#### Please Solve RD Sharma Class 12 Chapter 14 Inverse Trigonometric Functions Exercise Fill in the blanks Question 1 Maths Textbook Solution.

$\sqrt{3}$

Hint:

Use mean value theorem’s formula.

Given:

$f\left ( x \right )=x+\frac{1}{x},x\: \epsilon \left [ 1,3 \right ]$

Solution:

$f\left ( x \right )=x+\frac{1}{x}$

On differentiating we get,

$\Rightarrow$         ${f}'\left ( x \right )=1-\frac{1}{x^{2}}$

$\Rightarrow$         ${f}'\left ( c\right )=1-\frac{1}{c^{2}}$

Apply mean value theorem,

\begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \Rightarrow & 1-\frac{1}{c^{2}}=\frac{3+\frac{1}{3}-(1+1)}{3-1} \\ \Rightarrow & 1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{2} \\ \Rightarrow & 1-\frac{1}{c^{2}}=\frac{4}{6} \end{aligned}

$\Rightarrow \frac{1}{c^{2}}=1-\frac{4}{6}$

$\Rightarrow \frac{1}{c^{2}}=\frac{2}{6}=\frac{1}{3}$

$\Rightarrow c^{2}=3$

$c=\pm \sqrt{3}$

$\Rightarrow c=\sqrt{3}$                                                                                                                             $\left [ \because x\: \epsilon \left [ 1,3 \right ] \right ]$