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

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

Hint:

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

Given:

$f(x)=\frac{\sin x}{e^{x}}$ on$0 \leq x \leq \pi$

Explanation:

We have

$f(x)=\frac{\sin x}{e^{x}}$on$[0, \pi]$

This can be written as

$\Rightarrow \quad f(x)=e^{-x} \sin x$ on$[0, \pi]$

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

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

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)

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

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

$\\ \Rightarrow \quad e^{-c}(-\sin c+\cos c)=0\\\\ \Rightarrow \quad-\sin c+\cos c=0\\\\ \Rightarrow \quad-\frac{1}{\sqrt{2}} \sin c+\frac{1}{\sqrt{2}} \cos c=0$           (Multiplying both the sides by$\frac{1}{\sqrt2}$)

$\\ \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 \ \ \ \ \ \quad[\because \cos (A+B)=\cos A \cos B-\sin A \sin B]$$\\ \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{\pi}{4} \in(0, \pi)$

Hence, Rolle’s Theorem is verified.