#### Explain Solution RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Very Short Answer Question 5 Maths.

$c=\sqrt{5}$

Hint:

Find differentiability of $f\left ( x \right )$ and then apply the formula of ${f}'\left ( c \right )$.

Given:

$f\left ( x \right )=\sqrt{x^{2}-4},x\: \epsilon \left [ 2,3 \right ]$

Solution:

$f\left ( x \right )=\sqrt{x^{2}-4}$

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

Differentiate,

$\Rightarrow {f}'\left ( x \right )=\frac{1}{2}\left ( x^{2}-4 \right )^{\frac{1}{2}-1}\cdot 2x$                                                                     $\left [ \because \frac{d}{dx}\left ( x^{n} \right ) =n\left ( x \right )^{n-1}\right ]$

\begin{aligned} \Rightarrow &f^{\prime}(x)=\frac{1}{2}\left(x^{2}-4\right)^{\frac{-1}{2}} \cdot 2 x \\\\ \Rightarrow &f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x \\\\ \Rightarrow &f^{\prime}(x)=\frac{x}{\sqrt{x^{2}-4}} \\\\ \Rightarrow &f^{\prime}(c)=\frac{c}{\sqrt{c^{2}-4}} \end{aligned}

Use mean value theorem,

\begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\\\ \Rightarrow & \frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{5}-0}{3-2} \\\\ \Rightarrow & \frac{c}{\sqrt{c^{2}-4}}=\sqrt{5} \\\\ \Rightarrow & c=\sqrt{5}\left(\sqrt{c^{2}-4}\right) \\\\ \Rightarrow & c^{2}=5\left(c^{2}-4\right) \end{aligned}

$\Rightarrow c^{2}=5c^{2}-20$

$\Rightarrow4 c^{2}=20$

$\Rightarrow c^{2}=5$

$\Rightarrow c=\pm \sqrt{5}$                                                                                                                                    $\left [ \because x\: \epsilon \left [ 2,3 \right ] \right ]$

$\Rightarrow c=\sqrt{5}$