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

Answer:  $\sqrt{\frac{4}{\pi}-1}$

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

Given: $f(x)=\tan ^{-1} x \text { on }[0,1]$

Solution:

$f(x)=\tan ^{-1} x$

$f(x)$  is a polynomial function.

It is continuous in [0, 1]

$f^{\prime}(x)=\frac{1}{\left(1+x^{2}\right)}$

(Which is defined in [0, 1])

$\therefore f(x)$ is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \\ &\frac{1}{1+c^{2}}=\frac{\tan ^{-1}(1)-\tan ^{-1}(0)}{1} \end{aligned}

\begin{aligned} &\frac{1}{1+c^{2}}=\frac{\frac{\pi}{4}-0}{1} \\ &\frac{1}{1+c^{2}}=\frac{\pi}{4} \end{aligned}

\begin{aligned} &\pi\left(1+c^{2}\right)=4 \\ &\pi+\pi c^{2}=4 \\ &\pi c^{2}=4-\pi \end{aligned}

\begin{aligned} &c^{2}=\frac{4-\pi}{\pi} \\ &c=\sqrt{\frac{4-\pi}{\pi}} \end{aligned}

$c=\sqrt{\frac{4}{\pi}-1}$