Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (ii)

$c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

Hint:

$f(0)=f\left(\frac{\pi}{2}\right)$ , so there exists$c \in\left(0, \frac{\pi}{2}\right)$

Given:

$f(x)=\sin 2 x$ on $\left[0, \frac{\pi}{2}\right]$

Explanation:

We have

$f(x)=\sin 2 x \text { on }\left[0, \frac{\pi}{2}\right]$

We know that sine function is continuous and differentiable on

$\begin{array}{l} f(0)=\sin 2(0) \\\\ f(0)=\sin (0) \\\\ f(0)=0 \\\\ f\left(\frac{\pi}{2}\right)=\sin 2\left(\frac{\pi}{2}\right) \\\\ f\left(\frac{\pi}{2}\right)=\sin \pi \end{array}$.

$f\left(\frac{\pi}{2}\right)=0$

We have$f(0)=f\left(\frac{\pi}{2}\right)$

So there exists$c \in\left(0, \frac{\pi}{2}\right)$

Such that$f^{\prime}(c)=0$

Let find the derivative of$f(x)$

$\\ f^{\prime}(x)=\frac{d(\sin 2 x)}{d x} \\\\ f^{\prime}(x)=\cos 2 x \frac{d(2 x)}{d x} \\\\ \qquad f^{\prime}(x)=2 \cos 2 x \\\\$

We have

$\\f^{\prime}(c)=0 \\\\ 2 \cos 2 c=0 \\\\ 2 c=\frac{\pi}{2} \\\\ c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

Hence, Rolle’s Theorem is verified.