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

$c=\cos ^{-1}\left(\frac{-1 \pm \sqrt{33}}{-8}\right)$

Hint:

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

Given:

$f(x)=\sin x-\sin 2 x$ on$c \in[0, \pi]$

Explanation:

$f(x)=\sin x-\sin 2 x$ on$c \in[0, \pi]$

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

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

$\\ \Rightarrow \quad f(0)=\sin (0)-\sin 2(0)\\\\ \Rightarrow \quad f(0)=0-\sin (0)\\\\ \Rightarrow \quad f(0)=0\\\\ \Rightarrow \quad f(\pi)=\sin (\pi)-\sin 2(\pi)\\\\ \Rightarrow \quad f(\pi)=0-\sin (2 \pi)\\\\ \Rightarrow \quad f(\pi)=0$

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)

$\Rightarrow \quad f^{\prime}(x)=\frac{d(\sin x-\sin 2 x)}{d x}\\\\ \Rightarrow f^{\prime}(x)=\cos x-\cos 2 x \frac{d(2 x)}{d x} \ \ \ \ \ \left[\because \frac{d(\sin x)}{d x}=\cos x\right]\\\\ \Rightarrow \quad f^{\prime}(x)=\cos x-2 \cos 2 x\\\\ \Rightarrow \quad f^{\prime}(x)=\cos x-2\left(2 \cos ^{2} x-1\right) \ \ \ \ \left[\because \cos 2 x=2 \cos ^{2} x-1\right]\\\\ \Rightarrow \quad f^{\prime}(x)=\cos x-4 \cos ^{2} x+2$

We have$f' (c) = 0$,

$\Rightarrow \quad \cos c-4 \cos ^{2} c+2=0\\\\ \cos c=\frac{-1 \pm \sqrt{(1)^{2}-(4 \times-4 \times 2)}}{2 \times-4}\\\\ \Rightarrow \quad \cos c=\frac{-1 \pm \sqrt{1+32}}{-8} \\\\ \Rightarrow c=\cos ^{-1}\left(\frac{-1 \pm \sqrt{33}}{-8}\right)$so there exists at$c \in (0, \pi )$

Hence Rolle’s Theorem is verified.