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

$c=\frac{\pi}{6} \in(0, \pi)$

Hint:

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

Given:

$f(x)=\sin 3 x$ on $[0, \pi]$

Explanation:

We have

$f(x)=\sin 3 x$ on $[0, \pi]$

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

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

$\\ \Rightarrow \quad f(0)=\sin 3(0) \\\\ \Rightarrow \quad f(0)=\sin (0)\\\\ \therefore f(0)=0 \\\\ \Rightarrow \quad f(\pi)=\sin 3(\pi) \\\\ f (\pi) = 0$

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)

$\\ \Rightarrow f^{\prime}(x)=\frac{d(\sin 3 x)}{d x} \\\\ \Rightarrow f^{\prime}(x)=\cos 3 x \frac{d(3 x)}{d x}\\\\ \Rightarrow \quad f^{\prime}(x)=3 \cos 3 x$

We have $\\ f^{\prime}(c)=0$

$\\\\ \Rightarrow \quad 3 \cos 3 c=0\\\\ \Rightarrow 3c = \frac{\pi}{2}\\\\ \Rightarrow c=\frac{\pi}{6} \in(0, \pi)$

Hence, Rolle’s Theorem is verified.