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

$c=\frac{\pi}{3} \in(0, \pi)$

Hint:

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

Given:

$f(x)=2 \sin x+\sin 2 x$ on$[0, \pi ]$

Explanation:

We have

$f(x)=2 \sin x+\sin 2 x$ on$[0, \pi ]$

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

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

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

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)

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

$\\ f^{\prime}(c)=0 \\\\ \Rightarrow \quad 4 \cos ^{2} c+2 \cos c-2=0\\\\ \Rightarrow \quad 2 \cos ^{2} c+\cos c-1=0\\\\ \Rightarrow \quad 2 \cos ^{2} c+2 \cos c-\cos c-1=0\\\\ \Rightarrow \quad 2 \cos c(\cos c+1)-(\cos c+1)=0 \\\\ \cos c = \frac{1}{2} \ or \ \cos c = -1 \\\\ c = \frac{\pi}{3} \in (0, \pi )$

Hence, Rolle’s Theorem is verified.