RD Sharma Solutions Class 12 Mathematics Chapter 14 MCQ

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RD Sharma Class 12 Solutions Chapter 14MCQ Mean Value Theorems - Other Exercise

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RD Sharma Class 12 Solutions Chapter 14MCQ Mean Value Theorems - Other Exercise
Mean Value Theorems questions:MCQ
RD Sharma Chapter-wise Solutions

Mean Value Theorems questions:MCQ

Mean Value Theoram exercise multiple choice question 1

Answer:
Option (c)
Hint:
You must know about the concept of roots of the equation.
Given:
Polynomial equation,
$a_{n}x^{n}+a_{n-1\; }x^{n-1}+a_{n-2}\; x^{n-2}+....+a_{2}x^{2}+a_{1}x+a_{0}=0$
n being a positive integer, has two different real roots $\alpha$ and $\beta$ .
Solution:
$f(x)=a_{n}x^{n}+a_{n-1\; }x^{n-1}+a_{n-2}\; x^{n-2}+....+a_{2}x^{2}+a_{1}x+a_{0}=0$
Now, $f(\alpha )=0$ [ $\because \alpha$ is root of equation]
$f(\beta )=0$ [ $\because \beta$ is root of equation]

Now, $f^{'}(x)=na_{n}x^{n-1}+(n-1)a_{n-1\; }x^{n-2}+....+a_{1}=0$ has at least one root in $\left [ \alpha ,\beta \right ]$ .
Hence, option (c) is correct.

Mean Value Theoram exercise multiple choice question 2

Answer:
Option (c)
Hint:
Use Rolle’s Theorem.
Given:
$4a+2b+c=0$ and $3ax^{2}+2bx+c=0$ has at least one real root.
Solution:
Let, $f(x)=ax^{3}+bx^{2}+cx+d$
$f(0)=d$
$f(2)=a\left ( 2 \right )^{3}+b\left ( 2 \right )^{2}+c\left ( 2 \right )+d$
$=8a+4b+2c+d$
$=2(4a+2b+c)+d$
$=0+d$ $\left [ \because 4a+2b+c=0 \right ]$
$f(2)=d$
$\therefore f$ is continuous in closed interval [0,2]
$f(0)=f(2)$
As per Rolle’s Theorem,
$f^{'}(x)=3ax^{2}+2bx+c$
$f^{'}(\alpha )=3a\alpha ^{2}+2b\alpha +c$
$3a\alpha ^{2}+2b\alpha +c=0$
Hence, equation $f(x)$ has at least one root in the interval (0,2)
$\therefore f(x)$ must have one root in the interval (0,2) .
So, option (c) is correct.

Mean Value Theoram exercise multiple choice question 3

Answer:
Option (b)
Hint:
You must know about Lagrange’s mean value theorem.
Given:
$f(x)=x+\frac{1}{x},\; x\in \left [ 1,3 \right ]$
Solution:
$f(x)=x+\frac{1}{x},\; x\in \left [ 1,3 \right ]$
$f(x)=\frac{x^{2}+1}{x}$
Using mean value theorem,
$f^{'}(c)=\frac{f\left ( b \right )-f\left ( a \right )}{b-a}$
$f^{'}(c)=\frac{f\left ( 3 \right )-f\left ( 1 \right )}{3-1}$ $\left [ \because x=c,\; b=3,\; a=1 \right ]$
$f^{'}(c)=\frac{\left ( 3+\frac{1}{3} \right )-\left ( 1+\frac{1}{1} \right )}{3-1}$
$\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{\frac{9+1-6}{3}}{2}$
$\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{4}{3}\times \frac{1}{2}$
$\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{2}{3}$
$\Rightarrow \; 1-\frac{2}{3}=\frac{1}{c^{2}}$
$\Rightarrow \; 1-\frac{2}{3}=\frac{1}{c^{2}}$
$\Rightarrow \; \frac{1}{3}=\frac{1}{c^{2}}$
$\Rightarrow \; c^{2}=3$
$\Rightarrow \;c=\pm \sqrt{3}$
but only $c=\sqrt{3}\in \left [ 1,3 \right ]$
Hence option (b) is correct.

Mean Value Theoram exercise multiple choice question 4 maths

Answer:
Option (c)
Hint:
Prove mean value theorem
Given: $f^{'}(x_{1})=\frac{f(b)-f(a)}{b-a}$
Solution:
Mean value theorem,
$f(x)$ is continuous in $\left [ a,b \right ]$
$f(x)$ is differentiable in $\left ( a,b \right )$
$f^{'}(x_{1})=\frac{f(b)-f(a)}{b-a}$
$a< x_{1}< b$
$\therefore f^{'}(x_{1})=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow a< x_{1}< b$
Hence, option (c) is correct.

Mean Value Theoram exercise multiple choice question 5

Answer:
Option (b)
Hint:
We will use the concept of Rolle’s Theorem.
Given:
$\phi (x)=a^{\sin x},a> 0$ and Rolle’s theorem is applicable to it.
Solution:
$\phi (x)=a^{\sin x},a> 0$
Differentiate it with respect to ' $x$ '
$\phi ^{'}(x)=\log a\left ( \cos x.a^{\sin x} \right )$
$\Rightarrow \; \; \phi ^{'}(c)=\log a\left ( \cos c.a^{\sin c} \right )$
Let, $\phi (c)=0$
$\Rightarrow \; \;\log a\left ( \cos c.\; a^{\sin c} \right )=0$
$\Rightarrow \; \;\cos c.\; a^{\sin c}=0$
$\Rightarrow \; \;\cos c=0$
$\Rightarrow \; \; \cos c=\cos \frac{\pi }{2}$ $\left [ \because \cos \frac{\pi }{2} =0\right ]$
$\Rightarrow \; \; c=\frac{\pi }{2}\in \left [ 0,\pi \right ]$
Hence, Rolle’s Theorem is applicable in the interval $\left [ 0,\pi \right ]$
Option (b) is correct.

Mean Value Theoram exercise multiple choice question 6

Answer:
option (a) 2
Hint:
Use Rolle’s Theorem formula
Given:
$f(x)=2x^{3}-5x^{2}-4x+3,\; x\in \left [ \frac{1}{3},3 \right ]$
Solution:
We know that,
$f(x)=ax^{3}+bx^{2}+cx+d,x\in \left [ a,b \right ]$
$f(0)=d$
$f^{'}(x)=6x^{2}-10x-4$
As per Rolle’s Theorem,
$f^{'}(x)=6x^{2}-10x-4$
$f^{'}(c)=6c^{2}-10c-4$
$f^{'}(c)=0$
$\Rightarrow 6c^{2}-10c-4=0$
$\Rightarrow 2(3c^{2}-5c-2)=0$
$\Rightarrow 3c^{2}-5c-2=0$
$\Rightarrow 3c^{2}-6c+c-2=0$
$\Rightarrow (3c+1)(c-2)=0$
$\Rightarrow c=2,\frac{-1}{3}$
$\Rightarrow c=2$ $\left [ \because \frac{-1}{3}\notin \left ( \frac{1}{3},3 \right ) \right ]$
Hence, Option (a) is correct.

Mean Value Theoram exercise multiple choice question 7

Answer:
Option (a)
Hint:
Find the slope of points.
Given:
$y=x\log x$ parallel to the chord joining the points (1,0) and (e,e) .
Solution:
Slope of points $=\frac{e-0}{e-1}$
$y=x\log x$
$\frac{dy}{dx}=1+log x$
Now, $1+log\; x=\frac{e}{e-1}$ [? slope of tangent $=\frac{dy}{dx}$ ]
$\Rightarrow \; \; \; \; \; \;\; \; \log x=\frac{e}{e-1}-1$
$\Rightarrow \; \; \; \; \; \;\; \; \log x=\frac{e-(e-1)}{e-1}$
$\Rightarrow \; \; \; \; \; \;\; \; \log x=\frac{1}{e-1}$
$\Rightarrow \; \; \; \; \; \;\; \; x=e^{\frac{1}{e-1}}$
Hence option (a) is correct.

Mean Value Theoram exercise multiple choice question 8 maths

Answer:
Option (c)
Hint:
You must know about Rolle’s Theorem.
Given: $f(x)=\frac{x(x+1)}{e^{x}}$ defined on [ -1, 0]
Solution:
$f(x)=\frac{x(x+1)}{e^{x}}$
$\Rightarrow \; \; \; \; \; f(x)=\frac{x^{2}+x}{e^{x}}$
$\Rightarrow \; \; \; \; \; f^{'}(x)=\frac{e^{x}\left ( 2x+1 \right )-\left ( x^{2}+x \right )\left ( e^{x} \right )}{e^{2x}}$
$\Rightarrow \; \; \; \; \; f^{'}(x)=\frac{\left ( 2x+1 \right )-\left ( x^{2}+x \right )}{e^{x}}$
$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{\left ( 2c+1 \right )-\left ( c^{2}+c \right )}{e^{c}}$
Using Rolle’s Theorem,
$f^{'}(c)=0$
$\Rightarrow \; \; \; \; \; \frac{\left ( 2c+1 \right )-\left ( c^{2}+c \right )}{e^{c}}=0$
$\Rightarrow \; \; \; \; \; 2c+1-c^{2}-c=0$
$\Rightarrow \; \; \; \; \; c^{2}+c+1=0$
$\Rightarrow \; \; \; \; \; c^{2}-c-1=0$
$\Rightarrow \; \; \; \; \; c=\frac{-(-1)\pm \sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}$ $\left [ \because c=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \right ]$
$\Rightarrow \; \; \; \; \; c=\frac{1\pm \sqrt{1+4}}{2}$
$\Rightarrow \; \; \; \; \; c=\frac{1\pm \sqrt{5}}{2}$
but $\; \; \; \; \; c=\frac{1- \sqrt{5}}{2}\in \left [ -1,0 \right ]$
Hence option (c) is correct.

Mean Value Theoram exercise multiple choice question 9

Answer:

Option (d)

Hint:

Find the derivative ofand then apply the mean value theorem.

Given:

$f(x)=x(x-2),x\in \left [ 1,2 \right ]$

Solution:

$f(x)=x(x-2)$

$\Rightarrow \; \; \; \; \; f(x)=x^{2}-2x$

$\Rightarrow \; \; \; \; \; f^{'}(x)=2x-2$

$\Rightarrow \; \; \; \; \; f^{'}(c)=2c-2$

Using mean value theorem,

$f^{'}(c)=\frac{f(b)-f(a)}{b-a}$

$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{0-(-1)}{2-1}$ $\left [ \because b=2,a=1 \right ]$

$\Rightarrow \; \; \; \; \; f^{'}(c)=\frac{1}{1}$

$\Rightarrow \; \; \; \; \; 2c-2=1$ $\Rightarrow \; \; \; \; \; \left [ \because f^{'}(c)=2c-2 \right ]$

$\Rightarrow \; \; \; \; \; 2c=1+2$

$\Rightarrow \; \; \; \; \; c=\frac{3}{2}$

Hence option (d) is correct.

Mean Value Theoram exercise multiple choice question 10

Answer:
Option (a)
Hint:
Differentiate the given function and then apply Rolle’s Theorem.
Given:
$f(x)=x^{3}-3x,x\in \left [ 0,\sqrt{3} \right ]$
Solution:
$f(x)=x^{3}-3x,x\in \left [ 0,\sqrt{3} \right ]$
$f(x)=x^{3}-3x$
$\Rightarrow \; \; \; \; \; f^{'}(x)=3x^{2}-3$
$\Rightarrow \; \; \; \; \; f^{'}(c)=3c^{2}-3$
Applying Rolle’s Theorem,
$\Rightarrow \; \; \; \; \; f^{'}(c)=0$
$\Rightarrow \; \; \; \; \; 3c^{2}-3=0$
$\Rightarrow \; \; \; \; \; 3c^{2}=3$
$\Rightarrow \; \; \; \; \; c^{2}=1$
$\Rightarrow \; \; \; \; \; c=\pm 1$
$\Rightarrow \; \; \; \; \; c=1$ $\left [ \because c\in \left [ 0,\sqrt{3} \right ] \right ]$
Hence option (a) is correct.

Mean Value Theoram exercise multiple choice question 11

Answer:
Option (b)
Hint:
You must know about the concept of Rolle’s Theorem.
Given:
$f(x)=e^{x}\sin x,x\in \left [ 0,\pi \right ]$
Solution:
$f(x)=e^{x}\sin x$
$\Rightarrow f^{'}(x)=e^{x}\cos x-\sin x\left ( e^{x} \right )$
$\Rightarrow f^{'}(x)=e^{x}(\cos x -\sin x )$
$\Rightarrow f^{'}(c)=e^{c}(\cos c -\sin c )$
As Rolle’s Theorem,
$\Rightarrow f^{'}(c)=0$
$\Rightarrow e^{c}(\cos c -\sin c )=0$
$\Rightarrow \cos c -\sin c =0$
$\Rightarrow \cos c=\sin c$
$\Rightarrow \sin \left ( \frac{\pi }{2}-c \right )=\sin c$
$\Rightarrow c=\frac{\pi }{2}-c$
$\Rightarrow 2c=\frac{\pi }{2}$
$\Rightarrow c=\frac{\pi }{4}$
Option (b) is correct.

RD Sharma Class 12th Chapter 14 MCQ contains the chapter ‘Mean Value Theorem.’ This particular exercise has 11 questions that are quite fundamental and easy to answer. The concepts that students will learn from these questions are:

1. Mean value theorem at a closed interval

2. Lagrange’s theorem

3. Rolle’s theorem

4. Logarithmic and algebraic expressions with limits

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