#### need solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (xiii)

Answer:  $\sqrt{6}$

Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=\sqrt{x^{2}-4} \text { on }[2,4]$

Solution:

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

$f(x)$  is a polynomial function.

It is continuous in [2, 4]

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

(Which is defined in [2, 4])

$\therefore f(x)$ is differentiable in [2, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[2,4] \\ &f^{\prime}(c)=\frac{f(4)-f(2)}{4-2} \end{aligned}

\begin{aligned} &\frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{(4)^{2}-4}-\sqrt{(2)^{2}-4}}{2} \\ &\frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{16-4}-\sqrt{4-4}}{2} \end{aligned}

\begin{aligned} &\frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{12}-0}{2} \\ &\frac{c}{\sqrt{c^{2}-4}}=\frac{2 \sqrt{3}}{2} \end{aligned}

\begin{aligned} &\frac{c}{\sqrt{c^{2}-4}}=\sqrt{3} \\ &c=\sqrt{3\left(c^{2}-4\right)} \end{aligned}

Squaring on both sides,

\begin{aligned} &c^{2}=3\left(c^{2}-4\right) \\ &c^{2}=3 c^{2}-12 \\ &3 c^{2}-c^{2}=12 \\ &2 c^{2}=12 \\ &c^{2}=6 \\ &c=\sqrt{6} \end{aligned}