#### need solution for rd sharma maths class 12 chapter 14 Mean Value Theoram exercise multiple choice question 3

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.