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

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

Hint:

$f(0)=f(\pi)$ , so there exists c belongs to

Given:

$f(x)=\cos 2 x$ on $[0, \pi]$

Explanation:

We have

$f(x)=\cos 2 x$ on $[0, \pi]$

We know that exponential and cosine functions are continuous and differentiable on R.

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

$\\ \Rightarrow \quad f(0)=\cos 2(0) \\\\ \Rightarrow \quad f(0)=\cos (0)\\\\ \therefore \quad f(0)=1\\\\ \Rightarrow \quad f(\pi)=\cos 2(\pi)\\\\ \therefore f(\pi)=1$

We have$f(0)=f(\pi)$, so there exists at$c \in(0, \pi)$ , such that $f^{\prime}(c)=0$

Let’s find the derivative of f(x)

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

We have$f^{\prime}(c)=0$

$\\ \Rightarrow \quad-2 \sin 2 c=0\\\\ \Rightarrow \quad \sin 2 c=0\\\\ \Rightarrow \quad 2 c=0 \ or \ \pi \\\\$

$\\ \Rightarrow c = 0 \ or \ \frac{\pi}{2} \\\\ \Rightarrow c = \frac{\pi}{2} \in(0, \pi)$

Hence, Rolle’s Theorem is verified.