Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (v) maths textbook solution

$c=\frac{\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

Hint:

$f\left(\frac{-\pi}{2}\right)=f\left(\frac{\pi}{2}\right)$ , so there exists at$c \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

Given:

$f(x)=e^{x} \cos x$ on $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Explanation:

We have

$f(x)=e^{x} \cos x$ on $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

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

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

$\\ \Rightarrow \quad f\left(\frac{-\pi}{2}\right)=e^{\frac{-\pi}{2}} \cos \left(\frac{-\pi}{2}\right)\\\\ \Rightarrow \quad f\left(\frac{-\pi}{2}\right)=e^{\frac{-\pi}{2}} \times 0 \\\\ \therefore \quad f\left(\frac{-\pi}{2}\right)=0 \\\\ \Rightarrow \quad f\left(\frac{\pi}{2}\right) =e^{\frac{\pi}{2}} \cos \left(\frac{\pi}{2}\right)\) \\\\\Rightarrow \quad f\left(\frac{\pi}{2}\right)=e^{\frac{\pi}{2}} \times 0 \\\\ \therefore \quad f\left(\frac{\pi}{2}\right)=0$

We have$f\left(\frac{-\pi}{2}\right)=f\left(\frac{\pi}{2}\right)$ , so there exists $c \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ , such that$f^{\prime}(c)=0$

Let’s find the derivative of f(x)

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

We have

$\\ f^{\prime}(c)=0 \\\\ \Rightarrow \quad e^{c}(-\sin c+\cos c)=0 \\\\ \ \Rightarrow \quad-\sin c+\cos c=0 \\\\ \Rightarrow \quad \frac{-1}{\sqrt{2}} \sin c+\frac{1}{\sqrt{2}} \cos c=0 \\\\ \Rightarrow \quad-\sin \left(\frac{\pi}{4}\right) \sin c+\cos \left(\frac{\pi}{4}\right) \cos c=0 \\\\ \Rightarrow \quad \cos \left(c+\frac{\pi}{4}\right)=0 \\\\ \Rightarrow \quad c+\frac{\pi}{4}=\frac{\pi}{2} \\\\ \Rightarrow c=\frac{\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \\\\$

$\therefore$   Rolle’s Theorem is verified.