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

$c=\frac{3 \pi}{4} \in(0, \pi)$

Hint:

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

Given:

$f(x)=e^{x} \sin x$ on $[0, \pi]$

Explanation:

We have

$f(x)=e^{x} \sin x$ on $[0, \pi]$

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

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

ù$\\ \Rightarrow \quad f(0)=e^{0} \sin (0)\\\\ \Rightarrow \quad f(0)=1 \times 0\\\\ \therefore \quad f(0)=0\\\\ \Rightarrow \quad f(\pi)=e^{\pi} \sin (\pi)\\\\ \Rightarrow \quad f(\pi)=e^{\pi} \times 0\\\\ \therefore \quad f(\pi)=0$

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

Let’s find the derivative of f(x)

$\\ \Rightarrow f^{\prime}(x)=\frac{d\left(e^{x} \sin x\right)}{d x}\) \\\\ \Rightarrow \quad f^{\prime}(x)=\sin x \frac{d\left(e^{x}\right)}{d x}+\frac{e^{x} d(\sin x)}{d x} \\\\ \Rightarrow \quad f^{\prime}(x)=e^{x}(\sin x+\cos x)$

We have$f'(c) = 0$

$\begin{array}{l} \Rightarrow \quad e^{c}(\sin c+\cos c)=0 \\\\ \Rightarrow \quad \sin c+\cos c=0 \\\\ \Rightarrow \quad \frac{1}{\sqrt{2}} \sin c+\frac{1}{\sqrt{2}} \cos c=0 \\\\ \Rightarrow \quad \sin \left(\frac{\pi}{4}\right) \sin c+\cos \left(\frac{\pi}{4}\right) \cos c=0 \\\\ \Rightarrow \quad \cos \left(c-\frac{\pi}{4}\right)=0 \\\\ \Rightarrow \quad c-\frac{\pi}{4}=\cos ^{-1}(0) \\\\ \Rightarrow \quad c-\frac{\pi}{4}=\frac{\pi}{2} \\\\\left [ \because \cos ^{-1} (0)= \frac{\pi}{2} \right ] \\\\ \Rightarrow \quad c=\frac{3 \pi}{4} \in(0, \pi) \end{array}$

$\therefore$          Rolle’s Theorem is verified.