mathematics-class-11-unit-05-chapter-02-quadratic-equations-l-2

Quadratic equation lecture-2

Question 7

Let $a \in \mathbb{R}$ and the equation $-3(x-[x])^2+2(x-[x])+a^2=0$ has no integral solutions. Here $[x]$ denotes the greatest integer $\leq x$ . Then all the possible values of a lie in :

$(1) (-2,1)$
$(2) (-\infty,-2) \cup(2, \infty)$
$(3) (-1,0) \cup(0,1)$
$(4) (1,2)$

Quadratic equation lecture-2

Solution

$a \neq 0$

$[\text { if } a=0$

$x \in \mathbb{Z}$

$x-[x]=\{x\}=0]$

$x-[x]=\{x\}$

Quadratic equation lecture-2

Solution

$-3{x}^2+2{x}+a^2=0 \rArr 3{x}^2-2{x}-a^2=0$

$\frac{2 \pm \sqrt{4+12 a^2}}{6}$ $=\frac{1 \pm \sqrt{1+3 a^2}}{3}$

$1+3 a^2>1 \text { as } a \neq 0$

$\frac{1-\sqrt{1+3 a^2}}{3}<0$

Not a possible choice for $\{x\}$

${x} =\frac{1+\sqrt{1+3 a^2}}{3}>0$

Quadratic equation lecture-2

Solution

$0< 1+\sqrt{1+3 a^2}< 3$

$\Rightarrow-1<\sqrt{1+3 a^2}< 2$

$1< 1+3 a^2< 4$

$\Rightarrow 0< a^2< 1$

$\therefore \quad a \in(-1,0) \cup(0,1)$

$(-1,0) \cup(0,1) \subseteq(-2,1)$

Quadratic equation lecture-2

Question 8

For $x \geq 0$ , let $g(x)=\cos \left(x^2\right)$ and $f(x)=\sqrt{x}$ . Let $\alpha<\beta$ be the solutions of the equation $18 x^2-9 \pi x+\pi^2=0$ . Then the area (in sq. units) bounded by the curve $y=(g \circ f)(x)$ and the lines $x=\alpha, x=\beta$ and $y=0$ is :

$(1) \frac{\sqrt{3}+1}{2}$

$(2) \frac{\sqrt{3}-\sqrt{2}}{2}$

$(3) \frac{\sqrt{2}-1}{2}$

$(4) \frac{\sqrt{3}-1}{2}$

Quadratic equation lecture-2

Solution

$\frac{9 \pi \pm \sqrt{81 \pi^2-72 \pi^2}}{36}=\frac{9 \pi \pm 3 \pi}{36}$

$\alpha=\frac{9 \pi-3 \pi}{36}=\frac{\pi}{6}, \beta=\frac{9 \pi+3 \pi}{36}=\frac{\pi}{3}$

Quadratic equation lecture-2

Solution

$(g \circ f)(x)= \cos x$

$\int_{\pi / 6}^{\pi / 3} \cos x d x =[\sin x]_{\pi / 6}^{\pi / 3}$

$=\sin \frac{\pi}{3}-\sin \frac{\pi}{6}$

$=\frac{\sqrt{3}}{2}-\frac{1}{2}$

$=\frac{\sqrt{3}-1}{2}$

alt text

Quadratic equation lecture-2

Question 9

For $0 \leq \alpha \leq 1$ , if the line $x=\alpha$ divides the area of the region

${R}=\{{(x, y)} \in \mathbb{R}^2: {x^3 \leq y \leq x} \text { and } 0 \leq x \leq 1 \}$

into two equal parts, then

(1) $0<\alpha \leq \frac{1}{2}$

(2) $\frac{1}{2}<\alpha< 1$

(3) $2 \alpha^4-4 \alpha^2+1=0$

(4) $\alpha^4+4 \alpha^2-1=0$

Quadratic equation lecture-2

Solution

Area of A

$=\int_\alpha^\alpha x d x-\int_0^\alpha x^3 d x$

$=\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^\alpha$

$=\frac{\alpha^2}{2}-\frac{\alpha^4}{4}$

alt text

Quadratic equation lecture-2

Solution

$\text { Area of } B$

$=\int_\alpha^1 x d x-\int_\alpha^1 x^3 d x$ $= {\left[\frac{x^2}{2}-\frac{x^4}{4}\right]^{1} _\alpha }$

$=\frac{1}{4}-\left(\frac{\alpha^2}{2}-\frac{\alpha^4}{4}\right)$

$\rArr \frac{\alpha^2}{2}-\frac{\alpha^4}{4}$ $=\frac{1}{4}-\frac{\alpha^2}{2}+\frac{\alpha^4}{4}$

$\frac{\alpha^4}{2}-\alpha^2+\frac{1}{4}=0, \quad 2 \alpha^4-4 \alpha^2+1=0$

Quadratic equation lecture-2

Solution

$2 \alpha^4-4 \alpha^2+1=0$

$\frac{4 \pm \sqrt{16-8}}{4}=1 \pm \frac{\sqrt{2}}{2}$

$1+\frac{\sqrt{2}}{2}>1$

Not a possible choice for $\alpha^2$

as $\alpha \leqslant 1$

Quadratic equation lecture-2

Solution

$\alpha^2=1-\frac{\sqrt{2}}{2}$

If $\alpha \leqslant \frac{1}{2} \quad$ then $\quad \alpha^2 \leqslant \frac{1}{4}$

$1-\frac{\sqrt{2}}{2} \leqslant \frac{1}{4}$

$\Rightarrow \frac{3}{4} \leqslant \frac{\sqrt{2}}{2}$ $\Rightarrow \frac{9}{16} \leqslant \frac{2}{4}=\frac{1}{9}$ Not possible

$\alpha^2 \rightarrow \frac{-4 \pm \sqrt{16+4}}{2}$

$=-2 \pm \sqrt{5}$

Quadratic equation lecture-2

Question 10

If, for a positive integer $n$ , the quadratic equation

${x(x+1)+(x+1)(x+2)+\cdots+(x+n-1)(x+n)=10 n}$

has two consecutive integral solutions, then $n$ is equal to

(1) $9$
(2) $10$
(3) $11$
(4) $12$

Quadratic equation lecture-2

Solution

coefficient of $x^2$ is $n$

$x^2+x \quad x^2+x+2 x+2=x^2+3 x+2$

$=x^2+(n-1) x+n x+n(n-1)$

$= x^2+(2 n-1) x+n(n-1)$

Quadratic equation lecture-2

Solution

co-efficient of $x$

$= 1+3+\cdots+(2 n-1)$

$= 1+2+3+4+\cdots+(2 n-1)+2 n$ $-(2+4+\cdots+2 n)$

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

$= n(2 n+1)-n(n+1)$

$= 2 n^2+n-n^2-n=n^2$

Quadratic equation lecture-2

Solution

$x(x+1) \longrightarrow 0$

$(x+1)(x+2) \longrightarrow 2$

$(x+2)(x+3) \longrightarrow 6$

$\vdots$

$(x+n-1)(x+n) \longrightarrow(n-1) n$

$10 n$

Quadratic equation lecture-2

Solution

constant term

$=0+2+6+\cdots+n(n-1)-10 n$

$=\sum_{k=1}^n k(k-1)-10 n =\sum_{k=1}^n k^2-\sum_{k=1}^n k-10 n$

$=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}-10 n$

$\frac{n(n+1)}{2} \{\frac{2 n+1}{3}-1\}-10 n$

$= \frac{n(n+1)(n-1)}{3}-10 n = \frac{n\left(n^2-1\right)}{3}-10 n$

$n x^2 + n^2 x + \frac{n(n^2-1)}{3}-10 n=0$

Quadratic equation lecture-2

Solution

$x^2+n x+\frac{n^2-1}{3}-10=0$

$x^2+n x+\frac{n^2-31}{3}=0$

The two consecutive integer solutions, are

say $\quad m, m+1$

$m+m+1=-n \qquad$ $\biggl[\therefore \quad 2 m=-(n+1)\biggl]$

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

$m(m+1)$ $=\frac{n^2 -31}{3}$

Quadratic equation lecture-2

Solution

$\rArr -\frac{(n+1)}{2}\left(1-\frac{n+1}{2}\right)=\frac{n^2-31}{3}$

$\rArr -\frac{(n+1)(1-n)}{4}=\frac{n^2-31}{3}$

$\Rightarrow \frac{n^2-1}{4}=\frac{n^2-31}{3}$

$\Rightarrow 3 n^2-3=4 n^2-124$

$\Rightarrow n^2=121 \Rightarrow n=11$

Quadratic equation lecture-2

Question 11

Suppose the quadratic equation with real coefficients

p(x)=0

has purely imaginary solutions (ie. the real part of the solutions are zero). Then the equation

p(p(x))=0

has

$i \alpha \quad \alpha \in \mathbb{R}$

(1) one purely imaginary solution

(2) all real solutions

(3) two real and two purely imaginary solutions

(4) neither real nor purely imaginary solutions

Quadratic equation lecture-2

Solution

$a x^2+b x+c=0 \quad a>0$

$\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$

$-4 a c<0$

$b^2-4 a c<0$

$b=0$

$\Rightarrow a c>0$

Quadratic equation lecture-2

Solution

$a, c \longrightarrow$ both have same signs

$P(x)=0$

$a x^2+c=0$

$\Rightarrow x^2+\frac{c}{a}=0 \quad \quad e{\prime}>0$

$\begin{aligned} & P(P(x))=0 \\\ & \left(x^2+e^{\prime}\right)^2+e^{\prime}=0\end{aligned}$

$x^4+2 x^2 c^{\prime}+\left(c^{{\prime}^2}+c^{\prime}\right)=0\quad\biggl[$ quadratic equation $\text { in } x^2 \biggl]$

Quadratic equation lecture-2

Solution

$\rArr ~ \frac{-2 c^{\prime} \pm \sqrt{4 c^2-4\left(c^{\prime 2}+c^{\prime}\right)}}{2}$

$\rArr ~ \beta^2 \text { is a solution of } P(P(x))=0$

$\rArr ~ -c^{\prime} \pm i \sqrt{c^{\prime}}$

Quadratic equation lecture-2

Solution

$\text { if } \beta=i \alpha \text { or } \beta=\alpha \quad \alpha \in \mathbb{R}$

$\text { then } \beta^2=-\alpha^2 \text { or } \beta^2=\alpha^2$

$\Rightarrow \beta^2 \rightarrow \text { is not real }$

Quadratic equation lecture-2

Question 12

Let $a, b, c, d$ be four distinct numbers. Let $a, b$ be the solutions of $x^2-10 c x-11 d=0$ and $c, d$ be the solutions of $x^2-10 a x-11 b=0$ . Then the value of $a+b+{c+d}$ is :

(1) 0

(2) 11

(3) 121

(4) 1210

Quadratic equation lecture-2

Solution

$a, b$ solutions of $x^2-10 e x-11 d=0$ ,

a+b=10 c

$c, d$ solutions of $x^2-10 a x-11 b=0$ , $c+d=10$ a

a+b+c+d=10(a+c)

$a^2-10 c a-11 d=0$

$c^2-10 a c-11 b=0$

$a^2-c^2-11 d+11 b=0$

Quadratic equation lecture-2

Solution

$(a+c)(a-c)=11(d-b)\quad \left\{ \underset{\bold{c+d=10 a}}{a+b=10 c}\right\}$

$a+b-c-d=10(c-a)$

$b-d=11(c-a)$
$d-b=11(a-c)$

$(a+c)(a-c)={121}{(a-c)}$

${a \neq c},\qquad a-c \neq 0$

$a+c=121$

$a+b+c+d=10(a+c)=10 \times 121$

Quadratic equation lecture-2

Question 13

Let $S$ be the set of all nonzero real numbers $\alpha$ such that the quadratic equation $\alpha x^2-x+\alpha=0$ has two distinct real solutions $x_1$ and $x_2$ with $\left|x_1-x_2\right|<1$ . Identify the possible subsets of $S$ from the ${\text { list given below }}$

(1) $\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right)$

(2) $\left(-\frac{1}{\sqrt{5}}, 0\right)$

(3) $\left(0, \frac{1}{\sqrt{5}}\right)$

(4) $\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$

Quadratic equation lecture-2

Solution

$a x^2+b x+c=0$

$b^2-4 a c>0$

$1-4 \alpha^2>0$

$\Rightarrow 4 \alpha^2< 1$

$\Rightarrow \alpha^2< 1 / 4$

Quadratic equation lecture-2

Solution

$\alpha \in\left(-\frac{1}{2}, \frac{1}{2}\right) \backslash\{0\}$

$\left|x_ 1-x_2\right|< 1$

$\Leftrightarrow \underbrace{\left(x_ 1-x_ 2\right)^2}_ {\left(x_ 1+x_ 2\right)^2-4 x_ 1 x_2 < 1}< 1$

$\alpha x^2-x+\alpha=0$

$x_ 1+x_2=\frac{1}{\alpha}$

$x_ 1 x_2=1$

Quadratic equation lecture-2

Solution

$\frac{1}{\alpha^2}-4 < 1 \Rightarrow ~ \frac{1}{\alpha^2} < 5$

$\Rightarrow ~ \alpha^2 > \frac{1}{5}$

$\Rightarrow ~ \alpha > \frac{1}{\sqrt{5}} \quad \text { or } \quad \alpha < -\frac{1}{\sqrt{5}}$

$S=\left(-\frac{1}{2},-\frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$

Quadratic equation lecture-2

Question 14

For $p \neq 0$ , let $\alpha, \beta$ be the solutions of $p x^2+q x+r=0$ . If $p, q, r$ are in arithmetic progression and $\frac{1}{\alpha}+\frac{1}{\beta}=4$ , then the value of $|\alpha-\beta|$ is :

(1) $\frac{\sqrt{34}}{9}$

(2) $\frac{2 \sqrt{ } 13}{9}$

(3) $\frac{\sqrt{61}}{9}$

(4) $\frac{2 \sqrt{34}}{9}$

Quadratic equation lecture-2

Solution

$\alpha+\beta=-\frac{q}{p} \ldots (i)$

$\alpha \beta=\frac{r}{p}$

$\frac{1}{\alpha}+\frac{1}{\beta}=4$

$\alpha+\beta=4 \alpha \beta$

$=\frac{4 r}{p} \ldots (ii)$

from (i) and (ii)

$-\frac{q}{p}=\frac{4 r}{p}$

Quadratic equation lecture-2

Solution

$\operatorname{as} ~ {p \neq 0}$

$[{q}=-{4 p}]$

$q=\frac{p+r}{2} \quad$

$\Rightarrow-4 r=\frac{b+r}{2} \quad$

$\Rightarrow p=-9 r \quad$

$\frac{r}{p}=-\frac{1}{9}$

Quadratic equation lecture-2

Solution

$\alpha+\beta = -\frac{4}{9}$

${(\alpha-\beta)}^2 ={\alpha+\beta}^2 -4\alpha\beta$

$=\frac{16}{81}-4\left(-\frac{1}{9}\right) ~ =\frac{16}{81}+\frac{4}{9} ~ =\frac{52}{81}$

$\alpha-\beta ~ = \pm \frac{2 \sqrt{13}}{9}$

$\therefore|\alpha-\beta| = \frac{2 \sqrt{13}}{9}$

Quadratic equation lecture-2

Question 15

Let $a, b, c$ be real numbers and $a \neq 0$ . If $\alpha$ is a solution of $a^2 x^2+b x+c=0, \beta$ is a solution of $a^2 x^2-b x-c=0$ and $0<\alpha<\beta$ , then the equation $a^2 x^2+2 b x+2 c={0}$ has a solution $\gamma$ which always satisfies :

(1) $\gamma=\frac{\alpha+\beta}{2}$

(2) $\gamma=\alpha+\frac{\beta}{2}$

(3) $\gamma=\alpha$

(4) $\alpha<\gamma<\beta$

Quadratic equation lecture-2

Solution

$a^2 \alpha^2+b \alpha+c=0$

$a^2 \beta^2-b \beta-c=0$

$f(x)=a^2 x^2+2 b x+2 c$

Quadratic equation lecture-2

Solution

$f(\alpha)=a^2 \alpha^2+2 b \alpha+2 c$

$=\underbrace{a^2 \alpha^2+b \alpha+e}+(b \alpha+e)$

$=b \alpha+c$

$=-a^2 \alpha^2<0$

$\gamma \neq \alpha$

Quadratic equation lecture-2

Solution

$f(\beta)={a^2 \beta^2}+2 b \beta+2 c$

$= 3 b \beta + 3 c ~ =3 a^2 \beta^2 > 0$

$b \alpha+c=-a^2 \alpha^2$

$a^2 \beta^2=b \beta+c$

$0 < \alpha < \beta$

$\exists \alpha<\gamma<\beta$

$\text { so that } f(\gamma)=0$

Quadratic equation lecture-2

Solution

$b \alpha+c=-a^2 \alpha^2$

$a^2 \beta^2=b \beta+c$

$f\left(\frac{\alpha+\beta}{2}\right) =a^2\left(\frac{\alpha+\beta}{2}\right)^2+2 b\left(\frac{\alpha+\beta}{2}\right)+2 c$

$=\underbrace{a^2\left(\frac{\alpha+\beta}{2}\right)^2}+\underbrace{(b \alpha+e)}+\underbrace{(b \beta+e)}$

$>a^2\left(\frac{2 \alpha}{2}\right)^2 \quad \quad -a^2 \alpha^2 \quad \quad +a^2 \beta^2$

$=a^2 \beta^2>0$

Quadratic equation lecture-2

Solution

$f(\alpha)<0$

$f(\beta)>0$

$f\left(\frac{\alpha+\beta}{2}\right)>0$

$\alpha+\frac{\beta}{2}>\frac{\alpha+\beta}{2} \quad \text { as } \quad\alpha>0$

$f\left(\alpha+\frac{\beta}{2}\right)>0$

alt text

Thank you