Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14 point 1 question 3 sub question (iii)

$c \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$

Hint:

$f\left(-\frac{\pi}{4}\right)=f\left(\frac{\pi}{4}\right)$, so there exists$c \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$

Given:

$f(x)=\cos 2 x$ on $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$

Explanation:

We have

$f(x)=\cos 2 x$ on $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$

We know that cosine function is continuous and differentiable on R.

Let’s find the value of the function at an extreme

$\begin{array}{l} f\left(\frac{-\pi}{4}\right)=\cos 2\left(\frac{-\pi}{4}\right) \\\\ f(0)=\cos \left(\frac{-\pi}{2}\right) \end{array}$

We know that

$\cos (-x)=\cos x$

$\begin{array}{l} f(0)=0 \\\\ f\left(\frac{\pi}{4}\right)=\cos 2\left(\frac{\pi}{4}\right) \\\\ f\left(\frac{\pi}{2}\right)=\cos \left(\frac{\pi}{2}\right) \\\\ f\left(\frac{\pi}{2}\right)=0 \end{array}$

We have$f\left(\frac{-\pi}{4}\right)=f\left(\frac{\pi}{4}\right)$ , so there exists$c \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$

Such that$f '(c) = 0$

Let’s find the derivative of f(x)

$\begin{array}{l} f^{\prime}(x)=\frac{d(\cos 2 x)}{d x} \\\\ f^{\prime}(x)=-\sin 2 x \frac{d(2 x)}{d x} \\\\ f^{\prime}(x)=-2 \sin 2 x \\\\ \end{array}$

We have$f '(c) = 0$

$\begin{array}{l} -2 \sin 2 c=0 \\\\ \sin 2 c=0 \\\\ 2 c=0 \\\\ c=0_{\mathrm{as}} \quad c \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \end{array}$

Hence, Rolle’s Theorem is verified