2, E_3, E_4$ and $A$ be defined as follows:

$E_1:$ Event that the person goes to office by car

$E_2:$ Event that the person goes to office by scooter

$E_3:$ Event that the person goes to office by bus

$E_4:$ Event that the person goes to office by train.

$A$ : Event that the person reaches office in time.

Then, $P(E_1)=\frac{1}{7}, P(E_2)=\frac{3}{7}, P(E_3)=\frac{2}{7}, P(E_4)=\frac{1}{7}$

$P(A / E_1)=P($ Person reaches office in time if he goes by car $)$

$=1-P($ Person reaches office late if he goes by car $)$

$=1-\frac{2}{9}=\frac{7}{9}$

$P(A / E_2)=P($ Person reaches office in time if he goes by scooter $)$

$=1-P($ Person reaches office late if are goes by scooter $)$

$=1-\frac{1}{9}=\frac{8}{9}$

$P(A / E_3)=P($ Person reaches office in time if he goes by bus $)$

$=1-P($ Person reaches office late if he goes by bus)

$=1-\frac{4}{9}=\frac{5}{9}$ $P(A / E_4)=P($ person reaches office in time if he goes by train)

$=1-P$ (person reaches office late if he goes by train)

$=1-\frac{1}{9}=\frac{8}{9}$

Now, $P$ (person travelled by car if he reached office in time)

$ \begin{aligned} P(E_1 / A) & =\frac{P(E_1) \times P(A / E_1)}{P(E_1) \times P(A / E_1)+P(E_2) \times P(A / E_2)+P(E_3)+P(A / E_3)+P(E_4) \times P(A / E_4)} \\ & =\frac{\frac{1}{7} \times \frac{7}{9}}{\frac{1}{7} \times \frac{7}{9}+\frac{3}{7} \times \frac{8}{9}+\frac{2}{7} \times \frac{5}{9}+\frac{1}{7} \times \frac{8}{9}}=\frac{\frac{7}{63}}{\frac{7}{63}+\frac{24}{63}+\frac{10}{63}+\frac{8}{63}}=\frac{\frac{7}{63}}{\frac{49}{63}}=\frac{7}{49}=\frac{\mathbf{1}}{\mathbf{7}} \end{aligned} $

Ex. 6 . Box $A$ contains 2 black and 3 red balls, while box $B$ contains 3 black and 4 red balls. Out of these two boxes one is selected at random; and the probability of choosing box $A$ is double that of box $B$. If a red ball is drawn from the selected box, then find the probability that it has come from box $B$.

(EAMCET 2005)

Sol. Let the events be defined as:

$A$ : Selection of box $A$

$B$ : Selection of box $B$

$R$ : Drawing a red ball.

Let $P(B)=p$. Then, according to given condition $P(A)=2 P(B)=2 p$

$P(\frac{R}{A})=\frac{{ }^{3} C_1}{{ }^{5} C_1}=\frac{3}{5}, P(\frac{R}{B})=\frac{{ }^{4} C_1}{{ }^{7} C_1}=\frac{4}{7} $

$\therefore \quad P(\frac{B}{R})=\frac{P(B) \cdot P(\frac{R}{B})}{P(A) \cdot P(\frac{R}{A})+P(B) \cdot P(\frac{R}{B})}=\frac{p \cdot \frac{4}{7}}{2 p \cdot \frac{3}{5}+p \cdot \frac{4}{7}}=\frac{4 / 7}{6 / 5+4 / 7} =\frac{4 / 7}{\frac{42+20}{35}}$

$=\frac{4 / 7}{62 / 35}=\frac{\mathbf{1 0}}{\mathbf{3 1}} .$

Ex. 7 . $A$ and $B$ are two independent witnesses (i.e., there is no collision between them) in a case. The probability that $A$ will speak the truth is $x$ and the probability that $B$ will speak the truth is $y . A$ and $B$ agree in a certain statement. What is the probability that the statement is true?

(BITSAT 2004)

Sol. $A$ and $B$ will agree in a certain statement if both speak truth or both tell a lie. Now, let us define the events as follows:

$E_1: A$ and $B$ both speak the truth $\Rightarrow P(E_1)=x y$

$E_2: A$ and $B$ both tell a lie $\Rightarrow P(E_2)=(1-x)(1-y)$

$E: A$ and $B$ agree on a certain statement.

Clearly, $P(E / E_1)=P(E / E_2)=1$

Now, we are required to find $P(E_1 / E)$.

$ P(E_1 / E)=\frac{P(E_1) \cdot P(E / E_1)}{P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2)}=\frac{x y \cdot 1}{x y \cdot 1+(1-x)(1-y) \cdot 1}=\frac{\boldsymbol{x y}}{\mathbf{1}-\boldsymbol{x}-\boldsymbol{y}+\mathbf{2} \boldsymbol{x} \boldsymbol{y}} . $

PRACTICE SHEET

~~ 1. If $A$ and $B$ are events such that $P(A \cup B)=0.5, P(\bar{B})=0.8$ and $P(A / B)=0.4$, then what is $P(A \cap B)$ equal to ?

(a) 0.08

(b) 0.02

(d) 0.8

(NDA/NA 2011)

~~ 2. If $P(S)=0.3, P(T)=0.4$ and $S$ and $T$ are independent events, then $P(S / T)$ is equal to

(a) 0.12

(b) 0.2

(d) 0.4

(Orissa JEE 2011)

~~ 3. It is given that the events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A / B)=\frac{1}{2}$ and $P(B / A)=\frac{2}{3}$. Then $P(B)$ is

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

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

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

(AIEEE 2008)

~~ 4. Let $X$ and $Y$ be two events such that $P(X / Y)=\frac{1}{2}$, $P(Y \mid X)=\frac{1}{3}$ and $P(X \cap Y)=\frac{1}{6}$. Which of the following is/are correct?

(a) $P(X \cup Y)=2 / 3$

(b) $X$ and $Y$ are independent

(d) $P(X^{C} \cap Y)=\frac{1}{3}$

(IITJEE 2012) 5. Given, $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$, then $P(\frac{A^{\prime}}{B^{\prime}})$ is equal to

(a) $\frac{1}{3}$

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

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

(MHCET 2009)

~~ 6. A die is rolled. If the outcome is an odd number, what is the probability that it is a prime number?

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

(b) $\frac{7}{9}$

(d) $\frac{5}{6}$ ~~ 7. Two dice are thrown. Find the probability that the sum is 8 or greater than 8 , if 3 appears on the first die.

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

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

(d) $\frac{7}{8}$ ~~ 8. One ticket is selected at random from 50 tickets numbered $00,01,02, \ldots .49$. Then the probability that the sum of the digits selected is 8 , given that the product of these digits is zero is equal to

(a) $1 / 14$

(b) $1 / 7$

(d) $1 / 50$

(AIEEE 2009)

~~ 9. One Indian and four American men and their wives are to be seated randomly around a circular table. Then, the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

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

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

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

(IIT JEE 2007)

~~ 10 A bag contains 6 red and 9 blue balls. Two successive drawing of four balls are made such that the balls are not replaced before the second draw. Find the probability that the first draw gives 4 red balls and the second draw gives 4 blue balls.

(a) $\frac{3}{715}$

(b) $\frac{7}{715}$

(d) None of these ~~ 11 Two numbers are selected at random from the integer 1 through 9. If the sum is even, find the probability that both numbers are odd.

(a) $5 / 8$

(b) $3 / 8$

(d) None of these ~~ 12 A card is drawn from a well shuffled deck of cards. What is the probability of getting a king, given that the card drawn is black.

(a) $\frac{1}{13}$

(b) $\frac{4}{13}$

(d) $\frac{7}{13}$ ~~ 13 A bag $A$ contains 2 white and 2 red balls and another bag $B$ contains 4 white and 5 red balls. $A$ ball is drawn and is found to be red. The probability that it was drawn from bag $B$ is:

(a) $\frac{5}{19}$

(b) $\frac{21}{52}$

(d) $\frac{25}{52}$

(AMU 2010)

~~ 14 A bag $A$ contains 4 green and 3 red balls and bag $B$ contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted to be green. The probability that it comes from bag $B$ is

(a) $\frac{2}{7}$

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

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

(DCE 2005)

~~ 15 In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct. The probability that a student knows the answer to a question is $90 %$. If he gets the correct answer to a question, then the probability that he was guessing is

(a) $\frac{37}{40}$

(b) $\frac{1}{37}$

(d) $\frac{1}{9}$

(Kerala CEE 2004, EAMCET 2012)

~~ 16 In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct given that he copied it is $\frac{1}{8}$. The probability that his answer is correct, given that he guessed it is $\frac{1}{4}$. The probability that they knew the answer to the questions given that he correctly answered it is

(a) $\frac{24}{31}$

(b) $\frac{31}{24}$

(d) $\frac{29}{24}$

(J&K CET 2004, IIT)

~~ 17 In four schools $B_1, B_2, B_3$ and $B_4$ the percentage of girl students is 12, 20, 13 and 17 respectively. From a school selected at random, one student is picked up at random and it is found that the student is a girl. The probability that the girl selected from school $B_2$ is

(a) $\frac{6}{31}$

(b) $\frac{10}{31}$

(d) $\frac{17}{62}$

(UPSEE 2000)

~~ 18 An architecture company built 200 bridges, 400 hospitals and 600 hotels. The probability of damage due to an earthquake of a bridge, a hospital and a hotel is $0.01,0.15$, 0.03 respectively. One of the construction gets damaged in an earthquake. What is the probability that it is a hotel?

(a) $\frac{1}{26}$

(b) $\frac{1}{40}$

(d) $\frac{9}{40}$ ~~ 19 A box $B_1$ contains 1 white ball, 3 red balls and 2 black balls. Another box $B_2$ contains 2 white balls, 3 red balls and 4 black balls. $A$ third box $B_3$ contains 3 white balls, 4 red balls and 5 black balls. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $B_2$ is

(a) $\frac{116}{181}$

(b) $\frac{126}{181}$

(d) $\frac{55}{181}$

(IIT JEE 2013)

~~ 20 Let $U_1$ and $U_2$ be two urns such that $U_1$ contains 3 white and 2 red balls, and $U_2$ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from $U_1$ and put into $U_2$. However, if tail appears then 2 balls are drawn at random from $U_1$ and put into $U_2$. Now 1 ball is drawn at random from $U_2$. Given that the drawn ball from $U_2$ is white, the probability that head appeared on the coin is

(a) $\frac{17}{23}$

(b) $\frac{11}{23}$

(d) $\frac{12}{23}$

(IIT JEE 2011)

ANSWERS

1. $(a)$	2. (c)	3. $(d)$	4. $(a)$ and $(b)$	5. $(c)$	6. $(c)$	7. $(c)$	8. $(a)$	9. $(c)$	10. $(a)$
11. $(a)$	12. $(a)$	13. $(c)$	14. $(c)$	15. (b)	16. $(c)$	17. $(b)$	18. $(d)$	19. $(d)$	20. $(d)$

HINTS AND SOLUTIONS

~~ 1. $P(A \cup B)=0.5, P(B)=0.8, P(A / B)=0.4$

$ \begin{aligned} & P(A / B)=\frac{P(A \cap B)}{P(B)} \Rightarrow P(A \cap B)=P(A / B) \times P(B) \\ & \Rightarrow P(A \cap B)=(1-P(\bar{B})) \times P(A / B) \\ & =(1-0.8) \times 0.4=0.2 \times 0.4=\mathbf{0 . 8} . \end{aligned} $

~~ 2. $S$ and $T$ being independent events,

$P(S \cap T)=P(S) \times P(T)=0.3 \times 0.4$

$P(S / T)=\frac{P(S \cap T)}{P(T)}=\frac{0.3 \times 0.4}{0.4}=\mathbf{0 . 3}$.

~~ 3. Given, $\quad P(B / A)=\frac{P(A \cap B)}{P(A)}$

$\Rightarrow \quad P(A \cap B)=P(B / A) \times P(A)=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}$

Now $P(A / B)=\frac{P(A \cap B)}{P(B)}$

$\Rightarrow \quad \frac{1}{2}=\frac{1}{6} \times \frac{1}{P(B)} \Rightarrow P(B)=\frac{2}{6}=\frac{1}{\mathbf{3}}$.

~~ 4. $P(X / Y)=\frac{1}{2}, P(Y / X)=\frac{1}{3}, P(X \cap Y)=\frac{1}{6}$

$\therefore P(\frac{X}{Y})=\frac{P(X \cap Y)}{P(Y)}$ $\Rightarrow \frac{1}{2}=\frac{1 / 6}{P(Y)} \Rightarrow P(Y)=\frac{1}{3}$

Now $P(\frac{Y}{X})=\frac{P(Y \cap X)}{P(X)}=\frac{P(X \cap Y)}{P(X)}$

$\Rightarrow \quad \frac{1}{3}=\frac{1}{6} \times \frac{1}{P(X)} \Rightarrow P(X)=\frac{1}{2}$

$\therefore \quad P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$

$P(X \cap Y)=\frac{1}{6} \quad$ and $\quad P(X) \cdot P(Y)=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}$

$\Rightarrow P(X \cap Y)=P(X) . P(Y)$

$\Rightarrow X$ and $Y$ are independent events

Now $P(X^{C} \cap Y)=P(Y)-P(X \cap Y)=\frac{1}{3}-\frac{1}{6}=\frac{\mathbf{1}}{\mathbf{6}}$.

~~ 5. $P(\frac{A^{\prime}}{B^{\prime}})=\frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}=\frac{P(A \cup B)^{\prime}}{P(B^{\prime})}$

$ \begin{aligned} & =\frac{1-P(A \cup B)}{1-P(B)}=\frac{1-{P(A)+P(B)-P(A \cap B)}}{1-P(B)} \\ & =\frac{1-{0.5+0.4-0.3}}{1-0.4}=\frac{0.4}{0.6}=\frac{\mathbf{2}}{\mathbf{3}} . \end{aligned} $

~~ 6. Let $S$ be the sample space of rolling a dice. Then,

$S={1,2,3,4,5,6} \Rightarrow n(S)=6$

Let $A$ : Event of rolling an odd number and

$B:$ Event of rolling a prime number.

Then, $A={1,3,5} \Rightarrow n(A)=3$

$ B={2,3,5} \quad \Rightarrow n(B)=3 $

$ A \cap B={3,5} \Rightarrow n(A \cap B)=2 $

$\therefore \quad P(A)=\frac{3}{6}=\frac{1}{2}, P(B)=\frac{3}{6}=\frac{1}{2}, P(A \cap B)=\frac{2}{6}=\frac{1}{3}$

Now, $P($ Rolling a prime number, if the outcome is an odd number)

$ =P(\frac{B}{A})=\frac{P(A \cap B)}{P(A)}=\frac{1 / 3}{1 / 2}=\frac{\mathbf{2}}{\mathbf{3}} . $

~~ 7. Let $A$ : Event of getting a sum of 8 or greater than 8 in a throw of two dice

$B$ : Event of getting 3 on the first die.

Then,A=$\lbrace(2,6),(3,5),(4,4),(5,3),(6,2),(3,6),(4,5)$

$(5,4),(6,3),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\rbrace$.

$\Rightarrow \quad n(A)=13$

$B={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)}$.

$\Rightarrow n(B)=6$

$A \cap B={(3,5),(3,6)} \quad \Rightarrow n(A \cap B)=2$.

$\therefore \quad P($ Event of getting sum $=8$ or $>8$ when 3 appears on first die)

$\Rightarrow$ Occurence of event $A$ on the satisfaction of condition $B$

$=P(A / B)=\frac{P(A \cap B)}{P(A)}=\frac{\frac{n(A \cap B)}{n(S)}}{\frac{n(B)}{n(S)}}=\frac{n(A \cap B)}{n(B)}=\frac{2}{6}=\frac{\mathbf{1}}{\mathbf{3}}$.

~~ 8. Let $S={00,01,02, \ldots, 48,49}$.

$n(S)=50$

Let $A$ be the event that sum of the digits on the selected ticket is 8 , then

$A={08,17,26,35,44} \Rightarrow n(A)=5$

Let $B$ be the event that the product of the digits is zero.

Then,

$B={00,01,02, \ldots, 08,09,10,20,30,40} \Rightarrow n(B)=14$

$\therefore A \cap B={08} \Rightarrow n(A \cap B)=1$

$\therefore$ Required probability $=P(\frac{A}{B})$

$=\frac{P(A \cap B)}{P(B)}=\frac{1 / 50}{14 / 50}=\mathbf{1} / \mathbf{1 4}$.

~~ 9. Let $A$ : Event that Indian man is seated adjacent to his wife.

Let $B$ : Event that each American is seated adjacent to his wife.

Consider each couple as one entity. Thus, there are 5 entities to be arranged and husbands and wife can interchange their seats in 2 ! ways.

$\therefore P(A \cap B)=\frac{4 !(2 !)^{5}}{9 !}$

Next consider each American couple as an entity. Thus, there are 6 entities to be arranged including the Indian and his wife.

$ \begin{matrix} \therefore & P(B)=\frac{5 !(2 !)^{4}}{9 !} \\ \therefore & P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{4 !(2 !)^{5}}{5 !(2 !)^{4}}=\frac{\mathbf{2}}{\mathbf{5}} . \end{matrix} $

~~ 10 Let

$A$ : Event of drawing 4 red balls in the first draw and

$B$ : Event of drawing 4 blue balls in the second draw without replacement of the balls drawn in $A$.

Then, $P(A)=\frac{n(A)}{n(S)}$

$=\frac{\text{ No. of ways of drawing } 4 \text{ red balls out of } 6 \text{ red balls }}{\text{ No. of ways of drawing } 4 \text{ balls out of } 15 \text{ balls }}$

$=\frac{{ }^{6} C_4}{{ }^{15} C_4}=\frac{\frac{16}{\underline{24}}}{\frac{\underline{15}}{\underline{11} 4}}=\frac{\frac{6 \times 5}{2}}{\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}}=\frac{15}{1365}=\frac{1}{91}$

No. of ways of drawing 4 blue

$P(\frac{B}{A})=\frac{n(B)}{n(S)}=\frac{\text{ balls out of } 9 \text{ blue balls }}{\begin{matrix} \text{ No. of ways of drawing } 4 \text{ balls out of } \\ 11 \text{ balls remaining after the } 1 \text{ st draw }\end{matrix} }$

$=\frac{{ }^{9} C_4}{{ }^{11} C_4}=\frac{\frac{\underline{9}}{\underline{5}\lfloor 4}}{\frac{\lfloor 11}{\lfloor\lfloor 4}}=\frac{9 \times 8 \times 7 \times 6}{11 \times 10 \times 9 \times 8}=\frac{42}{110}=\frac{21}{55}$

Note: This second event is denoted by $B / A$ as it depends on condition $A$.

$\therefore \quad$ Required probability $=P(A) \times P(B / A)$

$ =\frac{1}{91} \times \frac{21}{55}=\frac{\mathbf{3}}{\mathbf{7 1 5}} . $

~~ 11 In the set of integers from 1 to 9 , there are four even numbers $2,4,6,8$ and 5 odd numbers $1,3,5,7,9$.

Let $A$ : Event of choosing odd numbers

$\Rightarrow n(A)={ }^{5} C_2$

( $\because 2$ numbers are chosen from 5 odd numbers)

$B:$ Event of getting the sum as even number.

$\Rightarrow \quad n(B)={ }^{4} C_2+{ }^{5} C_2 \quad(\because$ The sum is even if both the numbers chosen are even or both are odd)

$\therefore \quad n(A \cap B)={ }^{5} C_2$

(Event of getting sum as even if both the numbers are odd)

$\therefore \quad P$ (Selecting both odd numbers or getting an even sum)

$=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{n(A \cap B)}{n(B)}=\frac{{ }^{5} C_2}{{ }^{4} C_2+{ }^{5} C_2}=\frac{10}{16}=\frac{\mathbf{5}}{\mathbf{8}}$.

~~ 12 Let $S$ be the sample space of drawing a card out of 52 cards.

Then,

$ n(S)=52 $

Let $A$ : Event of drawing a king $\Rightarrow n(A)=4$

(A pack has 4 kings)

$B$ : Event of drawing a black card $\Rightarrow n(B)=26$

(A pack has 26 black cards)

$A \cap B$ : Event of drawing a king of a black card

$\Rightarrow n(A \cap B)=2$

(A pack has 2 black kings)

$\therefore P(A)=\frac{4}{52}=\frac{1}{13}, P(B)=\frac{26}{52}=\frac{1}{2}, P(A \cap B)=\frac{2}{52}=\frac{1}{26}$

$P($ Getting a king, given card drawn is black

$ =P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 26}{1 / 2}=\frac{2}{26}=\frac{\mathbf{1}}{\mathbf{1 3}} . $

$(\because$ Event $A$ depends on $B$ )

~~ 13 Let the events $E_1, E_2$ and $A$ be defined as follows:

$E_1=$ Choosing bag $A$

$E_2=$ Choosing bag $B$

$A=$ Choosing red ball.

Then, $P(E_1)=P(E_2)=\frac{1}{2}$

( $\because$ There are two bags that have an equally likely chance of being chosen)

$P(A / E_1)=P($ Drawing red ball from bag $A)=\frac{2}{4}=\frac{1}{2}$

( $\because 2$ red out of 4 balls)

$P(A / E_2)=P($ Drawing red ball from bag $B)=\frac{5}{9}$

( 5 red out of 9 balls)

$\therefore P($ Red balls drawn from bag $B)$

$=P(E_2 / A)=\frac{P(E_2) \times P(A / E_2)}{P(E_1) \times P(A / E_1)+P(E_2) \times P(A / E_2)}$

$=\frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{5}{9}}=\frac{\frac{5}{18}}{\frac{1}{4}+\frac{5}{18}}=\frac{5 / 18}{19 / 36}=\frac{\mathbf{1 0}}{\mathbf{1 9}}$.

~~ 14 $P($ Drawing a green ball from bag $A)=P(\frac{G}{A})=\frac{4}{7}$

$P($ Drawing a green ball from bag $B)=P(\frac{G}{B})=\frac{3}{7}$

$\therefore$ Required probability $P(\frac{B}{G})$

$ \begin{aligned} & =\frac{P(B) \cdot P(\frac{G}{B})}{P(A) \cdot P(\frac{G}{A})+P(B) \cdot P(\frac{G}{B})} \\ & =\frac{\frac{1}{2} \cdot \frac{3}{7}}{\frac{1}{2} \cdot \frac{4}{7}+\frac{1}{2} \cdot \frac{3}{7}}=\frac{3 / 14}{4 / 14+3 / 14}=\frac{3 / 14}{7 / 14}=\frac{\mathbf{3}}{7} . \end{aligned} $

~~ 15 We define the given events as:

$A_1$ : Student knows the answer $A_2:$ Student does not know the answer

$E:$ He gets the correct answer.

$P(A_1)=\frac{9}{10}, P(A_2)=1-\frac{9}{10}=\frac{1}{10}$

$\therefore P(E / A_1)=P($ Student gets the correct answer when he

knows the answer) $=1$

$P(E / A_2)=P($ Student gets the correct answer when he does not know the correct answer) $=1 / 4$

$\therefore$ Required probability

$ \begin{aligned} & =P(A_2 / E)=\frac{P(A_2) \cdot P(E / A_2)}{P(A_1) \cdot P(E / A_1)+P(A_2) \cdot P(E / A_2)} \\ & =\frac{\frac{1}{10} \cdot \frac{1}{4}}{\frac{9}{10} \cdot 1+\frac{1}{10} \cdot \frac{1}{4}}=\frac{\frac{1}{40}}{\frac{37}{40}}=\frac{\mathbf{1}}{\mathbf{3 7}} . \end{aligned} $

~~ 16 Let,

$E_1:$ Examinee guesses the answer

$E_2$ : Examinee copies the answer

$E_3$ : Examinee knows the answer

$E$ : Event examinee answers correctly.

Given, $P(E_1)=\frac{1}{3}, P(E_2)=\frac{1}{6}$

$\therefore P(E_3)=1-(P(E_1)+P(E_3))=1-(\frac{1}{3}+\frac{1}{6})=\frac{1}{2}$

Given, $P(E / E_1)=\frac{1}{4}, P(E / E_2)=\frac{1}{8}$ and $P(E / E_3)=1$

$\therefore$ Required probability $=P(E_3 / E)$

$=\frac{P(E_3) \cdot P(E / E_3)}{P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2)+P(E_3) \cdot P(E / E_3)}$

$=\frac{\frac{1}{2} \cdot 1}{\frac{1}{3} \times \frac{1}{4}+\frac{1}{6} \times \frac{1}{8}+\frac{1}{2} \times 1}=\frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}$

$=\frac{\frac{1}{2}}{\frac{4+1+24}{48}}=\frac{\frac{1}{2}}{\frac{29}{48}}=\frac{\mathbf{2 4}}{\mathbf{2 9}}$.

~~ 17 Let $E_1, E_2, E_3, E_4$ and $A$ be the events defined as follows:

$E_1=$ Event of selecting school $B_1$

$E_2=$ Event of selecting school $B_2$

$E_3=$ Event of selecting school $B_3$

$E_4=$ Event of selecting school $B_4$

$A=$ Event of selecting a girl.

Since there are four schools and each school has an equal chance of being chosen, $P(E_1)=P(E_2)=P(E_3)=P(E_4)=\frac{1}{4}$

Now, $P(.$ Girl chosen is from school $.B_1)=P(\frac{A}{E_1})=\frac{12}{100}$

Similarly $P(\frac{A}{E_2})=\frac{20}{100}, P(\frac{A}{E_3})=\frac{13}{100}, P(\frac{A}{E_4})=\frac{17}{100}$

$\therefore P(.$ Girl chosen is from school $.B_2)$

$=\frac{P(E_2) \times P(A / E_2)}{P(E_1) \times P(A / E_1)+P(E_2) \times P(A / E_2)+P(E_3) \times P(A / E_3)+P(E_4) \times P(A / E_4)}$

$=\frac{\frac{1}{4} \times \frac{20}{100}}{\frac{1}{4} \times \frac{12}{100}+\frac{1}{4} \times \frac{20}{100}+\frac{1}{4} \times \frac{13}{100}+\frac{1}{4} \times \frac{17}{100}}$

(Using Bayes Th.)

$=\frac{\frac{1}{4} \times \frac{20}{100}}{\frac{1}{4} \times \frac{62}{100}}=\frac{20}{62}=\frac{10}{\mathbf{3 1}}$.

~~ 18 Let $E_1, E_2, E_3$ and $A$ be the events defined as follows:

$E_1=$ Construction chosen is a bridge.

$E_2=$ Construction chosen is a hospital.

$E_3=$ Construction chosen is a hotel.

$A=$ Construction gets damaged.

Since there are $(200+400+600)=1200$ constructions,

$P(E_1)=\frac{200}{1200}=\frac{1}{6}, P(E_2)=\frac{400}{1200}=\frac{1}{3}, P(E_3)=\frac{600}{1200}=\frac{1}{2}$

Given, Probability that the construction that gets damaged is a bridge $=P(A / E_1)=0.01$

Similarly, $P(A / E_2)=0.15$ and $P(A / E_3)=0.03$

$\therefore$ Probability that a hotel gets damaged in an earthquake

$=P(\frac{E_3}{A})$

$=\frac{P(E_3) \times P(A / E_3)}{P(E_1) \times P(A / E_1)+P(E_2) \times P(A / E_2)+P(E_3) \times P(A / E_3)}$

$=\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times 0.01+\frac{1}{3} \times 0.15 \times \frac{1}{2} \times 0.03}$

(Using Bayes Th.)

$=\frac{\frac{1}{2} \times 0.03}{\frac{1}{6} \times(0.01+0.3 \times 0.09)}=\frac{6}{2} \times \frac{0.03}{0.4}=\frac{6 \times 3 \times 10}{2 \times 4 \times 100}=\frac{9}{40}$. ~~ 19 Let $E$ : Event of selecting red and one white ball

Probability of selecting a box $=P(B_1)=P(B_2)=P(B_2)=\frac{1}{3}$

Probability of selecting 1 Red and 1 White ball from box $B_1$

$ \begin{aligned} & =P(\frac{E}{B_1})=\frac{{ }^{1} C_1 \times{ }^{3} C_1}{{ }^{6} C_2}=\frac{1 \times 3 \times 2}{6 \times 5}=\frac{1}{5} \\ & P(\frac{E}{B_2})=\frac{{ }^{2} C_1 \times{ }^{3} C_1}{{ }^{9} C_2}=\frac{2 \times 3 \times 3}{9 \times 8}=\frac{1}{6} \\ & P(\frac{E}{B_3})=\frac{{ }^{3} C_1 \times{ }^{4} C_1}{{ }^{12} C_2}=\frac{3 \times 4 \times 2}{12 \times 11}=\frac{2}{11} \\ & \therefore P(\frac{B_2}{E})=\frac{P(B_2) \times P(\frac{E}{B_2})}{P(B_1) \times P(\frac{E}{B_1})+P(B_2) \times P(\frac{E}{B_2})+P(B_3) \times P(\frac{E}{B_3})} \\ & =\frac{\frac{1}{3} \times \frac{1}{6}}{\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{1}{6}+\frac{1}{3} \times \frac{2}{11}}=\frac{\frac{1}{6}}{\frac{66+55+30}{330}}=\frac{\frac{1}{6}}{\frac{181}{330}}=\frac{\mathbf{5 5}}{\mathbf{1 8 1}} . \end{aligned} $

~~ 20 The movement of balls from Urn 1 to Urn 2 on the condition that head or tail appears on the coin can be shown as:

Head appears U $_1 \begin{vmatrix} 2 W \\ 2 R\end{vmatrix} \xrightarrow[U_2]{1 \text{ White }}$ or

Tail appears $\underbrace{ \begin{vmatrix} 1 W \\ 2 R\end{vmatrix} } _{U_1} \xrightarrow{2 \text{ White }}$ or $3 \mid$

$ \begin{aligned} & \underbrace{ \begin{vmatrix} 3 W \\ 0 R \end{vmatrix} } _{U_1} \xrightarrow{\text{ 2 Red }} \xrightarrow[U_2]{ \begin{vmatrix} 1 W \\ 2 R \end{vmatrix} } \text{ or } \\ & \underbrace{ \begin{vmatrix} 2 W \\ 1 R \end{vmatrix} } _{U_1} \xrightarrow[U_2]{1 R 1 W} \end{aligned} $

Let the events be defined as:

$W$ : Drawing a white ball from $U_2$

$H$ : Tossing a head

$T$ : Tossing a tail. $\therefore P(H)=P(T)=\frac{1}{2}$

$P(W / H)=$ Probability of drawing a white ball from Urn 2, when head is tossed

$=\frac{{ }^{3} C_1}{{ }^{5} C_1} \times \frac{{ }^{2} C_1}{{ }^{2} C_1}+\frac{{ }^{2} C_1}{{ }^{5} C_1} \times \frac{{ }^{1} C_1}{{ }^{2} C_1}$

From (i)

$=\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}$

$P(W / T)=$ Probability of drawing a white ball from Urn 2, when tail is tossed

$=\frac{{ }^{3} C_2}{{ }^{5} C_2} \times \frac{{ }^{3} C_1}{{ }^{3} C_2}+\frac{{ }^{2} C_2}{{ }^{5} C_2} \times \frac{{ }^{1} C_1}{{ }^{3} C_1}+\frac{{ }^{3} C_1 \times{ }^{2} C_1}{{ }^{5} C_2} \times \frac{{ }^{2} C_1}{{ }^{3} C_2}$ (see diag. (ii))

$ \begin{gathered} =\frac{3 \times 2}{5 \times 4} \times 1+\frac{1 \times 2}{5 \times 4} \times \frac{1}{3}+\frac{3 \times 2 \times 2}{5 \times 4} \times \frac{2}{3} \\ =\frac{3}{10}+\frac{1}{30}+\frac{2}{5}=\frac{9+1+12}{30}=\frac{22}{30} \\ P(\frac{H}{W})=\frac{P(H) \cdot P(W / H)}{P(H) \cdot P(W / H)+P(T) \cdot P(W / T)} \\ =\frac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5}+\frac{1}{2} \times \frac{22}{30}} \\ =\frac{4 / 5}{4 / 5+22 / 30}=\frac{4 / 5}{\frac{24+22}{30}}=\frac{4 / 5}{46 / 30}=\frac{4}{5} \times \frac{30}{46}=\frac{\mathbf{1 2}}{\mathbf{2 3}} . \end{gathered} $

Trigonometric Equations

KEY FACTS

~~ 1. A trigonometric equation is an equation involving the trigonometric function or functions of unknown angles, e.g. $\cos x=0, \sin ^{2} x=\frac{1}{2}, \cos (\theta+\frac{\pi}{4})=\frac{1}{2}$, etc. ~~ 2. The solution of a trigonometric equation is a value of the unknown angle that satisfies the equation.

A trigonometric equation may have an unlimited number of solutions.

For example, if $\sin x=0$, then $x=0, \pi, 2 \pi, 3 \pi, \ldots$.

A solution lying between $0^{\circ}$ and $360^{\circ}$ is called the principal solution.
Since the trigonometric functions are periodic, a solution generalized by means of periodicity is known as the general solution.

Every equation will have a principal solution as well as a general solution.

3. Solutions of equations of type $\sin \theta=0, \cos \theta=0$ and $\tan \theta=0$

(a) $\sin \theta=0$

$\because \sin \theta=\sin 0=\sin \pi=\sin 2 \pi=\sin (-2 \pi)=\ldots .=0$

Therefore $\sin \theta=0$ is satisfied by the following values of $\theta$.

$\therefore \theta=0, \pm \pi, \pm 2 \pi, \pm 3 \pi, \pm 4 \pi, \ldots$.

$\Rightarrow$ The general solution of $\sin \boldsymbol{\theta}=\mathbf{0}$ is $\boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}, \boldsymbol{n} \in \boldsymbol{I}$

(b) $\cos \theta=0$

$\theta= \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \pm \frac{5 \pi}{2}, \ldots$ satisfy the equation $\cos \theta=0$ as

$\cos \theta=\cos ( \pm \frac{\pi}{2})=\cos ( \pm \frac{3 \pi}{2})=\ldots .=0$

$\Rightarrow$ The general solution of $\cos \theta=0$ is $\theta=(2 n+1) \frac{\pi}{2}, n \in I$

This is satisfied by $\theta=0, \pm \pi, \pm 2 \pi, \ldots$.

$\Rightarrow$ The general solution of $\boldsymbol{\operatorname { t a n }} \boldsymbol{\theta}=\mathbf{0}$ is $\boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}, \boldsymbol{n} \in \boldsymbol{I}$

~~ 4. Solution of equations of type $\sin \theta=\sin \alpha$

$\sin \theta=\sin \alpha \Rightarrow \sin \theta-\sin \alpha=0$

$\Rightarrow 2 \cos \frac{\theta+\alpha}{2} \sin \frac{\theta-\alpha}{2}=0$

$ \begin{matrix} \Rightarrow \text{ Either } \cos \frac{\theta+\alpha}{2}=0 & \text{ or } & \sin \frac{\theta-\alpha}{2}=0 \\ \Rightarrow \frac{\theta+\alpha}{2}=(2 m+1) \frac{\pi}{2}, m \in I & \frac{\theta-\alpha}{2}=m \pi, m \in I \\ \Rightarrow \theta+\alpha=(2 m+1) \pi & & \theta-\alpha=2 m \pi, m \in I \\ \Rightarrow \theta=(2 m+1) \pi-\alpha & \ldots .(i) & \Rightarrow \theta=2 m \pi+\alpha, m \in I \end{matrix} $

Thus $\theta=(-\alpha+$ odd multiple of $\pi)$ or $(\alpha+$ even multiple of $\pi)$

$\therefore \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}+(-1)^{n} \alpha$ where $\boldsymbol{n} \in I$.

5. Solution of equations of type $\cos \theta=\cos \alpha$

$\cos \theta=\cos \alpha \Rightarrow \cos \theta-\cos \alpha=0$

$\Rightarrow 2 \sin \frac{\theta+\alpha}{2} \sin \frac{\alpha-\theta}{2}=0 \Rightarrow-\sin \frac{\theta+\alpha}{2} \sin \frac{\theta-\alpha}{2}=0 \Rightarrow-\sin \frac{\theta+\alpha}{2}=0$ or $\sin \frac{\theta-\alpha}{2}=0$

$\Rightarrow \sin \frac{\theta+\alpha}{2}=0 \quad$ or $\quad \sin \frac{\theta-\alpha}{2}=0$

$\Rightarrow \frac{\theta+\alpha}{2}=n \pi$

$\Rightarrow \theta+\alpha=2 n \pi$

$\Rightarrow \theta=2 n \pi-\alpha$

$ \frac{\theta-\alpha}{2}=n \pi $

$ \theta-\alpha=2 n \pi $

$ \therefore \theta=2 n \pi \pm \alpha, n \in I $

6. Solution of equations of $type \tan \theta=\tan \alpha$

$\tan \theta=\tan \alpha$

$\Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \theta \cos \alpha-\cos \theta \sin \alpha=0$

$\Rightarrow \sin (\theta-\alpha)=0 \quad \Rightarrow \quad \theta-\alpha=n \pi \quad \Rightarrow \quad \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}+\boldsymbol{\alpha}, \boldsymbol{n} \in \boldsymbol{I}$.

Note: 1. In the above results, $\alpha$ is numerically the least angle which should be expressed in radians as far as possible. ~~ 2. $cosec \theta=cosec \alpha$ follows from $\sin \theta=\sin \alpha$ $\sec \theta=\sec \alpha$ follows from $\cos \theta=\cos \alpha$ $\cot \theta=\cot \alpha$ follows from $\tan \theta=\tan \alpha$.

~~ 7. Solution of the equations $\sin ^{2} \theta=\sin ^{2} \alpha, \cos ^{2} \theta=\cos ^{2} \alpha, \tan ^{2} \theta=\tan ^{2} \alpha$

$\begin{matrix}{l|l} \text{ (a) } \sin ^{2} \theta=\sin ^{2} \alpha & \text{ (b) } \cos ^{2} \theta=\cos ^{2} \alpha \\ \Rightarrow \frac{1-\cos 2 \theta}{2}=\frac{1-\cos 2 \alpha}{2} & \Rightarrow \frac{1+\cos 2 \theta}{2}=\frac{1+\cos 2 \alpha}{2} \\ \Rightarrow \cos 2 \theta=\cos 2 \alpha & \Rightarrow \cos 2 \theta=\cos 2 \alpha \\ \Rightarrow 2 \theta=2 n \pi \pm 2 \alpha & \Rightarrow 2 \theta=2 n \pi \pm 2 \alpha \\ \Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi} \pm \boldsymbol{\alpha}, \boldsymbol{n} \in \boldsymbol{I} & \Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi} \pm \boldsymbol{\alpha}, \boldsymbol{n} \in \boldsymbol{I} \end{matrix}$

$\Rightarrow \frac{1}{\tan ^{2} \theta}=\frac{1}{\tan ^{2} \alpha}$

$\Rightarrow \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}$

$\Rightarrow \cos 2 \theta=\cos 2 \alpha$

$\Rightarrow 2 \theta=2 n \pi \pm 2 \alpha$

$\Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi} \pm \boldsymbol{\alpha}, \boldsymbol{n} \in \boldsymbol{I}$

SOLVED EXAMPLES

Ex. 1 . Solve for $x, \sin x=-\frac{\sqrt{3}}{2},(0<x<2 \pi)$

Sol. $\sin x=-\frac{\sqrt{3}}{2}=-\sin 60^{\circ}=\sin (180^{\circ}+60^{\circ})=\sin (360^{\circ}-60^{\circ})$

$\Rightarrow x=240^{\circ}, 300^{\circ}$.

Ex. 2 . Solve $4 \cos ^{2} \theta=3(0^{\circ} \leq \theta \leq 360^{\circ})$

Sol. $4 \cos ^{2} \theta=3 \Rightarrow \cos ^{2} \theta=\frac{3}{4} \Rightarrow \cos \theta= \pm \frac{\sqrt{3}}{2}$

$ \begin{aligned} & \cos \theta=\frac{\sqrt{3}}{2} \quad \Rightarrow \quad \theta=30^{\circ}, 330^{\circ} \quad(\because \cos \theta \text{ is }+ \text{ ve and so } \theta \text{ lies in } 1 \text{ st and } 4 \text{ th quad.) } \\ & \cos \theta=-\frac{\sqrt{3}}{2} \Rightarrow \theta=150^{\circ}, 210^{\circ} \quad(\because \cos \theta \text{ is }- \text{ ve and so } \theta \text{ lies in } 2 \text{ nd and 3rd quad. }) \\ & \Rightarrow \theta=\mathbf{3 0}, \mathbf{1 5 0}^{\circ}, \mathbf{2 1 0}^{\circ}, \mathbf{3 3 0 ^ { \circ } .} \end{aligned} $

Note: Recall that if the value of $\theta$ is $\alpha$ when $\theta$ lies in Ist quadrant then it is $180^{\circ}-\alpha, 180^{\circ}+\alpha$ and $360^{\circ}-\alpha$. If $\theta$ lies in 2nd, 3rd and 4th quadrant respectively.

Ex. 3 . Solve $\sin ^{2} x+\sin x-2=0 .(0^{\circ}<\theta<360^{\circ})$

Sol. $\sin ^{2} x+\sin x-2=0 \Rightarrow(\sin x+2)(\sin x-1)=0$

$\Rightarrow \sin x+2=0$

$\Rightarrow \sin x=-2$

$\sin x-1=0$

This value of $x$ being numerically $>1$ is inadmissible as $\sin x$ cannot be numerically greater than 1 .

$\therefore \boldsymbol{x}=\mathbf{9 0}$.

Ex. 4 . Solve $\cos ^{2} \theta-\sin \theta-\frac{1}{4}=0(0^{\circ}<\theta<360^{\circ})$

Sol. $\cos ^{2} \theta-\sin \theta-\frac{1}{4}=0 \Rightarrow 1-\sin ^{2} \theta-\sin \theta-\frac{1}{4}=0$

$\Rightarrow 4 \sin ^{2} \theta+4 \sin \theta-3=0 \Rightarrow(2 \sin \theta+3)(2 \sin \theta-1)=0$

$\Rightarrow 2 \sin \theta+3=0$ or $2 \sin \theta-1=0$

$\Rightarrow \sin \theta=-\frac{3}{2} \quad$ or $\quad \sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}, 150^{\circ}$.

Since $|\sin \theta|=\frac{3}{2}$ is $>1$, the value $\sin \theta=-\frac{3}{2}$ is inadmissible.

$\therefore \theta=30^{\circ}, \mathbf{1 5 0}$.

Ex. 5 . Solve $2 \sin \theta \cos \theta=\cos \theta(0^{\circ}<\theta<360^{\circ})$

Sol. The given equation may be written as:

$2 \sin \theta \cos \theta-\cos \theta=0 \Rightarrow \cos \theta(2 \sin \theta-1)=0$

$\Rightarrow \cos \theta=0 \quad$ or $2 \sin \theta-1=0$

$\Rightarrow \theta=90^{\circ}, 270^{\circ} \quad \sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}, 150^{\circ}$

$\therefore \theta=30^{\circ}, 0^{\circ}, 150^{\circ} 270^{\circ}$.

Ex. 6 . Solve $\tan ^{2} \alpha+\sec \alpha-1=0,0 \leq \alpha \leq 2 \pi$

Sol. $\tan ^{2} \alpha+\sec \alpha-1=0 \Rightarrow \sec ^{2} \alpha-1+\sec \alpha-1=0$

$\Rightarrow \sec ^{2} \alpha+\sec \alpha-2=0 \Rightarrow(\sec \alpha+2)(\sec \alpha-1)=0$

$ \begin{aligned} & \Rightarrow \sec \alpha+2=0 \quad \text{ or } \quad \sec \alpha-1=0 \\ & \Rightarrow \sec \alpha=-2 \quad \sec \alpha=1 \\ & \Rightarrow \cos \alpha=-\frac{1}{2} \quad \cos \alpha=1 \\ & \Rightarrow \quad \alpha=\frac{2 \pi}{3}, \frac{4 \pi}{3} \quad \alpha=0 . \\ & \therefore \quad \alpha=0, \frac{2 \pi}{3}, \frac{4 \pi}{3} . \end{aligned} $

Ex. 7 . Find the general values of $\theta$ if (i) $2 \sin \theta-1=0$ (ii) $\cos \theta=-\frac{1}{2}$ (iii) $4 \sin ^{2} \theta=1$ (iv) $\tan 2 x-\sqrt{3}=0$ (v) $2 \cot ^{2} \theta=cosec^{2} \theta$

Sol. (i) $2 \sin \theta-1=0 \Rightarrow \sin \theta=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}+(-\mathbf{1})^{\boldsymbol{n}} \frac{\boldsymbol{\pi}}{\mathbf{6}}, \boldsymbol{n} \in \boldsymbol{I}$.

(ii) $\cos \theta=-\frac{1}{2}=\cos (180^{\circ}-60^{\circ})=\cos (\pi-\frac{\pi}{3})=\cos \frac{2 \pi}{3} \quad \therefore \boldsymbol{\theta}=\mathbf{2 n \pi} \pm \frac{\mathbf{2 \pi}}{\mathbf{3}}, \boldsymbol{n} \in \boldsymbol{I}$.

(iii) $4 \sin ^{2} \theta=1 \Rightarrow \frac{4(1-\cos 2 \theta)}{2}=1 \Rightarrow 2(1-\cos 2 \theta)=1 \Rightarrow 1-\cos 2 \theta=\frac{1}{2} \Rightarrow \cos 2 \theta=\frac{1}{2}=\cos \frac{\pi}{3}$

$\therefore 2 \theta=2 n \pi \pm \frac{\pi}{3} \Rightarrow \theta=n \pi \pm \frac{\pi}{6}$.

(iv) $\tan 2 x-\sqrt{3}=0 \Rightarrow \tan 2 x=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow 2 x=n \pi+\frac{\pi}{3} \Rightarrow \boldsymbol{x}=\frac{\boldsymbol{n} \pi}{\mathbf{2}}+\frac{\boldsymbol{\pi}}{\mathbf{6}}, \boldsymbol{n} \in \boldsymbol{I}$

(v) $2 \cot ^{2} \theta=cosec^{2} \theta \Rightarrow 2 \cot ^{2} \theta=1+\cot ^{2} \theta \Rightarrow \cot ^{2} \theta=1$

$\Rightarrow \cot \theta= \pm 1=\cot ( \pm \frac{\pi}{4}) \Rightarrow \theta=\boldsymbol{n} \pi \pm \frac{\pi}{4}, \boldsymbol{n} \in \boldsymbol{I}$.

Ex. 8 . If $3 \cos x \neq 2 \sin x$, then find the general solution of $\sin ^{2} x-\cos 2 x=2-\sin 2 x$.

(EAMCET 2009)

Sol. $\sin ^{2} x-\cos 2 x=2-\sin 2 x$

$\Rightarrow 1-\cos ^{2} x-(2 \cos ^{2} x-1)=2-2 \sin x \cos x$

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

$\Rightarrow \cos x(2 \sin x-3 \cos x)=0 \quad \Rightarrow \cos x=0 \quad(\because 2 \sin x \neq 3 \cos x)$

$\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \Rightarrow x=(4 n \pm 1) \frac{\pi}{2}$.

Ex. 9 . Find the general solution of $|\sin x|=\cos x, n \in I$.

Sol. $|\sin x|=\cos x \Rightarrow \sin ^{2} x=\cos ^{2} x \Rightarrow 1-\cos ^{2} x=\cos ^{2} x \Rightarrow 2 \cos ^{2} x=1$

$ \begin{matrix} \Rightarrow \cos x=+\frac{1}{\sqrt{2}} & \\ \Rightarrow \cos x=\cos \frac{\pi}{4} \Rightarrow \boldsymbol{x}=\mathbf{2 n \pi} \pm \frac{\pi}{4} . \end{matrix} $

Ex. 10 . Find the general solution of $\tan 2 \theta \tan \theta=1$.

(Gujarat CET 2007)

Sol. Given, $\tan 2 \theta \tan \theta=1$

$ \begin{aligned} & \Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta} \cdot \tan \theta=1 \Rightarrow \frac{2 \tan ^{2} \theta}{1-\tan ^{2} \theta}=1 \Rightarrow 2 \tan ^{2} \theta=1-\tan ^{2} \theta \\ & \Rightarrow 3 \tan ^{2} \theta=1 \Rightarrow \tan ^{2} \theta=\frac{1}{3} \Rightarrow \tan ^{2} \theta=(\frac{1}{\sqrt{3}})^{2}=\tan ^{2} \frac{\pi}{6} \Rightarrow \theta=\boldsymbol{n} \pi \pm \frac{\pi}{6} \end{aligned} $

Ex. 11 . Find the number of solutions of the equation $\tan x+\sec x=2 \cos x, x \in[0, \pi]$.

Sol. $\tan x+\sec x=2 \cos x$

$ \Rightarrow \quad \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x $

$ \begin{aligned} & \Rightarrow 1+\sin x=2 \cos ^{2} x \quad \Rightarrow 1+\sin x=2(1-\sin ^{2} x)=2-2 \sin ^{2} x \\ & \Rightarrow 2 \sin ^{2} x+\sin x-1=0 \quad \Rightarrow \quad(\sin x+2)(2 \sin x-1)=0 \\ & \Rightarrow(\sin x+2)=0 \quad \text{ or } \quad 2 \sin x=1 \Rightarrow \sin x=-2 \text{ or } \sin x=\frac{1}{2} \end{aligned} $

Since $\sin x=-2$ is inadmissible, therefore, $\sin x=\frac{1}{2}$

$\Rightarrow \quad x=30^{\circ}, 150^{\circ}$, i.e. $x=\frac{\pi}{\mathbf{6}}, \frac{\mathbf{5 \pi}}{\mathbf{6}}$.

$\therefore \quad$ The number of solutions $x \in[0, \pi]$ are 2 .

Ex. 12 . Find the general solution of $\sin 9 \theta=\sin \theta$.

Sol. $\sin 9 \theta=\sin \theta$

$ \begin{aligned} & \Rightarrow 9 \theta=2 n \pi+\theta \quad \text{ or } 9 \theta=(2 n+1) \pi-\theta, n \in I \quad(\because \sin \theta=\sin \alpha \Rightarrow \theta=n \pi+(-1)^{n} \alpha, \text{ where } n \in I) \\ & \Rightarrow \theta=\frac{2 n \pi}{8} \text{ or } \theta=\frac{(2 n+1) \pi}{10} \Rightarrow \theta=\frac{n \pi}{\mathbf{4}} \text{ or } \frac{(\mathbf{2 n + 1 ) \pi}}{\mathbf{1 0}} . \end{aligned} $

Ex. 13 . Solve $\tan 3 x=\cot 5 x(0<x<2 \pi)$.

Sol. $\tan 3 x=\tan (\frac{\pi}{2}-5 x)$

$ \begin{aligned} & \Rightarrow 3 x=n \pi+(\frac{\pi}{2}-5 x) \\ & \Rightarrow 8 x=(2 n+1) \frac{\pi}{2} \Rightarrow x=(2 n+1) \frac{\pi}{\mathbf{1 6}} \end{aligned} \quad(\because \tan \theta=\tan \alpha \Rightarrow \theta=n \pi+\alpha) $

$\therefore \quad$ Putting $n=0,1,2, \ldots .15$, we see that the values of $x$ between 0 and $2 \pi$ are

$ x=\frac{\pi}{16}, \frac{3 \pi}{16}, \frac{5 \pi}{16}, \ldots . ., \frac{31 \pi}{16} $

Ex. 14 . Solve $\cos 3 x+\cos 2 x=\sin \frac{3}{2} x+\sin \frac{1}{2} x, 0<x \leq \pi$.

Sol. $\cos 3 x+\cos 2 x=\sin \frac{3}{2} x+\sin \frac{1}{2} x \Rightarrow 2 \cos \frac{5}{2} x \cos \frac{x}{2}=2 \sin x \cos \frac{x}{2}$

$ \Rightarrow \quad \cos \frac{x}{2}[\cos \frac{5 x}{2}-\sin x]=0 \Rightarrow \cos \frac{x}{2}=0 \quad \text{ or } \quad \cos \frac{5 x}{2}-\sin x=0 $

Now, $\cos \frac{x}{2}=0 \Rightarrow \frac{x}{2}=\frac{\pi}{2} \Rightarrow x=\pi$

and $\cos \frac{5 x}{2}-\sin x=0 \Rightarrow \cos \frac{5 x}{2}=\sin x \Rightarrow \cos \frac{5 x}{2}=\cos (\frac{\pi}{2}-x)$ or $\sin (2 \pi+\frac{\pi}{2}-x)$

$\Rightarrow \frac{5 x}{2}=\frac{\pi}{2}-x \quad$ or $\quad \frac{5 x}{2}=2 \pi+\frac{\pi}{2}-x \quad \Rightarrow \quad \frac{7}{2} x=\frac{\pi}{2} \quad$ or $\quad \frac{7 x}{2}=\frac{5 \pi}{2} \quad \Rightarrow \quad x=\frac{\pi}{7} \quad$ or $\quad \frac{5 \pi}{7}$

$\therefore x=\frac{\pi}{7}, \frac{5 \pi}{7}$ or $\pi$.

Ex. 15 . Solve $\sqrt{3} \sin \theta-\cos \theta=\sqrt{2}$.

Sol. Dividing both sides of the equation by $\sqrt{(\sqrt{3})^{2}+(-1)^{2}}=\sqrt{4}=2$, we get $\frac{\sqrt{3}}{2} \sin \theta-\frac{1}{2} \cos \theta=\frac{1}{\sqrt{2}}$

$ \begin{aligned} & \Rightarrow \cos 30^{\circ} \sin \theta-\sin 30^{\circ} \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \sin (\theta-30^{\circ})=\sin (\theta-\frac{\pi}{6})=\frac{1}{\sqrt{2}} \\ & \Rightarrow \sin (\theta-\frac{\pi}{6})=\sin \frac{\pi}{4} \Rightarrow \theta-\frac{\pi}{6}=n \pi+(-1)^{n} \frac{\pi}{4}, n \in I \Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}+\frac{\pi}{\mathbf{6}}+(-\mathbf{1})^{n} \frac{\pi}{\mathbf{4}}, \boldsymbol{n} \in \boldsymbol{I} . \end{aligned} $

Ex. 16 . Solve $\cos x-\sqrt{3} \sin x=1,0^{\circ} \leq x \leq 360^{\circ}$.

Sol. Dividing both the sides of the equation $\cos x-\sqrt{3} \sin x=1$ by $\sqrt{(1)^{2}+(-\sqrt{3})^{2}}=2$, we get

$ \begin{aligned} & \frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x=\frac{1}{2} \Rightarrow \cos 60^{\circ} \cos x-\sin 60^{\circ} \sin x=\frac{1}{2} \Rightarrow \cos (x+60^{\circ})=\cos 60^{\circ} \\ \Rightarrow & \cos (x+60^{\circ})=\cos 60^{\circ}=\cos (360^{\circ}-60^{\circ})=\cos (360^{\circ}+60^{\circ}) \\ \Rightarrow & x+60^{\circ}=60^{\circ} \text{ or } 300^{\circ} \text{ or } 420^{\circ} \Rightarrow \boldsymbol{x}=\mathbf{0}^{\circ}, \mathbf{2 4 0}, \mathbf{3 6 0 ^ { \circ } .} \end{aligned} $

Ex. 17 . Show that the equation $k \sin x+\cos 2 x=2 k-7$ has a solution only if $2 \leq k \leq 6$.

(Kerala PET 2011)

Sol. $k \sin x+\cos 2 x=2 k-7 \quad \Rightarrow \quad k \sin x+1-2 \sin ^{2} x=2 k-7$

$ \begin{aligned} & \Rightarrow 2 \sin ^{2} x-k \sin x-1+2 k-7=0 \Rightarrow 2 \sin ^{2} x-k \sin x+2 k-8=0, \text{ which is a quadratic in } \sin x . \\ & \therefore \sin x=\frac{k \pm \sqrt{k^{2}-4(2)(2 k-8)}}{4}=\frac{k \pm \sqrt{k^{2}-16 k+64}}{4}=\frac{k \pm(k-8)}{4} \\ & \quad=\frac{k+k-8}{4} \text{ or } \frac{k-k+8}{4}=\frac{2 k-8}{4} \text{ or } 2=\frac{k-4}{2} \text{ or } 2 \\ & \because \quad-1 \leq \sin x \leq 1, \therefore \sin x=2 \text{ is inadmissible. } \\ & \therefore \quad-1 \leq \frac{k-4}{2} \leq 1 \Rightarrow-2 \leq k-4 \leq 2 \Rightarrow \mathbf{2} \leq \boldsymbol{k} \leq \mathbf{6} . \end{aligned} $

Ex. 18 . Find the total number of solutions of the equation $\sin ^{4} x+\cos ^{4} x=\sin x \cos x$ in $[0,2 \pi]$.

Sol. $\sin ^{4} x+\cos ^{4} x=\sin x \cos x$

$ \begin{aligned} & \Rightarrow(\sin ^{2} x+\cos ^{2} x)^{2}-2 \sin ^{2} x \cos ^{2} x=\sin x \cos x \Rightarrow 1-\frac{(2 \sin x \cos x)^{2}}{2}=\frac{2 \sin x \cos x}{2} \\ & \Rightarrow 1-\frac{\sin ^{2} 2 x}{2}=\frac{\sin 2 x}{2} \Rightarrow \sin ^{2} 2 x+\sin 2 x-2=0 \\ & \Rightarrow(\sin 2 x+2)(\sin 2 x-1)=0 \Rightarrow \sin 2 x=1 \Rightarrow \sin 2 x=\sin \frac{\pi}{2}=\sin (2 \pi+\frac{\pi}{2}) \quad(\because \sin 2 x \neq-2 \text{ is indivisible }) \\ & \Rightarrow 2 x=\frac{\pi}{2} \text{ or }(2 \pi+\frac{\pi}{2}) \Rightarrow 2 x=\frac{\pi}{2} \text{ or } \frac{5 \pi}{2} \Rightarrow x=\frac{\pi}{4} \text{ or } \frac{5 \pi}{4} . \end{aligned} $

Ex. 19 . If

$ \begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix} =0$, then what is the number of distinct real roots of this equation in the interval

$-\pi / 2<x<\pi / 2$ ?

(AMU 2011)

Sol. Let $A= \begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix} \quad \therefore$ Applying $R_1 \to R_1+R_2+R_3$, we have

$ A= \begin{vmatrix} \sin x+2 \cos x & 2 \cos x+\sin x & 2 \cos x+\sin x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} =(\sin x+2 \cos x) \begin{vmatrix} 1 & 1 & 1 \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} $

Applying $C_2 \to C_2-C_1$ and $C_3 \to C_3-C_1$, we have

$ \begin{aligned} & =(\sin x+2 \cos x) \begin{vmatrix} 1 & 0 & 0 \\ \cos x & \sin x-\cos x & 0 \\ \cos x & 0 & \sin x-\cos x \end{vmatrix} \\ & =(\sin x+2 \cos x) \times(\sin x-\cos x)^{2}(\text{ Expanding along } R_1) \end{aligned} $

$\therefore$ The given equation reduces to

$ (\sin x+2 \cos x)(\sin x-\cos x)^{2}=0 \Rightarrow(\tan x+2)(\tan x-1)^{2}=0 $

$\Rightarrow(\tan x+2)=0$ or $(\tan x-1)=0 \quad \Rightarrow \tan x=-2$ or $\tan x=1$

$\Rightarrow x=\tan ^{-1}(-2)$ or $\frac{\pi}{4}$

$(\because-\infty<\tan x<\infty)$.

$\therefore$ There are two possible values of $x$.

Ex. 20 . Find the smallest positive value of $x$ satisfying the equation $\log _{\cos x} \sin x+\log _{\sin x} \cos x=2$. (EAMCET)

Sol. Let $\log _{\cos x} \sin x=p$. Then, $\log _{\sin x} \cos x=\frac{1}{p}$.

$\therefore \log _{\cos x} \sin x+\log _{\sin x} \cos x=2$

$\Rightarrow p+\frac{1}{p}=2 \Rightarrow p^{2}-2 p+1=0 \Rightarrow \quad(p-1)^{2}=0 \Rightarrow p=1$

$\Rightarrow \log _{\cos x} \sin x=1 \Rightarrow \sin x=\cos x$

The smallest positive value of $x$ for which $\sin x=\cos x$ is $x=\frac{\pi}{4}$.

PRACTICE SHEET

~~ 1. The value of $\theta(0<\theta<2 \pi)$ satisfying $cosec \theta+2=0$ are

(a) $210^{\circ}, 330^{\circ}$

(b) $210^{\circ}, 240^{\circ}$

(d) $240^{\circ}, 300^{\circ}$

(EAMCET)

~~ 2. The value of $x$ in $(0, \frac{\pi}{2})$ satisfying the equation $\sin x \cos x=\frac{1}{4}$ is

(a) $\frac{\pi}{6}$

(b) $\frac{\pi}{8}$

(d) $\frac{\pi}{4}$

(Kerala CEE 2010)

~~ 3. If the equation $\tan \theta+\tan 2 \theta+\tan \theta \cdot \tan 2 \theta=1, \theta=$

(a) $\frac{n \pi}{6}+\frac{\pi}{6}$

(b) $\frac{n \pi}{2}+6$

(d) $\frac{n \pi}{2}+\frac{\pi}{12}$

(Odisha JEE 2011)

~~ 4. The general value of $\theta$ obtained from the equation $\cos 2 \theta$ $=\sin \alpha$ is

(a) $\theta=2 n \pi \pm(\frac{\pi}{2}-\alpha)$

(b) $\theta=\frac{n \pi+(-1)^{n} \alpha}{2}$

(d) $2 \theta=\frac{\pi}{2}-\alpha$

(MPPET, AMU 2002)

~~ 5. The general value of $\theta$ for which $\frac{\tan \theta-1}{\tan \theta+1}=\sqrt{3}$ is

(a) $\frac{n \pi}{3}+\frac{\pi}{12}$

(b) $\frac{n \pi}{3}+\frac{7 \pi}{36}$

(d) $n \pi+\frac{7 \pi}{12}$

(MPPET 2012) ~~ 6. The solution of the equation $4 \cos ^{2} x+6 \sin ^{2} x=5$ are

(a) $x=n \pi \pm \frac{\pi}{4}$

(b) $x=n \pi \pm \frac{\pi}{3}$

(d) $x=n \pi \pm \frac{2 \pi}{3}$

(MPPET 2009)

~~ 7. The root of the equation $1-\cos \theta=\sin \theta \cdot \sin \frac{\theta}{2}$ is

(a) $k \pi, k \in I$

(b) $2 k \pi, k \in I$

(d) None of these (DCE 2008)

~~ 8. The most general value of $\theta$ satisfying the equation $\sin \theta=\sin \alpha$ and $\cos \theta=\cos \alpha$ is

(a) $2 n \pi+\alpha$

(b) $2 n \pi-\alpha$

(d) $n \pi-\alpha$

(UPSEE 2008) ~~ 9. $x \in R: \cos 2 x+2 \cos ^{2} x=2$ is equal to

(a) $2 n \pi+\frac{\pi}{3}, n \in Z$

(b) $n \pi \pm \frac{\pi}{6}, n \in Z$

(d) $2 n \pi-\frac{\pi}{3}, n \in Z$

(EAMCET 2008)

~~ 10 If $2 \sec 2 \alpha=\tan \beta+\cot \beta$, then one of the values of $\alpha+\beta$ is

(a) $\frac{\pi}{4}$

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

(d) $n \pi-\frac{\pi}{4}, n \in I$ ~~ 11 If $\cos x \neq-\frac{1}{2}$, then the solutions of $\cos x+\cos 2 x+\cos 3 x=0$ are

(a) $2 n \pi \pm \frac{\pi}{4}, n \in Z$

(b) $2 n \pi \pm \frac{\pi}{3}, n \in Z$

(d) $2 n \pi \pm \frac{\pi}{2}, n \in Z$

(J&K CET 2008)

~~ 12 The most general solution of the equation $\sec ^{2} x=\sqrt{2}(1-\tan ^{2} x)$ are given by

(a) $n \pi \pm \frac{\pi}{4}$

(b) $2 n \pi+\frac{\pi}{4}$

(d) None of these ~~ 13 If $\sin 2 x=4 \cos x$, then $x$ is equal to

(BITSAT 2013)

(a) $\frac{n \pi}{2} \pm \frac{\pi}{4}, n \in Z$

(b) No value

(d) $2 n \pi \pm \frac{\pi}{2}, n \in Z$

(Karnataka CET 2012)

~~ 14 If $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$, then the general value of $\theta$ is

(a) $\frac{n \pi}{4}, n \pi \pm \frac{\pi}{3}$

(b) $\frac{n \pi}{4}, n \pi \pm \frac{\pi}{6}$

(d) $\frac{n \pi}{4}, 2 n \pi \pm \frac{\pi}{6}$

(WBJEE 2010)

~~ 15 If $2 \sin ^{2} \theta+\sqrt{3} \cos \theta+1=0$, then the value of $\theta$ is

(a) $\frac{\pi}{6}$

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

(d) $\pi$

(Odisha JEE 2008)

~~ 16 If $1+\sin \theta+\sin ^{2} \theta+\ldots \infty=4+2 \sqrt{3}, 0<\theta<\pi, \theta \neq \frac{\pi}{2}$, then

(a) $\theta=\frac{\pi}{3}$

(b) $\theta=\frac{\pi}{6}$

(d) $\theta=\frac{\pi}{3}$ or $\theta=\frac{2 \pi}{3}$

(Manipal2008)

~~ 17 The equation $3 \sin ^{2} x+10 \cos x-6=0$ is satisfied, if

(a) $x=n \pi \pm \cos ^{-1}(\frac{1}{3})$

(b) $x=2 n \pi \pm \cos ^{-1}(\frac{1}{3})$

(d) $x=2 n \pi \pm \cos ^{-1}(\frac{1}{6})$

(WBJEE 2007)

~~ 18 If $\tan \theta+\sec \theta=\sqrt{3}$, then the principal value of $\theta+\frac{\pi}{6}$ is equal to

(a) $\frac{\pi}{4}$

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

(d) $\frac{3 \pi}{4}$

(EAMCET 2000)

~~ 19 The solution of equation $\cos ^{2} \theta+\sin \theta+1=0$ lies in the interval:

(a) $(-\frac{\pi}{4}, \frac{\pi}{4})(b)(\frac{\pi}{4}, \frac{3 \pi}{4})$

(MPPET2006)

~~ 20 If $12 \cot ^{2} \theta-31 cosec \theta+32=0$, then the value of $\sin \theta$ is

(a) $\frac{3}{5}$ or 1

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

(d) $\pm \frac{1}{2}$

(Karnataka CET 2005)

~~ 21 Which one of the following equations has no solution?

(a) $\cos \theta+\sin \theta=\sqrt{2}$

(b) $cosec \theta \sec \theta=1$

(d) $cosec \theta-\sec \theta=cosec \theta \sec \theta$

(KCET 2006)

~~ 22 The number of real roots of the equation $\sin ^{4} \theta-2 \sin ^{2} \theta-1$ $=0$ in the interval $(0,2 \pi)$ is

(a) 0

(b) 1

(d) 4

(IIT 1983)

~~ 23 The general solution of $\sin x-\cos x=\sqrt{2}$ for any integer $n$ is

(a) $n \pi$

(b) $(2 n+1) \pi$

(d) $2 n \pi+\frac{3 \pi}{4}$

(BITSAT2006)

~~ 24 The number of values of $x$ in the interval $[0,5 \pi]$ satisfying the equation $3 \sin ^{2} x-7 \sin x+2=0$ is

(a) 0

(b) 5

(d) 10 (IIT) ~~ 25 One of the solutions of the equation $4 \sin ^{4} x+\cos ^{4} x=1$ is

(a) $n \pi$

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

(d) $(2 n+1) \frac{\pi}{2}$

(Odisha JEE 2008)

~~ 26 The solution set of $\sin (x+\frac{\pi}{4})=\sin 2 x$ equals

(a) $\frac{n \pi+\pi / 4}{1-(-1)^{n} 2}$

(b) $\frac{n \pi-\pi / 4}{1-(-1)^{n} 2}$

(d) $\frac{n \pi-\pi / 4}{1+(-1)^{n} 2}$

(IIT)

~~ 27 If $\tan (\frac{\alpha \pi}{4})=\cot (\frac{\beta \pi}{4})$, then

(a) $\alpha+\beta=0$

(b) $\alpha+\beta=2 n, n \in I$

(d) $\alpha+\beta=2(2 n+1), n \in I$ ~~ 28 The solution of the equation $(\sin x+\cos x)^{1+\sin 2 x}=2$, $-\pi \leq x \leq \pi$ is

(a) $\pi / 2$

(b) $\pi$

(d) $\frac{3 \pi}{4}$ ~~ 29 The general solution of $\sin 3 x+\sin x-3 \sin 2 x=\cos 3 x$ $+\cos x-3 \cos 2 x$ is

(a) $\frac{n \pi}{2}+\frac{\pi}{8}, n \in I$

(b) $\frac{n \pi}{2}-\frac{\pi}{8}, n \in I$

(d) $n \pi-\frac{\pi}{8}, n \in I$

(AMU 2013) ~~ 30 Solve:

$\cot (\theta+\pi / 4)+\cot (\theta-\pi / 4)=2 \tan \theta \cdot \cot (\theta-\pi / 4)$ . $\cot (\theta+\pi / 4)$ for the general value of $\theta$.

(a) $\theta=n \pi$

(b) $\theta=n \pi+(-1)^{n} \pi / 4$

(d) $\theta=2 n \pi \pm \pi / 4$

ANSWERS

1. (a)	2. $(c)$	3. $(c)$	4. $(c)$	5. $(b)$	6. $(a)$	7. $(b)$	8. (c)	9. (b)	10. (a)
1. (a)	12. $(c)$	13. $(d)$	14. $(a)$	15. $(c)$	16. $(d)$	17. $(b)$	18. $(b)$	19. $(d)$	20. (c)
1. (b)	22. (a)	23. $(d)$	24. (c)	25. (a)	26. (b)	27. $(d)$	28. (c)	29. (a)	30. (a)

HINTS AND SOLUTIONS

~~ 1. Given, $cosec \theta+2=0$

$\Rightarrow cosec \theta=-2 \Rightarrow \sin \theta=-\frac{1}{2}=-\sin 30^{\circ}$

$\Rightarrow \sin \theta=-\sin 30^{\circ}=\sin (180^{\circ}+30^{\circ})$ or $\sin (360^{\circ}-30^{\circ})$

$\Rightarrow \sin \theta=\sin 210^{\circ}$ or $\sin 330^{\circ}$

$\Rightarrow \boldsymbol{\theta}=\mathbf{2 1 0 ^ { \circ } , \mathbf { 3 3 0 }}$.

~~ 2. $\sin x \cos x=\frac{1}{4} \Rightarrow 2 \sin x \cos x=\frac{1}{2}$

$\Rightarrow \sin 2 x=\frac{1}{2} \Rightarrow \sin 2 x=\sin \frac{\pi}{6}$

$\Rightarrow 2 x=\frac{\pi}{6}, x \in(0, \frac{\pi}{2}) \Rightarrow x=\frac{\pi}{\mathbf{1 2}}$.

~~ 3. $\tan \theta+\tan 2 \theta=1-\tan \theta \cdot \tan 2 \theta$

$\Rightarrow \frac{\tan \theta+\tan 2 \theta}{1-\tan \theta \tan 2 \theta}=1$

$\Rightarrow \tan (\theta+2 \theta)=1 \quad(\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B})$

$\Rightarrow \tan 3 \theta=\tan \frac{\pi}{4} \Rightarrow 3 \theta=n \pi+\frac{\pi}{4} \Rightarrow \theta=\frac{n \pi}{\mathbf{3}}+\frac{\pi}{\mathbf{1 2}}$.

~~ 4. $\cos 2 \theta=\sin \alpha \Rightarrow \cos 2 \theta=\cos (\frac{\pi}{2}-\alpha)$

$\Rightarrow 2 \theta=2 n \pi \pm(\frac{\pi}{2}-\alpha)(\because \cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha)$

$\Rightarrow \theta=n \pi \pm(\frac{\pi}{4}-\frac{\alpha}{2})$.

~~ 5. $\frac{\tan 3 \theta-1}{\tan 3 \theta+1}=\sqrt{3} \Rightarrow \frac{\tan 3 \theta-\tan \pi / 4}{1+\tan \frac{\pi}{4} \cdot \tan 3 \theta}=\sqrt{3}=\tan \frac{\pi}{3}$

$\Rightarrow \tan (3 \theta-\pi / 4)=\tan \pi / 3 \Rightarrow 3 \theta-\pi / 4=n \pi+\frac{\pi}{3}$

$\Rightarrow 3 \theta=n \pi+\frac{\pi}{3}+\frac{\pi}{4} \Rightarrow 3 \theta=n \pi+\frac{7 \pi}{12} \Rightarrow \theta=\frac{n \pi}{3}+\frac{7 \pi}{36}$.

~~ 6. $4 \cos ^{2} x+6 \sin ^{2} x=5$

$\Rightarrow 4(\cos ^{2} x+\sin ^{2} x)+2 \sin ^{2} x=5$ $\Rightarrow 2 \sin ^{2} x=5-4=1 \Rightarrow \sin ^{2} x=\frac{1}{2}$

$\Rightarrow \sin x= \pm \frac{1}{\sqrt{2}} \Rightarrow \boldsymbol{x}=\boldsymbol{n} \boldsymbol{\pi} \pm \frac{\boldsymbol{\pi}}{\mathbf{4}}$.

~~ 7. $1-\cos \theta=\sin \theta$. $\sin \theta / 2$

$\Rightarrow 2 \sin ^{2} \theta / 2=2 \sin \theta / 2 \cdot \cos \theta / 2 \cdot \sin \theta / 2$

$\Rightarrow 2 \sin ^{2} \theta / 2(1-\cos \theta / 2)=0$

$\Rightarrow 2 \sin ^{2} \theta / 2=0$ or $1-\cos \theta / 2=0 \Rightarrow \sin \theta / 2=0$

or $2 \sin ^{2} \theta / 4=0$

(Using, $1-\cos 2 \theta=2 \sin ^{2} \theta$ )

$\Rightarrow \frac{\theta}{2}=k \pi \quad$ or $\quad \frac{\theta}{4}=k \pi$, where $\boldsymbol{k} \in \boldsymbol{I}$.

$\Rightarrow \theta=2 k \pi \quad$ or $\quad \theta=4 k \pi, k \in I$.

$\Rightarrow \boldsymbol{\theta}=\mathbf{2 k} \boldsymbol{\pi}, \boldsymbol{k} \in \mathbf{I}$.

~~ 8. Given, $\sin \theta=\sin \alpha$

$ \cos \theta=\cos \alpha $

$\therefore$ Dividing eqn (i) by eqn (ii), we have $\tan \theta=\tan \alpha$

$\Rightarrow \boldsymbol{\theta}=\boldsymbol{n} \boldsymbol{\pi}+\boldsymbol{\alpha}$.

~~ 9. $\cos 2 x+2 \cos ^{2} x=2 \Rightarrow 2 \cos ^{2} x-1+2 \cos ^{2} x=2$

$ \begin{aligned} & \Rightarrow \quad 4 \cos ^{2} x=3 \Rightarrow \cos x= \pm \frac{\sqrt{3}}{2} \\ & \Rightarrow x=n \pi \pm \frac{\pi}{6}, \boldsymbol{n} \in \boldsymbol{Z} . \end{aligned} $

~~ 10 $2 \sec 2 \alpha=\tan \beta+\cot \beta$

$ \begin{aligned} & \Rightarrow 2 \sec 2 \alpha=\frac{1+\tan ^{2} \beta}{\tan \beta}=\frac{\sec ^{2} \beta}{\tan \beta} \\ &=\frac{2}{2 \cos \beta \sin \beta}=2 cosec 2 \beta \\ & \Rightarrow \sec 2 \alpha=\sec (\frac{\pi}{2}-2 \beta) \Rightarrow 2 \alpha=2 n \pi \pm(\pi / 2-2 \beta) \end{aligned} $

For $(\alpha+\beta)$, taking positive sign, we have

$ 2(\alpha+\beta)=2 n \pi+\pi / 2 $

$\Rightarrow \alpha+\beta=n \pi+\pi / 4, n \in I$

$\therefore$ For $n=0$

$\alpha+\beta=\frac{\pi}{4}$.

~~ 11 $\cos x+\cos 2 x+\cos 3 x=0$

$\Rightarrow \cos x+\cos 3 x+\cos 2 x=0$

$\Rightarrow 2 \cos (\frac{x+3 x}{2}) \cos (\frac{3 x-x}{2})+\cos 2 x=0$

$ (\because \cos C+\cos D=2 \cos (\frac{C+D}{2}) \cos (\frac{D-C}{2})) \text{. } $

$\Rightarrow 2 \cos 2 x \cos x+\cos 2 x=0$

$\Rightarrow \cos 2 x(2 \cos x+1)=0$

$\Rightarrow \cos 2 x=0$ or $2 \cos x=-1 / 2$

(which is not possible, as given)

$\Rightarrow \cos 2 x=0=\cos \frac{\pi}{2} \Rightarrow 2 x=\pi / 2 \quad \Rightarrow \quad x=\pi / 4$

$\therefore$ The general solution of the given equation are

$ 2 n \pi \pm \frac{\pi}{4}, n \in I $

~~ 12 $\sec ^{2} x=\sqrt{2}(1-\tan ^{2} \alpha) \Rightarrow \tan ^{2} \alpha+1=\sqrt{2}(1-\tan ^{2} \alpha)$

$\Rightarrow \tan ^{2} \alpha(1+\sqrt{2})=\sqrt{2}-1$

$\Rightarrow \tan ^{2} \alpha=\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2}+1)(\sqrt{2}-1)}=(\sqrt{2}-1)^{2}=\tan ^{2} \frac{\pi}{8}$

$\therefore \tan \alpha=\tan ( \pm \frac{\pi}{8})$

$\therefore \quad \boldsymbol{\alpha}=\boldsymbol{n} \pi \pm \frac{\pi}{\mathbf{8}}$.

~~ 13 $\sin 2 x=4 \cos x$

$\Rightarrow 2 \sin x \cos x=4 \cos x \Rightarrow 2 \cos x(\sin x-2)=0$

$\Rightarrow \sin x-2=0 \quad$ or $\quad \cos x=0$

$\Rightarrow \sin x=2 \quad$ or $\cos x=\cos \pi / 2$

which is not possible $\Rightarrow x=2 n \pi \pm \pi / 2, n \in Z$.

~~ 14 $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$

$\Rightarrow(\sin 6 \theta+\sin 2 \theta)+\sin 4 \theta=0$

$\Rightarrow 2 \sin 4 \theta \cos 2 \theta+\sin 4 \theta=0$

$\Rightarrow \sin 4 \theta(2 \cos 2 \theta+1)=0$

$\Rightarrow$ Either $\sin 4 \theta=0$ or $(2 \cos 2 \theta+1)=0$

$\Rightarrow \sin 4 \theta=0 \quad \cos 2 \theta=-\frac{1}{2}=\cos 2 \pi / 3$

$\Rightarrow 4 \theta=n \pi \quad \Rightarrow 2 \theta=2 n \pi \pm 2 \pi / 3$

$\Rightarrow \theta=n \pi / 4 \quad \Rightarrow \theta=n \pi \pm \pi / 3$.

$\therefore \quad \theta=\frac{n \pi}{4}$ or $n \pi \pm \pi / 3, n \in I$.

~~ 15 $2 \sin ^{2} \theta+\sqrt{3} \cos \theta+1=0$

$\Rightarrow 2(1-\cos ^{2} \theta)+\sqrt{3} \cos \theta+1=0$

$\Rightarrow 2 \cos ^{2} \theta-\sqrt{3} \cos \theta-3=0$

$\Rightarrow \cos \theta=\frac{\sqrt{3} \pm \sqrt{3-4 \times 2 \times(-3)}}{2 \times 2}$

$\Rightarrow \cos \theta=\frac{\sqrt{3} \pm 3 \sqrt{3}}{4}=\frac{4 \sqrt{3}}{4}$ or $-\frac{2 \sqrt{3}}{4}=\sqrt{3}$ or $-\frac{\sqrt{3}}{2}$

Since $\cos \theta=\sqrt{3}$, is inadmissible, therefore

$\cos \theta=-\frac{\sqrt{3}}{2}=\cos \frac{5 \pi}{6} \Rightarrow \theta=\frac{5 \pi}{\mathbf{6}}$. ~~ 16 $1+\sin \theta+\sin ^{2} \theta+\ldots .+\infty=4+2 \sqrt{3}$

$\Rightarrow \frac{1}{1-\sin \theta}=4+2 \sqrt{3}$

( $\because$ Sum of an infinite G.P $=\frac{a}{1-r}$, where $r<1$. This is an infinite G.P with common ratio $\sin \theta$ and $0<\sin \alpha<1$ )

$\Rightarrow 1-\sin \theta=\frac{1}{4+2 \sqrt{3}}$

$\Rightarrow 1-\sin \theta=\frac{1}{4+2 \sqrt{3}} \times \frac{4-2 \sqrt{3}}{4-2 \sqrt{3}}$

$\Rightarrow 1-\sin \theta=\frac{4-2 \sqrt{3}}{16-12}=\frac{4-2 \sqrt{3}}{4}=1-\frac{\sqrt{3}}{2}$

$\Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3}$ or $\frac{2 \pi}{3}$.

~~ 17 $3 \sin ^{2} x+10 \cos x-6=0$

$\Rightarrow 3(1-\cos ^{2} x)+10 \cos x-6=0$

$\Rightarrow-3 \cos ^{2} x+10 \cos x-3=0$

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

Since $\cos x=3$ is inadmissible, therefore, $\cos x=\frac{1}{3}$

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

$\therefore \quad$ The general solution is $x=2 \boldsymbol{n} \pi \pm \cos ^{-1}(\frac{1}{3})$.

~~ 18 $\tan \theta+\sec \theta=\sqrt{3}$

$\Rightarrow \frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=\sqrt{3} \Rightarrow \sin \theta+1=\sqrt{3} \cos \theta$

$\Rightarrow \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=\frac{1}{2}$

$\Rightarrow \cos \frac{\pi}{6} \cos \theta-\sin \frac{\pi}{6} \sin \theta=\frac{1}{2}$

$\Rightarrow \cos (\theta+\pi / 6)=\cos \pi / 3$

$\Rightarrow$ Principal value of $\theta+\pi / 6=\pi / 3$.

~~ 19 $\cos ^{2} \theta+\sin \theta+1=0 \Rightarrow(1-\sin ^{2} \theta)+\sin \theta+1=0$

$\Rightarrow \sin ^{2} \theta-\sin \theta-2=0 \Rightarrow(\sin \theta+1)(\sin \theta-2)=0$

$\Rightarrow(\sin \theta+1)=0$ or $(\sin \theta-2)=0$

$\Rightarrow \sin \theta=-1 \quad(\because \sin \theta=2$ is inadmissible $)$

$\Rightarrow \sin \theta=\sin \frac{3 \pi}{2} \Rightarrow \theta=\frac{3 \pi}{2} \in(\frac{5 \pi}{4}, \frac{7 \pi}{4}.)$

~~ 20 $12 \cot ^{2} \theta-31 cosec \theta+32=0$

$ \begin{gathered} \Rightarrow 12 \frac{\cos ^{2} \theta}{\sin ^{2} \theta}-\frac{31}{\sin \theta}+32=0 \\ \Rightarrow 12 \cos ^{2} \theta-31 \sin \theta+32 \sin ^{2} \theta=0 \\ \Rightarrow 12(1-\sin ^{2} \theta)-31 \sin \theta+32 \sin ^{2} \theta=0 \\ \Rightarrow 20 \sin ^{2} \theta-31 \sin \theta+12=0 \\ \therefore \quad \sin \theta=\frac{31 \pm \sqrt{31^{2}-4 \times 20 \times 12}}{2 \times 20} \\ \quad=\frac{31 \pm \sqrt{961-960}}{40}=\frac{31 \pm 1}{40} \end{gathered} $

$ \begin{aligned} & =\frac{32}{40} \text{ or } \frac{30}{40}=\frac{4}{5} \text{ or } \frac{3}{4} \\ \therefore \quad \sin \theta & =\frac{4}{5} \text{ or } \frac{3}{4} . \end{aligned} $

~~ 21 (a) $\cos \theta+\sin \theta=\sqrt{2} \Rightarrow \frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta=1$

$\Rightarrow \cos \pi / 4 \cos \theta+\sin \pi / 4 \sin \theta=1$

$\Rightarrow \cos (\theta-\pi / 4)=\cos 0 \Rightarrow \theta-\pi / 4=2 n \pi$

$\Rightarrow \theta=2 n \pi+\pi / 4 \Rightarrow$ Solutions exist

(b) $cosec \theta \cdot \sec \theta=1 \Rightarrow \frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta}=1$

$\Rightarrow \sin \theta \cdot \cos \theta=1 \Rightarrow 2 \sin \theta \cos \theta=2$

$\Rightarrow \sin 2 \theta=2$ which is not possible as $-\mathbf{1} \leq \boldsymbol{\operatorname { s i n }} \boldsymbol{x} \leq \mathbf{1}$.

Hence no solutions exist.

Dividing throughout by $\sqrt{(\sqrt{3})^{2}+(-1)^{2}}=2$, we get $\frac{\sqrt{3}}{2} \sin \theta-\frac{1}{2} \cos \theta=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \frac{\pi}{6} \sin \theta-\sin \frac{\pi}{6} \cos \theta=\frac{1}{\sqrt{2}}$

$\Rightarrow \sin (\theta-\pi / 6)=\sin \pi / 4$

$\Rightarrow \theta-\pi / 6=n \pi+(-1)^{n} \pi / 4, n \in I$

$\Rightarrow \theta=n \pi+(-1)^{n} \pi / 4+\pi / 6, n \in I$

$\Rightarrow$ Solution exist.

(d) $cosec \theta-\sec \theta=cosec \theta$. $\sec \theta$

$\Rightarrow \frac{1}{\sin \theta}-\frac{1}{\cos \theta}=\frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} \Rightarrow \cos \theta-\sin \theta=1$

Dividing throughout by $\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$, we get $\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \theta \cos \pi / 4-\sin \theta \sin \pi / 4=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos (\theta+\pi / 4)=\cos (\pi / 4) \Rightarrow \theta+\pi / 4=2 n \pi \pm \pi / 4, n \in I$

$\Rightarrow \theta=2 n \pi$ or $\theta=2 n \pi-\pi / 2, n \in I$.

$\Rightarrow$ Solutions exist.

~~ 22 $\sin ^{4} \theta-2 \sin ^{2} \theta-1=0 \Rightarrow(\sin ^{2} \theta)^{2}-2 \sin ^{2} \theta-1=0$

$\begin{aligned} \Rightarrow \sin ^{2} \theta & =\frac{2 \pm \sqrt{4-4 \times 1 \times(-1)}}{2 \times 1} \\ & =\frac{2 \pm \sqrt{8}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1-\sqrt{2} \text{ or } 1+\sqrt{2}\end{aligned}$

$\sin ^{2} \theta=1-\sqrt{2}$ is inadmissible as it is not real.

$\because \quad-1 \leq \sin \theta \leq 1 \Rightarrow 0 \leq \sin ^{2} \theta \leq 1$

$\Rightarrow \sin ^{2} \theta=1+\sqrt{2}$ is not possible.

Hence the given equation has no real root.

~~ 23 $\sin x-\cos x=\sqrt{2} \Rightarrow \frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x=1$

$\Rightarrow \sin \pi / 4 \sin x-\cos \pi / 4 \cos x=1$

$\Rightarrow-(\cos \pi / 4 \cos x-\sin \pi / 4 \sin x)=1$ $\Rightarrow-\cos (x+\pi / 4)=1 \Rightarrow \cos (x+\pi / 4)=-1=\cos \pi$

$\Rightarrow x+\pi / 4=2 n \pi+\pi \Rightarrow x=2 n \pi+\frac{3 \pi}{4}$.

~~ 24 $3 \sin ^{2} x-7 \sin x+2=0 \Rightarrow(3 \sin x-1)(\sin x-2)=0$

$\Rightarrow(3 \sin x-1)=0$ or $(\sin x-2)=0$

$\because \quad \sin x=2$ is inadmissible, therefore, $\sin x=\frac{1}{3}$ Since, $\sin x=\sin \alpha$ where $\sin \alpha=\frac{1}{3}$, so $\alpha$ lies in the 1 st quadrant

$\Rightarrow x=n \pi+(-1)^{n} \alpha, n \in I$, where $0<\alpha<\pi / 2$

Since $x$ lies in the interval $[0,5 \pi]$, so we have one value of $x$ corresponding to each of the values $0,1,2,3,4,5$ or $n$.

$\therefore$ The number of values of $x$ in the interval $[0,5 \pi]$ is 6 .

~~ 25 $4 \sin ^{4} x+\cos ^{4} x=1 \Rightarrow 4 \sin ^{4} x+(1-\sin ^{2} x)^{2}=1$

$\Rightarrow 4 \sin ^{4} x+1+\sin ^{4} x-2 \sin ^{2} x=1$

$\Rightarrow 5 \sin ^{4} x-2 \sin ^{2} x=0 \Rightarrow \sin ^{2} x(5 \sin ^{2} x-2)=0$

$\Rightarrow \sin ^{2} x=0$ or $5 \sin ^{2} x-2=0$

$\Rightarrow \sin x=0$ or $\sin x=\sqrt{\frac{2}{5}}$

$\Rightarrow x=n \pi$ or $x=\sin ^{-1}(\sqrt{\frac{2}{5}})$

$ =n \pi+(-1)^{n} \sin ^{-1}(\sqrt{\frac{2}{5}}), n \in I $

$\therefore \quad x=n \pi$ is one of the solutions of the given equation.

~~ 26 $\sin (x+\pi / 4)=\sin 2 x \Rightarrow x+\pi / 4=n \pi+(-1)^{n} 2 x, n \in I$

$\Rightarrow x-(-1)^{n} 2 x=n \pi-\pi / 4, n \in I$

$\Rightarrow x{1-(-1)^{n} .2}=n \pi-\pi / 4 \Rightarrow x=\frac{\boldsymbol{n} \boldsymbol{\pi}-\boldsymbol{\pi} / \mathbf{4}}{\mathbf{1}-(-\mathbf{1})^{\boldsymbol{n}} . \mathbf{2}}, n \in I$.

~~ 27 $\tan (\frac{\alpha \pi}{4})=\cot (\frac{\beta \pi}{4}) \Rightarrow \tan (\frac{\alpha \pi}{4})=\tan (\frac{\pi}{2}-\frac{\beta \pi}{4})$

$\Rightarrow \frac{\alpha \pi}{4}=n \pi+\frac{\pi}{2}-\frac{\beta \pi}{4}, n \in I$

$\Rightarrow(\alpha+\beta) \pi / 4=n \pi+\pi / 2, n \in I$

$\Rightarrow(\alpha+\beta) \pi / 4=(2 n+1) \pi / 2, n \in I$

$\Rightarrow \alpha+\beta=2(2 n+1), \boldsymbol{n} \in \mathbf{I}$.

~~ 28 $[\sin x+\cos x]^{1+2 \sin x \cos x}=2$

$\Rightarrow[\sin x+\cos x]^{(\sin x+\cos x)^{2}}=2$

$\Rightarrow[\sin x+\cos x]^{(\sin x+\cos x)^{2}}=(\sqrt{2})^{(\sqrt{2})^{2}}$

Comparing both the sides, we have

$\sin x+\cos x=\sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x=1$

$\Rightarrow \sin \pi / 4 \sin x+\cos \pi / 4 \cos x=1$

$\Rightarrow \sin (x+\pi / 4)=1=\sin \pi / 2$

$\Rightarrow x+\pi / 4=n \pi+(-1)^{n} \pi / 2 \Rightarrow x=n \pi+(-1)^{n} \pi / 2-\pi / 4$

For $-\pi \leq x \leq \pi$, put $n=0$.

Then $x=\pi / 4$.

~~ 29 $(\sin 3 x+\sin x)-3 \sin 2 x=(\cos 3 x+\cos x)-3 \cos 2 x$

$\Rightarrow 2 \sin 2 x \cos x-3 \sin 2 x=2 \cos 2 x \cdot \cos x-3 \cos 2 x$

$\Rightarrow \sin 2 x(2 \cos x-3)=\cos 2 x(2 \cos x-3)$

$\Rightarrow(2 \cos x-3)(\sin 2 x-\cos 2 x)=0$

$\Rightarrow(2 \cos x-3)=0$ or $(\sin 2 x-\cos 2 x)=0$

$\Rightarrow \cos x=\frac{3}{2}$ or $\sin 2 x=\cos 2 x$

$\because-1 \leq \cos x \leq 1, \cos x \neq \frac{3}{2}$.

$\therefore \sin 2 x=\cos 2 x \Rightarrow \tan 2 x=1$

$\Rightarrow \tan 2 x=\tan (\pi / 4) \Rightarrow 2 x=n \pi+\pi / 4$

$\Rightarrow x=\frac{\boldsymbol{n} \pi}{\mathbf{2}}+\frac{\pi}{\mathbf{8}}, \boldsymbol{n} \in \boldsymbol{I}$.

~~ 30 $\cot (\theta-\pi / 4)+\cot (\theta+\pi / 4)$

$ =2 \tan \theta \cdot \cot (\theta-\pi / 4) \cdot \cot (\theta+\pi / 4) $

Dividing both sides by $\cot (\theta-\pi / 4) \cdot \cot (\theta+\pi / 4)$, we have $\frac{1}{\cot (\theta+\pi / 4)}+\frac{1}{\cot (\theta-\pi / 4)}=2 \tan \theta$

$\Rightarrow \tan (\theta+\pi / 4)+\tan (\theta-\pi / 4)=2 \tan \theta$

$\Rightarrow \frac{\tan \theta+1}{1-\tan \theta}+\frac{\tan \theta-1}{1+\tan \theta}=2 \tan \theta$

$\Rightarrow(1+\tan \theta)^{2}-(1-\tan \theta)^{2}=2 \tan \theta(1-\tan ^{2} \theta)$

$\Rightarrow 1+2 \tan \theta+\tan ^{2} \theta-1+2 \tan \theta-\tan ^{2} \theta$

$ =2 \tan \theta-2 \tan ^{3} \theta $

$\Rightarrow 4 \tan \theta=2 \tan \theta-2 \tan ^{3} \theta$

$\Rightarrow 2 \tan \theta+2 \tan ^{3} \theta=0$

$\Rightarrow 2 \tan \theta(1+\tan ^{2} \theta)=0$

$\Rightarrow 2 \tan \theta=0$

$(\because 1+\tan ^{2} \theta \neq 0)$

$\Rightarrow \tan \theta=0 \Rightarrow \theta=\boldsymbol{n} \pi, n \in I$.

SELF ASSESSMENT SHEET

~~ 1. The general solution of $\tan 5 \theta=\cot 2 \theta$ is

(a) $\theta=\frac{n \pi}{7}+\frac{\pi}{14}$

(b) $\theta=\frac{n \pi}{7}+\frac{\pi}{5}$

(d) $\theta=\frac{n \pi}{7}+\frac{\pi}{2}$

(AIEEE 2002)

~~ 2. The most general solution of $\tan \theta=-1$ and $\cos \theta=\frac{1}{\sqrt{2}}$ is

(a) $n \pi+(-1)^{n} \frac{\pi}{4}$

(b) $2 n \pi+\frac{3 \pi}{4}$

(d) $2 n \pi+\frac{7 \pi}{4}$

(EAMCET 2011)

~~ 3. If $\sqrt{3} \cos \theta-\sin \theta=1$, then $\theta$ is

(d) $\frac{(2 r+1) \pi}{2(m-n)}$

(Roorkee)

~~ 6. The number of values of $x$ in $(0,2 \pi)$ satisfying the equation $\sin 3 \theta=\sin \theta$ are

(a) 8

(b) 9

(d) 7

(Karnataka CET 2007)

~~ 7. The number of solutions of the equation $\sin x \cos 3 x$ $=\sin 3 x \cos 5 x$ in $[0, \pi / 2]$ is

(a) 3

(b) 4

(d) 6

(J&K CET 2009)

~~ 8. The most general solutions of the equation $\sec x-1=(\sqrt{2}-1) \tan x$ are given by

(a) $n \pi+\pi / 8$

(b) $2 n \pi, 2 n \pi+\pi / 4$

(d) None of these

(WBJEE 2007)

(a) $\frac{-5 \pi}{6}$

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

(d) $\frac{5 \pi}{6}$

~~ 5. If $\tan m \theta+\cot n \theta=0$, then the general value of $\theta$ is

(a) $\frac{r \pi}{m+n}$

(b) $\frac{r \pi}{m-n}$ ~~ 6. If $0 \leq x \leq \pi$ and $81^{\sin ^{2} x}+81^{\cos ^{2} x}=30$, then $x$ is equal to

(a) $\pi / 6$

(b) $\pi / 2$

(d) $3 \pi / 4$ ~~ 7. If $3 \tan (\theta-15^{\circ})=\tan (\theta+15^{\circ}), 0<\theta<\pi$, then $\theta$ is equal to

(a) $\pi$

(b) $\pi / 2$

(d) $2 \pi$

(EAMCET)

ANSWERS

~~ 1. (a) ~~ 2. $($ d) ~~ 3. $(d)$ ~~ 4. (b) ~~ 5. $(d)$ ~~ 6. (c) ~~ 7. $(c)$ ~~ 8. (b) ~~ 9. (a) ~~ 10 (c)

HINTS AND SOLUTIONS

~~ 1. $\tan 5 \theta=\cot 2 \theta \Rightarrow \tan 5 \theta=\tan (\pi / 2-2 \theta)$

$\Rightarrow 5 \theta=n \pi+(\pi / 2-2 \theta), n \in I \Rightarrow 7 \theta=n \pi+\pi / 2, n \in I$

$\Rightarrow \theta=\frac{n \pi}{7}+\frac{\pi}{14}, n \in I$.

~~ 2. $\tan \theta=-1$

$\Rightarrow \tan \theta=-\tan \pi / 4=\tan (\pi-\pi / 4)=\tan (2 \pi-\pi / 4)$

$ \begin{aligned} \Rightarrow \tan \theta & =\tan \frac{3 \pi}{4} \text{ or } \tan \frac{7 \pi}{4} \Rightarrow \theta=\frac{3 \pi}{4}, \frac{7 \pi}{4} \\ \cos \theta & =\frac{1}{\sqrt{2}} \\ \Rightarrow \cos \theta & =\cos \frac{\pi}{4}=\cos (2 \pi-\pi / 4) \end{aligned} $

$\Rightarrow \cos \theta=\cos \frac{\pi}{4}=\cos \frac{7 \pi}{4} \Rightarrow \theta=\frac{\pi}{4}, \frac{7 \pi}{4}$

$\therefore$ From (i) and (ii) $\theta=\frac{7 \pi}{4}$ is the only value of $\theta$ in $[0,2 \pi[$ which satisfies both the equations.

$\therefore$ The general value of $\theta$ satisfying both the equations is $\theta=2 n \pi+\frac{7 \pi}{4}, n \in I$.

~~ 3. $\sqrt{3} \cos \theta-\sin \theta=1$

Dividing throughout by $\sqrt{(\sqrt{3})^{2}+(-1)^{2}}=2$, we have

$\frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=\frac{1}{2}$

$\Rightarrow \sin \pi / 3 \cos \theta-\cos \pi / 3 \sin \theta=\frac{1}{2}$

$\Rightarrow \sin (\pi / 3-\theta)=\sin \pi / 6 \Rightarrow \pi / 3-\theta=\pi / 6 \Rightarrow \theta=\pi / 6$.

~~ 4. $\sin \theta=\sqrt{3} \cos \theta \Rightarrow \tan \theta=\sqrt{3}$

$\Rightarrow \tan \theta=\tan \pi / 3 \Rightarrow \theta=n \pi+\pi / 3, n \in I$

Putting $n=-1$, we get $\theta=-\pi+\frac{\pi}{3}=-\frac{2 \pi}{3}$, which lies between $-\pi$ and 0 .

~~ 5. $\tan m \theta+\cot n \theta=0 \Rightarrow \tan m \theta=-\cot n \theta$

$\Rightarrow \tan m \theta=\tan (\pi / 2+n \theta) \quad(\because \tan (\pi / 2+x)=-\cot x)$

$\Rightarrow m \theta=r \pi+\pi / 2+n \theta, r \in I$

$\Rightarrow(m-n) \theta=(2 r+1) \pi / 2, r \in I$

$\Rightarrow \theta=\frac{(2 r+1) \pi}{2(m-n)}, r \in I$.

~~ 6. $\sin 3 \theta-\sin \theta=0$

$\Rightarrow 2 \cos (\frac{3 \theta+\theta}{2}) \sin (\frac{3 \theta-\theta}{2})=0$

$\Rightarrow \cos 2 \theta \cdot \cos \theta=0 \quad \Rightarrow \quad \cos 2 \theta=0$ or $\cos \theta=0$

$\Rightarrow 2 \theta=\pi / 2,3 \pi / 2,5 \pi / 2,7 \pi / 2$ or $\theta=0, \pi, 2 \pi$

$\Rightarrow \theta=\pi / 4,3 \pi / 4,5 \pi / 4,7 \pi / 4$ or $\theta=\pi$

Hence the total number of solutions is $\mathbf{5}$.

~~ 7. $\sin x \cos 3 x=\sin 3 x \cos 5 x$

$\Rightarrow 2 \sin x \cos 3 x-2 \sin 3 x \cos 5 x=0$

$\Rightarrow[\sin (3 x+x)-\sin (3 x-x)]$

$ -[\sin (3 x+5 x)-\sin (5 x-3 x)]=0 $

$\Rightarrow \sin 4 x-\sin 2 x-\sin 8 x+\sin 2 x=0$

$\Rightarrow \sin 4 x-\sin 8 x=0$

$\Rightarrow 2 \cos (\frac{4 x+8 x}{2}) \sin (\frac{8 x-4 x}{2})=0$

$\Rightarrow 2 \cos 6 x \sin 2 x=0 \Rightarrow \cos 6 x=0$ or $\sin 2 x=0$

$\Rightarrow 6 x=(2 n+1) \pi / 2$ or $2 x=n \pi$ $\Rightarrow x=(2 n+1) \pi / 2$ or $x=\frac{n \pi}{2}$

$\Rightarrow x=0, \frac{\pi}{2}, \frac{\pi}{12}, \frac{3 \pi}{12}, \frac{5 \pi}{12}$ in $[0, \frac{\pi}{2}]$.

~~ 8. $\sec x-1=(\sqrt{2}-1) \tan x$

$\Rightarrow \frac{1}{\cos x}-1=(\sqrt{2}-1) \frac{\sin x}{\cos x}$

$\Rightarrow \frac{1-\cos x}{\cos x}=(\sqrt{2}-1) \frac{\sin x}{\cos x}$

$\Rightarrow 1-\cos x=(\sqrt{2}-1) \sin x$

$\Rightarrow 2 \sin ^{2} x / 2-(\sqrt{2}-1) \cdot 2 \sin x / 2 \cos x / 2=0$

$\Rightarrow \sin x / 2[\sin x / 2-(\sqrt{2}-1) \cos x / 2]=0$

$\Rightarrow \sin x / 2=0$ or $\sin x / 2-(\sqrt{2}-1) \cos x / 2=0$

$\Rightarrow x / 2=n \pi$ or $\tan x / 2=\sqrt{2}-1=\tan \frac{45^{\circ}}{2}=\tan \pi / 8$

$\Rightarrow x=2 n \pi$ or $\frac{x}{2}=n \pi+\pi / 8$

$\Rightarrow x=\mathbf{2 n \pi}$ or $\mathbf{2 n \pi}+\boldsymbol{\pi} / \mathbf{4}$.

~~ 9. $81^{\sin ^{2} x}+81^{\cos ^{2} x}=30$

$\Rightarrow 81^{\sin ^{2} x}+81^{1-\sin ^{2} x}=30$

$\Rightarrow 81^{\sin ^{2} x}+\frac{81}{81^{\sin ^{2} x}}=30$

$\Rightarrow y+\frac{81}{y}=30 \quad$ (putting $y=81^{\sin ^{2} x}$ )

$\Rightarrow y^{2}-30 y+81=0 \Rightarrow(y-27)(y-3)=0$

$\Rightarrow 81^{\sin ^{2} x}=27$ or $81^{\sin ^{2} x}=3$

$\Rightarrow 3^{4 \sin ^{2} x}=3^{3}$ or $3^{4 \sin ^{2} x}=3^{1}$

$\Rightarrow 4 \sin ^{2} x=3$ or $4 \sin ^{2} x=1$

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

$\Rightarrow \sin x=\frac{\sqrt{3}}{2}$ or $\sin x=\frac{1}{4}$

$\Rightarrow x=\pi / 3,2 \pi / 3$ or $x=\pi / 6,5 \pi / 6$.

~~ 10 $3 \tan (\theta-15^{\circ})=\tan (\theta+15^{\circ})$

$\Rightarrow \frac{3 \sin (\theta-15^{\circ})}{\cos (\theta-15^{\circ})}=\frac{\sin (\theta+15^{\circ})}{\cos (\theta+15^{\circ})}$

$\Rightarrow 3 \cos (\theta+15^{\circ}) \sin (\theta-15^{\circ})=\sin (\theta+15^{\circ}) \cos (\theta-15^{\circ})$

$\Rightarrow 3[\sin (2 \theta)-\sin 30^{\circ}]=\sin 2 \theta+\sin 30^{\circ}$

$\Rightarrow 3(\sin 2 \theta-\frac{1}{2})=\sin 2 \theta+\frac{1}{2}$

$\Rightarrow 3 \sin 2 \theta=\sin 2 \theta+\frac{1}{2}+\frac{3}{2}$

$\Rightarrow 2 \sin 2 \theta=2$

$\Rightarrow \sin 2 \theta=1 \Rightarrow 2 \theta=\pi / 2 \Rightarrow \theta=\pi / 4$.

9 Heights and Distances

KEY FACTS

Angles of Elevation and Depression

Suppose a person $P$ is looking up towards an object $O$, say an aeroplane. $O$ is obviously at a higher level than $P$. Let $P H$ be the horizontal line drawn through the point $P$ to meet in $M$ the vertical line drawn through $O$. Then $\angle O P H$ is called the angle of elevation of $O$ from $P$.

Thus, the angle of elevation of an object $O$ from $P$ is the angle between the line of sight $P O$ and a particular horizontal line $P M H$ through the observer $P$.

Conversely, if the observer $P$ is at a higher level, say on the top of a tower and the object is at a lower level than himself, then the angle $O P H$ between the line of sight $P O$ and a particular horizontal line $P M H$ through the observer $P$ is called the angle of depression of the object $O$ from $P$, where $O M$ is perpendicular to $P H$.

It should be noted that the angle of elevation of one position as seen from the other is equal to the angle of depression of the latter, as seen from the former.

SOLVED EXAMPLES

Ex. 1 . The angle of elevation of the top of a tower at a distance of 150 metres from its foot on a horizontal plane is found to be $30^{\circ}$. Find the height of the tower correct to one place of decimal.

Sol. Let $T M$, the tower, be $h$ metres. Then, $\frac{h}{150}=\tan 30^{\circ}$

$ \begin{aligned} \therefore \quad h & =150 \tan 30^{\circ}=150 \times \frac{1}{\sqrt{3}}=\frac{150}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{150 \sqrt{3}}{3} \\ & =50 \sqrt{3}=50 \times 1.732=\mathbf{8 6 . 6} \mathbf{~ m .} \end{aligned} $

Ex. 3 . An observer $1.5 m$ tall is $28.5 m$ away from a tower $30 m$ high. Determine the angle of elevation from his eye to the top of the tower.

Sol. $B C$ is the tower $30 m$ high, and the observer is at $A$. $E$ denotes his eyes’ position. Let the angle of elevation be $\theta$.

$ \begin{aligned} F C & =B C-A E=30-1.5=28.5 m \\ E F & =A B=28.5 m \\ \text{ In } \triangle E F C, \tan \theta & =\frac{F C}{E F}=\frac{28.5}{28.5}=1 . \\ \therefore \theta & =\mathbf{4 5}^{\circ} . \end{aligned} $

Ex. 4 . The angle of depression of a boat $B$ from the top $K$ of cliff $H K, 300$ metres high, is $40^{\circ}$. Find the distance of the boat from the foot $\boldsymbol{H}$ of the cliff.

Sol. Let the required distance $B H$ be equal to $x$ metres. $H K$ is the cliff and $\angle L K B$ is the angle of depression of $B$ from $K$. Then,

$ \begin{matrix} \angle K B H=\angle L K B & =40^{\circ} \quad \\ \therefore \quad \angle B K H=90^{\circ}-40^{\circ} & =50^{\circ} \\ & \angle & \\ \text{ From } \triangle B K H, \text{ we get, } \frac{x}{300} & =\tan 50^{\circ} \\ \therefore \quad & x & =300 \times \tan 50^{\circ}=300 \times 1.1918=357.54 \\ & & =\mathbf{3 5 8} \mathbf{m}, \text{ correct to the nearest metre. } \end{matrix} $

Ex. 5 . From a light house the angles of depression of two ships on opposite side of the light house are observed to be $30^{\circ}$ and $45^{\circ}$. If the height of light house be 300 metres, find the distance between the ships if the line joining them passes through the foot of the light house.

Sol. Let $H K$ be the light house $300 m$ in height. Let $A$ and $B$ be the two ships, such that angles of depression $L H A$ and $M H B$ are respectively $30^{\circ}$ and $45^{\circ}$. It follows that $\angle H A K=30^{\circ}$ and $\angle H B K=45^{\circ}$.

Also, $\quad \angle K H B=45^{\circ}$, and therefore, $K B=H K=300 m$

Also, $\angle A H K=60^{\circ}$

From $\triangle H A K$, we get, $\frac{A K}{H K}=\tan 60^{\circ}$

$\therefore \quad A K=H K \times \tan 60^{\circ}=300 \times 1.7321=519.63 m$

$\therefore \quad A B=A K+K B=519.63 m+300 m=\mathbf{8 1 9 . 6 3} \mathbf{~ m .}$

Ex. 6 A man sitting in an aeroplane observes that the angles of depression of two temples $2 \mathbf{~ k m}$ apart are $60^{\circ}$. If the plane is exactly above the middle point of the line joining the temples, calculate its height.

Sol. $A$ and $B$ represent the temples and $P$ the aeroplane

$ \begin{aligned} \frac{P M}{A M} & =\tan 60^{\circ} \quad \Rightarrow \frac{P M}{1}=\tan 60^{\circ} \\ \Rightarrow \quad P M & =1 \tan 60^{\circ}=\sqrt{3} \end{aligned} $

Hence the plane is at a height of $\sqrt{\mathbf{3}} \mathbf{~ k m}$.

Ex. 7 . The angle of depression of a $37 m$ high building from the top of a tower $117 m$ high is $30^{\circ}$. Calculate the distance between the building and the tower.

Sol. $A B$ and $C D$ represent the tower and the building respectively, $\angle X B D=30^{\circ}$ is the angle of depression. We have to calculate the measure of $A C$. Let it be $x$ metres.

In right $\triangle B D E$,

$ \begin{aligned} D E & =C A=x \\ B E & =117-37=80 m \\ \angle B D E & =\angle X B D=30^{\circ} \quad(\text{ alt. } \angle s) \end{aligned} $

Now, from right $\triangle B E D$, we have $\frac{D E}{E B}=\cot B D E$

$ \begin{matrix} \Rightarrow & \frac{x}{80}=\cot 30^{\circ} \Rightarrow x=80 \cot 30^{\circ} \\ \Rightarrow & x=80 \sqrt{3}=80 \times 1.732=\mathbf{1 3 8 . 5 6} \mathbf{~ m .} \end{matrix} $

Ex. 8 . The angular elevation of a tower from a point is $30^{\circ}$, at a point in a horizontal line to the foot of the tower and 100 metres nearer it is $60^{\circ}$, find the height of the tower. Find also the distance of the first point from the tower.

Sol. Let $A B$ represent the tower and $P, Q$ the two points of observation. It is given that $P Q=100 m$. Angles of elevation at $P$ and $Q$ are given to be $30^{\circ}$ and $60^{\circ}$ respectively. Let the required height of the tower be $h$ metres. Let $A Q=x$ metres. Then,

From right $\triangle B A Q, \quad \frac{h}{x}=\tan 60^{\circ} \Rightarrow h=x \tan 60^{\circ}$

From right $\triangle B A P, \frac{h}{x+100}=\tan 30^{\circ} \Rightarrow h=(x+100) \tan 30^{\circ}$

Eq. (i) and $(i i) \Rightarrow x \tan 60^{\circ}=(x+100) \tan 30^{\circ}$

$ x \sqrt{3}=\frac{x+100}{\sqrt{3}} \Rightarrow 3 x=x+100 \Rightarrow 2 x=100 \Rightarrow x=50 $

$\therefore$ From $(i) \quad h=x \tan 60^{\circ}=50 \sqrt{3}=50 \times 1.732=\mathbf{8 6 . 6} \mathbf{~ m}$ (nearly)

Also,

$ A P=x+100=50+100=\mathbf{1 5 0} \mathbf{~ m .} $

Ex. 9 . From the top of a cliff, 200 metres high, the angle of depression of the top and bottom of a tower are observed to be $30^{\circ}$ and $60^{\circ}$, find the height of the tower.

Sol. Let $A B$ represent the tower and $P$ the top of the cliff $L P$. If $P X$ be the horizontal line through $P$, then $\angle X P A=30^{\circ}$ and $\angle X P B=60^{\circ}$. Let the height of the tower be $h$ metres. From $A$ draw $A M$ prep. to $L P$.

$ L M = A B = h $

$\therefore M P = (200-h) m $

$\text{ Again, } \angle P B L = \angle X P B = 60^{\circ} $

$\angle P A M = \angle X P A = 30^{\circ}$

$\therefore$ From right $\triangle P L B, \frac{B L}{200}=\cot 60^{\circ}=\frac{1}{\sqrt{3}} \Rightarrow B L=\frac{200}{\sqrt{3}}$

From right $\triangle P M A, \frac{A M}{M P}=\cot 30^{\circ}$

$ \therefore \quad A M=M P \cot 30^{\circ}=(200-h) \sqrt{3} $

But $\quad A M=B L \therefore(200-h) \sqrt{3}=\frac{200}{\sqrt{3}} \Rightarrow(200-h) \times 3=200 \Rightarrow 3 h=400 \Rightarrow h=\frac{400}{3}=\mathbf{1 3 3} \frac{\mathbf{1}}{\mathbf{3}} \mathbf{m}$.

Ex. 10 . A man on the roof of a house, which is $10 m$ high, observes the angle of elevation of the building as $42^{\circ}$ and angle of depression of the base of the building as $40^{\circ}$. Find the height of the building and its distance from the house.

Sol. Let $A B$ be the house and $C D$ be the building. Let the height of the building be $h m$ and let $x m$ be the distance between the house and the building. Let $A E$ be $\perp C D$. Then, $C D=h m, C E=(h-10) m$

In $\triangle A B D, \frac{B D}{A B}=\tan 50^{\circ} \Rightarrow \frac{x}{10}=1.1918 \Rightarrow x=11.918$

In $\triangle C A E, \frac{C E}{A E}=\tan 42^{\circ} \Rightarrow \frac{h-10}{x}=0.9004 \quad$ [Using $t-$ tables]

$ \begin{aligned} & \Rightarrow h-10=0.9004 x=0.9004 \times 11.9 .18=10.731 \\ & \Rightarrow h=10.731+10=20.73 \end{aligned} $

Hence, height of the building $=\mathbf{2 0 . 7 3} \mathbf{~ m}$

and distance between house and the building $=\mathbf{1 1 . 9 2} \mathbf{~ m}$ (approx).

Ex. 11 . The angle of elevation of a jet plane from a point $A$ on the ground is $60^{\circ}$. After a flight of 15 seconds, the angle of elevation changes to $30^{\circ}$. If the jet plane is flying at a constant height of $1500 \sqrt{3} m$, find the speed of the jet plane.

Sol. Height of the jet plane $=1500 \sqrt{3} m$

Initially the jet plane is at $A$ and after 15 seconds flight at $B$.

$E D=x, E C=y$, where $D$ is right below $A$ and $C$ is right below $B$

(the position of the plane after 15 secs).

From $\triangle E B C, \quad \cot 30^{\circ}=\frac{y}{1500 \sqrt{3}}$

$\Rightarrow \frac{y}{1500 \sqrt{3}}=\sqrt{3} \Rightarrow y=1500 \sqrt{3} \times \sqrt{3}=4500 m$

From $\triangle E D A, \cot 60^{\circ}=\frac{x}{1500 \sqrt{3}} \Rightarrow \frac{x}{1500 \sqrt{3}}=\frac{1}{\sqrt{3}} \Rightarrow x=1500 m$

$\therefore A B=D C=$ Distance travelled in 15 seconds $=y-x=(4500-1500) m=3000 m$.

$\Rightarrow$ Distance travelled in 1 second $=\frac{3000}{15} m=200 m$

Hence speed of the jet plane $=200 m / sec=\frac{200}{1000} \times 60 \times 60 km / hr=\mathbf{7 2 0} \mathbf{~ k m} / \mathbf{h r}$.

Ex. 12 . Two pillars are of equal height and on either sides of a road, which is $100 m$ wide. The angles of elevation of the top of the pillars are $60^{\circ}$ and $30^{\circ}$ at a point on the road between the pillars. Find the position of the point between the pillars and the height of each pillar.

Sol. Let $A B$ and $E D$ be two pillars each of height $h$ metres. Let $C$ be a point on the road such that $B C=x$ metres. Then $C D=(100-x)$ metres. Given, $\angle A C B=60^{\circ}$ and $\angle E C D=30^{\circ}$

In $\triangle A B C$, we have $\tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x} \Rightarrow h=\sqrt{3} x$

In $\triangle E C D$, we have $\tan 30^{\circ}=\frac{E D}{C D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{100-x} \Rightarrow h \sqrt{3}=100-x$

From eq. (i) and (ii) on eliminating $h$, we have

$ 3 x=100-x \Rightarrow 4 x=100 \Rightarrow x=25 $

Substituting $x=25$ in $(i)$, we have $h=25 \sqrt{3}=25 \times 1.732=43.3 m$.

Thus the required point is at a distance of $\mathbf{2 5 ~} \mathbf{~ m}$ from the first pillar and $\mathbf{7 5} \mathbf{~ m}$ from the second pillar. The height of the pillars in $43.3 m$.

Ex. 13 . A man on a cliff observes a boat at an angle of depression of $30^{\circ}$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be $60^{\circ}$. Find the time taken by the boat to reach the shore.

Sol. Let $P C$ be the cliff and $A$ the initial position of the boat, where the angle of depression is $30^{\circ}$. The boat reaches point $B$ after 6 minutes, where the angle of depression is $60^{\circ}$.

In $\triangle s A P C$ and $B P C$,

$\tan 30^{\circ}=\frac{P C}{A C}$ and $\tan 60^{\circ}=\frac{P C}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{P C}{A C}$ and $\sqrt{3}=\frac{P C}{B C}$

$\Rightarrow \quad P C=\frac{A C}{\sqrt{3}}$ and $P C=\sqrt{3} B C \Rightarrow \frac{A C}{\sqrt{3}}=\sqrt{3} B C \Rightarrow A C=3 B C$.

Now $A B=A C-B C=A C-\frac{1}{3} A C=\frac{2}{3} A C \Rightarrow A B=\frac{2}{3} A C$

Let the speed of the boat be $x$ metres/minutes. Then $A B=$ Distance travelled by the boat in 6 minutes $=6 x$

From eq. (i), we have $\frac{2}{3} A C=6 x \Rightarrow A C=9 x$

$\therefore$ Time taken by the boat to reach the shore $=\frac{A C}{x} \quad[T=\frac{D}{S}]$

$ =\frac{9 x}{x} \text{ minutes }=9 \text{ minutes. } $

Ex. 14 . A person observes the angle of elevation of the top of a building as $30^{\circ}$. He proceeds towards the building with a speed of $25(\sqrt{3}-1) m /$ hour. After two hours, he observes the angle of elevation as $45^{\circ}$. What is the height of the building in metres?

(UPSEAT 2003)

Sol. Let the height of the building $P Q$ be $h$ metres.

Let $R$ be the first point of observation such that $\angle P R Q=30^{\circ}$.

Let $S$ be the second point of observation such that $\angle P S Q=45^{\circ}$

$R S=$ Distance moved in two hours with a speed of $25(\sqrt{3}-1) m / hour$

$ =2 \times 25(\sqrt{3}-1) m=50(\sqrt{3}-1) m $

Let $S Q=x$ metres

Then, in $\triangle P Q S, \tan 45^{\circ}=\frac{P Q}{Q S} \Rightarrow \frac{h}{x}=1 \Rightarrow h=x$

In $\triangle P Q R$, we have $\tan 30^{\circ}=\frac{P Q}{R Q} \Rightarrow \frac{h}{x+50(\sqrt{3}-1)}=\frac{1}{\sqrt{3}}$

$\Rightarrow \sqrt{3} h=x+50(\sqrt{3}-1) \Rightarrow \sqrt{3} h=h+50(\sqrt{3}-1) \quad($ From $(i) x=h)$

$\Rightarrow(\sqrt{3}-1) h=50(\sqrt{3}-1) \Rightarrow h=\mathbf{5 0} \mathbf{~ m}$.

$\therefore$ Height of the building $=\mathbf{5 0}$ metres.

Ex. 15 . If the angles of depression of the upper and lower ends of a lamp post from the top of a hill of height $h$ metres are $\alpha$ and $\beta$ respectively, then show that the height of the lamp post (in metres) is $\frac{h \sin (\beta-\alpha)}{\cos \alpha \sin \beta}$.

(EAMCET)

Sol. Let $A B$ be the hill of height $h$ metres and $P Q$ be the lamp post of height $x$ metres.

The $A B=h m, P Q=x m$.

Let $B Q=y$ metres

Given $\angle T A P=\alpha$ and $\angle T A Q=\beta$

Draw $a$ line $P C|B Q| T A$, such that $C$ lies on $A B$.

Now $\angle A P C=\angle T A P=\alpha \quad(A T | P C$, alt. $\angle S)$

$ \angle A Q B=\angle T A Q=\beta \quad(A T | B Q, \text{ alt. } \angle s) $

Also, $P C | B Q, C B=P Q=x m$ and $C P=B Q=y m$.

$\therefore$ In rt. $\triangle A C P, \tan \alpha=\frac{A C}{C P}=\frac{A B-B C}{C P}=\frac{h-x}{y} \Rightarrow y=(h-x) \cot \alpha$

In rt. $\triangle A B Q, \quad \tan \beta=\frac{A B}{B Q}=\frac{h}{y} \Rightarrow y=h \cot \beta$

$\therefore$ From $(i)$ and $(i i) \quad(h-x) \cot \alpha=h \cot \beta$

$\Rightarrow h \cot \alpha-h \cot \beta=x \cot \alpha$

$\Rightarrow h{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos \beta}{\sin \beta}}=x \frac{\cos \alpha}{\sin \alpha}$

$\Rightarrow h{\frac{\cos \alpha \sin \beta-\cos \beta \sin \alpha}{\sin \alpha \sin \beta}}=x \frac{\cos \alpha}{\sin \alpha} \Rightarrow h{\frac{\sin (\beta-\alpha)}{\sin \alpha \sin \beta}}=x \frac{\cos \alpha}{\sin \alpha} \Rightarrow x=\frac{\boldsymbol{h} \sin (\beta-\alpha)}{\cos \alpha \sin \beta}$.

Ex. 16 . A tower, $x$ metres, has a flagstaff at its top. The tower and the flagstaff subtend equal angles at a point distant $y$ metres from the foot of the tower. Then find the length of the flagstaff (in metres).

(EAMCET 2005)

Sol. Let the height of the tower $A Q$ be $x$ metres

Let the flagstaff $P A=h$ metres

Let the point $R$ be at $a$ distance $y$ metres from the foot of the tower $Q$, i.e., $R Q=y$ metres

Given, $\angle P R A=\angle A R Q=\alpha$ (say)

Then, in $\triangle A R Q, \tan \alpha=\frac{x}{y}$

In $\triangle P R Q, \tan 2 \alpha=\frac{h+x}{y} \Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\frac{h+x}{y} \Rightarrow \frac{2 x / y}{1-x^{2} / y^{2}}=\frac{h+x}{y}($ From (i))

$\frac{2 x y}{y^{2}-x^{2}}=\frac{h+x}{y} \Rightarrow 2 x y^{2}=h(y^{2}-x^{2})+x y^{2}-x^{3} \Rightarrow \frac{x y^{2}+x^{3}}{y^{2}-x^{2}}=h \Rightarrow h=\frac{x(y^{2}+x^{2})}{y^{2}-x^{2}} m$

$\therefore A P=$ Length of flagstaff $=\frac{\boldsymbol{x}(\boldsymbol{y}^{2}+\boldsymbol{x}^{2})}{\boldsymbol{y}^{2}-\boldsymbol{x}^{2}} \mathbf{m}$.

Ex. 17 . Two vertical poles $A L$ and $B M$ of heights $20 m$ and $80 m$ respectively stand apart on a horizontal plane. If $A, B$ be the feet of the poles and $A M$ and $B L$ intersect at $P$. Find the height of $P$ from the horizontal plane.

Sol. As shown in the figure, $A L=20 m, B H=80 m$.

Let the height $P=P Q=h$ metres (say)

Let $\angle L B A=\phi$ and $\angle M A B=\theta$. Then,

In $\triangle M A B=\tan \theta=\frac{M B}{A B}=\frac{80}{A B}$

In $\triangle P Q A,=\tan \theta=\frac{P Q}{Q A}=\frac{h}{Q A}$

$\therefore$ From (i) and (ii), $\frac{80}{A B}=\frac{h}{Q A} \Rightarrow Q A=\frac{h}{80} \cdot A B$

In $\triangle L B A, \tan \phi, \frac{L A}{A B}=\frac{20}{A B}$

In $\triangle P Q B, \tan \phi=\frac{P Q}{B Q}=\frac{h}{B Q}$

$\therefore$ From (iv) and (v), $\frac{20}{A B}=\frac{h}{B Q} \Rightarrow B Q=\frac{h}{20} A B$

from (iii) and (vi) $B Q+Q A=\frac{h}{20} A B+\frac{h}{80} A B$

$\Rightarrow A B=h(\frac{1}{20}+\frac{1}{80}) A B \Rightarrow 1=(\frac{1+4}{80}) h \Rightarrow h=\frac{80}{5}=\mathbf{1 6} \mathbf{~ m}$.

$\therefore$ Height of $P Q$ from horizontal plane $=\mathbf{1 6} \mathbf{~ m}$.

Ex. 18 . The angle of elevation of a cloud from a point $h$ metres above the level of water in a lake is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$. Show that the height of the cloud above the water level is $\frac{h \sin (\beta+\alpha)}{\sin (\beta-\alpha)}$.

Sol. Let $C D$ be the water level.

Let the height of the cloud above the water level be $x$ metres, i.e., $P D=x$ metres.

As $R$ is the reflection of cloud at $P$, therefore $D R=P D=x$ metres.

Let $A$ be the point $h$ metres above the water level, such that, $A C=B D=h$ metres.

$\therefore B P=P D-B D=(x-h)$ metres $\quad B R=D R+B D=(x+h)$ metres.

Given, $\angle P A B=\alpha$ and $\angle B A R=\beta$.

Let $A B=C D=y$ metres.

$\therefore$ In $\triangle P A B, \tan \alpha=\frac{P B}{A B}=\frac{x-h}{y} \Rightarrow x-h=y \tan \alpha$

In $\triangle B A R, \tan \beta=\frac{B R}{A B}=\frac{x+h}{y} \Rightarrow x+h=y \tan \beta$

$\therefore \frac{x+h}{x-h}=\frac{y \tan \beta}{y \tan \alpha}$

(On dividing eqn. (ii) by eqn. (i))

$\Rightarrow \frac{2 x}{2 h}=\frac{y(\tan \beta+\tan \alpha)}{y(\tan \beta-\tan \alpha)}$

(Applying componendo and dividendo)

$\Rightarrow \frac{x}{h}=\frac{\frac{\sin \beta}{\cos \beta}+\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin \beta}{\cos \beta}-\frac{\sin \alpha}{\cos \alpha}}=\frac{\sin \beta \cos \alpha+\sin \alpha \cos \beta}{\sin \beta \cos \alpha-\sin \alpha \cos \beta}=\frac{\sin (\beta+\alpha)}{\sin (\beta-\alpha)} \Rightarrow x=\frac{h \sin (\beta+\alpha)}{\sin (\beta-\alpha)}$.

Ex. 19 . A tower $P Q$ of height $h$ subtends an angle of $45^{\circ}$ at a point $A$ on the horizontal plane. At another point $B$ on $A B$ inclined to the horizontal at an angle $30^{\circ}$, the elevation of the top of the tower is found to be $60^{\circ}$. If $A B=a$, then show that $a=h(\sqrt{3}-1)$.

Sol. Given, $P Q=h$ metres, $A B=a$ metres, $\angle P A Q=45^{\circ}, \angle B A Q=30^{\circ}, \angle P B M=60^{\circ}$

In rt. $\triangle A P Q, \frac{P Q}{A Q}=\tan 45^{\circ}$

$\Rightarrow \frac{P Q}{A Q}=1 \Rightarrow A Q=P Q=h$

$\Rightarrow \triangle A P Q$ is an isosceles rt. angled $\triangle$

$\Rightarrow \angle A P Q=\angle P A Q=45^{\circ}$

$\therefore$ In rt. $\triangle P B M, \angle P M B=90^{\circ}$ and $\angle P B M=60^{\circ} \Rightarrow \angle B P M=30^{\circ} \Rightarrow \angle A P B=45^{\circ}-30^{\circ}=15^{\circ}$.

Also $\angle P A B=15^{\circ}$.

$\therefore$ In $\triangle A B P, \angle A B P=180^{\circ}-(\angle A P B+\angle P A B)=180^{\circ}-(15^{\circ}+15^{\circ})=180^{\circ}-30^{\circ}=150^{\circ}$

Also, in $\triangle A P Q, A P=\sqrt{A Q^{2}+P Q^{2}}=\sqrt{h^{2}+h^{2}}=h \sqrt{2}$

$\therefore$ By sine rule on $\triangle A B P, \frac{A P}{\sin 150^{\circ}}=\frac{A B}{\sin 15^{\circ}} \Rightarrow \frac{h \sqrt{2}}{\frac{1}{2}}=\frac{a \sqrt{2}}{\frac{\sqrt{3}-1}{2 \sqrt{2}}} \quad(\begin{matrix} \text{ Calculate } \sin 15^{\circ}=\sin (60^{\circ}-45^{\circ}) \\ =\sin 60^{\circ} \cos 45^{\circ}-\sin 45^{\circ} \cos 60^{\circ}\end{matrix} )$

$\Rightarrow h \sqrt{2}=\frac{a \sqrt{2}}{\sqrt{3}-1} \quad \Rightarrow \boldsymbol{a}=\boldsymbol{h}(\sqrt{\mathbf{3}}-\mathbf{1})$

Ex. 20 . The angle of elevation of the top of a tower observed from each of the three point $A, B$ and $C$ on the ground forming a triangle is the same angle $\alpha_0$ If $R$ is the circumradius of triangle $A B C$, then show that the height of the tower is $R \tan \alpha$.

Sol. Let the triangle formed on the ground be $A B C, P Q$ be the tower with its base $Q$ on the ground. $P Q$ is perpendicular to the horizontal plane on which the three points $A, B, C$ lie. $\therefore$ Angle of elevation of point $P$ (top of tower) from each of the points $A, B, C$ is the same, let $\angle P A Q=\angle P B Q=\angle P C Q=\alpha$.

Also, $\angle P Q A=\angle P Q C=\angle P Q B=90^{\circ}$

Let $P Q=h$ metres

(Note: Students are advised to visualise the given diagram in 3- $D$ for better understanding.)

$\therefore$ In rt $\angle d \triangle A P Q, \tan \alpha=\frac{P Q}{A Q} \Rightarrow \tan \alpha=\frac{h}{A Q} \Rightarrow A Q=\frac{h}{\tan \alpha}$

Similarly from rt. $\angle d \triangle s B P Q$ and $C P Q$, we have $B Q=\frac{h}{\tan \alpha}, C Q=\frac{h}{\tan \alpha}$

$\therefore A Q=B Q=C Q \Rightarrow Q$ is the circumcentre of $\triangle A B C$

$\because R$ is the circumradius of $\triangle A B C, \therefore A Q=B Q=C Q=R$

$\therefore$ From (i) $R=\frac{h}{\tan \alpha} \Rightarrow \boldsymbol{h}=\boldsymbol{R} \tan \boldsymbol{\alpha} \Rightarrow$ Height of the tower $=\boldsymbol{R} \tan \boldsymbol{\alpha}$.

PRACTICE SHEET

~~ 1. A ladder rests against a wall so that its top touches the roof of the house. If the ladder makes an angle of $60^{\circ}$ with the horizontal and the height of the house be $6 \sqrt{3} m$, then the length of the ladder is

(a) $12 \sqrt{3} m$

(b) $12 m$

(d) None of these

(BITSAT 2011)

~~ 2. From the top of a tower, the angle of depression of a point on the ground is $60^{\circ}$. If the distance of this point from the tower is $\frac{1}{\sqrt{3}+1}$ metres, then the height of the tower is

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

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

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

(Kerala PET 2009)

~~ 3. From the top of a light house, the angle of depression of a boat is $15^{\circ}$. If the light house is $60 m$ high and its base is at sea level, the distance of the boat from the light house is

(a) $(\frac{\sqrt{3}-1}{\sqrt{3}+1}) 60 m$

(b) $(\frac{\sqrt{3}+1}{\sqrt{3}-1}) 60 m$

(d) $(\frac{\sqrt{3}+1}{\sqrt{3}-1}) 60 m$

(IIT)

~~ 4. A tree is broken by wind, its upper part touches the ground at a point 10 metres from the foot of the tree and makes an angle of $60^{\circ}$ with the ground. The entire length of the tree is

(a) $15 m$

(b) $(10+\frac{\sqrt{3}}{2}) m$

(d) $\sqrt{3}(10+\frac{20}{\sqrt{3}}) m$

(Rajasthan PET 2006)

~~ 5. When the length of the shadow of the pole is equal to the height of the pole, then the elevation of the source of light is

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $75^{\circ}($ BVP 2006) ~~ 6. $20 m$ high flagpole is fixed on a $80 m$ high pillar. $50 m$ away from the base of the pillar, the flag pole along with the pillar make an angle $\alpha$, then the value of $\tan \alpha$ is

(a) $2 / 11$

(b) 2

(d) $2^{1 / 4}$

(MPPET 2003)

~~ 7. $A B$ is a vertical pole with $B$ at the ground level and $A$ at the top. A man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $60^{\circ}$. He moves away from the pole along the line $B C$ to a point $D$ such that $C D=7 m$. From $D$, the angle of elevation of point $A$ is $45^{\circ}$. Then, the height of the pole is

(a) $(\frac{7 \sqrt{3}}{2}-\frac{1}{\sqrt{3}+1}) m$

(b) $(\frac{7 \sqrt{3}}{2}-\frac{1}{\sqrt{3}-1}) m$

(d) $\frac{7 \sqrt{3}}{2}(\sqrt{3}-1) m$

(AIEEE 2008)

~~ 8. The shadow of a tower standing on a level ground is $x$ metres long when the sun’s altitude is $30^{\circ}$, while it is $y$ metres long when the altitude is $60^{\circ}$. If the height of the tower is 45. $\frac{\sqrt{3}}{2} m$, then $x-y$ is

(a) $45 m$

(b) $45 \sqrt{3} m$

(d) $45 \cdot \frac{\sqrt{3}}{2} m$

(MPPET 2008)

~~ 9. The angle of elevation of an object from a point on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object, the angle of elevation is found to be $\beta$. Then the height of the object in metres is

(a) $d \tan \alpha$

(b) $d \cot \beta$

(d) $\frac{d}{\cot \alpha-\cot \beta}$

(EAMCET 2007)

~~ 10 A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of $30^{\circ}$. After sometime, the angle of depression becomes $60^{\circ}$. The distance (in metres) travelled by the car during this time is

(a) $200 \sqrt{3}$

(b) $\frac{100}{\sqrt{3}}$

(d) $100 \sqrt{3}$

(IIT 2001)

~~ 11 $P T$ is a tower of height $2^{x}$ metres, $P$ being the foot, $T$ being the top of the tower. $A$ and $B$ are points on the same line with $P$. If $A P=2^{x+1} m$ and $B P=192 m$ and if the angle of elevation of the tower as seen from $B$ is double the angle of elevation of the tower as seen from $A$, then what is the value of $x$ ?

(a) 6

(b) 7

(d) 9

(NDA/NA 2008)

~~ 12 The angles of elevation of the top of a tower from two points situated at distance $36 m$ and $64 m$ from its base and in the same straight line with it are complementary. What is the height of the tower?

(a) $50 m$

(b) $48 m$

(d) $24 m$

(CDS 2011)

~~ 13 A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of $45^{\circ}$ with the man’s eye when at a distance of $60 m$ from the bottom of the tower. After $5 s$, the angle of depression becomes $30^{\circ}$. What is the approximate speed of the boat assuming that it is running in still water?

(a) $31.5 km / h$

(b) $36.5 km / h$

(d) $40.5 km / h$

(CDS 2010)

~~ 14 A tower subtends angles $\alpha, 2 \alpha, 3 \alpha$ respectively at points $A$, $B$ and $C$ all lying on a horizontal line through the foot of the tower. Then, $\frac{A B}{B C}$ is equal to

(a) $1+2 \cos 2 \alpha$

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

(d) $\frac{\sin 2 \alpha}{\alpha}$

(EAMCET 2003)

~~ 15 From the top of a cliff $50 m$ high, the angles of depression of the top and bottom of a tower are observed to be $30^{\circ}$ and $45^{\circ}$. The height of the tower is

(a) $50 m$

(b) $50 \sqrt{3} m$

(d) $50(1-\frac{\sqrt{3}}{3}) m$

(UPSEE 2009)

~~ 16 A flagpole stands on a building and an observer on a level ground is $300 ft$ from the base of the building. The angle of elevation of the bottom of the flagpole is $30^{\circ}$ and the height of the flagpole is $50 ft$. If $\theta$ is the angle of elevation of the top of the flagpole, then $\tan \theta$ is equal to

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

(b) 1

(d) $\frac{2 \sqrt{3}+1}{6}$

(Kerala CEE 2008) ~~ 17 A flag is standing vertically on a tower of height $b$. On a point at a distance $a$ from the foot of the tower, the flag and the tower subtend equal angles. The height of the flag is

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

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

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

(UPSEE 2006)

~~ 18 $A B$ is a vertical pole. The end $A$ is on the level ground. $C$ is the middle point of $A B . P$ is a point on the level ground. The portion $B C$ subtends an angle $\beta$ at $P$. If $A P=n A B$, then $\tan \beta$ is equal to

(a) $\frac{n}{n^{2}-2}$

(b) $\frac{n}{n^{2}-1}$

(d) $\frac{n}{2 n^{2}+1}$

(AMU 2001)

~~ 19 The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance $100 m$ from its base is $45^{\circ}$. If the angle of elevation of the top of the complete pillar at the same point is to be $60^{\circ}$, then the height of the incomplete pillar is increased by

(a) $50 \sqrt{2} m$

(b) $100 m$

(d) $100(\sqrt{3}+1)$

( $A M U$ 2006)

~~ 20 What should be the height of a flag where a $20 ft$ long ladder reaches $20 ft$ below the flag. The angle of elevation of the top of the flag at the foot of the ladder is $60^{\circ}$.

(a) $20 ft$

(b) $30 ft$

(d) $20 \sqrt{2} ft$

(NDA/NA 2007)

~~ 21 From a light house, the angles of depression of two ships on opposite sides of the lighthouse are observed to be $30^{\circ}$ and $45^{\circ}$. If the height of the lighthouse is $h$. What is the distance between the ships?

(a) $(\sqrt{3}+1) h$

(b) $(\sqrt{3}-1) h$

(d) $(1+\frac{1}{\sqrt{3}}) h$

(CDS 2011)

~~ 22 Suppose the angle of elevation of the top of a tree at a point $E$ due east of the tree is $60^{\circ}$ and that at a point $F$ due west of the tree is $30^{\circ}$. If the distance between the points $E$ and $F$ is $160 ft$, then what is the height of the tree?

(a) $40 \sqrt{3} ft$

(b) $60 ft$

(d) $23 ft$

(CDS 2010)

~~ 23 From the top of a lighthouse $A B$, the angles of depression of two stations $C$ and $D$ on opposite sides at a distance $d$ apart are $\alpha$ and $\beta$. The height of the lighthouse is

(a) $\frac{d}{\cot \alpha \cot \beta}$

(b) $\frac{d}{\cot \alpha-\cot \beta}$

(d) $\frac{d \cot \alpha \cot \beta}{\cot \alpha+\cot \beta}$

(KCET)

~~ 24 The length of the shadows of a vertical pole of height $h$, thrown by the sun’s rays at three different moments are $h$, $2 h$ and $3 h$. The sum of the angles of elevation of the rays at these three moments is equal to:

(a) $\pi / 4$

(b) $\pi / 6$

(d) $\pi / 3$

(MPPET 2000)

~~ 25 An aeroplane flying with uniform speed horizontally 2 kilometer above the ground is observed at an elevation of $60^{\circ}$. After $10 s$, if the elevation from the same point is observed to be $45^{\circ}$, then the distance travelled by the aeroplane is

(a) $\frac{2(\sqrt{3}-1)}{\sqrt{3}} km$

(b) $2(\sqrt{3}+1) km$

(d) None of these

(MPPET 2013)

~~ 26 The angle of elevation of the top of a tower observed from each of 3 points $A, B, C$ on the ground are $\alpha, 2 \alpha$ and $3 \alpha$ respectively. If $A B=a, B C=b$, then the height of the tower is

(a) $\frac{a}{2 b} \sqrt{(3 b-a)(a+b)}$

(b) $a / b \sqrt{(3 b+a)(a+b)}$

(d) None of these

(EAMCET)

~~ 27 The angle of elevation of the top of a tower from the top and bottom of a building of height $a$ are $30^{\circ}$ and $45^{\circ}$ respectively. If the tower and the building stand at the same level, then the height of the tower is

(a) $a \sqrt{3}+1$

(b) $\frac{a \sqrt{3}}{\sqrt{3}+1}$

(d) $\frac{a+\sqrt{3}}{2(\sqrt{3}-1)}$

(KCET 2000)

~~ 28 The angle of elevation of the top of a tower from the bottom of a building is twice that from its top. What is the height of the building, if the height of the tower is $75 m$ and the angle of elevation of the top of the tower from the bottom of the building is $60^{\circ}$ ?

(a) $25 m$

(b) $37.5 m$

(d) $60 m$

(CDS 2010)

~~ 29 From the top of a hill $h$ metres high, the angles of depressions of the top and the bottom of a pillar are $\alpha$ and $\beta$ respectively. The height (in metres) of the pillar is

(a) $\frac{h(\tan \beta-\tan \alpha)}{\tan \beta}$

(b) $\frac{h(\tan \alpha-\tan \beta)}{\tan \alpha}$

(d) $\frac{h(\tan \beta+\tan \alpha)}{\tan \alpha}$

(BITSAT 2008)

~~ 30 A vertical pole subtends an angle $\tan ^{-1}(\frac{1}{2})$ at a point $P$ on the ground. If the angles subtended by the upper half and lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(\tan \alpha, \tan \beta)$ is equal to

(a) $(\frac{1}{4}, \frac{1}{5})$

(b) $(\frac{1}{5}, \frac{2}{9})$

(d) $(\frac{1}{4}, \frac{2}{9})$

(EAMCET 2012) ~~ 31 The angle of elevation of the cloud at a point $2500 m$ high from a lake is $15^{\circ}$ and from the same point the angle of depression of its reflection in the lake is $45^{\circ}$. The height (in metres) of the cloud above the lake (given $\cot 15^{\circ}=$ $2+\sqrt{3}$ ) is

(a) 2500

(b) $2500 \sqrt{2}$

(d) 5000

(MPPET 2003, EAMCET 2011)

~~ 32 An aeroplane flying at a height of $300 m$ above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. What is the height of the lower plane from the ground?

(a) $50 m$

(b) $\frac{100}{\sqrt{3}} m$

(d) $150(\sqrt{3}+1) m($ NDA/NA 2011)

~~ 33 The angle of depression of the top and the foot of the chimney as seen from the top of a second chimney which is $150 m$ high and standing on the same level as the first are $\theta$ and $\phi$ respectively. The distance between their tops when $\tan \theta=\frac{4}{3}$ and $\tan \phi=5 / 2$ is equal to

(a) $50 m$

(b) $100 m$

(d) $\frac{200}{\sqrt{2}} m$

(AMU 2007)

~~ 34 The angle of elevation of the top of the minar from the foot of the tower of height $h$ is $\alpha$ and the tower subtends and angle $\beta$ at the top of the minar. Then, the height of the minar is

(a) $\frac{h \cot (\alpha-\beta)}{\cot (\alpha-\beta)-\cot \alpha}$

(b) $\frac{h \tan (\alpha-\beta)}{\tan (\alpha-\beta)-\tan \alpha}$

(d) None of the above

(UPSEE 2010)

~~ 35 The angle of elevation of the top two vertical towers as seen from the middle point joining the feet of the towers are $60^{\circ}$ and $30^{\circ}$ respectively. The ratio of the heights of the towers is

(a) $2: 1$

(b) $\sqrt{3}: 1$

(d) $3: 1$

(EAMCET, MLNR, AMU)

~~ 36 Two pillars of equal height stand on either side of a road way which is $60 m$ wide. At a point in the roadway between the pillars, the elevation of the top of the pillars are $60^{\circ}$ and $30^{\circ}$. The height of the pillar is

(a) $\frac{15}{\sqrt{3}} m$

(b) $15 m$

(d) $20 m$

(UPSEE 2004)

~~ 37 $P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^{\circ}$ each. Then, the square of the distance between the tops of the poles is

(a) $\frac{a^{2}+b^{2}}{2}$

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

(d) $4(a^{2}+b^{2})$

(EAMCET 2009)

~~ 38 The base of a cliff is circular. From the extremities of a diameter of the base, angles of elevation of the top of the cliff are $30^{\circ}$ and $60^{\circ}$. If the height of the cliff be $500 m$, then the diameter of the base of the cliff is

(a) $2000 / \sqrt{3} m$

(b) $1000 / \sqrt{3} m$

(d) $1000 \sqrt{3} m$

(Manipal 2008)

~~ 39 $A B C D$ is a rectangle field. $A$ vertical lamp post of height $12 m$ stands at the corner $A$. If the angle of elevation of its top from $B$ is $60^{\circ}$ and from $C$ is $45^{\circ}$, then the area of the field is

(a) $48 \sqrt{2} sq m$

(b) $48 \sqrt{3} sq m$

(d) $12 \sqrt{3} sq m$

(KeralaCEE2005)

~~ 40 The upper $(\frac{3}{4})$ th portion of a vertical pole subtends an angle $\tan ^{-1}(\frac{3}{5})$ at a point in the horizontal plane through its foot and at a distance $40 m$ from the foot. A possible height of the vertical pole is

(a) $20 m$

(b) $40 m$

(d) $80 m$

(AIEEE 2003)

~~ 41 A spherical balloon of radius $r$ subtends an angle $\alpha$ at the eye of an observer, when the angular elevation of its centre is $\beta$. The height of the centre of the balloon is

(a) $r \sin \alpha / 2 cosec \beta$

(b) $r cosec \alpha \sin \beta / 2$

(d) $r \sin \alpha cosec \beta / 2$ ~~ 42 A ladder rests against a wall at an angle $\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance ’ $a$ ‘, so that it slides a distance ’ $b$ ’ down the wall, making an angle $\beta$ with the horizontal. The value of $a / b$ is

(a) $\tan (\frac{\alpha+\beta}{\alpha-\beta})$

(b) $\tan (\frac{\alpha-\beta}{\alpha+\beta})$

(d) $\tan (\frac{\alpha+\beta}{2})$

(IIT 1985)

~~ 43 Two towers stands on a horizontal plane. $P$ and $Q$, where $P Q=30 m$ are two points on the line joining their feet. As seen from $P$, the angle of elevation of the tops of the towers are $30^{\circ}$ and $60^{\circ}$ but as seen from $Q$ are $60^{\circ}$ and $45^{\circ}$. The distance between the towers is equal to

(a) $15(4+\sqrt{3})$

(b) $15(4-\sqrt{3})$

(d) $15(2+\sqrt{3})$

(UPSEAT 2000)

~~ 44 A pole stands vertically inside a triangle park $A B C$. If the angle of elevation of the top of the pole from each corner of the park is the same, then in triangle $A B C$, the foot of the pole is at the

(a) centroid

(b) in centre

(d) orthocentre

(UPSEAT 2008, IIT 2000)

~~ 45 $A B C D$ is a square plot. The angle of elevation of the top of the pole standing at $D$ from $A$ or $C$ is $30^{\circ}$ and that from $B$ is $\theta$. Then $\tan \theta$ is equal to

(a) $\sqrt{6}$

(b) $1 / \sqrt{6}$

(d) $\sqrt{2} / 3$

(Karnataka CET 2013)

ANSWERS

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

HINTS AND SOLUTIONS

~~ 1. Let $A B$ be the height of the house. Given $A B=6 \sqrt{3} m$ Let $A C$ be the ladder whose length $A C=x$ metres (say) Given $\angle A C B=60^{\circ}$

$\therefore$ In rt. $\triangle A B C, \frac{A B}{A C}=\sin 60^{\circ}$ $\Rightarrow \frac{6 \sqrt{3}}{x}=\frac{\sqrt{3}}{2} \Rightarrow x=12 m$.

~~ 2. Let $P Q$ be the tower and $R$ be the point on the ground which is at a distance $\frac{1}{\sqrt{3}+1}$ metres from the base of the tower, i.e., $Q R=\frac{1}{\sqrt{3}+1}$ metres.

Let the height of the tower $P Q=h$ metres.

Given, angle of depression $\angle A P R=60^{\circ}$,

$\therefore \angle P R Q=\angle A P R=60^{\circ} \quad(A P | Q R$, alt $\angle s)$

$\therefore$ In rt. $\triangle P Q R, \frac{P Q}{Q R}=\tan 60^{\circ}$

$\Rightarrow \frac{h}{\frac{1}{\sqrt{3}+1}}=\sqrt{3} \Rightarrow h(\sqrt{3}+1)=\sqrt{3}$

$\Rightarrow h=\frac{\sqrt{3}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\mathbf{3}-\sqrt{\mathbf{3}}}{\mathbf{2}} \mathbf{m}$.

~~ 3. Let $P Q$ be the lighthouse and $R$ the position of the boat.

Given, $P Q=60 m$

Angle of depression of boat from point $P=15^{\circ}$

$\therefore \angle A P R=15^{\circ} \Rightarrow \angle P R Q=15^{\circ}$

Distance of boat from the lighthouse $=R Q$

In rt. $\triangle P R Q, \frac{P Q}{R Q}=\tan 15^{\circ}$

$\Rightarrow \frac{60}{R Q}=\tan 15^{\circ} \Rightarrow \frac{60}{R Q}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

$[\tan 15^{\circ}=\tan (45^{\circ}-30^{\circ})=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \cdot \tan 30^{\circ}}.$

$=\frac{1-1 / \sqrt{3}}{1+1.1 / \sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

$\Rightarrow R Q=(\frac{\sqrt{3}+1}{\sqrt{3}-1}) .60$ metres.

~~ 4. Let $P Q$ be the total length of the tree. Let $A$ be the point at which the tree breaks and $B$ be the point where the top of the tree touches the ground. $\therefore A P=A B$

Given, $\angle A B Q=60^{\circ}, B Q=10 m$.

In rt. $\triangle A Q B, \frac{A Q}{B Q}=\tan 60^{\circ}$

$A Q=B Q \tan 60^{\circ}=10 \sqrt{3} m$

Also, $\frac{B Q}{A B}=\cos 60^{\circ}$

$\Rightarrow A B=B Q \sec 60^{\circ}=(10 \times 2) m=20 m$

$\therefore P Q=A Q+A P=A Q+A B=(10 \sqrt{3}+20) m(\therefore A P=A B)$

$=\sqrt{\mathbf{3}}(\mathbf{1 0}+\frac{\mathbf{2 0}}{\sqrt{\mathbf{3}}}) \mathbf{m}$.

~~ 5. Let $P Q$ be the pole, $Q R$ be its shadow. Let $L$ be the source of light and $\theta$ be the angle of elevation of the source of light, such that

$P Q=Q R \quad \therefore \angle P R Q=\theta$

$\therefore$ In rt. $\triangle P Q R$

$\frac{P Q}{Q R}=\tan \theta \Rightarrow \tan \theta=1$

$\Rightarrow \boldsymbol{\theta}=\mathbf{4 5}^{\circ}$.

$\therefore$ Required angle of elevation $=\mathbf{4 5}^{\circ}$.

~~ 6. In the given diagram,

$A C$ is the flag pole of height $20 m$

$C B$ is the pillar of height $80 m$

$\angle A D B=\alpha, D B=50$

$\therefore \tan \alpha=\frac{A B}{D B}=\frac{100}{50}=\mathbf{2}$.

~~ 7. Le the height of the pole $A B=h$ metres.

Given, $\angle A C B=60^{\circ}$,

$\angle A D B=45^{\circ}$ and $D C=7 m$. In $\triangle A B C, \frac{B C}{h}=\cot 60^{\circ}$

$\Rightarrow B C=h \cot 60^{\circ}$

In $\triangle A B D, \frac{D B}{h}=\cot 45^{\circ}$

$\Rightarrow D B=h \cot 45^{\circ}$

$\therefore D B-B C=C D$

$\Rightarrow h \cot 45^{\circ}-h \cot 60^{\circ}=7$

$ \begin{aligned} \Rightarrow h & =\frac{7}{\cot 45^{\circ}-\cot 60^{\circ}}=\frac{7}{1-\frac{1}{\sqrt{3}}}=\frac{7 \sqrt{3}}{\sqrt{3}-1} \\ & =\frac{7 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{\mathbf{7} \sqrt{\mathbf{3}}(\sqrt{\mathbf{3}}+\mathbf{1})}{\mathbf{2}} \mathbf{m} . \end{aligned} $

~~ 8. Let the given tower be $A C$

$ A C=\frac{45 \sqrt{3}}{2} m $

When angle of elevation is $30^{\circ}$, length of shadow is $x m$.

$ \Rightarrow \angle A B C=30^{\circ}, B C=x $

When angle of elevation is $60^{\circ}$, length of shadow is $y m$.

$\Rightarrow \angle A D C=60^{\circ}, D C=y$

In rt. $\triangle A D C, \frac{A C}{D C}=\tan 60^{\circ}$

$ \Rightarrow \frac{\frac{45 \sqrt{3}}{2}}{y}=\sqrt{3} \Rightarrow y=\frac{45}{2} $

In rt. $\triangle A B C, \frac{A C}{B C}=\tan 30^{\circ}$

$ \Rightarrow \frac{\frac{45 \sqrt{3}}{2}}{x}=\frac{1}{\sqrt{3}} \Rightarrow x=\frac{135}{2} $

$\therefore x-y=\frac{135}{2}-\frac{45}{2}=\frac{90}{2}=\mathbf{4 5} m . \quad$ (From (i) and (ii))

~~ 9. Let the height of the object $A B$ be $h$ metres.

Given, $\angle A C B=\alpha$,

$\angle A D B=\beta, C D=d$ metres

Let $D B=x$ metres.

Then, in rt. $\triangle A B D$,

$\frac{h}{x}=\tan \beta \Rightarrow x=h \cot \beta$

In rt. $\triangle A C B$,

$ \begin{aligned} \frac{h}{x+d}=\tan \alpha & \Rightarrow x+d=h \cot \alpha \\ \therefore(i i)-(i) \quad & \Rightarrow d=h \cot \alpha-h \cot \beta \\ & \Rightarrow h=\frac{\boldsymbol{d}}{\cot \alpha-\cot \beta} . \end{aligned} $

~~ 10 Let $P Q$ be the given tower of height 100 metres.

Let $P$ be the position of the man and $R$ and $S$ the positions of the car moving towards the tower. Given angle of depression at points $R$ and $S$ are $30^{\circ}$

and $60^{\circ}$ respectively.

Distance travelled by the car $=R S=Q R-Q S$.

In rt. $\triangle P Q R, \tan 30^{\circ}=\frac{P Q}{Q R}=\frac{100}{Q R}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{Q R} \Rightarrow Q R=100 \sqrt{3} m$

In rt. $\triangle P S Q, \tan 60^{\circ}=\frac{P Q}{Q S}=\frac{100}{Q S}$

$\Rightarrow \sqrt{3}=\frac{100}{Q S} \Rightarrow Q S=\frac{100}{\sqrt{3}} m$

$\therefore$ From (i), (ii) and (iii)

$ \begin{aligned} R S=100 \sqrt{3}-\frac{100}{\sqrt{3}} & =100 \sqrt{3}-\frac{100 \sqrt{3}}{3} \\ & =100 \sqrt{3}(1-\frac{1}{3})=\frac{200 \sqrt{3}}{3}=\frac{\mathbf{2 0 0}}{\sqrt{\mathbf{3}}} \mathbf{~ m .} \end{aligned} $

~~ 11 Given, $P T=2^{x} m$,

$A P=2^{x+1} m, B P=192 m$,

$\angle T A P=\theta, \angle T B P=2 \theta$

In rt. $\triangle T A P, \tan \theta=\frac{P T}{A P}$

$\Rightarrow \tan \theta=\frac{2^{x}}{2^{x+1}}=\frac{1}{2}$

In rt. $\triangle T B P, \tan 2 \theta=\frac{P T}{B P}=\frac{2^{x}}{192}$

$\Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{2^{x}}{192} \Rightarrow \frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}=\frac{2^{x}}{192}$

(from (i))

$\Rightarrow \frac{4}{3} \times 192=2^{x} \Rightarrow 2^{x}=256 \Rightarrow 2^{x}=2^{8} \Rightarrow \boldsymbol{x}=\mathbf{8}$.

~~ 12 Let the height of the tower $A B=h$ metres,

Given, $B D=36 m$ and $B C=64 m$

Let $\angle A D B=\alpha, \angle A C B=\beta$

In rt. $\triangle A B C$,

$\tan \beta=\frac{h}{64}$

In rt. $\triangle A B D$,

$\tan \alpha=\frac{h}{36}$

Also, given $\alpha+\beta=\pi / 2$

$ \begin{aligned} & \Rightarrow \tan (\alpha+\beta)=\tan \pi / 2 \\ & \Rightarrow \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{1}{0} \\ & \Rightarrow \tan \alpha \tan \beta=1 \Rightarrow \frac{h}{36} \times \frac{h}{64}=1 \\ & \Rightarrow h^{2}=36 \times 64 \Rightarrow h=6 \times 8 \Rightarrow h=\mathbf{4 8} \mathbf{~ m .} \end{aligned} $

~~ 13 Let the man be standing at the top of the tower $A B$ of height $h$ at $A$.

Let the original position of the boat be $C$, then, $\angle A C B=45^{\circ}$ and $B C=60 m$.

Let the position of the boat after 5 seconds be $D$ where $\angle A D B=30^{\circ}$ and $C D=x m$ (say)

In rt. $\triangle A B C, \tan 45^{\circ}=\frac{A B}{B C}=\frac{h}{60}$

$\Rightarrow \frac{h}{60}=1=\boldsymbol{h}=\mathbf{6 0}$.

In rt. $\triangle A D B, \tan 30^{\circ}=\frac{A B}{x+60}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60}{60+x} \Rightarrow x+60=60 \sqrt{3} \quad(\therefore h=60 m)$

$\Rightarrow x=60(\sqrt{3}-1)$

$\therefore x=60 \times 0.732=43.8 m$ (approx)

$\therefore$ Speed of boat $=\frac{\text{ Distance travelled }}{\text{ Time }} \times \frac{18}{5} km / hr$

$ =(\frac{43.8}{5} \times \frac{18}{5}) km / hr=\mathbf{3 1 . 5} \mathbf{~ k m} / \mathbf{h r} . $

~~ 14 Let the height of the tower $P Q=h$ metres.

Given, $\angle P A Q=\alpha, \angle P B Q=2 \alpha, \angle P C Q=3 \alpha$.

$\therefore$ In rt. $\triangle P A Q, \tan \alpha=\frac{P Q}{A Q}=\frac{h}{A Q}$

$\Rightarrow A Q=h \cot \alpha$

In rt. $\triangle P B Q, \tan 2 \alpha=\frac{P Q}{B Q}=\frac{h}{B Q}$

$\Rightarrow B Q=h \cot 2 \alpha$

In rt. $\triangle P C Q, \tan 3 \alpha=\frac{P Q}{C Q}=\frac{h}{C Q}$

$\Rightarrow C Q=h \cot 3 \alpha$

$\therefore$ From (i) and (ii), $A B=A Q-B Q$

$ =h \cot \alpha-h \cot 2 \alpha $

From (iv) and (iii), $B C=B Q-C Q$

$ =h \cot 2 \alpha-h \cot 3 \alpha $

$\therefore$ From (iv) and (v), we have

$ \begin{aligned} \frac{A B}{B C}= & \frac{h(\cot \alpha-\cot 2 \alpha)}{h(\cot 2 \alpha-\cot 3 \alpha)} \\ = & \frac{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha}-\frac{\cos 3 \alpha}{\sin 3 \alpha}} \end{aligned} $

$\cos \alpha \sin 2 \alpha-\cos 2 \alpha \sin \alpha$

$ =\frac{\sin \alpha \sin 2 \alpha}{\frac{\cos 2 \alpha \sin 3 \alpha-\cos 3 \alpha \sin 2 \alpha}{\sin 2 \alpha \sin 3 \alpha}} $

$ \begin{aligned} & =\frac{\frac{\sin (2 \alpha-\alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin (3 \alpha-2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}}=\frac{\frac{\sin \alpha}{\sin \alpha \sin 2 \alpha}}{\frac{\sin \alpha}{\sin 2 \alpha \sin 3 \alpha}}=\frac{\sin 3 \alpha}{\sin \alpha} \\ & =\frac{3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha \\ & =3-2(1-\cos 2 \alpha)=1+2 \cos 2 \alpha . \end{aligned} $

~~ 15 Let the height of the cliff $B D=50 m$ and the height of the tower $A E=h$ metres.

Given, $\angle D E C=30^{\circ}, \angle D A B=45^{\circ}$

Let $B A=C E=x$ metres

In rt. $\triangle D E C$,

$\tan 30^{\circ}=\frac{D C}{C E}=\frac{50-h}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50-h}{x} \Rightarrow x=(50-h) \sqrt{3}$

In rt. $\triangle B A D$,

$\tan 45^{\circ}=\frac{D B}{B A} \Rightarrow 1=\frac{50}{x} \Rightarrow x=50$

$\therefore$ From $(i)$ and $(i i)$

$ \begin{aligned} 50 & =(50-h) \sqrt{3} \Rightarrow 50-50 \sqrt{3}=-h \sqrt{3} \\ \Rightarrow & h=\frac{50(\sqrt{3}-1)}{\sqrt{3}}=\mathbf{5 0}(\mathbf{1}-\frac{\sqrt{\mathbf{3}}}{\mathbf{3}}) \mathbf{m} . \end{aligned} $

~~ 16 Let $A B$ be the building of height $h$ metres and $B C$ the flagpole of height $50 ft$.

Given, $D$ is the position of the observer $300 ft$ away from the base of the building, i.e., $A D=300 ft$.

Also $\angle B D A=30^{\circ}$ and $\angle C D A=\theta$

$\therefore$ In rt. $\triangle A B D, \tan 30^{\circ}=\frac{A B}{A D}$

$\Rightarrow \frac{h}{300}=\frac{1}{\sqrt{3}} \Rightarrow h=\frac{300}{\sqrt{3}}=100 \sqrt{3} ft$.

In rt. $\triangle A C D, \tan \theta=\frac{A C}{A D}$

$\Rightarrow \frac{50+h}{300}=\frac{50+100 \sqrt{3}}{300}=\frac{\mathbf{2} \sqrt{\mathbf{3}}+\mathbf{1}}{\mathbf{6}} \mathbf{~ m .}$ ~~ 17 Type Solved Example 16 .

~~ 18 Let the height of the vertical pole $A B=h$ metres.

Then, $A C=B C=h / 2, A P=n . A B=n h$

Let $\angle A P C=\alpha$. Given $\angle B P C=\beta$

$\therefore \angle A P B=\alpha+\beta$

In rt. $\triangle A C P$,

$\tan \alpha=\frac{A C}{A P}=\frac{h / 2}{n h}=\frac{1}{2 h}$

In rt. $\triangle B A P, \tan (\alpha+\beta)=\frac{A B}{A P}=\frac{h}{n h}=\frac{1}{n}$

Now $\tan \beta=\tan {(\alpha+\beta)-\alpha}$

$=\frac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \tan \alpha}$

$=\frac{\frac{1}{n}-\frac{1}{2 n}}{1+\frac{1}{n} \cdot \frac{1}{2 n}}=\frac{\boldsymbol{n}}{\mathbf{2 n}^{2}+\mathbf{1}}$.

(Using (i) and (ii))

~~ 19 Let $B C$ be the incomplete pillar and $B D$ be the complete pillar. In $\triangle A B C$,

$ \begin{aligned} & \tan 45^{\circ}=\frac{B C}{A B} \\ \Rightarrow & \frac{B C}{100}=1 \Rightarrow B C=100 \end{aligned} $

In $\triangle A B D$,

$ \begin{aligned} \tan 60^{\circ} & =\frac{B D}{A B} \Rightarrow \sqrt{3}=\frac{B C+x}{100} \Rightarrow x=100 \sqrt{3}-B C \\ & =100 \sqrt{3}-100=\mathbf{1 0 0}(\sqrt{\mathbf{3}}-\mathbf{1}) \mathbf{m} . \end{aligned} $

$(\because B C=100 m)$

~~ 20 Let $B D$ be the height of the flag. Given, $D C=20 ft$. and length of ladder $A C=20 ft$.

$ \angle D A B=60^{\circ} \text{. } $

Let $\quad B C=\boldsymbol{h}$ metres

In rt. $\triangle A B D, \tan 60^{\circ}=\frac{B D}{A B}$

$ \begin{matrix} \Rightarrow & \sqrt{3}=\frac{h+20}{A B} \\ \Rightarrow & A B=\frac{h+20}{\sqrt{3}}=\frac{(h+20) \sqrt{3}}{3} \end{matrix} $

In rt. $\triangle A B C, A C^{2}=A B^{2}+B C^{2}$

$ \Rightarrow 20^{2}=\frac{3(h+20)^{2}}{9}+h^{2} $

(From $(i))$

$\Rightarrow 400=\frac{(h+20)^{2}+3 h^{2}}{3} \Rightarrow 1200=h^{2}+40 h+400+3 h^{2}$

$\Rightarrow 4 h^{2}+40 h-800=0 \Rightarrow h^{2}+10 h-200=0$

$ (h+20)(h-10)=0 \Rightarrow h=10 \quad(\because h \neq-20) $

$\therefore \quad$ Height of the flag $=B C+C D=10+20=\mathbf{3 0} \mathbf{~ f t}$.

~~ 21 Let $P Q$ be the light house whose height $=\boldsymbol{h}$ metre. Let $A$ and $B$ be the position of the ships on opposite sides of the lighthouse such that angle of depression for $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$ respectively. Let

$A Q=\boldsymbol{x}$ metre, $Q B=\boldsymbol{y}$ metres. $\angle P A Q=30^{\circ}, \angle P B Q=45^{\circ}$. Required distance between the ships $=A B$

$ =A Q+Q B=x+y $

In rt. $\triangle P A Q$,

$\tan 30^{\circ}=\frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \Rightarrow x=h \sqrt{3}$

In rt. $\triangle P B Q$,

$\tan 45^{\circ}=\frac{h}{y} \Rightarrow 1=\frac{h}{y} \Rightarrow y=h$

$\therefore \quad x+y=h \sqrt{3}+h=\boldsymbol{h}(\sqrt{\mathbf{3}}+\mathbf{1}) \mathbf{m}$.

~~ 22 Let $h$ be the height of the tree $C D . \angle C E D=60^{\circ}$, $\angle C F D=30^{\circ}$ and $E F=160 ft$. Let $F D=x ft$, therefore $D E=(160-x) ft$

In rt. $\triangle D F C$,

$ \tan 30^{\circ}=\frac{h}{x} \Rightarrow \frac{h}{x}=\frac{1}{\sqrt{3}} \Rightarrow x=h \sqrt{3} $

In rt. $\triangle D E C$,

$ \begin{aligned} & \tan 60^{\circ}=\frac{h}{160-x} \Rightarrow \frac{h}{160-x}=\sqrt{3} \\ & \Rightarrow h=\sqrt{3}(160-x) \Rightarrow h=\sqrt{3}(160-h \sqrt{3}) \\ & \Rightarrow h=160 \sqrt{3}-3 h \Rightarrow 4 h=160 \sqrt{3} \Rightarrow h=\mathbf{4 0} \sqrt{\mathbf{3}} \mathbf{f t} . \end{aligned} $

~~ 23 Let the height of the light house $A B=h$ metres. Given $\angle A C B=\alpha, \angle A D B=\beta, C D=d$.

$ \begin{aligned} \tan \alpha & =\frac{A B}{B C} \\ \Rightarrow \quad \tan \alpha & =\frac{h}{B C} \Rightarrow B C=\frac{h}{\tan \alpha} \end{aligned} $

In rt. $\triangle A B D$,

$ \begin{aligned} \tan \beta & =\frac{A B}{B D} \Rightarrow \frac{h}{B D}=\tan \beta \Rightarrow B D=\frac{h}{\tan \beta} \\ \therefore \quad & C D=B C+B D=\frac{h}{\tan \alpha}+\frac{h}{\tan \beta} \end{aligned} $

(Adding eqn. (i) and (ii)

$ \Rightarrow \quad d=\frac{h(\tan \beta+\tan \alpha)}{\tan \alpha+\tan \beta} \Rightarrow h=\frac{\boldsymbol{d} \tan \alpha \tan \beta}{\tan \alpha+\tan \beta} $

~~ 24 Let $P Q$ be the vertical pole of height $h$.

If the angle of elevation of the sun’s ray is $\alpha$, then the length of the shadow $Q R=h \cot \alpha$

In $\triangle Q R P$, for the first moment, we take $\alpha=\alpha_1$

Given, $h \cot \alpha_1=h \Rightarrow \cot \alpha_1=1 \Rightarrow \tan \alpha_1=1$

For the second moment, $\alpha=\alpha_2$ (say)

Given, $h \cot \alpha_2=2 h \Rightarrow \tan \alpha_2=\frac{1}{2}$

Now, for the third moment, $\alpha=\alpha_3$ (say)

$ h \cot \alpha_3=3 h \Rightarrow \tan \alpha_3=\frac{1}{3} $

Now, we need to find the value of $\alpha_1+\alpha_2+\alpha_3$. $\tan (\alpha_1+\alpha_2+\alpha_3)$

$ \begin{aligned} & =\frac{\tan \alpha_1+\tan \alpha_2+\tan \alpha_3-\tan \alpha_1 \tan \alpha_2 \tan \alpha_3}{1-\tan \alpha_1 \tan \alpha_2-\tan \alpha_3 \tan \alpha_1-\tan \alpha_2 \tan \alpha_3} \\ & =\frac{(1+\frac{1}{2}+\frac{1}{3})-(1 \times \frac{1}{2} \times \frac{1}{3})}{1-(1 \times \frac{1}{2}+1 \times \frac{1}{3}+\frac{1}{2} \times \frac{1}{3})} \\ & =\frac{\frac{11}{6}-\frac{1}{6}}{1-(\frac{1}{2}+\frac{1}{3}+\frac{1}{6})}=\frac{\frac{10}{6}}{1-1}=\infty \\ & \tan (\alpha_1+\alpha_2+\alpha_3)=\tan \frac{\pi}{2} \Rightarrow \alpha_1+\alpha_2+\alpha_3=\frac{\pi}{2} . \end{aligned} $

~~ 25 Let $O$ be the position of the aeroplane and $A$ be the point on the ground. Then $\angle O A M=60^{\circ}$

$O^{\prime}$ is the position of the plane after $10 s$.

In $\triangle A M O$,

$ \tan 60^{\circ}=\frac{O M}{A M} \Rightarrow \sqrt{3}=\frac{2}{A M} \Rightarrow A M=\frac{2}{\sqrt{3}} km $

In $\triangle A O^{\prime} N, \tan 45^{\circ}=\frac{O^{\prime} N}{A N} \Rightarrow A N=O^{\prime} N=2 km$

$\therefore \quad$ Distance travelled by the plane $=O^{\prime} O=M N$

$ =A N-A M=2-\frac{2}{\sqrt{3}}=\frac{2 \sqrt{3}-2}{\sqrt{3}}=\frac{\mathbf{2}(\sqrt{\mathbf{3}}-\mathbf{1})}{\sqrt{\mathbf{3}}} \mathbf{k m} . $

~~ 26 Let the height of the tower $P Q=\boldsymbol{h}$ metres

Given, $\angle Q A P=\alpha, \angle Q B P=2 \alpha$, $\angle Q C P=3 \alpha, A B=a, B C=b, C P=x$ (say)

Now, in rt. $\triangle Q A P$,

$ \tan \alpha=\frac{Q P}{A P}=\frac{h}{a+b+x} $

$\Rightarrow a+b+x=h \cot \alpha$

In rt. $\triangle Q B P$,

$\tan 2 \alpha=\frac{Q P}{B P}=\frac{h}{b+x}$

$\Rightarrow b+x=h \cot 2 \alpha$

In rt. $\triangle Q C P$,

$ \tan 3 \alpha=\frac{Q P}{C P}=\frac{h}{x} \Rightarrow x=h \cot 3 \alpha $

Eq $(i)-$ Eq $(i i)$

$\Rightarrow \quad a=h(\cot \alpha-\cot 2 \alpha)=h[\frac{\cos \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}]$

$\Rightarrow \quad a=\frac{h[\cos \alpha \sin 2 \alpha-\sin \alpha \cos 2 \alpha}{\sin \alpha \sin 2 \alpha}$

$\Rightarrow \quad a=\frac{h \sin (2 \alpha-\alpha)}{\sin \alpha \sin 2 \alpha}=\frac{h \sin \alpha}{\sin \alpha \sin 2 \alpha}$

$\Rightarrow \quad h=a \sin 2 \alpha$

Eqn (ii) - Eqn (iii)

$\Rightarrow \quad b=h(\cot 2 \alpha-\cot 3 \alpha)=h[\frac{\cos 2 \alpha}{\sin 2 \alpha}-\frac{\cos 3 \alpha}{\sin 3 \alpha}]$

$\Rightarrow \quad b=\frac{h \cdot[\cos 2 \alpha \sin 3 \alpha-\cos 3 \alpha \sin 2 \alpha]}{\sin 2 \alpha \sin 3 \alpha}$

$\Rightarrow \quad b=\frac{h \sin (3 \alpha-2 \alpha)}{\sin 2 \alpha \sin 3 \alpha} \Rightarrow b=\frac{h \sin \alpha}{\sin 2 \alpha \sin 3 \alpha}$

$\Rightarrow \quad h=\frac{b \sin 2 \alpha \sin 3 \alpha}{\sin \alpha}$

$\therefore$ From eqn (iv) and (v), we have

$a \sin 2 \alpha=\frac{b \sin 2 \alpha \sin 3 \alpha}{\sin \alpha} \Rightarrow \sin \alpha=\frac{b}{a} \sin 3 \alpha$

$\Rightarrow \sin \alpha=\frac{b}{a}(3 \sin \alpha-4 \sin ^{3} \alpha) \Rightarrow 1=\frac{3 b}{a}-\frac{4 b}{a} \sin ^{2} \alpha$

$\Rightarrow \quad a=3 b-4 b \sin ^{2} \alpha \Rightarrow 4 b \sin ^{2} \alpha=3 b-a$

$\Rightarrow \sin \alpha=\sqrt{\frac{3 b-a}{4 b}} \Rightarrow \cos \alpha=\sqrt{1-\sin ^{2} \alpha}$

$=\sqrt{1-\frac{3 b-a}{4 b}}=\sqrt{\frac{\boldsymbol{b}+\boldsymbol{a}}{\mathbf{4 b}}}$

From eqn (iv) we have $h=a \sin 2 \alpha$

$ \begin{aligned} & h=2 a \sin \alpha \cos \alpha \\ & h=2 a \cdot \sqrt{\frac{3 b-a}{4 b}} \sqrt{\frac{b+a}{4 b}} \\ & h=\frac{2 a}{4 b} \sqrt{(3 b-a)(b+a)} \\ & h=\frac{\boldsymbol{a}}{\mathbf{2} \boldsymbol{b}} \sqrt{(\mathbf{3 b - a ) ( b + a )}} . \end{aligned} $

~~ 27 Let $A B$ be the tower of height $h$ metres (say).

Given, building $P Q=a$ metres. Draw $P R | B Q$ such that $R$ lies on $A B$.

$\Rightarrow \quad P R=B Q$

Given, $\angle A P R=30^{\circ}$ and $\angle A Q B=45^{\circ}$

$ R B=P Q=a \Rightarrow A R=(h-a) $

$\therefore \quad$ In rt. $\triangle A R P$,

$ \begin{aligned} \tan 30^{\circ} & =\frac{A R}{R P}=\frac{h-a}{R P} \\ \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{h-a}{R P} \Rightarrow R P=(h-a) \sqrt{3} \end{aligned} $

In rt. $\triangle A B Q$,

$ \begin{matrix} \tan 45^{\circ} & =\frac{A B}{B Q}=\frac{h}{B Q} \\ \Rightarrow & 1 & =\frac{h}{B Q} \Rightarrow B Q=h \\ \therefore & R P & =B Q, \text{ from }(i) \text{ and }(i i), \text{ we have } \\ \Rightarrow & h & =(h-a) \sqrt{3} \Rightarrow h(\sqrt{3}-1)=a \sqrt{3} \\ \Rightarrow & h & =\frac{a \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{\boldsymbol{a} \sqrt{\mathbf{3}}(\sqrt{\mathbf{3}}+\mathbf{1})}{\mathbf{2}} . \end{matrix} $

~~ 28 Let the height of the building $A B$ be $h$ metres and the distance between $A B$ and tower $E C$ be $\boldsymbol{x}$ metres.

Given, $\angle E A D=30^{\circ}, \angle E B C=60^{\circ}$

In rt. $\triangle A D E$,

$ \begin{aligned} \tan 30^{\circ} & =\frac{E D}{A D}=\frac{75-h}{x} \\ \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{75-h}{x} \Rightarrow x=(75-h) \sqrt{3} \end{aligned} $

In rt. $\triangle E B C$,

$ \tan 60^{\circ}=\frac{E C}{B C} \Rightarrow \sqrt{3}=\frac{75}{x} \Rightarrow x=\frac{75}{\sqrt{3}} $

$\therefore \quad$ From $(i)$ and (ii)

$ \begin{aligned} & \frac{75}{\sqrt{3}}=(75-h) \sqrt{3} \Rightarrow 75=(75-h) \times 3 \\ \Rightarrow \quad 75 & =225-3 h \quad \Rightarrow \quad 3 h=150 \quad \Rightarrow \quad h=\mathbf{5 0} \mathbf{~ m .} \end{aligned} $

~~ 29 Let $P Q$ be the hill of height $h$ metres.

Let $A B$ be the pillar of height $H$ metres.

Let $A C | B Q$ meet $P Q$ in $C$.

The angles of depression of the top and bottom of the pillar are given as $\alpha$ and $\beta$ respectively $\therefore \angle P A C=\alpha, \angle P B Q=\beta$

Let $B Q=A C=\boldsymbol{x}$ metres.

In rt. $\triangle P C A$,

$\tan \alpha=\frac{P C}{A C}=\frac{h-H}{x} \Rightarrow x=\frac{h-H}{\tan \alpha}$

In rt. $\triangle P Q B$,

$\tan \beta=\frac{P Q}{Q B}=\frac{h}{x} \Rightarrow x=\frac{h}{\tan \beta}$

$\therefore$ From (i) and (ii), we get

$ \frac{h-H}{\tan \alpha}=\frac{h}{\tan \beta} \Rightarrow h \tan \beta-H \tan \beta=h \tan \alpha $

$\Rightarrow h(\tan \beta-\tan \alpha)=H \tan \beta \Rightarrow H=\frac{\boldsymbol{h}(\boldsymbol{\operatorname { t a n }} \boldsymbol{\beta}-\boldsymbol{\operatorname { t a n }} \boldsymbol{\alpha})}{\boldsymbol{\operatorname { t a n }} \boldsymbol{\beta}}$.

~~ 30 Let $A C$ be the pole and point $P$ be the position on the ground. Then $C B$ subtends $\boldsymbol{\alpha}$ at $P, B A$ subtends $\boldsymbol{\beta}$ at $P$.

Given, $\angle C P A=\theta$.

Given, $\theta=\tan ^{-1} \frac{1}{2} \Rightarrow \tan \theta=\frac{1}{2}$

Also, $\quad \theta=\alpha+\beta$

$\Rightarrow \tan \theta=\tan (\alpha+\beta) \Rightarrow \frac{1}{2}=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$

Checking on all the given options,

(a) When $(\tan \alpha, \tan \beta)=(\frac{1}{4}, \frac{1}{5})$

$ \text{ R.H.S }=\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4} \times \frac{1}{5}}=\frac{\frac{9}{20}}{\frac{19}{20}}=\frac{9}{19} \neq \frac{1}{2} \text{, not true } $

(b) When $(\tan \alpha, \tan \beta)=(\frac{1}{5}, \frac{2}{9})$

$ \text{ R.H.S }=\frac{\frac{1}{5}+\frac{2}{9}}{1-\frac{1}{5} \times \frac{2}{9}}=\frac{\frac{19}{45}}{\frac{43}{45}}=\frac{19}{43} \neq \frac{1}{2} \text{, not true } $

$ \text{ R.H.S }=\frac{\frac{2}{9}+\frac{1}{4}}{1-\frac{2}{9} \times \frac{1}{4}}=\frac{\frac{\mathbf{1 7}}{\mathbf{3 6}}}{\frac{\mathbf{3 4}}{\mathbf{3 6}}}=\frac{\mathbf{1}}{\mathbf{2}}, \text{ true. } $

~~ 31 Let $F E$ be the level of the lake.

Let $A$ be a point $2500 m$ above the level of the lake from where the angle of elevation of cloud at $B$ is $15^{\circ}$.

Let $D$ be the reflection of the cloud in the lake. Given, $B E=E D=H$ metres (say)

Draw $A C | F E$ meeting $B E$ in $C$.

$ \angle B A C=15^{\circ}, \angle C A D=45^{\circ}, A F=C E=h=2500 m . $

Now, in rt. $\triangle A B C, \tan 15^{\circ}=\frac{B C}{A C}$

$\Rightarrow A C=B C \cot 15^{\circ} \Rightarrow A C=(H-h) \cot 15^{\circ}$

In rt. $\triangle A C D, \tan 45^{\circ}=\frac{C D}{A C}$

$\Rightarrow A C=C D \cot 45^{\circ} \Rightarrow A C=(H+h) \cot 45^{\circ}$

From eqns (i) and (ii),

$ \begin{aligned} & (H-h) \cot 15^{\circ}=(H+h) \cot 45^{\circ} \\ & \Rightarrow H(\cot 15^{\circ}-\cot 45^{\circ})=h(\cot 45^{\circ}+\cot 15^{\circ}) \\ & \Rightarrow \quad H=\frac{h(1+\cot 15^{\circ})}{\cot 15^{\circ}-1}=\frac{2500(2+\sqrt{3}+1)}{2+\sqrt{3}-1} \\ & =\frac{2500(3+\sqrt{3})}{\sqrt{3}+1} \times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}=2500 \sqrt{3} \mathbf{~ m .} \end{aligned} $

~~ 32 Let the positions of the two planes at the instant when they are vertically above each other be $C$ and $B$ respectively.

Let $A$ be the base level on the ground from where the height of the upper plane is $300 m$, i.e., $A C=300 m, A B=\boldsymbol{x}$ metres (say) Let angle of elevations of $P$ be the position from where the two planes are measured, i.e.,

$ \angle C P A=60^{\circ}, \angle B P A=45^{\circ} \text{, } $

$P A=\boldsymbol{y}$ m (say) $y$

In rt. $\triangle A B P$,

$\tan 45^{\circ}=\frac{A B}{P A} \Rightarrow 1=\frac{x}{y} \Rightarrow y=x$

In rt. $\triangle C P A$,

$\tan 60^{\circ}=\frac{A C}{A P} \Rightarrow \sqrt{3}=\frac{300}{y} \Rightarrow y=\frac{300}{\sqrt{3}}$

From (i) and (ii), $x=\frac{300}{\sqrt{3}}=\frac{300}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=100 \sqrt{\mathbf{3}} \mathbf{~ m}$

$\therefore$ The lower plane is flying at a height of $\mathbf{1 0 0} \sqrt{\mathbf{3}}$.

~~ 33 Let $P Q$ and $A B$ be the two chimneys such that the angles of depression of points $A$ and $B$ from point $P$ are $\theta$ and $\phi$ respectively. Also, $P Q=150 m$ Draw $A R | Q B$ meeting $P Q$ in $R$.

$\therefore \quad R A=Q B, R Q=A B=h$ (say)

Also, given, $\angle P A R=\theta, \angle P B Q=\phi$.

$\therefore$ In rt. $\triangle P R A$,

$ \tan \theta=\frac{P R}{R A} \Rightarrow \frac{4}{3}=\frac{P R}{R A} $

In rt. $\triangle P Q B, \tan \phi=\frac{P Q}{Q B} \Rightarrow \frac{5}{2}=\frac{150}{Q B}$

$\Rightarrow Q B=\frac{150 \times 2}{5}=60 m$

$\because \quad Q B=R A$, putting $R A=60$ in $(i)$, we get

$ \frac{4}{3}=\frac{P R}{60} \Rightarrow P R=80 m $

$\therefore$ In rt. $\triangle P R A$,

distance between the top of the two chimneys, i.e.,

$ \begin{aligned} A P & =\sqrt{P R^{2}+R A^{2}} \quad \text{ (Pythagoras’ Th.) } \\ & =\sqrt{80^{2}+60^{2}}=\sqrt{6400+3600}=\sqrt{10000}=\mathbf{1 0 0} \mathbf{~ m .} \end{aligned} $

~~ 34 Let $A B$ be the tower and $P Q$ be the minar.

Given $A B=h$, let $P Q=x$.

Draw $A C | B Q$ meeting $P Q$ in $C$

Then, $B Q=A C=y$ (say)

$\therefore C Q=A B=h$

$\Rightarrow \quad P C=P Q-C Q=x-h$

Given $\angle P B Q=\alpha, \angle A P B=\beta$.

Now $\angle P D C=\alpha$

$ (D C | B Q, \text{ corr. } \angle s) $

In $\triangle A D P, \angle P A D+\angle A P D=$ ext. $\angle P D C$

$\Rightarrow \angle P A D=$ ext. $\angle P D C-\angle A P D=\alpha-\beta$.

In rt. $\triangle P C A$

$\tan (\alpha-\beta)=\frac{P C}{A C}=\frac{x-h}{y} \Rightarrow y=\frac{x-h}{\tan (\alpha-\beta)}$

In rt. $\triangle P B Q$,

$\tan \alpha=\frac{P Q}{B Q}=\frac{x}{y} \Rightarrow y=\frac{x}{\tan \alpha}$

From (i) and (ii) $\frac{x-h}{\tan (\alpha-\beta)}=\frac{x}{\tan \alpha}$

$\Rightarrow \quad \frac{x-h}{x}=\frac{\tan (\alpha-\beta)}{\tan \alpha} \Rightarrow 1-\frac{h}{x}=\frac{\cot \alpha}{\cot (\alpha-\beta)}$

$\Rightarrow \quad \frac{h}{x}=1-\frac{\cot \alpha}{\cot (\alpha-\beta)}=\frac{\cot (\alpha-\beta)-\cot \alpha}{\cot (\alpha-\beta)}$

$\Rightarrow \quad x=\frac{h \cot (\alpha-\beta)}{\cot (\alpha-\beta)-\cot \alpha}$. ~~ 35 Let $P Q$ and $R S$ be the two given tower and $T$, the middle point of the line $Q S$, joining the feet of the towers.

Given, $\angle P T Q=60^{\circ}, \angle R T S=30^{\circ}$, $Q T=T S$.

In rt. $\triangle P Q T$,

$ \tan 60^{\circ}=\frac{P Q}{Q T} \Rightarrow \sqrt{3}=\frac{P Q}{Q T} \Rightarrow Q T=\frac{P Q}{\sqrt{3}} $

In rt. $\Delta R T S$,

$ \tan 30^{\circ}=\frac{R S}{S T} \Rightarrow \frac{1}{\sqrt{3}}=\frac{R S}{S T} \Rightarrow S T=R S \sqrt{3} $

From (i) and (ii)

$ Q T=S T \Rightarrow \frac{P Q}{\sqrt{3}}=R S \sqrt{3} \Rightarrow \frac{P Q}{R S}=\frac{\sqrt{3} \times \sqrt{3}}{1} $

$\Rightarrow P Q: R S=\mathbf{3}: \mathbf{1}$.

~~ 36 Let $P Q$ and $R S$ be the two pillar of equal height $=h$ metres (say)

Given, $T$ is a point on the line joining the bases of $P Q$ and $R S$ such that $Q T=\boldsymbol{x}$ m (say),

then $T S=(60-x) m$

$(\because Q S=60 m)$

$ \angle P T Q=60^{\circ}, \angle R T S=30^{\circ} $

In rt. $\triangle P Q T$,

$ \tan 60^{\circ}=\frac{P Q}{Q T} \Rightarrow \frac{h}{x}=\sqrt{3} \Rightarrow x=\frac{h}{\sqrt{3}} $

In rt. $\Delta$ RTS,

$ \tan 30^{\circ}=\frac{R S}{T S} \Rightarrow \frac{h}{60-x}=\frac{1}{\sqrt{3}} \Rightarrow x=60-\sqrt{3} h $

From (i) and (ii),

$ \frac{h}{\sqrt{3}}=60-\sqrt{3} h \Rightarrow 4 h=60 \sqrt{3} \Rightarrow \boldsymbol{h}=\mathbf{1 5} \sqrt{\mathbf{3}} \mathbf{~ m .} $

~~ 37 Let $A D$ and $B C$ be the two pillar of heights $\boldsymbol{a}$ and $\boldsymbol{b}$ respectively. $\angle D P A=45^{\circ}, \angle C P B=45^{\circ}$ Required distance $=C D$. Draw $D E | A B$ meeting $B C$ in $E$

$\therefore A B=D E$

In $\triangle A P D, \tan 45^{\circ}=\frac{a}{A P} \Rightarrow A P=a$

In $\triangle B P C, \tan 45^{\circ}=\frac{b}{P B} \Rightarrow P B=b$

$\therefore A B=A P+P B=a+b \Rightarrow D E=a+b$

$C E=B C-B E=B C-A D=b-a$

In $\triangle D E C, D C^{2}=D E^{2}+C E^{2}=(a+b)^{2}+(b-a)^{2}$

$ =2(a^{2}+b^{2}) . $

~~ 38 Let $A B$ be the diameter of the circular base of the cliff. Let $C$ be the top of the cliff.

Given, $C E=500 m, \angle C A E=60^{\circ}$, $\angle C B E=30^{\circ}, A E=d_1, B E=d_2$

In $\triangle A E C, \tan 60^{\circ}=\frac{500}{d_1}$

$\Rightarrow d_1=\frac{500}{\sqrt{3}} m$

In $\triangle B E C, \tan 30^{\circ}=\frac{500}{d_2} \Rightarrow d_2=500 \sqrt{3} m$

$\therefore$ Required diameter $=d_1+d_2=\frac{500}{\sqrt{3}}+500 \sqrt{3}=\frac{\mathbf{2 0 0 0}}{\sqrt{\mathbf{3}}} \mathbf{~ m}$

~~ 39 Let $A B C D$ be the rectangular field and $E A$ be the lamp post that stands at corner $A$.

Since $A E$ is a vertical pole, it is perpendicular to all lines in the plane of the rectangular field, i.e., $E A$ is perp. to $A B, B C, C D, D A$.

$\therefore \angle E A D=90^{\circ}$. Given, $E A=12 m$

Join $E B, E C$, and $A C$.

Given, $\angle E B A=60^{\circ}, \angle A C E=45^{\circ}$.

In rt. $\triangle A B E$,

$\tan 60^{\circ}=\frac{A E}{A B}=\frac{12}{A B} \Rightarrow \sqrt{3}=\frac{12}{A B} \Rightarrow A B=\frac{12}{\sqrt{3}}=4 \sqrt{3} m$

In rt. $\triangle A C E$,

$\tan 45^{\circ}=\frac{A E}{A C} \Rightarrow \frac{12}{A C}=1 \Rightarrow A C=12 m$.

In $\triangle A B C, B C=\sqrt{A C^{2}-A B^{2}} \quad$ (Pythagoras’ Th.)

$=\sqrt{144-48}=\sqrt{96} m=4 \sqrt{6} m$

$\therefore$ Area of the rectangular field $=A B \times B C=4 \sqrt{3} \times 4 \sqrt{6}$

$ =48 \sqrt{2} sq \cdot m \text{. } $

~~ 40 Let $P Q$ be the vertical pole of height $h$ metres. Let $A$ be a point on $P Q$ such that $P A=\frac{3}{4} h$, and $A Q=h-\frac{3}{4} h=\frac{1}{4} h$.

Let the portion $P A$ of the pole

subtend an angle $\theta_1$ at the observation point $O$ such that $Q O$ $=40 m$. Let $A Q$ subtend $\theta_2$ at 0 . Then, $\theta_1=\tan ^{-1}(\frac{3}{5}) \Rightarrow \tan \theta_1=\frac{3}{5}$

In rt. $\triangle A O Q$,

$\tan \theta_2=\frac{A Q}{Q O} \Rightarrow \tan \theta_2=\frac{h}{160}$

In rt. $\triangle P O Q, \tan (\theta_1+\theta_2)=\frac{P Q}{Q O}=\frac{h}{40}$

$\Rightarrow \frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}=\frac{h}{40} \quad$ (From eqns (i) and (ii))

$\Rightarrow \frac{5(h+96)}{800-3 h}=\frac{h}{40} \Rightarrow h^{2}-200 h+6400=0$

$\Rightarrow(h-160)(h-40)=0 \Rightarrow h=160$ or $h=40$.

According to the given options one of the possible heights $=\mathbf{4 0} \mathbf{~ m}$.

~~ 41 Let $O$ be the centre of the balloon of radius $r$ and $E$ the position of the eye of the observer.

Let $E A, E B$ be the tangents from $E$ to the balloon.

Then $\angle E A B=\alpha$. Let $O L$ be the perpendicular from $O$ on the

horizontal line $E X$. Given $\angle O E L=\beta$

$ \angle O L E=90^{\circ}, \angle O B E=90^{\circ}=\angle O A E, $

( $\because$ radius $\perp$ to the tangent at point of contact)

Also $\angle B E O=\angle A E O=\frac{\alpha}{2}$

$(\because \triangle A E O \cong \triangle B E O)$

In rt. $\triangle O A E, \sin \frac{\alpha}{2}=\frac{O A}{O E}$

$\Rightarrow O E=O A cosec \frac{\alpha}{2}=r cosec \frac{\alpha}{2}$

In rt. $\triangle O L E, \sin \beta=\frac{O L}{O E}$

$\Rightarrow O L=O E \sin \beta=\boldsymbol{r} \boldsymbol{\operatorname { c o s e c }} \frac{\boldsymbol{\alpha}}{\mathbf{2}} \sin \boldsymbol{\beta}$

(From (i))

~~ 42 Let $O Y$ be the wall and $O X$ the horizontal. Let $A B$ be the initial position of the ladder and $P Q$ the new position of the ladder after its foot is pulled away from the wall. Let the length of the ladder be $x$ metres.

Given, $Q B=a m, A P=b m$,

$A B=P Q=\boldsymbol{x} m, \angle A B O=\alpha, \angle P Q O=\beta$

In rt. $\triangle A O B, \cos \alpha=\frac{O B}{A B}$

$\Rightarrow O B=A B \cos \alpha=x \cos \alpha$

and $\quad \sin \alpha=\frac{O A}{A B} \Rightarrow O A=A B \sin \alpha=x \sin \alpha$

In rt. $\triangle P O Q, \cos \beta=\frac{O P}{P Q} \Rightarrow O P=P Q \cos \beta=x \cos \beta$

and

$ \sin \beta=\frac{O P}{P Q} \Rightarrow O P=P Q \sin \beta=x \cos \beta $

$ \begin{matrix} a=O Q-O B=x \cos \beta-x \cos \alpha=x(\cos \beta-\cos \alpha) \\ b=O A-O P=x \sin \alpha-x \sin \beta=x(\sin \alpha-\sin \beta) \\ \therefore \frac{a}{b}=\frac{\cos \beta-\cos \alpha}{\sin \alpha-\sin \beta}=\frac{2 \sin (\frac{\alpha+\beta}{2}) \sin (\frac{\alpha-\beta}{2})}{2 \cos (\frac{\alpha+\beta}{2}) \sin (\frac{\alpha-\beta}{2})} \\ =\tan (\frac{\alpha+\beta}{2}) . \end{matrix} $

~~ 43 Let the two towers be $A B$ and $C D$ whose heights are $H m$ and $h m$ respectively. Given $P Q=30 m$.

Also, $\angle A P B=60^{\circ}, \angle A Q B=45^{\circ}, \angle C P D=30^{\circ}, \angle C Q D=60^{\circ}$, From rt $\triangle s C P D$ and $C Q D$,

$ \begin{aligned} & P D=h \cot 30^{\circ}, Q D=h \cot 60^{\circ} \\ \therefore \quad & P Q=P D-Q D=h(\cot 30^{\circ}-\cot 60^{\circ}) \\ \Rightarrow \quad 30 & =h(\sqrt{3}-\frac{1}{\sqrt{3}}) \Rightarrow h=15 \sqrt{3} m . \end{aligned} $

From rt $\triangle s A Q B$ and $A P B$,

$ \begin{aligned} & B Q=H \cot 45^{\circ}, B P=H \cot 60^{\circ} \\ & \therefore \quad P Q=B Q-B P=H(\cot 45^{\circ}-\cot 60^{\circ}) \\ & \Rightarrow 30=H(1-\frac{1}{\sqrt{3}}) \Rightarrow H=\frac{30 \sqrt{3}}{\sqrt{3}-1} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)} \\ & =15 \sqrt{3}(\sqrt{3}+1) m . \end{aligned} $

$\therefore \quad$ Distance between the towers $=B D=B Q+Q D$

$ \begin{aligned} & =H \cot 45^{\circ}+h \cot 60^{\circ}=15 \sqrt{3}(\sqrt{3}+1)+15 \sqrt{3} \times \frac{1}{\sqrt{3}} \\ & =45+15 \sqrt{3}+15=60+15 \sqrt{3}=\mathbf{1 5}(\mathbf{4}+\sqrt{\mathbf{3}}) \mathbf{m} . \end{aligned} $

~~ 44 Let $P Q$ be the pole standing vertically in the centre of the triangular park $A B C$. Then $P Q$ being the pole standing vertically is perp. to all sides of the park, i.e. $Q B, Q C$ and $B C$. $\therefore \angle P Q B=90^{\circ}, \angle P Q A=90^{\circ}$, $\angle P Q C=90^{\circ}$.

Also, given $\angle P B Q=\angle P A Q$

$=\angle P C Q=\theta$ (say)

In rt. $\triangle P A Q, \tan \theta=\frac{P Q}{A Q} \Rightarrow A Q=P Q \cot \theta$

In rt. $\triangle P B Q, \tan \theta=\frac{P Q}{B Q} \Rightarrow B Q=P Q \cot \theta$

In rt. $\triangle P C Q, \tan \theta=\frac{P Q}{C Q} \Rightarrow C Q=P Q \cot \theta$

From (i), (ii), and (iii) $A Q=B Q=C Q$

$\Rightarrow Q$ is equidistant from the vertices $A, B$ and $C$

$\Rightarrow Q$ is the circumcentre of $\triangle A B C$.

~~ 45 Let $P D$ be the pole standing at the corner $D$ of the square plot $A B C D$. Let $P D=h m$ (say)

Given, $\angle P C D=30^{\circ}, \angle D A P=30^{\circ}$ and $\angle P B D=\theta$.

From rt. $\triangle P A D$ or $\triangle P C D$

$P D=A D \tan 30^{\circ}$ or $P D=C D \tan 30^{\circ}$

Let each side of the square $A B C D$ be $a$ metres.

(Then, $P D=a \tan 30^{\circ}=\frac{a}{\sqrt{3}}$ )

$B D$ being the diagonal of square $A B C D$,

$B D=a \sqrt{2}$

$\therefore$ In rt. $\triangle P B D, \tan \theta$

$ =\frac{P D}{B D}=\frac{\frac{a}{\sqrt{3}}}{\frac{a}{\sqrt{2}}}=\frac{\mathbf{1}}{\sqrt{\mathbf{6}}} \text{. } $

SELF ASSESSMENT SHEET

~~ 1. From the top of a building of height ’ $h$ ’ metres, the angle of depression of an object on the ground is $\alpha$. The distance (in metres) of the object from the foot of the building is

(a) $h \sin \alpha$

(b) $h \cos \alpha$

(d) $h \cot \alpha$ ~~ 2. Angles of elevation of the top of a tower from three points (collinear) $A, B$ and $C$ on a road leading to the foot of the tower are $30^{\circ}, 45^{\circ}$ and $60^{\circ}$ respectively. The ratio of $A B$ to $B C$ is

(a) $\sqrt{3}: 1$

(b) $\sqrt{3}: 2$

(d) $2: \sqrt{3}$

(Karnataka CET 2011)

~~ 3. The angles of elevation of the top of a tower at two point, which are at distances $a$ and $b$ from the foot in the same horizontal line and on the same sides of the tower are complementary. The height of the tower is

(a) $a b$

(b) $\sqrt{a b}$

(d) $\sqrt{\frac{b}{a}}$ ~~ 4. An aeroplane flying with uniform speed horizontally one kilometre above the ground is observed at an elevation of $60^{\circ}$. After 10 seconds if the elevation is observed to be $30^{\circ}$, then the speed of the plane (in $km / hr$ ) is

(a) $\frac{120}{\sqrt{3}}$

(b) $120 \sqrt{3}$

(d) $240 \sqrt{3}$.

(EAMCET 2004)

~~ 5. The angle of elevation of the top of a $T V$ tower from three points $A, B, C$ in a straight line (in a horizontal plane) through the foot of the tower are $\alpha, 2 \alpha, 3 \alpha$ respectively. If $A B=a$, the height of the tower is

(a) $a \sin \alpha$

(b) $a \sin 2 \alpha$

(d) $a \sin (\alpha^{2})$

(KCET 2008)

~~ 6. The top of a hill observed from the top and bottom of a building of height $h$ is at angles of elevation $p$ and $q$ respectively. The height of the hill is

(a) $\frac{h \tan p}{\tan p-\tan q}$

(b) $\frac{h \tan q}{\cot q-\cot p}$

(d) $\frac{h \cot q}{\cot q-\cot p}$

(UPSEAT 2001)

~~ 7. A tower subtends an angle $\alpha$ at a point $A$ in the plane of its base and the angle of depression of the foot of the tower at a point $b$ feet just above $A$ is $\beta$. The height of the tower is

(a) $b \tan \alpha \tan \beta$

(b) $b \tan \alpha \cot \beta$

(d) $b \cot \alpha \cot \beta$

(Kerala PET 2006)

~~ 8. The lower $24 m$ portion of a $50 m$ tall tower is painted green and the remaining portion red. What is the distance of a point on the ground from the base of the tower where two different portions of the tower subtend equal angles?

(a) $60 m$

(b) $72 m$

(d) $120 m$

(NDA/NA 2007)

~~ 9. The angle of depression of vertex of a regular hexagon lying in a horizontal plane, from the top of a tower of height $75 m$ located at the centre of the hexagon is $60^{\circ}$. What is the length of each side of the hexagon?

(a) $50 \sqrt{3} m$

(b) $75 m$

(d) $25 m$

(CDS 2007)

~~ 10 From a point on a horizontal plane, the elevation of the top of the hill is $45^{\circ}$. The elevation becomes $75^{\circ}$ after walking a distance of $500 m$ up a slope inclined at an angle of $15^{\circ}$ to the horizon. The height of the hill is:

(a) $500 \sqrt{6} m$

(b) $500 \sqrt{3} m$

(d) $250 \sqrt{3} m$

(UPSEAT 2007)

ANSWERS

~~ 1. (d) ~~ 2. (a) ~~ 3. (b) ~~ 4. $($ d) ~~ 5. (b) ~~ 6. (c) ~~ 7. (b) ~~ 8. (d) ~~ 9. (c) ~~ 10 (c)

HINTS AND SOLUTIONS

~~ 1. $P A$ is the building of height $h$.

$\angle P B A=\alpha$.

$\therefore \quad$ In rt. $\triangle P A B$,

$ \tan \alpha=\frac{P A}{B A} $

$\Rightarrow B A=P A \cot \alpha=\boldsymbol{h} \cot \boldsymbol{\alpha}$.

~~ 2. $\frac{A B}{B C}=\frac{A P-B P}{B P-C P}$

Let the height of the given tower $O P=h$

Then, in rt. $\triangle O A P$,

$ \begin{aligned} & \tan 30^{\circ}=\frac{O P}{A P} \\ \Rightarrow & A P=O P \cot 30^{\circ}=\sqrt{3} . h \end{aligned} $

In rt. $\triangle O B P$,

$ \tan 45^{\circ}=\frac{O P}{B P} \Rightarrow B P=O P \cot 45^{\circ}=h $

In rt. $\triangle O C P$,

$ \tan 60^{\circ}=\frac{O P}{C P} \Rightarrow C P=O P \cot 60^{\circ}=\frac{h}{\sqrt{3}} $

$\therefore$ Reqd. ratio $=\frac{\sqrt{3} h-h}{h-\frac{h}{\sqrt{3}}}=\frac{h(\sqrt{3}-1) \sqrt{3}}{h(\sqrt{3}-1)}=\frac{\sqrt{3}}{1}$

$\Rightarrow A B: B C=\sqrt{3}: 1$.

~~ 3. Let $A B$ be the tower of height and $C$ and $D$ be the two given points of observation such that

$ \begin{aligned} & B C=b, B D=a, \angle A C B=\beta, \\ & \angle A D B=\alpha . \end{aligned} $

In rt. $\triangle A B C$,

$ \tan \beta=\frac{A B}{b} \Rightarrow A B=b \tan \beta $

In rt. $\triangle A D B$,

$ \begin{aligned} & \tan \alpha=\frac{A B}{a} \\ \Rightarrow A B=a \tan \alpha & =a \tan (90^{\circ}-\beta) \quad(\because \alpha+\beta=90^{\circ}) \\ & =a \cot \beta \quad \end{aligned} $

Multiplying eqns (i) and (ii), we get

$ A B^{2}=b \cdot \tan \beta \times a \cot \beta \Rightarrow A B^{2}=a b \Rightarrow A B=\sqrt{\boldsymbol{a b}} . $

~~ 4. Type Solved Example 11 . ~~ 5. Let $E F$ be the $T V$ tower such that $\angle E A F=\alpha, \angle E B F=2 \alpha$, $\angle E C F=3 \alpha$ and $A B=a$

In $\triangle E B A$,

ext. $\angle E B F=\angle B E A+\angle B A E$

$\Rightarrow 2 \alpha=\angle B E A+\alpha \Rightarrow \angle B E A=\alpha$

$\Rightarrow \triangle B E A$ is isosceles $\Rightarrow B E=a$.

In rt. $\triangle E B F, \sin 2 \alpha=\frac{E F}{E B}$

$\Rightarrow E F=E B \sin 2 \alpha=\boldsymbol{a} \sin 2 \alpha$

~~ 6. Type Question No. 29 (practrice sheet)

You are required to find $H$ here.

~~ 7. Let $C D$ be the tower of height $h ft$.

Given, $\angle C A D=\alpha . B$ is a point $b$ feet above $A$, such that $\angle B D A=\beta$.

Now in rt. $\triangle C D A$,

$ \tan \alpha=\frac{C D}{D A} $

$\Rightarrow D A=C D \cot \alpha=h \cot \alpha$

In rt. $\triangle B D A$

$ \tan \beta=\frac{B A}{D A} \Rightarrow D A=a \cot \beta $

$\therefore \quad$ From $(i)$ and $(i i)$

$h \cot \alpha=b \cot \beta \Rightarrow h=\boldsymbol{b} \tan \boldsymbol{\alpha} \cot \boldsymbol{\beta}$.

~~ 8. Let $A C$ be the tower whose length is $50 m$.

Let $A B=24 m$ and $B C=26 m$ In rt. $\triangle A B D$, $\tan \theta=\frac{A B}{A D} \Rightarrow \tan \theta=\frac{24}{x}$

In rt. $\triangle A C D$,

$\tan 2 \theta =\frac{A C}{A D} \Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{50}{x}$

$\Rightarrow \frac{2 \cdot \frac{24}{x}}{1-\frac{576}{x^{2}}} =\frac{50}{x} \Rightarrow \frac{24 x^{2}}{x^{2}-576}=25$

$\begin{matrix} \Rightarrow & 24 x^{2} =25 x^{2}-576 \times 25 \\ \Rightarrow & x^{2} =576 \times 25 \\ \Rightarrow & x =24 \times 5=\mathbf{1 2 0} \mathbf{~ m .}\end{matrix}$

~~ 9. Let $O P$ be the tower of height $75 m$.

Angle of elevation $=$ Angle of depression

$\therefore$ In $\triangle F O P$,

$\tan 60^{\circ}=\frac{75}{x}$

$\Rightarrow x=\frac{75}{\sqrt{3}}=25 \sqrt{3} m$.

In a regular hexagon, six equilateral $\Delta s$ are formed, i.e.,

$O F=O E=O D=O C=O B=O A=$ side of hexagon

$\Rightarrow$ length of hexagon $=25 \sqrt{3} m$.

~~ 10 Type Solved Example 19 .

From rt. $\triangle P A Q$,

$ \begin{matrix} \tan 45^{\circ} & =\frac{P Q}{A Q} \\ \Rightarrow \quad A Q & =P Q=h \\ \therefore \quad A P & =h \sqrt{2} \\ & \Rightarrow \quad \angle B P M & =90^{\circ}-75^{\circ}=15^{\circ} \\ \Rightarrow \quad \angle A P B & =45^{\circ}-15^{\circ}=30^{\circ} \end{matrix} $

By sine rule on $\triangle A B P$, we have

$ \begin{aligned} \frac{h \sqrt{2}}{\sin 120^{\circ}} & =\frac{500}{\sin 30^{\circ}} \Rightarrow h \sqrt{2}=\frac{\sin 120^{\circ} \times 500}{\sin 30^{\circ}} \\ h \sqrt{2} & =\frac{\sqrt{3}}{2} \times 500 \times 2 \Rightarrow h \sqrt{2}=500 \sqrt{3} \\ h & =\frac{500 \sqrt{6}}{2}=\mathbf{2 5 0} \sqrt{\mathbf{6}} \mathbf{~ m .} \end{aligned} $

Circles: Area and

Perimeter

KEY FACTS

Circles: For a circle of radius $r$, and diameter $d$.

Circumference or perimeter of a circle $=2 \pi r=\pi d$
Area of circle $=\pi r^{2}=\frac{\pi d^{2}}{4}=\frac{1}{2} \times$ Circumference $\times$ Radius
Radius of circle $=\sqrt{\frac{\text{ Area }}{\pi}}=\frac{\text{ Perimeter or Circumference }}{2 \pi}$
Area of sector
Length of $arc \mathbf{A B}(l)=\frac{\theta}{360} \times 2 \pi r$
Area of sector $A C B O=\frac{\theta}{360} \times \pi r^{2}=\frac{\mathbf{1}}{\mathbf{2}} \times \boldsymbol{l} \times \boldsymbol{r}$.
Area of a segment
Area of segment $A C B=$ Area of sector $A C B O$ - Area of $\triangle A O B$
Area of segment $A D B=$ Area of circle - Area of segment $A C B$.
$\frac{\text{ Area of the circle circumscribing a square }}{\text{ Area of circle inscribed in the square }}=\frac{2}{1}$
$\frac{\text{ Area of square circumscribing a circle }}{\text{ Area of square inscribed in the circle }}=\frac{2}{1}$

Area of annulus or ring $=\pi(R^{2}-r^{2})$, where radius of outer circle $=R$, radius of inner circle $=r$
Angle described by minute hand in one minute $=\frac{360^{\circ}}{60^{\circ}}=6^{\circ}$
Angle described by the hour hand in one minute $=(\frac{360^{\circ}}{12 \times 60^{\circ}})=(\frac{\mathbf{1}}{\mathbf{2}})^{\circ}$

SOLVED EXAMPLES

Ex. 1 . Find the area of the shaded region in the figure, if $P Q=24 cm, P R=7 cm$ and $O$ is the centre of the circle.

(NCERT)

Sol. Join $R P$. As angle in a semi-circle is a right angle, $\angle Q P R=90^{\circ}$

$ \begin{matrix} \therefore & R Q^{2}=P R^{2}+P Q^{2} \\ \Rightarrow & R Q=\sqrt{49+576}=\sqrt{625}=25 cm . \end{matrix} $

$\therefore \quad$ Area of $\triangle R P Q=\frac{1}{2} \times R P \times P Q=\frac{1}{2} \times 7 \times 24=84 cm^{2}$

Area of semi-circle $=\frac{\pi r^{2}}{2}=\frac{1}{2} \times \frac{22}{7} \times(\frac{25}{2})^{2}=\frac{6875}{28} cm^{2}$

$\therefore \quad$ Area of the shaded region $=$ Area of semi-circle - Area of $\triangle R P Q$

$ =(\frac{6875}{28}-84) cm^{2}=(\frac{6875-2352}{28}) cm^{2}=\frac{\mathbf{4 5 2 3}}{\mathbf{2 8}} \mathbf{c m}^{2} $

Ex. 2 . In the given figure, are shown sectors of two concentric circles of radii $7 cm$ and $3.5 cm$. Find the area of the shaded region. (Use $\pi=\frac{22}{7}$ )

Sol. Required area $=$ Area of sector $O A B-$ Area of sector $O C D$

$ \begin{aligned} & =(\frac{30}{360} \times \frac{22}{7} \times 7^{2}) cm^{2}-(\frac{30}{360} \times \frac{22}{7} \times 3.5^{2}) cm^{2} \\ & =\frac{77}{6} cm^{2}-\frac{77}{24} cm^{2}=\frac{77}{24} \times(4-1) cm^{2}=\frac{77}{8} cm^{2}=\mathbf{9 . 6 2 5} cm^{2} \end{aligned} $

Ex. 3 . A pendulum swings through an angle of $30^{\circ}$ and describes an arc $8.8 cm$ in length. Find the length of the pendulum. $(.$ Use $.\pi=\frac{22}{7})$

Sol. Here $\theta=30^{\circ}$, arc length $=l=8.8$, length of the pendulum $=$ radius $=r$

$ \begin{aligned} & \because \quad l=\frac{\theta}{360^{\circ}} \times 2 \pi r \\ & \therefore \quad 8.8=\frac{30}{360} \times 2 \times \frac{22}{7} \times r \Rightarrow r=\frac{8.8 \times 6 \times 22}{22}=\mathbf{1 6 . 8} \mathbf{c m} . \end{aligned} $

Ex. 4 . In the figure given alongside $P Q R S$ is a square with four semi-circles drawn inside it in such a way that they meet each other at point $O$. Sides $P Q, Q R, R S$ and $P S$ are the respective diameters of the four semi-circles. Each side of the square is $8 cm$. What is the area of the shaded region?

Sol. Let the area of each shaded region be $x cm^{2}$, and that of each non shaded region be $y cm^{2}$. Given, Total area of square $=64 cm^{2}$

Also, Area of square $=4(x+y)$

$\Rightarrow 4(x+y)=64 \Rightarrow x+y=16$

Area of semicircle $A O B=x+y+x=\frac{1}{2} \times \pi \times(4)^{2}$

$\Rightarrow 2 x+y=8 \pi$

$\therefore \quad$ On subtracting equ. (i) from equ. (ii), we get $x=8 \pi-16$

$\therefore \quad$ Total area of shaded region $=4 x=4(8 \pi-16)=\mathbf{3 2}(\pi-2) \mathbf{c m}^{2}$.

Ex. 5 . The diagram given alongside represents three circular garbage cans each of unit radius. The three cans are touching each other as shown. Find, in metres, the perimeter of the rope encompassing the three cans and hence the area of the circumscribing circle.

Sol. $A B=C D=E F=$ Distance between two radii $=2 cm$

$ \angle A O F=\angle B P C=\angle D Q E=120^{\circ} . $

$\therefore \quad$ Perimeter of the circumscribing figure

$ =A B+C D+E F+\text{ Circumferences of sectors }(F O A+B P C+D Q E) $

Since, three equal sectors of $120^{\circ}=1$ full circle of same radius.

$\therefore \quad$ Perimeter of the circumscribing figure $=2 \pi r+2+2+2=(2 \pi r+6) cm=(2 \pi+6) cm$

If $R$ is the radius of the circumscribing circle, then,

$2 \pi R=2 \pi+6=(2 \times 3.14+6) cm=12.28 cm$

$\Rightarrow \quad R=\frac{12.28}{2 \times 3.14}=\mathbf{2} \mathbf{~ c m .}$

$\therefore \quad$ Area of circumscribing circle $=\pi R^{2}=(3.14 \times 4) cm^{2}=\mathbf{1 2 . 5 6} \mathbf{~ c m}^{2}$.

Ex. 6 . Let $S_1$ be a square of side $a$. Another square $S_2$ is formed by joining the mid-points of the sides of $S_1$. The same process is applied to $S_2$ to form yet another square $S_3$ and so on. If $A_1, A_2, A_3, \ldots . . .$. be the areas and $P_1, P_2, P_3, \ldots . . . .$. be the perimeters of $S_1, S_2, S_3, \ldots . . . . .$. respectively, then what does the ratio $\frac{P_1+P_2+P_3+\ldots . .}{A_1+A_2+A_3+\ldots . .}$ equal to?

(CAT 2004)

Sol. Each side of $S_1=a$ units

$\therefore$ Each side of $S_2=\sqrt{(\frac{a}{2})^{2}+(\frac{a}{2})^{2}}=\frac{a}{\sqrt{2}}$ units

$\Rightarrow$ Each side of $S_3=\sqrt{(\frac{a}{2 \sqrt{2}})^{2}+(\frac{a}{2 \sqrt{2}})^{2}}=\sqrt{\frac{2 a^{2}}{8}}=\frac{a}{2}$ units

$\Rightarrow$ Each side of $S_4=\sqrt{(\frac{a}{4})^{2}+(\frac{a}{4})^{2}}=\sqrt{\frac{2 a^{2}}{16}}=\frac{a}{2 \sqrt{2}}$ unit and so on.

$\therefore \quad P_1+P_2+P_3+P_4+\ldots \ldots \ldots=4 a+\frac{4 a}{\sqrt{2}}+\frac{4 a}{2}+\frac{4 a}{2 \sqrt{2}}+\ldots \ldots$

This is an infinite $G P$ with 1 st term $4 a$ and common ratio $\frac{1}{\sqrt{2}}$

$\therefore \quad P_1+P_2+P_3+P_4+\ldots \ldots \ldots=\frac{4 a}{1-\frac{1}{\sqrt{2}}}=\frac{4 a \times \sqrt{2}}{(\sqrt{2}-1)}=\frac{4 \sqrt{2} a}{(\sqrt{2}-1)}$

(Sum of infinite $G P=\frac{\text{ First term }}{1-\text{ Common Ratio }}$ )

Similarly, $A_1+A_2+A_3+A_4+\ldots \ldots \ldots=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}+\frac{a^{2}}{8}+\ldots \ldots \ldots \ldots=\frac{a^{2}}{1-\frac{1}{2}}=2 a^{2}$

$\therefore \frac{P_1+P_2+P_3+P_4+\ldots}{A_1+A_2+A_3+A_4+\ldots}=\frac{4 \sqrt{2} a}{(\sqrt{2}-1) \times 2 a^{2}}=\frac{2 \sqrt{2}}{(\sqrt{2}-1) a}=\frac{2 \sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1) a}=\frac{4+2 \sqrt{2}}{a}=\frac{\mathbf{2}(\mathbf{2}+\sqrt{\mathbf{2}})}{\boldsymbol{a}}$.

Ex. 7 . In the adjoining figure, there are three semi-circles in which $B C=6 cm$ and $B D=6 \sqrt{3} cm$. What is the area of the shaded region in $cm^{2}$ ?

Sol. Join $D A$ and $D C$.

$\angle A D C=90^{\circ}$ (Angle in a semi-circle is a right angle).

$\therefore A B \cdot B C=B D^{2}$

[From Geometry, $B D$ is the mean proportional between $A B$ and $B C$.]

$\Rightarrow A B \cdot 6=(6 \sqrt{3})^{2} \Rightarrow A B=18 cm$

$\therefore \quad$ Diameter of largest circle $=A B+B C=(18+6) cm=24 cm$

$\therefore$ Area of shaded region

$=$ Area of largest semi-circle - Sum of the areas of two smaller semi-circles

$=\frac{\pi \times 12^{2}}{2}-(\frac{\pi \times 9^{2}}{2}+\frac{\pi \times 3^{2}}{2})=72 \pi-45 \pi=\mathbf{2 7} \pi \mathbf{~ c m}^{2}$.

Ex. 8 . Consider a circle with unit radius. There are seven adjacent sectors $S_1, S_2, S_3, \ldots \ldots . ., S_7$ in a circle such that their total area is $\frac{1}{8}$ of the area of the circle. Further, the area of the $j$ th sector is twice that of the $(j-1)$ th sector for $j=2, \ldots . . . .7$. What is the angle in radians subtended by the arc of $S_1$ at the centre of the circle?

(CAT 2000)

Sol. Let the area of sector $S_1$ be $x$ units. Then the areas of the next six adjacent sectors $S_2, S_3, S_4, S_5, S_6, S_7$ will be $2 x$, $4 x, 8 x, 16 x, 32 x$ and $64 x$ units respectively.

$\therefore \quad$ Combined area of sectors $=x+2 x+4 x+8 x+16 x+32 x+64 x=127 x$

Now we know that the area of sectors is directly proportional to the angle subtended by the arc at the centre for a given radius.

Angle subtended by the combined sector $=\frac{2 \pi}{8}=\frac{\pi}{4}$ radians.

$\therefore \quad \frac{\text{ Area of } S_1}{\text{ Combined area of sectors }}=\frac{\text{ Angle subtended by } S_1}{\text{ Angle subtended by combined sectors }}$

$\Rightarrow \frac{x}{127 x}=\frac{\text{ Required angle }}{\pi / 4} \Rightarrow$ Required angle $=\frac{\pi}{127 \times 4}=\frac{\pi}{\mathbf{5 0 8}}$.

PRACTICE SHEET

~~ 1. The radius of a circle is so increased, that its circumference is increased by $5 %$. The area of the circle, then increases by

(a) $12.5 %$

(b) $10.25 %$

(d) $11.25 %$

(SNAP 2009)

~~ 2. A $36^{\circ}$ sector of a circle has area $3.85 cm^{2}$. What is the length of the arc of the sector?

(a) $1.1 cm$

(b) $3.5 cm$

(d) $2.2 cm$ ~~ 3. A car has two wipers which do not overlap. Each wiper has a blade of length $24 cm$ sweeping through an angle of $120^{\circ}$. Find the total area cleaned at each sweep of the blades.

(a) $192 \pi cm^{2}$

(b) $224 \pi cm^{2}$

(d) $240 \pi cm^{2}$ ~~ 4. A square of side length $\boldsymbol{a}$ is inscribed in a circle as shown in the figure. What is the area of the shaded region?

(a) $\frac{(3 \pi-2) a^{2}}{4}$

(b) $\frac{(\pi+4) a^{2}}{8}$

(d) $\frac{(3 \pi+2) a^{2}}{8}$

(CDS 2006)

~~ 5. In the given figure, $A B C$ is a right angled triangle, right angled at $B . B C=21 cm$ and $A B=28 cm$. Width $A C$ as the diameter of a semi-circle and width $B C$ as the radius of a quarter circle are drawn. What is the area of the shaded portion?

(a) $425 cm^{2}$

(b) $425.75 cm^{2}$

(d) $428.75 cm^{2}$

(CDS 2008)

~~ 6. There are two circles intersecting each other. Another smaller circle with centre $O$, is lying between the common region of the two larger circles. Centres of

the circle (i.e., $A, O$ and $B$ ) are lying on a straight line. $A B$ $=16 cm$ and the radii of the larger circles are $10 cm$ each. What is the area of the smaller circle?

(a) $4 \pi cm^{2}$

(b) $2 \pi cm^{2}$

(d) $\frac{\pi}{4} cm^{2}$

~~ 7. $A B C D$ and $E F G A$ are the squares of side $4 cm$ each. In square $A B C D$, $D M B$ and $P M Q$ are the arcs of circles with centres at $A$ and $C$ respectively. In square $A E F G$, the shaded region is enclosed by two arcs of circles with centres at $A$ and $F$ respectively. What is the ratio of the shaded regions of

the squares $A B C D$ and $A E F G$ respectively.

(a) $\frac{2+\pi(\sqrt{2}-2)}{(\pi-2)}$

(b) $\frac{(\pi-2)}{2(\sqrt{2}+1-\pi)}$

(d) None of these

~~ 8. There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let $A$, $B$ and $C$ be three distinct points on the perimeter of the outer circle such that $A B$ and $A C$ are tangents to the inner circle. If the area of the outer circle is 12 square centimetres, then the area of $\triangle A B C$ (in sq cm) would be:

(a) $\pi \sqrt{12}$

(b) $\frac{9}{\pi}$

(d) $\frac{6 \sqrt{3}}{\pi}$

(CAT 2003)

~~ 9. Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side $1 cm$. The area in sq $cm$ of the portion that is common to the two circles is:

(a) $\frac{\pi}{4}$

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

(d) $\sqrt{2}-1$

(CAT 2005)

~~ 10 A punching machine is used to punch a circular hole of diameter two units form a square sheet of aluminium of width 2 units as shown in the diagram. The hole is punched such that the circular hole touches one corner $P$ of the square sheet and the diameter of the hole originating at $P$ is in line with a diagonal of the square. The proportion of the sheet area that remains after

punching is:

(a) $(\pi+2) / 8$

(b) $(6-\pi) / 8$

(d) $(\pi-2) / 4$ (e) $(\pi-2) / 6$

(CAT 2006) ~~ 11 In the given figure, the circle with centre $O$ has a radius of 4. If the area of the shaded region is $14 \pi$, then what is the value of $x$ ?

(a) 90

(b) 60

(d) 45

~~ 12 $P Q R S$ is the diameter of a circle of radius $6 cm$. The lengths $P Q$, $Q R$ and $R S$ are equal. Semicircles are drawn with $P Q$ and $Q S$ as diameters as shown in the figure alongside. Find the ratio of the area of the shaded region to that of the unshaded region.

(a) $1: 2$

(b) $25: 121$

(d) $5: 13$

~~ 13 Two goats are tethered to diagonally opposite vertices of a field formed by joining the mid-points of the adjacent sides of another square field of side $20 \sqrt{2} m$. What is the total grazing area of the two goats?

(a) $100 \pi m^{2}$

(b) $50(\sqrt{2}-1) \pi m^{2}$

(d) $200 \pi(2-\sqrt{2}) m^{2}$ ~~ 14 In the given figure, the centre of the circle is $A$ and $A B C D E F$ is a regular hexagon of side $6 cm$. What is the approximate area of segment $B P F$ ? (Take $\pi=3.14$ )

(a) $25 cm^{2}$

(b) $22 cm^{2}$

(d) $30 cm^{2}$

~~ 15 In the given figure, crescent is formed by two circles which touch at point $A$. $O$ is the centre of the bigger circle. If $C B=9 cm$ and $D E=5 cm$, the area of the shaded portion is

(a) $625 cm^{2}$

(b) $614.25 cm^{2}$

(d) $671.35 cm^{2}$

~~ 16 $M$ is the centre of the circle. $Q S=10 \sqrt{2} . P R=R S$ and $P R | Q S$. Find the approximate area of the shaded region. (Take $\pi=3.14)$

(a) 100 sq. units

(b) 114 sq. units

(d) 200 sq. units

~~ 17 $A B C D$ is a square. A circle is inscribed in the square. Also taking $A, B, C, D$ as the centres, four quadrants are drawn inside the circle touching each other at the mid-points of the sides of the square. Area of the square is 4

$cm^{2}$. What is the area of the shaded region?

(a) $(4-\frac{3 \pi}{2}) cm^{2}$

(b) $(2 \pi-4) cm^{2}$

(d) $(4-\frac{\pi}{2}) cm^{2}$

~~ 18 The given diagram comprises of two sectors with angles $30^{\circ}$ and $60^{\circ}$, their radii are $7 cm$ and $14 cm$ respectively. Both have centre $O$. The perimeter of the

shaded region is

(a) $54 \frac{1}{3} cm$

(b) $105 cm$

(d) $77 cm$

~~ 19 A circle is circumscribed by the rhombus which in turn is made up by joining the mid-points of a rectangle whose sides are $12 cm$ and $16 cm$ respectively. What is the area of the circle?

(a) $\frac{625 \pi}{26}$

(b) $\frac{676 \pi}{25}$

(d) Can’t be determined ~~ 20 In the figure given alongside, $\triangle A B C$ is equilateral with area $S . M$ is the mid-point of $B C$ and $P$ is a point on $A M$ extended such that $M P=B M$. If the semi-circle on $A P$ intersects $C B$ extended at $Q$ and the area of a square with $M Q$ as a side is $T$, which of the following is true?

(a) $T=\sqrt{2} S$

(b) $T=S$

(d) $T=2 S$

(CAT 2011)

ANSWERS

~~ 1. (b) ~~ 2. $(d)$ ~~ 3. $(c)$ ~~ 4. $(d)$ ~~ 5. $(d)$ ~~ 6. (a) ~~ 7. (a) ~~ 8. (c) ~~ 9. (b) ~~ 10 (b) ~~ 11 $(d)$ ~~ 12 $(d)$ ~~ 13 $(a)$ ~~ 14 $(b)$ ~~ 15 $(c)$ ~~ 16 (c) ~~ 17 (b) ~~ 18 (c) ~~ 19 $(c)$ ~~ 20 $(b)$

HINTS AND SOLUTIONS

~~ 1. Let $r$ be the radius of the circle. Then, Increased circumference $=2 \pi r+2 \pi r \times \frac{5}{100}=\frac{21 \pi r}{10}$

$\therefore$ If $R$ be the radius of the increased circle, then

$2 \pi R=\frac{21 \pi r}{10} \Rightarrow R=\frac{21 r}{20}$

$\therefore$ Radius of increased circle $=\frac{21 r}{20}$

$\therefore$ Increased area $=\pi(\frac{21 r}{20})^{2}-\pi r^{2}=\frac{441}{400} \pi r^{2}-\pi r^{2}$

$ =\frac{41 \pi r^{2}}{400} $

$\therefore$ Percentage increase in area $=\frac{\frac{41 \pi r^{2}}{400}}{\pi r^{2}} \times 100 %=\mathbf{1 0 . 2 5 %}$

~~ 2. Area of a sector $=\frac{\theta}{360^{\circ}} \times \pi r^{2}$

$\Rightarrow \frac{36}{360} \times \frac{22}{7} \times r^{2}=3.85$

$\Rightarrow r^{2}=\frac{3.85 \times 10 \times 7}{22}=12.25$

$\Rightarrow r=3.5 cm$.

$\therefore$ Length of $arc=\frac{\theta}{360^{\circ}} \times 2 \pi r=\frac{36}{360} \times 2 \times \frac{22}{7} \times 3.5=\mathbf{2 . 2} \mathbf{~ c m}$.

~~ 3. Total area cleaned at each sweep of the blades

$=2 \times$ Area of sector cleaned by the sweep of one blade

$=2 \times\begin{bmatrix}\frac{120}{360} \times \pi \times(24)^{2}\end{bmatrix}$ $\begin{bmatrix} \therefore \text{ radius of sector } \\ \text{= length of blade } \end{bmatrix} $

$=384 \pi cm^{2}$. ~~ 4. Let the side of the square $=\boldsymbol{a}$

Then diameter of circle $=$ diagonal of square $=a \sqrt{2}$

$\Rightarrow$ Radius of circle $\frac{a \sqrt{2}}{2}=\frac{a}{\sqrt{2}}$

$\therefore$ Area of circle $=\pi(\frac{a}{\sqrt{2}})^{2}=\frac{\pi a^{2}}{2}$

As diagonals of a square bisect each

other at right angles,

$\angle A O B=90^{\circ}$ and $O A=O B=\frac{a}{\sqrt{2}}$

$\therefore$ Area of $\triangle O A B=\frac{1}{2} \cdot \frac{a}{\sqrt{2}} \cdot \frac{a}{\sqrt{2}}=\frac{a^{2}}{4}$

Hence area of unshaded portion $=\frac{90^{\circ}}{360^{\circ}} \times \pi \times(\frac{a}{\sqrt{2}})^{2}-\frac{a^{2}}{4}$

$ =\frac{\pi a^{2}}{8}-\frac{a^{2}}{4} $

$\therefore$ Area of shaded region

$=$ Area of circle - Area of unshaded region

$ \begin{aligned} & =\frac{\pi a^{2}}{2}-(\frac{\pi a^{2}}{8}-\frac{a^{2}}{4}) \\ & =\frac{4 \pi a^{2}-\pi a^{2}+2 a^{2}}{8}=\frac{3 \pi a^{2}+2 a^{2}}{8}=\frac{(3 \pi+2) a^{2}}{8} . \end{aligned} $

~~ 5. In $\triangle A B C$,

$ \begin{aligned} A C & =\sqrt{A B^{2}+B C^{2}} \\ & =\sqrt{28^{2}+21^{2}} cm \\ & =\sqrt{784+441} cm=\sqrt{1225} cm \\ & =35 cm \end{aligned} $

$\therefore$ Area of shaded portion $=$ Area of semi-circle $A C E$ + Area of $\triangle A B C$ - Area of quadrant circle $B C D$.

$=\frac{\pi}{2}(\frac{A C}{2})^{2}+\frac{1}{2} \times B C \times B A-\frac{\pi}{4}(B C)^{2}$

$=\frac{1}{2} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}+\frac{1}{2} \times 21 \times 28-\frac{1}{4} \times \frac{22}{7} \times 21 \times 21$

$=\frac{5 \times 11 \times 35}{4}+\frac{21}{2}[28-33]$

$=\frac{1925}{4}-\frac{105}{2}=\frac{1925-210}{4}=\frac{1715}{4}=\mathbf{4 2 8 . 7 5} cm^{2}$.

~~ 6. $A B=16 cm$

$A Q=10 cm$

$A O=\frac{A B}{2}=8 cm$.

$\therefore O Q=A Q-A O=10 cm-8 cm=2 cm$.

$\therefore$ Area of smaller circle with centre $O=\pi \times(2)^{2}=\mathbf{4} \pi cm^{2}$.

~~ 7. Area of square $A B C D=$ Area of square $A E F G=16 cm^{2}$

Area of quadrant $A D M B=\frac{1}{4} \times \pi \times(4)^{2}=4 \pi cm^{2}$

Radius of the smaller quadrant $C P M Q=C M=A C-A M$

$ =4 \sqrt{2}-4=4(\sqrt{2}-1) cm . $

$\therefore$ Area of the smaller quadrant

$ \begin{aligned} & =\frac{1}{4} \times \pi \times[4(\sqrt{2}-1)]^{2} \\ & =4 \pi(3-2 \sqrt{2}) \end{aligned} $

Area of shaded region inside the square $A B C D$

$=$ Area of sq. $A B C D-($ Area of quadrant $A D M B+$ Area of quadrant $C P M Q$ )

$ \begin{aligned} & =16-[4 \pi+4 \pi(3-2 \sqrt{2})] \\ & =16-[4 \pi(1+3-2 \sqrt{2})] \\ & =16-[4 \pi(4-2 \sqrt{2})] \\ & =8[2-2 \pi+\sqrt{2} \pi] \end{aligned} $

Now area of quadrants $A E G$ and $E F G$

$=2$ area of quadrant $A E G=2 \times \frac{1}{4} \times \pi \times(4)^{2}=8 \pi$.

$\therefore$ Area of shaded region inside the square $A E F G$

$=$ Sum of the areas of quadrants $A E G$ and $E F G-$ Area of square $A E F G$

$=8 \pi-16=8(\pi-2)$

$\therefore$ Required ratio $=\frac{8(2-2 \pi+\sqrt{2} \pi)}{8(\pi-2)}=\frac{[\mathbf{2}+\pi(\sqrt{2}-2)]}{(\pi-2)}$. ~~ 8. As a line through the centre of a circle is perpendicular to the tangent at the point of contact $\angle O D A=90^{\circ}$

$\frac{\text{ Area of outer circle }}{\text{ Area of inner circle }}=\frac{4}{1}$ (given)

$\Rightarrow \frac{(\text{ Radius of outer circle })^{2}}{(\text{ Radius of inner circle })^{2}}=\frac{4}{1}$

$\Rightarrow \frac{O A}{O D}=\frac{2}{1}$

$\therefore \sin \angle O A B=\frac{O D}{O A}=\frac{1}{2}$

$\therefore \angle O A D=30^{\circ} \Rightarrow \angle B A C=60^{\circ} \Rightarrow \triangle A B C$ is equilateral As, area of outer circle $(r)=12 cm^{2}$, so

$\pi r^{2}=12 \Rightarrow r=O B=\sqrt{\frac{12}{\pi}} \Rightarrow O D=\frac{1}{2} \sqrt{\frac{12}{\pi}}$

$\therefore$ In $\triangle O D B, D B^{2}=O B^{2}-O D^{2}=\frac{12}{\pi}-\frac{1}{4} \times \frac{12}{\pi}=\frac{9}{\pi}$

$\Rightarrow D B=\frac{3}{\sqrt{\pi}}$

$\therefore$ Side $A B=2 \times \frac{3}{\sqrt{\pi}}=\frac{6}{\sqrt{\pi}}$

$\therefore$ Area of $\triangle A B C=\frac{\sqrt{3}}{4} \times side^{2}=\frac{\sqrt{3}}{4} \times \frac{36}{\pi}=\frac{\mathbf{9} \sqrt{3}}{\pi}$.

~~ 9. Since each side of the square $A O B O^{\prime}=1 cm$, so radius $O A=O^{\prime} A=1 cm$

$\therefore$ Area of each circle $=\pi r^{2}$

$ =\pi \text{ sq. } cm $

Area of sector $A O B$

$ =\frac{90^{\circ}}{360^{\circ}} \times \pi=\frac{\pi}{4} $

$\Rightarrow$ Area of sector $A O^{\prime} B=\frac{\pi}{4}$

$\therefore$ Common area $=$ Area of sector $A O B+$ Area of sector

$A O^{\prime} B-$ Area of square $=\frac{\pi}{4}+\frac{\pi}{4}-1=\frac{\pi}{\mathbf{2}}-\mathbf{1}$.

~~ 10 In $\triangle F A E, \angle F A E=90^{\circ}$, as it is an angle of a square. $E F$ will be the diameter of the circle as an angle subtended by a diameter of the circumference of a circle $=90^{\circ}$. So $E F$ will pass through centre O.

In $\triangle A O F$ and $A O E$,

$O F=O E$ (radii of circle)

$A O=O A$ (common)

$\angle O A F=\angle O A E=45^{\circ}$

$\therefore \triangle A O F \sim \triangle A O E \Rightarrow A F=A E$

Also, area of $\triangle A F E=\frac{1}{2} \times A F \times A E=\frac{1}{2} A F^{2}$

Given $F E=2$

$\Rightarrow A F^{2}+A E^{2}=F E^{2}=2^{2}=4 \Rightarrow A F^{2}=2$ $\therefore$ Area of $\triangle A F E=\frac{1}{2} \times 2=1$.

Area of shaded region 2

$=2 \times($ Area of sector $A O E-$ Area of $\triangle A O E)$

$=2(\frac{\pi r^{2}}{4}-\frac{1}{2} \times 1 \times 1)=2(\frac{\pi}{4}-\frac{1}{2})$ as $r=1=\frac{\pi}{2}-1$

$\therefore$ Area of unshaded region of the circle

$=$ Area of circle - Area of shaded region 2

$=\pi r^{2}-\frac{\pi-2}{2}=\frac{2 \pi-(\pi-2)}{2}=\frac{\pi+2}{2}($ as $r=1)$

$\therefore$ Area of shaded region $1=$ Area of square - Area of unshaded region of circle.

$ =2^{2}-\frac{\pi+2}{2}=\frac{6-\pi}{2} $

$\therefore$ Proportion of the sheet that remains after punching

$=\frac{\text{ Area of region } 1}{\text{ Total area of the square }}=\frac{6-\pi}{\frac{2}{4}}=\frac{6-\pi}{8}$.

~~ 11 Area of circle $=\pi r^{2}=16 \pi \quad(\because r=4)$

$\angle O B A=90^{\circ}$ as $O B \perp$ tangent at $B$.

$\Rightarrow \angle A O B=90^{\circ}-x$

$\therefore$ Area of sector $O C B=$ Area of circle - Area of shaded region

$ =16 \pi-14 \pi=2 \pi cm^{2} $

$\therefore \frac{90^{\circ}-x}{360^{\circ}} \times 16 \pi=2 \pi$

$\Rightarrow \boldsymbol{x}=\mathbf{4 5}^{\circ}$

~~ 12 $P Q=Q R=R S=\frac{12}{3}=4 cm$

Area of unshaded region

$ \begin{aligned} & =\frac{\pi}{2} \times(6)^{2}+\frac{\pi}{2} \times(4)^{2} \\ & =18 \pi+8 \pi=26 \pi \end{aligned} $

Area of shaded region

$ \begin{aligned} & =\frac{\pi}{2} \times(6)^{2}-\frac{\pi}{2} \times(4)^{2} \\ & =18 \pi-8 \pi=10 \pi . \end{aligned} $

$\therefore$ Required ratio $=\frac{10 \pi}{26 \pi}=5: 13$.

~~ 13 $A B C D$ is original square, the mid-points of whose sides are joined to give the square $P Q R S$.

Let the goats be tethered at points $P$ and $R$ or $Q$ and $S$. Then, the length of the rope $=10 \sqrt{2} m$.

$\therefore$ The two goats together will graze an area equal to area of a semi-circle with radius $10 \sqrt{2} m$.

$\therefore$ Total area grazed $=\frac{\pi r^{2}}{2}=\frac{\pi \times(10 \sqrt{2})^{2}}{2}=\mathbf{1 0 0} \mathbf{\pi m}^{2}$.

~~ 14 As $\angle B A F$ is an angle of the regular hexagon, $\angle B A F=120^{\circ}$.

Draw $A K \perp B F$. The perpendicular from the centre of the circle to a chord bisects the chord, so $B K=K F$.

Also $\triangle A B F$ being an isosceles triangle, $\angle B A K=\frac{1}{2} \times \angle B A F=60^{\circ}$

$\therefore \angle A B K=30^{\circ}$

Also radius of the circle $=$ side $A B$ of hexagon $=6 cm$.

In $\triangle A B K, A K=A B \sin 30^{\circ}=6 \times \frac{1}{2}=3 cm$, and

$ B K=A B \cos 30^{\circ}=6 \times \frac{\sqrt{3}}{2}=3 \sqrt{3} cm . $

$\therefore B F=2 \times B K=6 \sqrt{3} cm$.

Now, area of segment $B P F=$ Area of sector $A B P F-$ Area of $\triangle A B F$

$ \begin{aligned} & =\frac{120}{360} \times 3.14 \times 6^{2}-\frac{1}{2} \times 6 \sqrt{3} \times 3 \\ & =37.68 cm^{2}-15.57 cm^{2} \\ & =\mathbf{2 2 . 1 1} cm^{2} \sim \mathbf{2 2} \mathbf{c m}^{2} \end{aligned} $

~~ 15 Let $r$ and $R$ be the radii of the inner and outer circle respectively.

Then, $A B-A C=C B$

$\Rightarrow 2 R-2 r=9 \Rightarrow 2(R-r)=9$

Join $A D$ and $D C$.

$\triangle A O D \sim \triangle D O C$

$\Rightarrow \frac{O D}{O A}=\frac{O C}{O D} \Rightarrow O D^{2}=O A \times O C$

$\Rightarrow(R-5)^{2}=R(R-9)$

$=O E-D E=R-5$

$(\because O D$

$\Rightarrow R^{2}-10 R+25=R^{2}-9 R$

$\Rightarrow R=25$

$\Rightarrow 2(25-r)=9$ from $(i)$

$\Rightarrow 2 r=41 \Rightarrow r=20.5$

$\therefore$ Area of shaded portion $=\pi(R^{2}-r^{2})$

$ \begin{aligned} & =\frac{22}{7}[(25)^{2}-(20.5)^{2}] \\ & =\frac{22}{7} \times(625-420.25) \\ & =\frac{22}{7} \times 204.75=643.5 cm^{2} \end{aligned} $

~~ 16 $\angle P Q S=\angle P R S=90^{\circ}$ (Angle in a semicircle)

$\because Q S | P R, \therefore \angle Q P R=\angle Q S R=90^{\circ}$

$\therefore P Q S R$ is a rectangle as a pair of opposite sides are parallel and all angles are $90^{\circ}$.

$\Rightarrow P R=Q S$ and $P Q=R S$

$\Rightarrow P Q=Q S=R S=P R=10 \sqrt{2} cm$. Hence $P Q S R$ is a square.

$\therefore$ Diagonal $P S$ of the square $=$ Diameter of the circle

$ =10 \sqrt{2} \times \sqrt{2}=20 cm \text{. } $

$\therefore$ Radius of the circle $=10 cm$.

Hence, Area of shaded portion

$=$ Area of semi-circle - Area of $\triangle P Q S$

$ \begin{aligned} & =\frac{\pi \times(10)^{2}}{2}-\frac{1}{2} \times 10 \sqrt{2} \times 10 \sqrt{2} \\ & =(50 \pi-100) cm^{2} . \\ & =(50 \times 3.14-100) cm^{2} \\ & =(157-100) cm^{2}=57 cm^{2} \simeq \mathbf{6 0} cm^{2} . \end{aligned} $

~~ 17 From the three figures we can see that,

Area of shaded region $=$ Area of square $-($ Total area shown by region $a+$ Total area shown by region $b$ )

Total area of region $a$

$=$ Area of square - Area of inscribed circle

$=[4-\pi(1)^{2}]=4-\pi(\because$ Each side of square $=\sqrt{4}=2 cm)$ Total area of region $b$

$=$ Area of square -4 (Area of a quadrant)

$ =4-4(\frac{1}{4} \pi \times(1)^{2})=(4-\pi) $

$\therefore \quad$ Required area $=4-(4-\pi+4-\pi)=(2 \pi-4) \mathbf{c m}^{2}$.

~~ 18 $\therefore$ Required perimeter $=A B$

arc length $B C+C D+arc$

length $D E+A E$

$=14+\frac{60}{360} \times 2 \times \frac{22}{7} \times 14$

$+7+\frac{30}{360} \times 2 \times \frac{22}{7} \times 7+7$

$=28+\frac{44}{3}+\frac{11}{3}=28+\frac{55}{3}=\frac{139}{3}=\mathbf{4 6} \frac{\mathbf{1}}{\mathbf{3}} \mathbf{c m}$.

~~ 19 Let $A B C D$ be the given rectangle, $P Q R S$, the rhombus obtained on joining the mid-points of $A B C D$ and circle with centre $O$, the circle inscribed in the rhombus.

$ \begin{aligned} & \therefore D S=\frac{A D}{2}=6 cm=R O \\ & D R= \frac{D C}{2}=8 cm=O S \end{aligned} $

$\therefore$ In right angled triangle $R O S, S R=\sqrt{O S^{2}+R O^{2}}$

$ =\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10 cm $

$\therefore$ Area of $\triangle O R S=\frac{1}{2} \times O S \times O R=\frac{1}{2} \times O T \times S R$

$\Rightarrow \quad \frac{8 \times 6}{2}=\frac{O T \times 10}{2} \Rightarrow O T=\frac{48}{10} cm$.

$\therefore$ Area of circle $=\pi r^{2}=\pi \times(\frac{48}{10})^{2}=\frac{\mathbf{5 7 6}}{\mathbf{2 5}} \boldsymbol{\pi}$.

~~ 20 $\angle P Q A=90^{\circ}$ (Angle in a semi-circle)

$\therefore(M P) \cdot(M A)=M Q^{2}$. (Mean

Proportional)

Let $M P=M B=1($ Given $M P=B M)$

Since $\triangle A B C$ is an equilateral $\triangle$,

$A M=\frac{\sqrt{3}}{2} \times B C=\sqrt{3}$.

$\therefore$ From (i) $1 \cdot \sqrt{3}=M Q^{2} \to M Q=3^{\frac{1}{4}}$

Area of square $=T=(M Q)^{2}=\sqrt{3}$

Area of $\triangle A B C=S=\frac{\sqrt{3}}{4}(B C)^{2}=\frac{\sqrt{3}}{4} \times 4=\sqrt{3}$

$\therefore \boldsymbol{T}=\boldsymbol{S}$.

SELF ASSESSMENT SHEET

~~ 1. The area of the shaded region in the figure given alongside is:

(a) $\frac{a^{2}}{2}(\frac{\pi}{2}-1)$ sq. units

(b) $a^{2}(\pi-1)$ sq. units

(d) $\frac{a^{2}}{2}(\pi-1)$ sq. units

(SSC 2007)

~~ 2. Three circles of diameter $10 cm$ each are bound together by a rubber band, as shown in the figure. The length of the rubber band in $cm$, if it is stretched as shown is:

(a) 30

(b) $30+10 \pi$

(d) $60+20 \pi$

(SSC 2011)

~~ 3. $C_1$ and $C_2$ are two concentric circles with centre at $O$. Their radii are $12 cm$ and $3 cm$ respectively. $B$ and $C$ are the points of contact of the two tangents drawn to $C_2$ from a point $A$ lying on the circle $C_1$. Then the area of quadrilateral $A B O C$ is:

(a) $\frac{9 \sqrt{15}}{2}$ sq. cm

(b) $12 \sqrt{15}$ sq. cm

(d) $6 \sqrt{15}$ sq. cm

(SSC 2013)

~~ 4. In the figure given alongside, the rectangle at the corner measures $10 cm$ $\times 20 cm$. The corner $A$ of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in $cm$ ?

(a) $10 cm$

(b) $40 cm$

(d) None of these

(CAT 2003)

~~ 5. Let $C$ be a circle with centre $P_0$ and $A B$ be the diameter of C. Suppose $P_1$ is the mid-point of the line segment $P_0 B$, $P_2$ is the mid-point of the line segment $P_1 B$ and so on. Let $C_1, C_2, C_3, \ldots \ldots$. be the circles with diameters $P_0 P_1, P_1 P_2$, $P_2 P_3, \ldots$. respectively. Suppose the circles $C_1, C_2, C_3 \ldots \ldots$ are all shaded. The ratio of the area of the unshaded portion of $C$ to the original circle $C$ is:

(a) $8: 9$

(b) $9: 10$

(d) $11: 12$

(CAT 2004)

~~ 6. Four identical coins are placed in a square. For each coins, the ratio of area to circumference is the same as the ratio of circumference to area. Then, find the area of the square that is not covered by the coins.

(a) $16(\pi-1)$

(b) $16(8-\pi)$

(d) $16(4-\pi / 2)$ ~~ 7. A regular hexagon is inscribed in a circle of radius $R$. Another circle is inscribed in the hexagon. Now another hexagon is inscribed in the second (smaller) circle. What is the ratio of the area of inner circle to the outer circle.

(a) $3: 4$

(b) $9: 16$

(d) None of these ~~ 8. If the sides $50 m$ and $130 m$ of a triangular field meet at an angle of $72^{\circ}$, then find the area in which wheat is cultivated

(shaded portion)

(a) $120 \pi m^{2}$

(b) $150 \pi m^{2}$

(d) $180 \pi m^{2}$

ANSWERS

~~ 1. (c) ~~ 2. (b) ~~ 3. (c) ~~ 4. (c) ~~ 5. (d) ~~ 6. (c) ~~ 7. (a) ~~ 8. (d)

HINTS AND SOLUTIONS

~~ 1. Since $C$ is the centre of the circle, radius of the circle $=a$ units

$\therefore$ Area of the semi-circle

$ =\frac{\pi a^{2}}{2} \text{ sq. units. } $

$A C=C D=C B=$ radii of the circle.

$\Rightarrow$ Both the triangles $\triangle A B C$ and $\triangle B C D$ are isosceles and are equal.

$\therefore$ Area of each triangle $=\frac{1}{2} a^{2}$.

$\therefore$ Total area of both the triangles $=2 \times \frac{1}{2} a^{2}=a^{2} \cdot$ sq. units.

$\therefore$ Area of shaded region $=\frac{\pi a^{2}}{2}-a^{2}=\boldsymbol{a}^{2}(\frac{\pi}{2}-1)$ sq. units.

~~ 2. Type Solved Ex. 5 .

Length of the rubber band $=3 d+2 \pi r$

$ \begin{aligned} & =3 \times 10+2 \times \pi \times 5 \\ & =(\mathbf{3 0}+\mathbf{1 0} \boldsymbol{\pi}) \mathbf{~ c m} . \end{aligned} $

~~ 3. Given, $O B=3 cm, O A=12 cm$.

As a line from the centre of the circle to

$\therefore A B=A C=\sqrt{12^{2}-3^{2}}$

$\therefore \sqrt{144-9}=\sqrt{135}=3 \sqrt{15} cm$.

$\therefore$ Area of $O B A C$

$=$ Area of $\triangle O B A+$ Area of $\triangle O C A$

$=2 \times \frac{1}{2} \times 3 \times 3 \sqrt{15} cm^{2}$

$=9 \sqrt{15} cm^{2}$.

~~ 4. Draw the perpendicular $O P$ and $A Q$ as shown in the figure and join the points $A$ and $O$, where $O$ is the centre of the circle.

$ \begin{aligned} & O A^{2}=(A Q^{2}+O Q^{2}) \\ &=(D Q-D A)^{2}+(O P-Q P)^{2} \\ & \Rightarrow r^{2}=(r-10)^{2}+(r-20)^{2} \\ & \quad(\because C P=D Q=O P=O A=\text{ radius } \end{aligned} $

of circle $=r$ )

$\Rightarrow r^{2}=r^{2}-20 r+100+r^{2}-40 r+400$

$\Rightarrow r^{2}-60 r+500=0$

$\Rightarrow(r-10)(r-50)=0$

$\Rightarrow r=10,50$ but $r>10 \quad \therefore \boldsymbol{r}=\mathbf{5 0} \mathbf{~ c m}$.

~~ 5. Area of $C=\pi r^{2}$, where $P_0 B=r$ is the radius of the circle $C$.

Then radius of circle $C_1=r / 4$ radius of circle $C_2=r / 8$ and so on.

$\therefore$ Area of shaded portion

$=$ Area of $C_1+$ Area of $C_2+$

$=\pi(r / 4)^{2}+\pi(r / 8)^{2}+\ldots .$.

$=\frac{\pi r^{2}}{16}+\frac{\pi r^{2}}{64}+\ldots$.

$=\pi r^{2}[\frac{1}{16}+\frac{1}{64}+\ldots ..]$

$=\pi r^{2}[\frac{1 / 16}{1-1 / 4}](\because.$ Infinite $G P$ with $.a=\frac{1}{16}, r=\frac{1}{4})$

$=\pi r^{2} \times \frac{1}{6} \times \frac{4}{3}=\frac{\pi r^{2}}{\mathbf{1 2}}$.

$\therefore$ Required ratio $=\frac{\text{ Area of unshaded portion }}{\text{ Area of shaded portion }}=\frac{\pi r^{2}-\frac{\pi r^{2}}{12}}{\frac{\pi r^{2}}{12}}$

~~ 6. Let ’ $r$ ’ be the radius of each circle. Then, = 11:12.

$\frac{\pi r^{2}}{2 \pi r}=\frac{2 \pi r}{\pi r^{2}} \Rightarrow \pi r^{2}=2 \pi r \Rightarrow r=\mathbf{2}$

$(\because r \neq 0)$

$\therefore$ Length of side of square $=4 r=8$

$\therefore$ Area of square not covered by coins

$=$ Area of square - Area of 4 circles

$=64-4 \pi(2)^{2}=\mathbf{1 6}(\mathbf{4}-\boldsymbol{\pi})$.

~~ 7. Each side of the outer (larger) hexagon is equal to the radius of the circle which is $R$.

Now $O C=O N=O D=$ radii of smaller circle.

But $\frac{O N}{O A}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\Rightarrow O N=\frac{\sqrt{3}}{2} O A=\frac{\sqrt{3}}{2} R$ $=$ each side of inner hexagon

$\therefore$ Required ratio

$=\frac{\text{ Area of inner circle }}{\text{ Area of outer circle }}$

$ =\frac{\pi(\frac{\sqrt{3}}{2} R)^{2}}{\pi(R)^{2}}=\frac{\mathbf{3}}{\mathbf{4}} . $

~~ 8. $\angle A B C=90^{\circ}(\because O Q \perp B C$ and $O D \perp A B$ Line through the centre of a circle is perpendicular to the tangent of the circle at the point of

contact.)

$\therefore$ Given, $\angle B A C=72^{\circ} \Rightarrow \angle A C B=90^{\circ}-72^{\circ}=18^{\circ}$ $B C=\sqrt{A C^{2}-A B^{2}}=\sqrt{130^{2}-50^{2}}=\sqrt{14400}=120 m$.

(Pythagoras’ Theorem)

Let $B Q=x m \Rightarrow Q C=(120-x) m$ and $A D=(50-x) m$.

$\Rightarrow B D=x m, C P=(120-x) m, A P=(50-x) m$ (Tangents from the same point are equal)

$\therefore B Q+Q C+C P+A P+A D+D B=(130+120+50) m$

$\Rightarrow x+(120-x)+(120-x)+(50-x)+(50-x)+x=300$

$\Rightarrow 340-2 x=300 \Rightarrow 2 x=40 \Rightarrow x=20 m$.

$\Rightarrow$ Radius of circle $=O Q=D B=20 m(\because O D B Q$ is a square $)$

$\therefore$ Area of circle $=\pi r^{2}=400 \pi m^{2}$

Now in cyclic quadrilateral $P O Q C, \angle P O Q=180^{\circ}-\angle A C B$

$\therefore$ Area of sector $P O Q=\frac{162^{\circ}}{360^{\circ}} \times$ Area of circle

$ =\frac{162}{360} \times 400 \pi=180 \pi \mathbf{m}^{2} $

Volume and Surface Area of Solids (Further Continued from Class IX)

KEY FACTS

~~ 1. Cuboid: For a cuboid of length $\boldsymbol{l}$, breadth $\boldsymbol{b}$ and height $\boldsymbol{h}$,

Volume of cuboid $=(l \times b \times h)$ cu. units.
Whole surface of cuboid $=2(l b+b h+l h)$ sq. units.
Diagonal of cuboid $=\sqrt{l^{2}+b^{2}+h^{2}}$ units.
Area of four walls $=2(l+b) h$ sq. units.

~~ 2. Cube: For a cube of edge length $\boldsymbol{a}$,

Volume $=a^{3}$ cu. units.
Whole surface $=6 a^{2}$ sq. units.
Diagonal $=\sqrt{3} a$ units.

~~ 3. 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,

Area of base (triangle) $=\frac{1}{2} \times b \times h$, if base $(b)$ and height $(h)$ of triangle are known.

Area of base (scalene triangle) $=\sqrt{s(s-a)(s-b)(s-c)}$, if all the sides $a, b, c$ are known where $s=\frac{a+b+c}{2}$
Area of base (equilateral triangle) $=\frac{\sqrt{3}}{4} a^{2}$, where ’ $a$ ’ is each side of the equilateral triangle.

For a hexagonal prism,

Area of base (regular hexagon) $=\frac{\sqrt{3}}{2}(\text{ edge })^{2}$.

~~ 4. (a) Right Circular Solid Cylinder: For a right circular cylinder of radius of base $(r)$ and perpendicular height $(h)$,

Curved Surface Area $(C S A)=\mathbf{2} \pi r h$ sq. units.
Total Surface Area $(T S A)=\mathbf{2} \pi \boldsymbol{r} \boldsymbol{h}+\mathbf{2} \pi \boldsymbol{r}^{2}=\mathbf{2} \pi \boldsymbol{r}(\boldsymbol{h}+\boldsymbol{r})$ sq. units

Volume $=\pi r^{2} h$ cu. units.

(b) Right Circular Hollow Cylinder: For a hollow cylinder, whose inner radius $=\boldsymbol{r}$, Outer radius $=\boldsymbol{R}$ and perpendicular height $=\boldsymbol{h}$,

Curved Surface Area $=$ External CSA + Internal CSA

$ =2 \pi R h+2 \pi r h=2 \pi \boldsymbol{h}(\boldsymbol{R}+\boldsymbol{r}) \text{ sq. units. } $

Total Surface Area $=$ Curved Surface Area + Area of bases

$ \begin{aligned} & =2 \pi h(R+r)+2 \pi(R^{2}-r^{2}) \\ & =2 \pi(R+r)(h+R-r) \text{ sq. units. } \end{aligned} $

Volume of material used in making the hollow cylinder $=\pi R^{2} h-\pi r^{2} h=\pi h(R^{2}-r^{2})$ cu. units.

~~ 5. Right Circular Cone: For a right circular cone of base radius ( $r$ ), perpendicular height

(h) and slant height $(l)$,

Curved Surface Area $=\pi r l=\pi r \sqrt{h^{2}+r^{2}}$ sq. units.
Total Surface Area $=\pi r l+\pi r^{2}=\pi r(l+r)=\pi r(\sqrt{h^{2}+r^{2}}+r)$ sq. units
Volume $=\frac{1}{3} \pi r^{2} h$ cu. units.
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 $\boldsymbol{r}$ the bounding radii

~~ 6. Sphere: For a sphere with radius $(r)$,

Volume $=\frac{4}{3} \pi r^{3}$ cu. units
Surface area $=4 \pi r^{2}$ sq. units

For a hollow sphere of external radius $(\boldsymbol{R})$ and internal radius $(\boldsymbol{r})$,

Volume $=\frac{4}{3} \pi(R^{3}-r^{3})$ cu. units

~~ 7. Hemisphere: For a hemisphere with radius ( $r$ ),

Volume $=\frac{2}{3} \pi r^{3}$ cu. units.
Curved surface area $=2 \pi r^{2}$ sq. units.

Total surface area $=3 \pi r^{2}$ sq. units.

For a hollow hemisphere of external radius $(\boldsymbol{R})$ and internal radius $(\boldsymbol{r})$,

Volume $=\frac{2}{3} \pi(R^{3}-r^{3})$ cu. units.
Curved surface area $=2 \pi(R^{2}+r^{2})$ sq. units.
Total surface area $=2 \pi(R^{2}+r^{2})+\pi(R^{2}-r^{2})$ sq. units.

~~ 8. Frustum of a cone: If a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of a cone.

If $\boldsymbol{R}$ and $\boldsymbol{r}$ are respectively the radii of the base and top of the frustum and $\boldsymbol{h}$, the height of the frustum, then

Volume of frustum of right circular cone $=\frac{\pi h}{3}[R^{2}+r^{2}+R r]$ cu. units.
Lateral curved surface area of frustum $=\pi(R+r) l$ sq. units,

where, $l=$ slant height of frustum $=\sqrt{h^{2}+(R-r)^{2}}$.

Total surface area of frustum $=$ Lateral surface area + Area of base + Area of top $=\pi(R+r) l+\pi R^{2}+\pi r^{2}=\pi[R^{2}+r^{2}+l(R+r)]$ sq. units.
Total surface area of bucket $=\pi(R+r) l+\pi r^{2}$ (as a bucket is open at the bigger end).

~~ 9. Pyramid: A solid figure with a polygonal base and triangular faces that meet at a common point (vertex) outside the plane of the base.

Types of Pyramids

Regular Pyramid: The base of the pyramid is a regular polygon and hence its lateral faces are equally sized.

Irregular Pyramid: The base of the pyramid is an irregular polygon and hence its lateral faces are not equally sized.

Right Pyramid: If the vertex or apex is directly above the centre of the base, it is a right pyramid, otherwise an oblique pyramid.

Pyramids can also be classified according to the shape of their bases :

Triangular Pyramid

Base: Triangle

Parts of a Pyramid:

Height: Perpendicular from vertex to base.
Lateral Edge: The edges through the vertex of a pyramid.
Slant Height: The height of a lateral face of a regular pyramid. It is the line segment joining the vertex to the mid-point of any one of the sides of the base.
Volume of a pyramid $=\frac{1}{3} \times$ Base Area $\times$ Height.
Surface area of a pyramid when all side faces are same $=$ Base Area $+\frac{1}{2} \times$ Perimeter of base $\times$ Slant height.
Surface area of a pyramid, when all side faces are different = Base Area + Lateral Area.
For a right pyramid with an equilateral triangle of side ’ $a$ ’ as base and height ’ $h$ ‘.
Lateral edge $=\sqrt{h^{2}+\frac{a^{2}}{3}}$
Slant height $=\sqrt{h^{2}+\frac{a^{2}}{12}}$
Lateral surface area $=\frac{1}{2} \times$ Perimeter of base $\times$ Slant height $=\frac{1}{2} \times 3 a \times \sqrt{h^{2}+\frac{a^{2}}{12}}$
Total surface area $=\frac{1}{3} \times 3 a \times \sqrt{h^{2}+\frac{a^{2}}{12}}+\frac{\sqrt{3}}{4} a^{2}$
Volume $=\frac{1}{3} \times \frac{\sqrt{3}}{4} a^{2} \times h=\frac{\sqrt{3}}{12} a^{2} h$
Area of one lateral face $=\frac{1}{2} \times$ Length of edge of base $\times$ Slant height $=\frac{1}{2} \times a \times \sqrt{h^{2}+\frac{a^{2}}{12}}$
For a regular tetrahedron (all edges are equal, all four faces including base are congurent equilateral triangles)
Height $=\sqrt{\frac{2}{3}}$ edge
Slant height $=\frac{\sqrt{3}}{2}$ edge
Lateral surface area $=\frac{3 \sqrt{3}}{4}(\text{ edge })^{2}$
Total surface area $=\sqrt{3}(\text{ edge })^{2}$
Volume $=\frac{\sqrt{2}}{12}(\text{ edge })^{3}$.

~~ 10 Immersion of Solids: If a solid or solids of given dimension is/are dropped in a vessel partly filled with water, and is submerged completely, then the rise in water level in the vessel can be calculated using the principle:

Volume of displaced water $=$ Volume of submerged solid, where the height of water displaced will be the required height.

~~ 11 When a solid is melted and converted to another solid, then volume of both the solids remain the same, assuming there is no wastage in conversions.

Also, Number of new solids obtained by recasting

$ =\frac{\text{ Volume of the solid that is melted }}{\text{ Volume of the solid that is made }} . $

~~ 12 Combination of Solids: When calculating the surface area of a combination of solids, you need to consider the visible surface of the component solids, which is normally the total curved surface area of the component solids. For volume, you need to calculate the total volume of the component solids.

13. Some common solids inscribed in a given solid or solid circumscribing other solid.

If a largest possible sphere is circumscribed by a cube of edge ’ $a$ ’ units, then radius of the sphere $=\frac{a}{2}$ units.

If a largest possible sphere is inscribed in a cylinder of radius ’ $a$ ’ units and height " $\boldsymbol{h}$ " units, then

(i) For $h>a$, radius of sphere $=a$ units

(ii) For $h<a$, radius $=\frac{h}{2}$ units.

If a largest possible cube is inscribed in a sphere of radius ’ $a$ ’ units, then the edge of the cube $=\frac{2 a}{\sqrt{3}}$ units.
If a largest possible cone is inscribed in a cylinder of radius ’ $a$ ’ units and height ’ $\boldsymbol{h}$ ’ units, then radius of the cone $=$ ’ $a$ ’ units, height of the cone $=$ ’ $h$ ’ units.

~~ 14 Volume of water that flows out through a pipe $=($ Cross-section area $\times$ Speed $\times$ Time $)$

SOLVED EXAMPLES

Combination of Solids

Ex. 1 . A toy is in the form of a cone mounted on a hemisphere such that the diameter of the base of the cone is equal to that of the hemisphere. If the diameter of the base of the cone is $6 \mathbf{~ c m}$ and its height is $4 cm$, what is the surface area of the toy in sq. $cm$. (Take $\pi=3.14$ )

Sol. For the conical part, $r=\frac{6}{2}=3 cm, h=4 cm, l=\sqrt{h^{2}+r^{2}}=\sqrt{3^{2}+4^{2}} cm=5 cm$ Surface area of the conical part $=\pi r l=(3.14 \times 3 \times 5) cm^{2}=47.1 cm^{2}$

For the hemispherical part, $r=\frac{6}{2}=3 cm$

Surface area of the hemispherical part $=2 \pi r^{2}=2 \times 3.14 \times 3 \times 3=56.52 cm^{2}$

$\therefore$ Surface area of the toy $=47.1 cm^{2}+56.52 cm^{2}=\mathbf{1 0 3 . 6 2} \mathbf{~ c m}^{2}$.

Ex. 2 . A storage tank consists of a circular cylinder with a hemisphere adjoined on either side. If the external diameter of the cylinder be $14 m$ and its length be $50 m$, then what will be the cost of painting it at the rate of ₹ 10 per sq. $m$ ?

(MAT 2003)

Sol. Let $r(=7) cm$ be the radius of the base of the cylinder, hence of hemispheres and $h$, the height of the cylinder

Surface area of the tank

$ \begin{aligned} & =\text{ Curved surface area of cylinder }+2 \times \text{ Curved surface area of hemisphere } \\ & =2 \pi r h+2 \times 2 \pi r^{2} \\ & =2 \times \frac{22}{7} \times 7 \times 50+4 \times \frac{22}{7} \times 7 \times 7 \\ & =2200+616=2816 m^{2} \end{aligned} $

$\therefore$ Required cost of painting $=2816 \times ₹ 10=₹ \mathbf{2 8 1 6 0}$.

Ex. 3 . A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of $₹ 2$ per square metre, if the radius of the base is 14 metres.

Sol. Total area to be painted $=$ Curved Surface Area of cylinder $+C S A$ of cone

$ \begin{aligned} C S A \text{ of cylinder } & =2 \pi r h=(2 \times \frac{22}{7} \times 14 \times 3) m^{2}=264 m^{2} \\ C S A \text{ of cone } & =\pi r l \text{ where } l=\text{ slant height } \\ l & =\sqrt{r^{2}+h^{2}}=\sqrt{14^{2}+(10.5)^{2}} m \\ & =\sqrt{196+110.25} m=\sqrt{306.25} m=17.5 m \\ \therefore \quad C S A \text{ of cone } & =(\frac{22}{7} \times 14 \times 17.5) m=770 m^{2} \end{aligned} $

$\therefore \quad$ Cost of painting $=₹ 2 \times$ Total CSA

$ =₹ 2 \times(264+770)=₹[2 \times 1034]=₹ \mathbf{2 0 6 8} . $

Immersion of Solids

Ex. 4 . In a cylindrical vessel of diameter $24 cm$ filled up with sufficient quantity of water, a solid spherical ball of radius $6 cm$ is completely immersed. What is the increase in height of water level?

(CDS 2012)

Sol. Let the height of the increased water level be $h cm$.

Then, Volume of water displaced $=$ Volume of the sphere

$\Rightarrow \pi r^{2} h=\frac{4}{3} \pi R^{3}$, (where, $r=$ radius of cylinder, $R=$ radius of sphere)

$\Rightarrow 12 \times 12 \times h=\frac{4}{3} \times 6 \times 6 \times 6 \Rightarrow h=\frac{4 \times 6 \times 6 \times 6}{3 \times 12 \times 12}=\mathbf{2} \mathbf{~ c m}$.

Ex. 5 . Marbles of diameter $1.4 cm$ are dropped into a cylindrical beaker containing some water and are fully submerged. The diameter of the beaker is $7 cm$. Find how many marbles have been dropped in it if the water rises by $5.6 cm$.

(SSC 2011)

Sol. Number of marbles $=\frac{\text{ Volume of raised water in the cylindrical beaker }}{\text{ Total Volume of marbles }}=\frac{\pi r^{2} h}{\frac{4}{3} \pi R^{3}}(\begin{matrix} \text{ where } r & =\text{ radius of beaker, } \\ R & =\text{ radius of marble }\end{matrix} )$

$ =\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 5.6}{\frac{4}{3} \times \frac{22}{7} \times 0.7 \times 0.7 \times 0.7}=\mathbf{1 5 0} . $

Frustum of a Cone

Ex. 6 . What is the capacity and surface area of a bucket, the radii of whose circular ends are $28 cm$ and $7 cm$ and height is $45 cm$ ?

Sol. Capacity of the bucket $=$ Volume of frustum of a cone $=\frac{\pi h}{3}[R^{2}+r^{2}+R r]$

$ \begin{aligned} & =\frac{22}{7} \times \frac{45}{3}[(28)^{2}+(7)^{2}+28 \times 7]=\frac{330}{7} \times[784+49+196] \\ & =\frac{330}{7} \times 1029=330 \times 147=48510 \text{ cu. cm. } \end{aligned} $

Surface area of the bucket = Lateral surface area of the frustum of the cone
$\hspace{38mm}$ + Area of the bottom

$\hspace {38mm}=\pi l(R+r)+\pi r^{2}$

(Slant height)l $ =\sqrt{h ^{2}+(R-r) ^{2}}=\sqrt{45 ^{2}+(28-7) ^{2}}=\sqrt{2025+441}=\sqrt{2466}=49.66 \ cm$ (approx)

$\therefore$ Surface area $=\frac{22}{7} \times 49.66 \times 35+\frac{22}{7} \times 7^{2}$

$ =(5462.6+154) \text{ sq. } cm=\mathbf{5 6 1 6 . 6} \text{ sq. } \mathbf{~ c m} . $

Ex. 7 . A right circular cone is divided into two portions by a plane parallel to the base and passing through a point which is at $\frac{1}{3}$ of the height from the top. The ratio of the volume of the smaller cone to that of the remaining frustum of the cone is:

(CDS 2001)

Sol. Let the height $A B$ of the cone be $h$ units. Then $A D=\frac{1}{3} h$

As $\triangle A D E \sim \triangle A B C($ similar $\triangle s) \frac{A D}{A B}=\frac{D E}{B C}=\frac{1}{3} \Rightarrow A D=\frac{1}{3} A B$ and $D E=\frac{1}{3} B C$

$ \begin{aligned} \therefore \quad \text{ Required ratio } & =\frac{\frac{1}{3} \pi(D E)^{2} \times A D}{\frac{1}{3} \pi(B C)^{2} \times A B-\frac{1}{3} \pi(D E)^{2} \times A D}=\frac{(D E)^{2} \times A D}{B C^{2} \times A B-D E^{2} \times A D} \\ & =\frac{\frac{1}{9} B C^{2} \times \frac{1}{3} A B}{B C^{2} \times A B-\frac{1}{9} B C^{2} \times \frac{1}{3} A B}=\frac{\frac{1}{27}}{1-\frac{1}{27}}=\frac{1 / 27}{26 / 27}=\mathbf{1 : 2 6 .} \end{aligned} $

Ex. 8 . A right circular cone is cut by two planes parallel to the base and trisecting the altitude. What is the ratio of the volumes of the three parts; top, middle and bottom respectively?

(CDS 2005)

Sol. The cone be divided into three parts by the two planes trisecting the cone. Let the height of each of the trisected portion be $h$ units.

It is obvious the $\triangle A D E, \triangle A F G$ and $\triangle A B C$ are similar. Let $DE=2 x$.

As the three given $\Delta s$ are similar, $\frac{F G}{D E}=\frac{A Q}{A P}=\frac{2 h}{h}$

$ \Rightarrow F G=2 D E=4 x $

Similarly $B C=3 D E=6 x$

$\therefore P E=x, Q G=2 x$ and $R C=3 x$

$ \begin{aligned} \text{ Now, Volume of cone } A B C & =\frac{1}{3} \times \pi \times(3 x)^{2} \times(3 h)=9 \pi x^{2} h \\ \text{ Volume of cone } A F G & =\frac{1}{3} \times \pi \times(2 x)^{2} \times(2 h)=\frac{8}{3} \pi x^{2} h \\ \text{ Volume of cone } A D E & =\frac{1}{3} \times \pi \times(x)^{2} \times h=\frac{1}{3} \pi x^{2} h \end{aligned} $

$\therefore \quad$ Volume of middle portion $D E F G=$ Vol. of cone $A F G-$ Vol. of cone $A D E=\frac{8}{3} \pi x^{2} h-\frac{\pi x^{2} h}{3}=\frac{7 \pi x^{2} h}{3} \ldots(i v)$ Volume of the lowermost portion $F G B C=$ Vol. of cone $A B C-$ Vol. of cone $A F G$

$ =9 \pi x^{2} h-\frac{8}{3} \pi x^{2} h=\frac{19 \pi x^{2} h}{3} $

$\therefore$ Reqd. ratio $=\frac{\pi x^{2} h}{3}: \frac{7 \pi x^{2} h}{3}: \frac{19 \pi x^{2} h}{3}=1: 7: 19$.

Ex. 9 . A container opened from the top and made up of a metal sheet is in the form of a frustum of a cone of height $16 cm$ with radii of its lower and upper ends as $8 cm$ and $20 cm$ respectively. Find the cost of the milk which can completely fill the container at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per $100 cm^{2}$. (Take $\pi=3.14$ )

(NCERT)

Sol. Reqd. cost of milk $=$ Vol. of the container $\times ₹ 20$. The container is in the shape of the frustum of a cone whose, height $(\boldsymbol{h})=16 cm$, radius of lower end $(\boldsymbol{r}_1)=8 cm$, radius of upper end $(\boldsymbol{r} _{\boldsymbol{2}})=20 cm$

$\therefore$ Volume of the container $=\frac{\pi}{3} \times h \times(r_1^{2}+r_2^{2}+r_1 r_2)$

[Using formula for frustum of a cone]

$ \begin{aligned} & =\frac{3.14}{3} \times 16 \times[(8)^{2}+(20)^{2}+160] \\ & =\frac{3.14}{3} \times 16 \times(64+400+160) \\ & =\frac{3.14 \times 16 \times 624}{3} cm^{3}=\frac{3.14 \times 16 \times 624}{3 \times 1000} L=10.45 L \text{ (approx) } \end{aligned} $

$\therefore$ Cost of milk $=₹(10.45 \times 20)=₹ 209$ (approx)

Area of the metal sheet used to make the container

$=$ Lateral surface area of the container + Area of the base $=\pi l(r_1+r_2)+\pi r_1{ }^{2}$

where, $l=\sqrt{h^{2}+(r_2-r_1)^{2}}=\sqrt{16^{2}+(20-8)^{2}}=\sqrt{256+144}=\sqrt{400}=20 cm$

$\therefore$ Required area $=3.14 \times 20 \times 28+3.14 \times 64=1758.4+200.96=1959.36 cm^{2}$

$\therefore$ Cost of metal sheet $=₹(\frac{8 \times 1959.36}{100})=₹ \mathbf{1 5 6 . 7 5}$ (approx).

Pyramids and Tetrahedron

Ex. 10 . A right pyramid $10 cm$ high has a square base of which the diagonal is $10 cm$. What is the whole surface area of the pyramid?

(CDS 1996)

Sol. Whole surface area of the pyramid

$=$ Lateral surface area + Area of the square base $+(\text{ Side })^{2}$

$=4 \times$ Area of triangular faces

Each side of the square $(a)=\frac{\text{ diagonal }}{\sqrt{2}}=\frac{10}{\sqrt{2}} cm$, Height $(h)=10 cm$

$E G=$ height of a triangular face

$A B=$ base of the triangular face

$\therefore$ Height of the triangle $=\sqrt{(10)^{2}+(\frac{1}{2} \times \frac{10}{\sqrt{2}})^{2}}=\sqrt{100+\frac{25}{2}}=\sqrt{112.5}=10.6 cm$ (approx)

$\therefore$ Whole surface area $=4 \times \frac{1}{2} \times A B \times E G+(A B)^{2}=4 \times \frac{1}{2} \times \frac{10}{\sqrt{2}} \times 10.6+(\frac{10}{\sqrt{2}})^{2}$

$ =149.91 cm^{2}+50 cm^{2}=\mathbf{2 0 0} \mathbf{~ c m}^{2} \text{ (approx). } $

Ex. 11 . The base of a right pyramid is an equilateral triangle each side of which is $4 m$ long. Every slant edge is $5 m$ long. Find the lateral surface area and volume of the pyramid?

Sol. Let ’ $h$ ’ be the height of the pyramid and ’ $a$ ’ the length of each side of the base of an equilateral triangle.

Then, slant edge $=\sqrt{h^{2}+\frac{a^{2}}{3}}$

$\Rightarrow 5=\sqrt{h^{2}+\frac{16}{3}} \Rightarrow 25-\frac{16}{3}=h^{2} \Rightarrow h^{2}=\frac{59}{3} \Rightarrow h=\sqrt{\frac{59}{3}} m$

Slant height $=\sqrt{h^{2}+\frac{a^{2}}{12}}=\sqrt{\frac{59}{3}+\frac{16}{12}}=\sqrt{21} m$

$\therefore$ Lateral surface area $=\frac{1}{2}($ Perimeter of base $\times$ Slant height $)=\frac{1}{2}(4+4+4) \times \sqrt{21} m^{2}=\mathbf{6} \sqrt{\mathbf{2 1}} \mathbf{~ m}^{\mathbf{2}}$

Volume of the pyramid $=\frac{1}{3}($ Area of base $\times$ height $)=\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 4^{2} \times \sqrt{\frac{59}{3}} m^{3}=\frac{\mathbf{4} \sqrt{\mathbf{5 9}}}{\mathbf{3}} m^{3}$.

Ex. 12 . A right pyramid is on a regular hexagonal base. Each side of the base is $10 m$ and its height is $30 m$. Find the volume of the pyramid?

Sol. Volume of a pyramid $=\frac{1}{3} \times($ Area of the base $) \times$ Height

$ =\frac{1}{3} \times \frac{3 \sqrt{3}}{2} \times(5)^{2} \times 30=649.52 m^{3} \approx \mathbf{6 5 0} \mathbf{m}^{3} .(\text{ Area of a regular hexagon of side ’ } a \text{ ’ }=\frac{3 \sqrt{3}}{2} a^{2}) $

Ex. 13 . Find the volume, lateral surface area and total surface area of a regular tetrahedron whose edge is $16 cm$.

Sol. Volume of a regular tetrahedron $=\frac{\sqrt{2}}{12}(\text{ edge })^{2}=\frac{\sqrt{2}}{12} \times 16^{2}=\frac{\sqrt{2} \times 256}{12} cm^{3}=\frac{\mathbf{6 4} \sqrt{\mathbf{2}}}{\mathbf{3}} cm^{3}$

Lateral surface area $=\frac{3 \sqrt{3}}{4}(\text{ edge })^{2}=\frac{3 \sqrt{3}}{4} \times 16^{2} cm^{2}=192 \sqrt{\mathbf{3}} cm^{2}$

Total surface area $=\sqrt{3}(\text{ edge })^{2}=\sqrt{3} \times 16^{2} cm^{2}=\mathbf{2 5 6} \sqrt{\mathbf{3}} \mathbf{c m}^{2}$.

Ex. 14 . If ’ $p$ ’ be the length of the perpendicular drawn from the vertex of a regular tetrahedron to its opposite face and each edge is of length $2 a$, show that $3 p^{2}=8 a^{2}$.

Sol. Height of tetrahedron $=\sqrt{\frac{2}{3}} \times($ length of an edge $)$

$\Rightarrow p=\sqrt{\frac{2}{3}} \times 2 a \Rightarrow p^{2}=\frac{8 a^{2}}{3} \Rightarrow \mathbf{3} p^{2}=\mathbf{8} \boldsymbol{a}^{\mathbf{2}}$.

PRACTICE SHEET-1

~~ 1. A vertical cone of Volume $V$ with vertex downwards is filled with water upto half its height. The volume of water is:

(a) $\frac{V}{16}$

(b) $\frac{V}{8}$

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

(CDS 2001)

~~ 2. If $h, C, V$ are respectively the height, the curved surface area and volume of a cone, then $3 \pi V^{3}-C^{2} h^{2}+9 V^{2}$ is equal to

(a) 0

(b) 1

(d) 3

(CDS 2003)

~~ 3. What is the semi-vertical angle of a cone whose lateral surface area is double the base area?

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $15^{\circ}$

(CDS 2004)

~~ 4. A square hole of cross-sectional area $4 cm^{2}$ is drilled across a cube with its length parallel to a side of the cube. If an edge of the cube measures $5 cm$, what is the total surface area of the body so formed?

(a) $140 cm^{2}$

(b) $142 cm^{2}$

(d) $182 cm^{2}$

(CDS 2004)

~~ 5. A closed right circular cone contains water upto a height $h / 2$ above the base, where $h$ is the height of the cone. To what height does the water rise if the cone is inverted?

(a) $h / 2$

(b) $3 h / 4$

(d) $(\frac{7}{8})^{1 / 3} h$

(CDS 2005)

~~ 6. Two concentric spheres $A$ and $B$, have radii $r$ and $2 r$ respectively. A cone is inscribed in the latter so as to circumscribe the former. What is the curved surface area of the cone?

(a) $2 \pi r^{2}$

(b) $4 \pi r^{2}$

(d) $8 \pi r^{2}$

(CDS 2005)

~~ 7. The radius and height of a right solid circular cone are $r$ and $h$ respectively. A conical cavity of radius $r / 2$ and height $h / 2$ is cut out of the cone. What is the whole surface of the rest of the portion?

(a) $\frac{\pi r}{4}(5 \sqrt{r^{2}+h^{2}}+3 r)$

(b) $\frac{5 \pi r}{4}(\sqrt{r^{2}+h^{2}})$

(d) $\frac{3 \pi r}{7}(\sqrt{r^{2}+h^{2}}+r)$

(CDS 2005)

~~ 8. A sphere of radius $13 cm$ is cut by a plane whose distance from the centre of the sphere is $5 cm$. What is the circumference of the plane circular section?

(a) $10 \pi cm$

(b) $12 \pi cm$

(d) $26 \pi cm$

(CDS 2005) ~~ 9. A double cone is formed by a complete revolution of the triangle $A B C$ about the side $A B$. The sides $B C=6.5 cm$, $C A=2 cm$ and the perpendicular from $C$ on $A B=1.6 cm$. The volume of the double cone is approximately:

(a) $25 cm^{3}$

(b) $24 cm^{3}$

(d) $20 cm^{3}$

(CDS 2001)

~~ 10 A sphere of radius $5 cm$ exactly fits into a cubical box. The ratio of the surface of the box and the surface of the sphere is:

(a) $19: 9$

(b) $21: 11$

(d) $25: 13$ ~~ 11 The diameter of a right conical tent is 6 metre. If a pole of length 2 metres can be fixed up in the tent at half the distance of the radius from the centre of the base, then the area of canvas required is

(a) $10 \pi$

(b) $12 \pi$

(d) $16 \pi$

~~ 2. A thin walled glass paper weight consists of a hollow cylinder with a hollow cone on top as shown in the given Fig. (a). The paper weight contains just enough sand to fill the cylinder. The paper weight is now turned upside down as shown in Fig. (b).

Fig. (a)

Fig. (b) Calculate the depth of the sand (marked $x$ in the diagram).

(a) $4 cm$

(b) $5 cm$

(d) $3 cm$

~~ 13 A solid cube has a square hole cut through horizontally and a circular hole cut through vertically. Both the holes are cut centrally in appropriate faces. The dimensions of the cube and the hole are shown in the diagram. Calculate the volume remaining after the holes

have been cut. (Take $\pi=3.14$ )

(a) $4995.2 cm^{3}$

(b) $5497.6 cm^{3}$

(d) $5994.2 cm^{3}$

~~ 14 Shown below in Fig. (a) is a closed box which is a prism of length $40 cm$. The cross-section of the box is shown in Fig. (b) with all right angles marked.

Fig. (a)

Fig. (b)

The approximate volume of the prism is

(a) $15200 m^{3}$

(b) $14600 m^{3}$

(d) $12500 m^{3}$

~~ 15 A child consumed an ice-cream of inverted right-circular conical shape form 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) $3 cm$

(d) $4 cm$.

(JMET 2009)

ANSWERS

HINTS AND SOLUTIONS

~~ 1. Let the water be filled in $A E F$ as shown in the given figure. Then, $A B=B D=h / 2$, where $h$ is the height of the cone.

Let $D E=r$ be the radius of the cone.

Then $V=\frac{1}{3} \pi r^{2} h$

$\triangle A B C \sim \triangle A D E$

$\therefore \quad \frac{A B}{A D}=\frac{B C}{D E}$

$\Rightarrow \frac{h / 2}{h}=\frac{B C}{r} \Rightarrow B C=\frac{1}{2} r$

$\therefore$ Volume of water $=$ Vol. of cone $A B C=\frac{1}{3} \pi .(B C)^{2} \times A B$

$ =\frac{1}{3} \pi \cdot(\frac{1}{2} r)^{2} \times \frac{1}{2} h=\frac{1}{24} \pi r^{2} h=\frac{1}{8}(\frac{1}{3} \pi r^{2} h)=\frac{\mathbf{1}}{\mathbf{8}} \boldsymbol{V} . $

~~ 2. $C=\pi r l=\pi r \sqrt{h^{2}+r^{2}}$ and $V=\frac{1}{3} \pi r^{2} h$ where, $r$ and $l$ are respectively the radius of the base and slant height of the cone.

$ \begin{aligned} \therefore 3 \pi V h^{3}-C^{2} h^{2}+9 V^{2}= & 3 \pi \times \frac{1}{3} \pi r^{2} h \times h^{3}-\pi^{2} r^{2}(h^{2}+r^{2}) h^{2} \\ & +9 \times \frac{1}{9} \pi^{2} r^{4} h^{2} \\ = & \pi^{2} r^{2} h^{4}-\pi^{2} r^{2}(h^{2}+r^{2}) h^{2}-\pi^{2} r^{4} h^{2}=\mathbf{0} . \end{aligned} $

~~ 3. Let the semi vertical angle of the cone be $\alpha$, the height $h$, radius of base $r$, and slant height $l$.

Then, Lateral (Curved) surface area of cone $=\pi r l=\pi r(r cosec \alpha)$ Base area of the cone $=\pi r^{2}$

$\therefore$ Given, $\pi r(r cosec \alpha)=2 \pi r^{2}$

$\Rightarrow cosec \alpha=2=cosec 30^{\circ}$

$\therefore \alpha=30^{\circ}$.

~~ 4. Total surface area of the cube having a hole of cross sectional area $4 cm^{2}$

$=T S A$ of cube $-2 \times 4 cm^{2}$

$ =6(5)^{2}-2 \times 4=(150-8) cm^{2}=142 cm^{2} $

Now, Area of four walls of the hole made parallel to a side of the cube $=4 \times 5 \times 2=40 cm^{2}$

$\therefore$ Total surface area of the body

$=142 cm^{2}+40 cm^{2}$

$=182 cm^{2}$.

~~ 5. Let $r$ and $h$ respectively be the radius of the base and height of the cone AFE (Fig. (a))

Then, Volume of water

$=$ Vol. of cone $A F E-$ Vol. of cone $A G C$

$ \begin{aligned} & =\frac{1}{3} \pi r^{2} h-\frac{1}{3} \pi(\frac{r}{2})^{2} \cdot \frac{h}{2} \\ & \quad(\because A B=\frac{h}{2} \text{ and } \frac{A B}{A D}=\frac{B C}{D E} \Rightarrow B C=\frac{1}{2} r) \\ & =\frac{1}{3} \pi r^{2} h-\frac{\pi r^{2} h}{24}=\frac{7}{24} \pi r^{2} h . \end{aligned} $

Fig. (a)

Fig. (b) Now as shown in Fig. (b), Let the height of water in the inverted cone be $H cm$ and radius of the surface of water $=R cm$.

Then, by similarity of $\Delta s, \frac{R}{H}=\frac{r}{h}$

$\Rightarrow R=\frac{H r}{h}$

$\therefore$ Volume of water in inverted cone $=\frac{1}{3} \pi R^{2} H$

$ =\frac{1}{3} \pi(\frac{r H}{h})^{2} H=\frac{1}{3} \pi \frac{r^{2}}{h^{2}} H^{3} $

But given, $\frac{1}{3} \pi \frac{r^{2} H^{3}}{h^{2}}=\frac{7}{24} \pi r^{2} h$

$\Rightarrow H^{3}=\frac{7}{8} h^{3}=\boldsymbol{H}=(\frac{\mathbf{7}}{\mathbf{8}})^{\mathbf{1 / 3}} \boldsymbol{h}$.

~~ 6. Let $A B C$ be the cone circumscribing the sphere with centre $O$ and radius $r$ and inscribed in the sphere with centre $O$ and radius $2 r$. By the properties of a circle,

$O D \perp A B$ and bisects $A B$,

$O F \perp A C$ and bisects $A C$,

$O E \perp B C$ and bisects $B C$,

Also, $\angle O D A=\angle O F A=\angle O E B=90^{\circ}$

Also $A B=B C=A C$

So, $A D^{2}+O D^{2}=A O^{2}$ (Pythagoras’ Th.)

$\Rightarrow A D^{2}=A O^{2}-O D^{2}=(2 r)^{2}-r^{2}=4 r^{2}-r^{2}=3 r^{2}$

$\Rightarrow A D=\sqrt{3} r$.

$\Rightarrow A B=2 \times A D=2 \sqrt{3} r=l$ (slant height of cone) and

$ B E=\frac{1}{2} B C=\frac{1}{2} A B=\frac{1}{2} \times 2 \sqrt{3} r=\sqrt{3} r $

$\therefore$ Curved surface area of cone $=\pi r l$

$ =\pi \times \sqrt{3} r \times 2 \sqrt{3} r=6 \pi r . $

~~ 7. Outer slant height $=\sqrt{h^{2}+r^{2}}$ Inner slant height

$=\sqrt{(\frac{h}{2})^{2}+(\frac{r}{2})^{2}}$

$=\frac{1}{2} \sqrt{h^{2}+r^{2}}$

$\therefore$ Total surface area $=$ Outer curved surface area + Inner curved surface area + Area of base

$ \begin{aligned} & =\pi r \sqrt{h^{2}+r^{2}}+\pi \frac{r}{2} \times \frac{\sqrt{h^{2}+r^{2}}}{2}+\pi[r^{2}-(\frac{r}{2})^{2}] \\ & =\pi r \sqrt{h^{2}+r^{2}}+\frac{\pi r}{4} \sqrt{h^{2}+r^{2}}+\frac{3 \pi r^{2}}{4} \\ & =\frac{5 \pi r}{4} \sqrt{h^{2}+r^{2}}+\frac{3 \pi r^{2}}{4}=\frac{\pi r}{4}[5 \sqrt{\boldsymbol{r}^{2}+\boldsymbol{h}^{2}}+3 r] . \end{aligned} $

~~ 8. Let $O$ be the centre of the sphere and let $A$ be the centre of the circular plane. Then radius $(r)$ of the plane

$ \begin{aligned} & =\sqrt{13^{2}-5^{2}}=\sqrt{169-25} \\ & =\sqrt{144}=12 cm \end{aligned} $

$\therefore$ Circumference of the plane circular section $=2 \pi r$ $=2 \times \pi \times 12 cm=\mathbf{2 4} \boldsymbol{\pi} \mathbf{~ c m}$.

~~ 9. Here, $C A B$ is the given triangle. Since $\triangle C A B$ revolves about the side $A B$, it forms two cones $C A D$ and $C B D$.

Radius of both the cones $=C E=E D=1.6 cm$ Height of cone $C A D$

$ =A E=\sqrt{2^{2}-1.6^{2}}=\sqrt{4-2.56}=\sqrt{1.44}=1.2 cm $

Height of cone $C B D=E B$

$=\sqrt{(6.5)^{2}-(1.6)^{2}}=\sqrt{42.25-2.56}=\sqrt{39.69}=6.3 cm$.

$\therefore$ Volume of double cone

$=$ Vol. of cone $C A D+$ Vol. of cone $C B D$

$ \begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times(1.6)^{2} \times 1.2+\frac{1}{3} \times \frac{22}{7} \times(1.6)^{2} \times 6.3 \\ & =\frac{22}{21} \times 2.56 \times(1.2+6.3)=\frac{22}{21} \times 2.56 \times 7.5 \\ & =20.11 cm^{3} \approx \mathbf{2 0} cm^{3} . \end{aligned} $

~~ 10 Radius of sphere $=5 cm$

$\therefore$ Each edge of cubical box $=10 cm$

$\therefore$ Required ratio $=\frac{\text{ Surface area of box }}{\text{ Surface area of sphere }}$

$ =\frac{6 \times(10)^{2}}{4 \times \frac{22}{7} \times(5)^{2}}=\frac{600 \times 7}{4 \times 25 \times 22}=\frac{21}{11}=\mathbf{2 1 : 1 1 .} $

~~ 11 Let $h$ be the height of the conical tent.

Radius $(r)$ of the conical tent $=3 cm$.

$D E=E B=\frac{1}{2} D B=1.5 cm$.

$\triangle C D B \sim \triangle F E B$

$\Rightarrow \frac{h}{2}=\frac{D B}{E B} \Rightarrow \frac{h}{2}=\frac{3}{1.5} \Rightarrow h=4 cm$

$\therefore$ Slant height of the cone $(l)=\sqrt{h^{2}+r^{2}}=\sqrt{4^{2}+3^{2}}=5 cm$.

$\therefore$ Curved surface area of the cone $=\pi r l$

~~ 12 Volume of sand in the cylinder $=\pi r^{2} h$

$ \begin{aligned} & =\pi \times 3 \times 5 cm^{2} \\ & =\mathbf{1 5} \pi cm^{2} \end{aligned} $

$ \begin{aligned} & =\pi \times(3)^{2} \times 4 cm^{3} \\ & =36 \pi cm^{3} \end{aligned} $

Volume of cone $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi \times(3)^{2} \times 2=6 \pi cm^{3}$

Now when the paper weight is turned upside down, vol. of sand in cone $=6 \pi cm^{3}$

Remaining volume of sand in cylinder $=36 \pi cm^{3}-6 \pi cm^{3}$

$ =30 \pi cm^{3} \text{. } $

Let ’ $h_1$ ’ be the height of the part of the cylinder filled with sand. Radius of base of cylinder $=3 cm$.

$\therefore \pi \times(3)^{2} \times h_1=30 \pi$

$\Rightarrow h_1=\frac{30}{9}=\frac{10}{3} cm$

$\therefore x=(2+\frac{10}{3}) cm=\frac{\mathbf{1 6}}{\mathbf{3}} \mathbf{~ c m}$.

~~ 13 Remaining volume $=$ Volume of cube

Volume of cuboid formed by cutting the square hole
Volume of the cylinder formed by cutting the circular hole

Common volume of cuboid and cylinder

$ \begin{aligned} & =(20)^{3}-(10)^{2} \times 20-\pi \times(4)^{2} \times 20+\pi \times(4)^{2} \times 10 \\ & =8000-2000-160 \pi=6000-160 \pi=5497.6 cm^{3} \end{aligned} $

Note: The common portion is a cylinder of diameter $8 cm$ and height $10 cm$ (side of square).

~~ 14 Volume of prism $=$ Area of base $\times$ height.

Area of base $=$ Area of crossection shown in Fig. (b)

$=$ Area of quadrant $A O B+$ Area of rectangle $B C F O+$ Area of rectangle $A F E D$

$ \begin{aligned} & =\frac{1}{4} \times \frac{22}{7} \times(12)^{2}+(10 \times 12)+(22 \times 6) \\ & =(\frac{792}{7}+120+132) cm^{2}=365 \frac{1}{7} cm^{2} \end{aligned} $

$\therefore$ Volume of prism $=(\frac{2556}{7} \times 40) cm^{3}=14605.71 cm^{3}$

$\approx 14600 cm^{3}$. ~~ 15 $A D F$ is the part of the cone that is filled with remaining ice cream. Now $\triangle A E F \sim \triangle A G C$, so

$\frac{A E}{E F}=\frac{A G}{G C} \Rightarrow E F=\frac{A E \times G C}{A G}$

Let $A E=h, G C=r$ (radius of the bigger cone), $A G=8 cm, E F=r_1$ (radius of smaller conical part)

Then, $r_1=\frac{h r}{8}$

So, Volume of bigger cone $=\frac{1}{3} \pi r^{2} .8$

Volume of smaller conical part filled with icecream

$=\frac{1}{3} \pi r_1^{2} \cdot h=\frac{1}{3} \pi \frac{h^{2} r^{2}}{64} \cdot h=\frac{1}{3} \times \frac{\pi h^{3} r^{2}}{64}$

Given, $\frac{1}{3} \times \frac{\pi h^{3} r^{2}}{64}=12.5 %$ of $\frac{1}{3} \pi r^{2} .8$

$\Rightarrow \quad \frac{h^{3}}{64}=\frac{125}{1000} \times 8$

$\Rightarrow \quad h^{3}=\frac{125 \times 8 \times 64}{1000} \Rightarrow h=\frac{5 \times 8}{10}=\mathbf{4} cm$.

PRACTICE SHEET-2

~~ 1. A container is in the form of a right circular cylinder surmounted by a hemisphere of the same radius 15 $cm$ as the cylinder. If the volume of the container is $32400 \pi cm^{3}$, then the height $h$ of the container satisfies which one of the following ?

(a) $135 cm<h<150 cm$

(b) $140 cm<h<147 cm$

(d) $139 cm<h<145 cm$

(CDS 2008)

~~ 2. A fountain having the shape of a right circular cone is fitted into a cylindrical tank of volume $V$, so that the base of the tank coincides with the base of the cone and the height of the tank is the same as that of the cone. The volume of water in the tank, when it is completely filled with water from the fountain is

(a) $\frac{V}{2}$

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

(d) $\frac{V}{4}$ ~~ 3. A vessel is in the form of a hemi-spherical bowl mounted by a hollow cylinder. The diameter of the sphere is $14 cm$ and the total height of the vessel is $13 cm$. Find its capacity. (Take $\pi=\frac{22}{7}$ )

(a) $1426.66 cm^{3}$

(b) $1264.66 cm^{3}$

(d) $1624.66 cm^{3}$ ~~ 4. A circus tent is cylindrical upto a height of $3 m$ and conical above it. If the diameter of the base is $105 m$ and the slant height of the conical part is $53 m$, find the total canvas used in the making the tent?

(a) $9735 m^{2}$

(b) $9537 m^{2}$

(d) $9753 m^{2}$ ~~ 5. A cylindrical vessel of base radius $14 cm$ is filled with water to some height. If a rectangular solid of dimensions $22 cm$ $\times 7 cm \times 5 cm$ is immersed in it, what is the rise in the water level?

(a) $0.5 cm$

(b) $1.0 cm$

(d) $1.5 cm$

(CDS 2009)

~~ 6. Half of a large cylindrical tank open at the top is filled with water and identical heavy spherical balls are to be dropped into the tank without spilling water out. If the radius and the height of the tank are equal and each is four times the radius of a ball, what is the maximum number of balls that can be dropped?

(a) 12

(b) 24

(d) 48

(CDS 2010)

~~ 7. A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is $3.5 cm$ and the height of the cone is $4 cm$. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylinder is $5 cm$ and its height is $10.5 cm$, find the volume of water left in the cylindrical tub. (Use $.\pi=\frac{22}{7})$

(a) $386.83 cm^{3}$

(b) $836.83 cm^{3}$

(d) $638.83 cm^{3}$ ~~ 8. A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is $2 cm$ and the diameter of the cone is $4 cm$. If a right circular cylinder circumscribes the solid, find how much more space will it cover.

(a) $6 \pi cm^{3}$

(b) $4 \pi cm^{3}$

(d) $10 \pi cm^{3}$ ~~ 9. In a rocket shape fire cracker, explosive powder is to be filled up inside the metallic enclosure. The metallic enclosure is made up of a cylindrical base and conical top with a base of radius $8 cm$. The ratio of the height of the cylinder and the cone is $5: 3$. A cylindrical hole is drilled through the metal solid with height one-third the height of metal solid. What should be the radius of the hole, so that the volume of the hole (in which the gun powder is filled up) is half of the volume of metal solid after drilling?

(a) $4 \sqrt{3} cm$

(b) $4 cm$

(d) None of the above

(IIFT 2010)

~~ 10 A spherical iron ball is dropped into a cylindrical vessel of base diameter $14 cm$, containing water. The water level is increased by $9 \frac{1}{3} cm$. What is the radius of the ball?

(a) $3.5 cm$

(b) $7 cm$

(d) $12 cm$

(CDS 2005)

~~ 11 A gulab jamun, contains sugar syrup upto about $30 %$ of its volume. Find approximately how much syrup would be found in 45 gulab jamuns each shaped like a cylinder with two hemispherical ends of length $5 cm$ and diameter $2.8 cm$.

(a) $383 cm^{3}$

(b) $833 cm^{3}$

(d) $388 cm^{3}$

(NCERT)

~~ 12 A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $41 \frac{19}{21} m^{3}$ of air. If the internal diameter of the dome is equal to its total height above the floor, find the height of the building?

(a) $2 m$

(b) $6 m$

(d) $8 m$

(NCERT)

~~ 13 A vessel is in the form of an inverted cone. Its height is $8 cm$ and the radius of its top which is open is $5 cm$. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius $0.5 cm$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

(a) 90

(b) 150

(d) 100

(NCERT)

~~ 14 A heavy sphere of maximum possible volume is to be completely immersed into a cylindrical jar of radius $a$ containing water upto to a height $2 a$. The minimum height of the jar so that no water spills out of it is:

(a) $\frac{10 a}{3}$

(b) $\frac{11 a}{3}$

(d) $\frac{13 a}{3}$

(CDS 2003)

~~ 15 A solid is hemispherical at the bottom and conical above it. If the surface areas of the two parts are equal, then the volumes of the two parts are in the ratio:

(a) $1: \sqrt{3}$

(b) $2: \sqrt{3}$

(d) $1: 1$ ~~ 16 The radii of the circular ends of a bucket of height $40 cm$ are of lengths $35 cm$ and $14 cm$. What is the volume of the bucket?

(a) 60060 cu. $cm$

(b) 70040 cu. $cm$

(d) 80160 cu. cm. (CDS 2011) ~~ 17 A right circular cone is cut by a plane parallel to its base in such a way that the slant heights of the original and the smaller cone thus obtained are in the ratio $2: 1$. If $V_1$ and $V_2$ are respectively the volumes of the original cone and the new cone, then what is $V_1: V_2$ ?

(a) $2: 1$

(b) $3: 1$

(d) $8: 1$ ~~ 18 A fez, the cap used by the turks is shaped like the frustum of a cone. If its radius on the open side is $10 cm$, radius at the upper base is $4 cm$ and its slant height is $15 cm$, find the area of the material used for making it.

(a) $210 \pi cm^{2}$

(b) $226 \pi cm^{2}$

(d) $341 \pi cm^{2}$

(CDS 2008)

~~ 19 A metallic right circular cone $20 cm$ high and whose vertica angle is $60^{\circ}$ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $\frac{1}{16} cm$, find the length of the wire.

(a) $6497.44 m$

(b) $4967.44 cm$

(d) $6794.44 cm$

(NCERT)

~~ 20 A tent is made in the form of a conic frustum surmounted by a cone. The diameters of the base and top of the frustum are $20 m$ and $6 m$ respectively and the height is $24 m$. If the height of the tent is $28 m$, find the quantity of canvas required.

(a) $280 \pi m^{2}$

(b) $325 \pi m^{2}$

(d) $340 \pi m^{2}$ ~~ 21 A cone $12 cm$ high is cut $8 cm$ from the vertex to form a frustum with a volume of $190 cu$. $cm$. Find the radius of the cone.

(a) $3.46 cm$

(b) $4.63 cm$

(d) $3.64 cm$ ~~ 22 There is a cone $2 m$ high which has a volume of $1.5 cu$. $m$. The cone is bisected by a horizontal plane, forming a smaller cone and a frustum of equal values. What is the height of the new cone?

(a) $\sqrt[3]{2} m$

(b) $\sqrt[3]{3} m$

(d) $\sqrt[3]{6} m$ ~~ 23 A right circular cone with radius to height ratio as $12: 5$ is cut parallel to its base to get a smaller cone and a frustum. If the height of the smaller cone to the height of the frustum are in the ratio $3: 1$, by what percentage is the combined total surface area of the smaller cone and frustum more with respect to the original cone.

(a) $22 %$

(b) $32 %$

(d) None of these ~~ 24 The base of a right pyramid is a square of side $40 cm$ long. If the volume of the pyramid is $800 cm^{3}$, then its height is:

(a) $5 cm$

(b) $10 cm$

(d) $20 cm$

(SSC 2011)

~~ 25 The base of a right pyramid is a square of side $16 cm$ long. If its height be $15 cm$, then the area of the lateral surface in square centimetre is:

(a) 136

(b) 544

(d) 1280

(SSC 2011)

~~ 26 If the length of each side of a regular tetrahedron is $12 cm$, then the volume of the tetrahedron is

(a) $144 \sqrt{2}$ cu. cm

(b) $72 \sqrt{2}$ cu.cm

(d) $12 \sqrt{2}$ cu. cm ~~ 27 If the slant height of a right pyramid with square base is 4 metre and the total slant surface of the pyramid is 12 square metre, then the ratio of total slant surface and area of the base is :

(a) $16: 3$

(b) $24: 5$

(d) $12: 3$

(SSC 2012)

~~ 28 A regular pyramid has a square base with side $10 cm$ and a vertical height of $20 cm$. If the height increases by $10 %$ of its original value and the volume is constant, the percentage change in the side of the square base with respect to its original value is approximately

(a) $+5 %$

(b) $+10 %$

(d) $-10 %$

(JMET 2009)

~~ 29 The base of a pyramid is an equilateral triangle of side 1 metre. If the height of the pyramid is 4 metres, then the volume is:

(a) $0.550 m^{3}$

(b) $0.577 m^{3}$

(d) $0.750 m^{3}$

(CDS 1999)

~~ 30 A right pyramid stands on an equilateral triangular base of area $36 \sqrt{3} cm^{2}$. If the area of one of the lateral faces is $42 cm^{2}$, find the volume of the pyramid.

(a) $\frac{360}{\sqrt{3}} cm^{3}$

(b) $12 \sqrt{471} cm^{3}$

(d) $12 \sqrt{111} cm^{3}$ ~~ 31 The surface area and volume respectively of a regular tetrahedron of height ’ $h$ ’ are:

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

(b) $\frac{3 \sqrt{3}}{2} h^{2}, \frac{3 \sqrt{3}}{8} h^{3}$

(d) $\frac{3 \sqrt{3}}{2} h^{2}, \frac{\sqrt{3}}{8} h^{3}$ ~~ 32 A right pyramid has an equilateral triangular base of side 4 units. If the number of square units of its whole surface area be three times the number of cubic units of its volume, find its height.

(a) 6 units

(b) 10 units

(d) 4 units ~~ 33 There is a pyramid on a base which is a regular hexagon of side ’ $2 a^{\prime}$ cm. If every slant height of this pyramid is of length $\frac{5 a}{2} cm$, then the volume of this pyramid is

(a) $3 a^{3} cm^{3}$

(b) $3 \sqrt{2} a^{3} cm^{3}$

(d) $6 a^{3} cm^{3}$

(SCC 2011)

~~ 34 The radius of a cone is $\sqrt{2}$ times the height of the cone. A cube of maximum possible volume is cut from the same cone. What is the ratio of volume of the cone to the volume of the cube?

(a) $3.18 \pi$

(b) $2.25 \pi$

(d) None of these ~~ 35 Ankit has a right circular cylinder which he inserted completely into a right circular cone of height $20 cm$. The vertical angle of the cone is $60^{\circ}$ and the diameter of the cylinder is $10 \sqrt{3} cm$. The volume of the cone is

(a) $\frac{4000}{7} \pi cm^{3}$

(b) $\frac{8000}{3} \pi cm^{3}$

(d) $\frac{3000}{7} \pi cm^{3}$

ANSWERS

1. (a)	2. (c)	3. (c)	4. $(a)$	5. $(c)$	6. $(b)$	7. (c)	8. (c)	9. (a)	10. $(b)$
11. (c)	12. $(c)$	13. (b)	14. (a)	15. $(b)$	16. $(c)$	17. $(d)$	18. $(b)$	19. $(c)$	20. $(d)$
21. (b)	22. $(c)$	23. $(d)$	24. (c)	25. (b)	26. (a)	27. (a)	28. $(c)$	29. (b)	30. $(b)$
31. (d)	32. $(c)$	33. $(c)$	34. (b)	35. (c)

HINTS AND SOLUTIONS

~~ 1. Let the height of the cylinder be $H cm$. Then, by given condition, Vol. of hemisphere + Vol. of cylinder $=$ Vol. of container, i.e, $\frac{2}{3} \pi r^{3}+\pi r^{2} H=32400 \pi \quad(\because r=15)$ $\Rightarrow \frac{2}{3} \times \pi \times 3375+\pi \times 225 \times H=32400 \pi$ $\Rightarrow 2 \times 1125+225 H=32400$ $\Rightarrow 10+H=144 \Rightarrow H=134 cm$

$\therefore$ Height of the container $=H+15=134 cm+15 cm=\mathbf{1 4 9} cm$ Hence option $(a)$ is the correct answer.

~~ 2. Volume of the cylindrical tank $=V=\pi r^{2} h$ Volume of the cone $=\frac{1}{3} \pi r^{2} h=\frac{V}{3}$

$\therefore$ Volume of water in the tank

$=V-\frac{V}{3}=\frac{2 V}{3}$.

~~ 3. Total capacity of the bowl

$=$ Vol. of the cylinder + Vol. of the hemisphere

$=(\pi r^{2} h+\frac{2}{3} \pi r^{3}) cm^{3}$

$=\pi r^{2}(h+\frac{2}{3} r) cm^{3}=\frac{22}{7} \times 7^{2} \times(6+\frac{2}{3} \times 7) cm^{3}$

$=\frac{22}{7} \times 7 \times \frac{32}{3} cm^{3}=\frac{4928}{7} cm^{3}=\mathbf{1 6 4 2 . 6 6} cm^{3}$.

~~ 4. Total canvas reqd. in making the tent

$=$ Curved surface area of the conical part + Curved surface area of cylindrical part

$=(\pi r l+2 \pi r h) m^{2}=\pi r(l+2 h) m^{2}$

$=(\frac{22}{7} \times 52.5 \times(53+6)) m^{2}$

$=(\frac{22}{7} \times 52.5 \times 59) m^{2}=\mathbf{9 7 3 5} \mathbf{m}^{2}$

~~ 5. Vol. of the solid $=22 cm \times 7 cm \times 5 cm=770 cu . cm$ Let the height of water rise in the cylinder be $h cm$. Then, $\pi r^{2} h=770$

$\Rightarrow \frac{22}{7} \times 14 \times 14 \times h=770 \Rightarrow h=\frac{770 \times 7}{14 \times 14 \times 22}=\frac{5}{4}=\mathbf{1 . 2 5} cm$

~~ 6. Let the radius of a ball $=r cm$.

$\therefore$ Radius of the base of the cylinder $=4 r$

Height of the cylinder $=4 r$

Vol. of spherical ball $=\frac{4}{3} \pi r^{3}$

Now, Vol. of water in the cylindrical tank $=\pi(4 r)^{2}(2 r)$ (height of water $=2 r$ )

$ =32 \pi r^{3} $

$\therefore$ Volume of remaining portion of the cylindrical tank

$ =32 \pi r^{3} \text{. } $

Now, if the number of spherical balls $=n$, then

$ n \times \frac{4}{3} \pi r^{3}=32 \pi r^{3} \Rightarrow n=\mathbf{2 4} $

~~ 7. In the given solid, radius of hemisphere

$=$ radius of base of cone $(r)=3.5 cm$ Height of cone $(h)=4 cm$.

$\therefore$ Volume of the solid $=$ Volume of the hemispherical part + Volume of the conical part

$ \begin{aligned} & =\frac{2}{3} \pi r^{3}+\frac{1}{3} \pi r^{2} h \\ & =\frac{1}{3} \pi r^{2}(2 r+h)=\frac{1}{3} \times \frac{22}{7} \times(3.5)^{2} \times(7+4) \\ & =\frac{1}{3} \times \frac{22}{7} \times 12.25 \times 11=141.17 cm^{3} \end{aligned} $

When the solid is submerged in the cylindrical tub, the volume of water that flows out of the cylinder is equal to the volume of the solid.

Therefore,

Volume of water left in the cylinder

$=$ Volume of cylinder - Volume of solid

$ \begin{aligned} & =(\frac{22}{7} \times 25 \times 10.5-141.17) cm^{3} \\ & =(825-141.17) cm^{3}=\mathbf{6 8 3 . 8 3} \mathbf{c m}^{3} . \end{aligned} $

~~ 8. Let $P Q R$ be the cone surmounted on the hemisphere $Q R S$ circumscribed by the cylinder $A B C D$. Then, Radius of the hemisphere $=$ radius of the base of the cylinder $=$ radius of the base of the cone $=\mathbf{2} \mathbf{~ c m}$.

$\Rightarrow E R=E S=S C=2 cm$

Height of the cone $=2 cm$

Height of the cylinder $P E+E S=2 cm+2 cm=4 cm$.

$\therefore$ Required space

$=$ Volume of the cylinder - Volume of the solid toy

$ =[\pi \times 2^{2} \times 4-(\frac{2}{3} \pi(2)^{3}+\frac{1}{3} \pi(2)^{2} 2)] cm^{3} $

$ =[16 \pi-8 \pi] cm^{3}=\mathbf{8} \pi cm^{3} . $

~~ 9. Given, the ratio of the height of the cylindrical base to height of the conical top $=5: 3$. Let their heights be $5 k$ and $3 k$ respectively. Let $R$ be the radius of the base of the cylinder as well as the cone,

Let $r$ be the radius of the cylindrical hole.

Height of the cylindrical hole

$=\frac{1}{3} \times$ Height of the solid $=\frac{8 K}{3}$.

Volume of the solid

$=$ Volume of cylindrical base + Volume of conical top

$=\pi R^{2} \times 5 K+\frac{1}{3} \pi R^{2} \times 3 K=6 \pi R^{2} K$.

Volume of the drilled cylindrical hole $=\pi r^{2} \frac{8 K}{3}=\frac{8 \pi r^{2} K}{3}$

Volume of the metal solid left after drilling $=6 \pi R^{2} k-\frac{8 \pi r^{2} k}{3}$

Given, Vol. of cylindrical hole $=\frac{1}{2} \times$ Vol. of metal solid left

$\Rightarrow \frac{8 \pi r^{2} k}{3}=\frac{1}{2}(6 \pi R^{2} k-\frac{8 \pi r^{2} k}{3})$

$\Rightarrow \frac{8 \pi r^{2} k}{3}+\frac{8 \pi r^{2} k}{6}=3 \pi R^{2} k$

$\Rightarrow 4 \pi r^{2} k=3 \pi R^{2} k \Rightarrow r^{2}=\frac{3 R^{2}}{4} \Rightarrow r=\frac{\sqrt{3}}{2} R$

Given, $R=8 \Rightarrow r=\frac{\sqrt{3}}{2} \times 8=\mathbf{4} \sqrt{\mathbf{3}} \mathbf{~ c m .}$

~~ 10 Let $R$ be the radius of the ball. Then,

Volume of water displaced $=$ Volume of iron ball

$\Rightarrow \pi \times(\frac{14}{2})^{2} \times \frac{28}{3}=\frac{4}{3} \times \pi \times R^{3}$

$\Rightarrow R^{3}=7^{3} \Rightarrow \boldsymbol{R}=7$.

~~ 11 The gulab jamun can be considered as the combination of three solids as shown in the given diagram.

Here, radius of the hemispherical ends $=$ radius of the cylinder $(r)=1.4 cm$

Height of the cylindrical part $(h)=5-(1.4+1.4) cm$

$ =2.2 cm \text{. } $

$\therefore$ Volume of one gulab jamun

$=2 \times($ Volume of hemisphere $)+$ Volume of cylinder

$=2 \times{\frac{2}{3} \pi r^{3}}+\pi r^{2} h=\pi r^{2}(\frac{4}{3} r+h)$

$=\frac{22}{7} \times 1.4 \times 1.4{\frac{4}{3} \times 1.4+2.2}$

$=\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}{\frac{4}{3} \times \frac{14}{10}+\frac{22}{10}}$

$ =\frac{154}{25} \times \frac{61}{15}=\frac{9394}{375} cm^{3} $

$\therefore \quad$ Quantity of syrup found in 45 gulab jamuns

$ \begin{aligned} & =30 % \text{ of }(45 \times \frac{9394}{375}) \\ & =\frac{3}{10} \times 45 \times \frac{9394}{375}=338.184 \approx \mathbf{3 3 8} \mathbf{c m}^{3} . \end{aligned} $

~~ 12 Let the height of the building

$=$ internal diameter of the dome $=\mathbf{2 r} m$.

$\therefore \quad$ Radius of the building $=$ radius of dome $=\frac{2 r}{2}=r m$

Height of cylindrical portion

$ =2 r-r=r m \text{. } $

Volume of the cylinder

$ =\pi r^{2}(r)=\pi r^{3} m^{3} $

Volume of hemispherical dome

$ =\frac{2}{3} \pi r^{3} m^{3} $

$\therefore$ Total volume of the building

$ =\pi r^{3}+\frac{2}{3} \pi r^{3}=\frac{5}{3} \pi r^{3} m^{3} $

Given, $\frac{5}{3} \pi r^{3}=41 \frac{19}{21}=\frac{880}{21} \Rightarrow r^{3}=\frac{880 \times 7 \times 3}{5 \times 22 \times 21}=8$

$\therefore \quad r=\sqrt[3]{8}=2 m$

Hence, height of the building $=2 r=\mathbf{4} \mathbf{~ m}$.

~~ 13 Volume of water in the vessel $=$ Volume of the inverted cone

$ =\frac{1}{3} \times \pi \times(5)^{2} \times 8 cm^{3} $

Let the number of lead shots $=n$

Volume of one lead shot $=\frac{4}{3} \times \pi \times(0.5)^{3}$

$\therefore$ Total volume of lead shots

$=$ Volume of water flowing out

$\Rightarrow n \times \frac{4}{3} \times \pi \times(0.5)^{3}=\frac{1}{4} \times \frac{1}{3} \times \pi \times(5)^{2} \times 8$

$\Rightarrow n=\frac{25 \times 8}{16 \times 0.125}=\mathbf{1 0 0}$.

~~ 14 Let the maximum height of the cylinder be $h$. Then,

Volume of cylinder of maximum height $=$ Volume of water filled

$\Rightarrow \pi a^{2} \cdot h=\pi \cdot a^{2} \cdot 2 a+\frac{4}{3} \pi a^{3}$

$\Rightarrow \pi a^{2} h=\pi a^{2}(2+\frac{4}{3}) \Rightarrow h=\frac{\mathbf{1 0 a}}{\mathbf{3}}$

~~ 15 Let the radius of the hemispherical as well as base of conical portion $=r cm$ Let the vertical height of the cone $=h cm$ Let the slant height of the cone $=l cm$. Given, $\pi r l=2 \pi r^{2}$

$\Rightarrow l=2 r$

$\therefore h=\sqrt{l^{2}-r^{2}}=\sqrt{4 r^{2}-r^{2}}=\sqrt{3 r^{2}}=r \sqrt{3}$

$\therefore \quad$ Required ratio $=$ Vol. of hemisphere $:$ Vol. of cone

$ =\frac{2}{3} \pi r^{3}: \frac{1}{3} \pi r^{2} \cdot r \sqrt{3}=\mathbf{2}: \sqrt{\mathbf{3}} . $

~~ 16 Bucket is a frustum of a right circular cone. Here,

$ R=35 cm, \quad r=14 cm, \quad h=40 cm \text{. } $

Volume of frustum of cone $=\frac{\pi h}{3}(R^{2}+r^{2}+R r)$

$=\frac{22}{7} \times \frac{40}{3}(35^{2}+14^{2}+35 \times 14)$

$=\frac{880}{21}(1225+196+490)=\frac{880}{21} \times 1911=\mathbf{8 0 0 8 0}$ cu. $\mathbf{~ c m}$.

~~ 17 Let $A B C$ be the original cone cut by the plane $D F E$ such that,

$A C: A D=2: 1$.

$\because \triangle A F E \sim \triangle A G C$

$\frac{F E}{G C}=\frac{A E}{A C}=\frac{1}{2}$

$\Rightarrow$ Radius of cone $A B C=G C=2 r$ (say),

then radius of cone $A D E=F E=r$.

So, required ratio $=\frac{\text{ Vol. of cone } A B C}{\text{ Vol. of cone } A D E}$

$\Rightarrow \frac{V_1}{V_2}=\frac{\frac{1}{3} \pi \times 4 r^{2} \times \sqrt{(2 l)^{2}-(2 r)^{2}}}{\frac{1}{3} \pi \times r^{2} \times \sqrt{l^{2}-r^{2}}}=\frac{\mathbf{8}}{\mathbf{1}}$.

~~ 18 For the fez, height $(h)=15 cm, r_1=4 cm, r_2=10 cm$

Total curved surface area of the fez $=$ Lateral surface area of the frustum + Area of upper closed end

$ \begin{aligned} & =\pi(r_1+r_2) l+\pi r_1^{2}=\pi{(4+10) \times 15}+\pi .16 \\ & =(210 \pi+16 \pi) cm^{2}=\mathbf{2 2 6} \pi \mathbf{c m}^{2} . \end{aligned} $

~~ 19 Given $\angle B A C=60^{\circ} \Rightarrow \angle D A C=30^{\circ}$.

$A D=20 cm \Rightarrow A E=E D=10 cm$.

In $\triangle A E F, \tan 30^{\circ}=\frac{E F}{A E}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{r_1}{10}$

$\Rightarrow r_1=\frac{10}{\sqrt{3}} cm$

In $\triangle A D C, \tan 30^{\circ}=\frac{D C}{A D}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{r_2}{20} \Rightarrow r_2=\frac{20}{\sqrt{3}} cm$

$\therefore$ Volume of the frustum cut by the plane

$ \begin{aligned} & =\frac{\pi h}{3}(r_1^{2}+r_2^{2}+r_1 r_2) \\ & =\frac{\pi \times 10}{3}(\frac{100}{3}+\frac{400}{3}+\frac{200}{3})=\frac{7000 \pi}{9} cm^{3} \end{aligned} $

Let the length of the wire be $h cm$.

Given, radius of wire $=\frac{1}{32} cm$.

$\therefore$ Volume of wire (cylinder) $=\pi \times(\frac{1}{32})^{2} \times h$

So, $\frac{7000 \pi}{9}=\frac{\pi h}{32 \times 32}$

$\Rightarrow h=\frac{7000 \times 32 \times 32}{9} cm=\frac{7000 \times 32 \times 32}{9 \times 100} m=\mathbf{7 9 6 4 . 4 4} m$.

~~ 20. Area of canvas required $=$ Lateral surface area of the conic frustum + Curved surface area of the cone. Height of conical part

$=28 m-24 m=4 m$

Slant height $(l_1)$ of the conical

part $=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5 m$.

For the furstum, $h=24 m, R=10 m, r=3 m$

Slant height $(l_2)$ of the frustum

$ \begin{aligned} & =\sqrt{h^{2}+(R-r)^{2}}=\sqrt{24^{2}+(10-3)^{2}} \\ & =\sqrt{576+49}=\sqrt{625}=25 m . \end{aligned} $

$\therefore \quad$ Area of canvas reqd $=\pi(R+r) l_2+\pi r l_1$

$ =\pi \times 13 \times 25+\pi \times 3 \times 5=(325 \pi+15 \pi) m^{2}=\mathbf{3 4 0} \pi \mathbf{m}^{2} . $

~~ 21. Given, Height of the cone $=A G=12 cm$

Height (h) of the frustum $=F G=4 cm$ and $A F=8 cm$

$\triangle A F C \sim \triangle A G E$, so

$ \frac{A F}{A G}=\frac{r_1}{r_2} \Rightarrow r_1=\frac{8}{12} r_2=\frac{2}{3} r_2 $

Volume of frustum $=\frac{\pi h}{3}(r_1^{2}+r_2^{2}+r_1 \times r_2)$

$\Rightarrow 190=\frac{22}{7} \times \frac{4}{3}(\frac{4}{9} r_2^{2}+r_2^{2}+\frac{2}{3} r_2^{2})$

$\Rightarrow 190=\frac{88}{21}(\frac{19}{9} r_2^{2})$

$\Rightarrow r_2^{2}=\frac{190 \times 9 \times 21}{88 \times 19}=21.47$ (approx)

$\Rightarrow r_2=\sqrt{21.47}=4.63 cm$ approx.

~~ 22. Volume of original cone

$V_1=\frac{\pi}{3} R^{2} H=\frac{3}{2}$

$\Rightarrow \frac{\pi}{3} R^{2} \times 2=\frac{3}{2}$

$\Rightarrow R^{2}=\frac{9}{4 \pi} \Rightarrow R=\frac{3}{2 \sqrt{\pi}}$

From similar $\triangle s A F C$ and $A G E$,

we have $\frac{r}{h}=\frac{R}{2} \Rightarrow r=\frac{h R}{2}$

$\Rightarrow r=\frac{h}{2} \times \frac{3}{2 \sqrt{\pi}}=\frac{3 h}{4 \sqrt{\pi}}$

Volume of smaller cone $=\frac{\pi}{3} r^{2} h=\frac{3}{4}$

$\Rightarrow \frac{\pi}{3}(\frac{3 h}{4 \sqrt{\pi}})^{2} \times h=\frac{3}{4}$

$\Rightarrow \frac{\pi}{3} \frac{9 h^{2}}{16 \pi} \times h=\frac{3}{4} \Rightarrow h^{3}=4 \Rightarrow h=\sqrt[3]{4} \mathbf{~ m}$.

~~ 23. Let the radius and height of the circular cone be $12 k$ and $5 k$ respectively.

Given : $\frac{\text{ Height }(h_1) \text{ of smaller cone }}{\text{ Height }(h_2) \text{ of frustum }}=\frac{3}{1}$

$\therefore \quad h_1=\frac{3}{4} \times 5 k$ and $h_2=\frac{1}{4} \times 5 k$

$\triangle A F E \sim \triangle A G C$, so

$\frac{F E}{G C}=\frac{A F}{A G}$

$\Rightarrow \frac{r_1}{r_2}=\frac{h_1}{A G}$

$\Rightarrow \frac{r_1}{12 k}=\frac{\frac{15 k}{4}}{5 k}$

$\Rightarrow r_1=\frac{3}{4} \times 12 k=9 k$.

Total surface area of original cone $=\pi r_2 l+\pi r_2^{2}$

$ \begin{aligned} & =\pi \times 12 k \times \sqrt{(12 k)^{2}+(5 k)^{2}}+\pi \times(12 k)^{2} \\ & =\pi \times 12 k \times 13 k+144 \pi k^{2} \\ & =156 \pi k^{2}+144 \pi k^{2}=300 \pi k^{2} \end{aligned} $

Total surface area of smaller cone

$ \begin{aligned} & =\pi \times 9 k \times \sqrt{(\frac{15 k}{4})^{2}+(9 k)^{2}}+\pi(9 k)^{2} \\ & =\pi \times 9 k \times \frac{39}{4} k+81 \pi k^{2}=\frac{351 \pi k^{2}}{4}+81 \pi k^{2}=\frac{675 \pi k^{2}}{4} \end{aligned} $

Total surface area of frustum

$ =\pi \times(9 k+12 k) \times \sqrt{(\frac{5 k}{4})^{2}+3 k^{2}}+\pi[(12 k)^{2}+(9 k)^{2}] $

$ \begin{aligned} & =\pi \times 21 k \times \frac{13}{4} k+\pi(144 k^{2}+81 k^{2}) \\ & =\frac{273 \pi k^{2}}{4}+225 \pi k^{2}=\frac{1173}{4} \pi k^{2} \end{aligned} $

$\therefore$ Reqd. difference $=$ Total surface area of (smaller cone + frustum) - Total surface area of original cone

$ \begin{aligned} & =(\frac{1173}{4} \pi k^{2}+\frac{675}{4} \pi k^{2})-300 \pi k^{2} \\ & =\frac{1848}{4} \pi k^{2}-300 \pi k^{2}=462 \pi k^{2}-300 \pi k^{2}=162 \pi k^{2} \end{aligned} $

$\therefore \quad$ Reqd. percent $=(\frac{162 \pi k^{2}}{300 \pi k^{2}} \times 100) %=\mathbf{5 4 %}$.

Area of the base $=40 cm \times 40 cm=1600$ sq. $cm$.

Volume of pyramid $=\frac{1}{3} \times$ Area of base $\times$ height

$\Rightarrow 8000=\frac{1}{3} \times 1600 \times h \Rightarrow h=\frac{8000 \times 3}{1000}=\mathbf{1 5} cm$.

~~ 25. Given, $O G=15 cm$

$G F=\frac{1}{2} \times$ side of square $=8 cm$

$\therefore \quad$ Height of triangle $=O F=\sqrt{15^{2}+8^{2}}$ $=\sqrt{225+64}=\sqrt{289}=17 cm$.

$\therefore \quad$ Area of the lateral surface of the pyramid

$=4 \times$ Area of one triangle edge

$=4 \times \frac{1}{2} \times$ base $\times$ height $=4 \times \frac{1}{2} \times 16 \times 17=544$ sq. $cm$.

~~ 26. Volume of a regular tetrahedron $=\frac{\sqrt{2}}{12}(\text{ edge })^{3}$

$ =\frac{\sqrt{2}}{12}(12)^{3} \text{ cu. } cm=\mathbf{1 4 4} \sqrt{2} \text{ cu. cm. } $

~~ 27. Slant surface or Lateral surface area of a pyramid

$ =\frac{1}{2} \times \text{ Perimeter of base } \times \text{ Slant height } $

Given, $12=\frac{1}{2} \times 4 \times a \times 4$

where, each side of the square $=\boldsymbol{a}$ metres

$\Rightarrow a=\frac{24}{16}=\frac{3}{2} m$.

$\therefore \quad$ Area of base $=(\frac{3}{2})^{2} m^{2}=\frac{9}{4} m^{2}$.

$\therefore$ Reqd. ratio $=12: \frac{9}{4}=\mathbf{1 6}: \mathbf{3}$.

~~ 28. Volume of the pyramid $=\frac{1}{3} \times$ Area of base $\times$ height

$ =\frac{1}{3} \times 100 \times 20 cm^{3} $

New height of the pyramid $=22 cm$ Let the new side of the square be $x cm$. Then,

New volume $=\frac{1}{3} \times x^{2} \times 22$

So, $\frac{1}{3} \times x^{2} \times 22=\frac{1}{3} \times 100 \times 20$

$\Rightarrow x^{2}=\frac{100 \times 20}{22}=90.90 \Rightarrow x \approx 9.5$

$\therefore \quad %$ change in the side of the square base

$ =(\frac{10-9.5}{10} \times 100) %=\mathbf{5 %} \text{ less. } $

~~ 29. Volume of the pyramid $=\frac{1}{3} \times$ Area of base $\times$ Height

$ =\frac{1}{3} \times \frac{\sqrt{3}}{4} \times(1)^{2} \times 4 m^{2}=\frac{1.732}{3}=\mathbf{0 . 5 7 7} \mathbf{m}^{2} \text{ (approx) } $

~~ 30. Let the length of each side of the base be $a cm$. Then, area of the base $=36 \sqrt{3} cm^{2}$

$\Rightarrow \frac{\sqrt{3}}{4} a^{2}=36 \sqrt{3} \Rightarrow a^{2}=36 \times 4 \Rightarrow a=12 cm$.

Let $h$ be the height of the pyramid and $l$ be its slant height.

Then, $l=\sqrt{h^{2}+\frac{a^{2}}{12}}$

$\Rightarrow l^{2}=h^{2}+\frac{a^{2}}{12} \Rightarrow l^{2}=h^{2}+\frac{144}{12} \Rightarrow l^{2}=h^{2}+12$

Area of a lateral face

$=\frac{1}{2}($ length of a edge of base $\times$ slant height $)$

$=\frac{1}{2}(a \times l)$

$\Rightarrow 42=\frac{1}{2} \times(12 \times l) \Rightarrow l=\frac{42 \times 2}{12}=7 cm$.

Putting the value of $l$ in $(i)$ we have $h^{2}=(7)^{2}-12=49-12=37 \Rightarrow h=\sqrt{37} cm$

$\therefore$ Volume of pyramid $=\frac{1}{3} \times$ Area of base $\times$ Height

$ =\frac{1}{3} \times 36 \sqrt{3} \times \sqrt{37} cm^{3}=\mathbf{1 2} \sqrt{\mathbf{1 1 1}} \mathbf{~ c m}^{3} . $

~~ 31. If the length of each edge of a regular tetrahedron $=a$ units, then height of the tetrahedron $=\sqrt{\frac{2}{3}} a$

$\Rightarrow h=\sqrt{\frac{2}{3}} a \Rightarrow a=\sqrt{\frac{3}{2}} h$

$\therefore \quad$ Surface area of the tetrahedron $=\sqrt{3}(\text{ edge })^{2}$

$ =\sqrt{3}(\sqrt{\frac{3}{2}} h)^{2}=\frac{3 \sqrt{3}}{2} h^{2} $

Volume of the tetrahedron $=\frac{\sqrt{2}}{12}(\text{ edge })^{3}$

$ =\frac{\sqrt{2}}{12}(\sqrt{\frac{3}{2}} h)^{3}=\frac{3 \sqrt{3}}{24} h^{3}=\frac{\sqrt{\mathbf{3}}}{\mathbf{8}} \boldsymbol{h}^{\mathbf{3}} . $

~~ 32. Let $a$ be the length of each side of the base, $h$ be the height and $l$ be the slant height of the pyramid. Here, $a=\mathbf{4} \mathbf{~ c m}$.

$\therefore$ Slant height $(l)=\sqrt{h^{2}+\frac{a^{2}}{12}}=\sqrt{h^{2}+\frac{16}{12}}=\sqrt{h^{2}+\frac{4}{3}}$

According to the given question,

Lateral surface area + Area of the base $=3$ (Volume)

$\Rightarrow \frac{1}{2} \times 12 \times \sqrt{h^{2}+\frac{4}{3}}+\frac{\sqrt{3}}{4} \times(4)^{2}=3 \times \frac{1}{3} \times(\frac{\sqrt{3}}{4} \times 4^{2} \times h)$

$\Rightarrow 6 \sqrt{h^{2}+\frac{4}{3}}+4 \sqrt{3}=4 \sqrt{3} h$

$\Rightarrow 6 \sqrt{h^{2}+\frac{4}{3}}=4 \sqrt{3}(h-1)$

$\Rightarrow 36(h^{2}+\frac{4}{3})=48(h-1)^{2}$

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

$\Rightarrow 3 h^{2}+4=4 h^{2}-8 h+4$

$\Rightarrow h^{2}-8 h=0 \Rightarrow h(h-8)=0 \Rightarrow \boldsymbol{h}=\mathbf{8}$ as $\boldsymbol{h} \neq \mathbf{0}$.

~~ 33. $A B C D E F$ is the regular hexagon with $A B=B C=C D=D E$ $=E F=F E=a cm$

Here, Height of pyramid $=O P$, Slant height of pyramid $=P Q$

So in right angled $\triangle P O Q, P Q^{2}=P O^{2}+O Q^{2}$.

Area of the hexagonal base $=6 \times \frac{\sqrt{3}}{4} \times(2 a)^{2}$

$ =6 \times \frac{\sqrt{3}}{4} \times 4 a^{2}=6 \sqrt{3} a^{2} cm^{2} . $

Height $=\sqrt{(\frac{5 a}{2})^{2}-(2 a)^{2}}$

$=\sqrt{\frac{25}{4} a^{2}-4 a^{2}}=\sqrt{\frac{9 a^{2}}{4}}=\frac{3}{2} a cm$.

$\therefore$ Volume of pyramid $=\frac{1}{3} \times$ area of base $\times$ height

$ =\frac{1}{3} \times 6 \sqrt{3} a^{2} \times \frac{3}{2} a=3 \sqrt{3} a^{3} cm^{3} . $

~~ 34. Let each side of the cube be $a cm$.

Then $D E=$ diagonal of square face $=\sqrt{2} a$

$\Rightarrow D G=\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}} cm$. Let the radius of the cone, i.e., $A F=F B=r cm$.

Height $F C=h cm$. Given, $r=h \sqrt{2}$

In similar triangles, $A F C$ and $D G C$

$\frac{A F}{F C}=\frac{D G}{G C} \Rightarrow \frac{r}{h}=\frac{\frac{a}{\sqrt{2}}}{(h-a)}$

$\Rightarrow \frac{a / \sqrt{2}}{h-a}=\sqrt{2} \Rightarrow a=2(h-a) \Rightarrow h=\frac{3 a}{2}, r=\frac{3 a}{2} * \sqrt{2}$

$\therefore$ Volume of cone: Volume of cube $=\frac{\frac{1}{3} \pi \times(\frac{3 a \sqrt{2}}{2})^{2} \times \frac{3 a}{2}}{a^{3}}$ $=\frac{9}{4} a^{3} \pi: a^{3}=\frac{9}{4} \boldsymbol{\pi}=\mathbf{2 . 2 5} \boldsymbol{\pi}$.

~~ 35. The cylinder inside the cone is shown in Fig. (a).

Here, $A Q=20 cm, P G=Q E=\frac{1}{2} D E=5 \sqrt{3} cm \angle C A B=60^{\circ}$.

Fig. (a)

Fig. (b) The cross-section as a plane figure is shown is Fig. (b).

$ \frac{P G}{A P}=\tan 30^{\circ} \Rightarrow \frac{5 \sqrt{3}}{A P}=\frac{1}{\sqrt{3}} \Rightarrow A P=15 cm $

Let $Q B=r$ be the radius of the cone.

Then, in similar triangles $A P G$ and $A Q B$,

$ \frac{A P}{P G}=\frac{A Q}{Q B} \Rightarrow Q B=\frac{A Q \cdot P G}{A P}=\frac{20 \times 5 \sqrt{3}}{15}=\frac{20}{\sqrt{3}} cm $

$\therefore$ Radius of cone $=\frac{20}{\sqrt{3}} cm$, Height of cone $=20 cm$.

$\therefore$ Volume of cone

$ =\frac{1}{3} \pi r^{2} h=\frac{1}{3} \times \pi \times(\frac{20}{\sqrt{3}})^{2} \times 20=\frac{\mathbf{8 0 0 0} \boldsymbol{\pi}}{\mathbf{9}} \mathbf{c m}^{3} . $

SELF ASSESSMENT SHEET

A solid iron pole consists of a cylinder of height $220 cm$ and base diameter $24 cm$, which is surmounted by another cylinder of height $60 cm$ and radius $8 cm$. Find the mass of the pole, given that $1 cm^{3}$ of iron has approximately $8 g$ mass. (Use $\pi=3.14$ ) (a) $982.62 kg$ (b) $892.26 kg$ (c) $829.26 kg$ (d) $928.62 kg$

~~ 2. A pen stand made of wood is in the shape of a cuboid with conical depressions to hold pens. The dimensions of the cuboid are $15 cm$ by $10 cm$ by $3.5 cm$. The radius of

each of the depressions is $0.5 cm$ and the depth is $1.4 cm$. Find the volume of wood in the entire stand. (a) $532.54 cm^{3}$ (b) $523.54 cm^{3}$ (c) $534.54 cm^{3}$ (d) $543.54 cm^{3}$

~~ 3. A solid consisting of a right circular cone of height $120 cm$ and radius $60 cm$ standing on a hemisphere of radius $60 cm$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60 cm$ and its height is $180 cm$. The answer upto $2 dp$ is : (a) $1.31 m^{3}$ (b) $1.13 m^{3}$ (c) $1.42 m^{3}$ (d) $1.24 m^{3}$

~~ 4. A solid cube of side $5.5 cm$ is dropped into a cylindrical vessel partly filled with water. The diameter of the vessel is $11 cm$. If the cube is wholly submerged, the level of water will rise by: (a) $3.75 cm$ (b) $0.75 cm$ (c) $1.75 cm$ (d) $2.85 cm$.

(CDS 2003)

~~ 5. A water tank is hemispherical at the bottom and cylindrical on top of it. The radius is $12 m$. If the total capacity is $3312 \pi m^{3}$, then the capacities of the two portions are in the ratio (a) $8: 9$ (b) $8: 11$ (c) $8: 13$ (d) $8: 15$

(CDS 1996)

~~ 6. A right circular cone and a right circular cylinder have equal bases and equal heights.

If the ratio $\frac{\text{ The lateral surface of cone }}{\text{ The lateral surface of the cylinder }}$ equals $\frac{5}{8}$, then the semi-vertical angle of the right circular cone must be

ANSWERS

(a) $22 \frac{1}2^{\circ}$ (b) $30^{\circ}$ (c) $\tan ^{-1}(\frac{3}{4})(d) 45^{\circ}$

~~ 7. If a right circular cone of a certain height has its upper part cut off by a plane passing through the mid-point of its axis and at right angles to it, then the volume of lower portion of the cone and that of the upper portion of the cone will be in the ratio (a) $7: 1$ (b) $1: 7$ (c) $1: 4$ (d) $4: 1$

~~ 8. Find the volume of a tetrahedron the sides of whose base are $9 cm, 12 cm$ and $15 cm$ and height $=15 cm$. (a) $225 cm^{3}$ (b) $270 cm^{3}$ (c) $360 cm^{3}$ (d) $200 cm^{3}$

~~ 9. The base of a right pyramid is an equilateral triangle of side $4 cm$. The height of the pyramid is half of its slant height. The volume of the pyramid is : (a) $\frac{8 \sqrt{3}}{9} cm^{3}$ (b) $\frac{4 \sqrt{3}}{9} cm^{3}$ (c) $\frac{7}{3} cm^{3}$ (d) $\frac{18}{\sqrt{3}} cm^{3}$

~~ 10. A container shaped like a right circular cylinder having diameter $12 cm$ and height $15 cm$ is full of icecream. The icecream is to be filled into cones of height $12 cm$ and diameter $6 cm$, having a hemispherical shape on the top. Find the number of such cones that can be filled with icecream. (a) 150 (b) 100 (c) 10 (d) 20 (NCERT) 11. (b) 12. (b) $3(b)$ 13. $(c)$ 14. $(d)$ 15. $(c)$ 16. (a) 17. (c) 18. (a) 19. $(c)$

HINTS AND SOLUTIONS

Mass of pole $=$ Volume of pole $\times 8 g$.

Volume of pole $=$ Vol. of base cylinder + Vol. of top cylinder

$ =\pi \times(12)^{2} \times 220+\pi \times(8)^{2} \times 60 $

Now calculate.

~~ 2. Volume of wood in the entire stand

$=$ Volume of cuboidal stand - Volume of 4 conical depressions

$ \begin{aligned} & =l \times b \times h-4 \times \frac{1}{3} \pi r^{2} h \\ & =15 \times 10 \times 3.5-4 \times \frac{1}{3} \times \frac{22}{7} \times(0.5)^{2} \times 1.4 \end{aligned} $

~~ 3. Volume of water left in the cylinder

$=$ Volume of water filled in th right circular cylinder - Volume of the solid

$=$ Vol. of right circular cylinder

(Vol. of cone + Vol. of hemisphere)

$=\frac{22}{7} \times 0.6 \times 0.6 \times 1.8-(\frac{1}{3} \times \frac{22}{7} \times(0.6)^{2} \times 1.2+\frac{2}{3} \times \frac{22}{7} \times(0.6)^{3})$

~~ 4. Let the level of water rise by $h cm$. Then,

Volume of displaced water $=$ Vol. of cube

$ \Rightarrow \quad \frac{22}{7} \times(5.5)^{2} \times h=(5.5)^{3} \text{. } $

~~ 5. Total capacity of the tank = Capacity of hemispherical portion + Capacity of cylindrical portion

$\Rightarrow \frac{2}{3} \pi r^{3}+\pi r^{2} h=3312 \pi$

$\Rightarrow \pi r^{2}(\frac{2}{3} r+h)=3312 \pi$

$\Rightarrow 144(\frac{2}{3} \times 12+h)=3312$

$\Rightarrow 144(8+h)=3312 \Rightarrow 8+h=23 \Rightarrow h=15 cm$.

$\therefore$ Reqd. ratio $=\frac{\frac{2}{3} \pi r^{3}}{\pi r^{2} h}=\frac{\frac{2}{3} \times 12}{15}=\mathbf{8}: \mathbf{1 5}$.

~~ 6. Let $r$ be the radius of the cone and the cylinder and $h$ be the height of both. Then,

$\frac{\text{ Lateral surface of the cone }}{\text{ Lateral surface of the cylinder }}=\frac{5}{8}$

$\Rightarrow \frac{\pi r l}{2 \pi r h}=\frac{5}{8} \Rightarrow \frac{h}{l}=\frac{4}{5}$

If $\theta$ is the semi-vertical angle of the cone, then $\cos \theta=\frac{4}{5}$

$\Rightarrow \tan \theta=\frac{3}{4}$ or $\theta=\tan ^{-1}(\frac{3}{4})$

Note : $r^{2}=\sqrt{l^{2}-h^{2}}=3$

~~ 7. Let $O A=h, O C=r$. Then,

$A F=h / 2$ and by similarity

of triangles $A F E$ and $A O C$,

$F E=r / 2$

$\therefore \frac{\text{ Vol. of portion } D E C B}{\text{ Vol. of cone } A D E}$

$=\frac{\text{ Vol. of cone } A B C-\text{ Vol. of cone } A D E}{\text{ Vol. of cone } A D E}$

$=\frac{\frac{1}{3} \pi r^{2} h-\frac{1}{24} \pi r^{2} h}{\frac{1}{24} \pi r^{2} h}=\frac{\frac{7}{24} \pi r^{2} h}{\frac{1}{24} \pi r^{2} h}=7: \mathbf{1}$.

~~ 8. Let $a=9 cm, b=12 cm, c=15 cm$. Then,

$ s=\frac{a+b+c}{2}=\frac{9+12+15}{2}=18 cm $

$\therefore$ Area of the base $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{18(18-9)(18-12)(18-15)} \\ & =\sqrt{18 \times 9 \times 6 \times 3}=54 cm^{2} \end{aligned} $

$\therefore$ Volume of the tetrahedron

$=\frac{1}{3} \times($ Area of the base $\times$ height $)$

$=\frac{1}{3} \times 54 \times 20 cm^{3}=\mathbf{3 6 0} \mathbf{c m}^{3}$.

~~ 9. Let $h$ be the height of the pyramid and $l$ its slant height, $a$ the length of each side of the base.

Given, $h=\frac{l}{2} \quad(\because l=\sqrt{h^{2}+\frac{a^{2}}{12}})$

$\Rightarrow h=\frac{1}{2} \sqrt{h^{2}+\frac{a^{2}}{12}} \Rightarrow 4 h^{2}=h^{2}+\frac{a^{2}}{12}$

$\Rightarrow 3 h^{2}=\frac{a^{2}}{12} \Rightarrow h^{2}=\frac{a^{2}}{36}=\frac{16}{36}=\frac{4}{9} \Rightarrow h=\frac{2}{3}$

$\therefore$ Volume of the pyramid $=\frac{1}{3} \times$ Area of the base $\times$ height

$ =\frac{1}{3} \times \frac{\sqrt{3}}{4} \times(4)^{2} \times \frac{2}{3} cm^{3}=\frac{8 \sqrt{3}}{9} cm^{3} . $

~~ 10. Number of cones

$=\frac{\text{ Volume of cylindrical icecream container }}{\text{ Volume of one cone filled with icecream }}$

$=\frac{\pi r_1^{2} h_1}{(\text{ Vol. of conical part }+ \text{ Vol. of hemispherical part })}$

$=\frac{\pi r_1^{2} h_1}{\frac{1}{3} \pi r_2^{2} h_2+\frac{2}{3} \pi r_2^{3}}$

$=\frac{\pi r_1^{2} h_1}{\frac{1}{3} \pi r_2^{2}(h_2+2 r_2)}$

$=\frac{3 \times 6 \times 6 \times 15}{3 \times 3 \times(12+6)}=\frac{1620}{162}=\mathbf{1 0}$.

Disclaimer : While the authors of this book have made every effort to avoid any mistakes or omissions and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information, the authors and S. Chand do not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S. Chand and the authors expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication.

Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual. ↩︎

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