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

$c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

Hint:

$f(0)=f\left(\frac{\pi}{2}\right)$ , so there exists at$c \in\left(0, \frac{\pi}{2}\right)$

Given:

$f(x)=\sin ^{4} x+\cos ^{4} x$ on $\left[0, \frac{\pi}{2}\right]$

Explanation:

$f(x)=\sin ^{4} x+\cos ^{4} x$ on $\left[0, \frac{\pi}{2}\right]$

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

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

$\begin{array}{l} \Rightarrow \quad f(0)=\sin ^{4}(0)+\cos ^{4}(0) \\\\ \Rightarrow \quad f(0)=0+1 \\\\ \Rightarrow f(0)=1 \\\\ \Rightarrow f\left(\frac{\pi}{2}\right)=\sin ^{4}\left(\frac{\pi}{2}\right)+\cos ^{4}\left(\frac{\pi}{2}\right) \\\\ f\left(\frac{\pi}{2}\right)=1+0 \\\\ \quad f\left(\frac{\pi}{2}\right)=1 \end{array}$

We have$f (0) = f ( \frac{\pi}{2})$ , so there exists at$c \in ( 0 , \frac{\pi}{2})$ , such that$f ' (c) = 0$

Let’s find the derivative of f (x)

$\Rightarrow \quad f^{\prime}(x)=\frac{d\left(\sin ^{4} x+\cos ^{4} x\right)}{d x}\\$

$\Rightarrow \quad f^{\prime}(x)=4 \sin ^{3} x \frac{d(\sin x)}{d x}+4 \cos ^{3} x \frac{d(\cos x)}{d x} \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]\\$

$\left[ \because \frac{d(\sin x)}{d x}=\cos x \right]$

$\left[ \because \frac{d(\cos x)}{d x}=-\sin x \right]$

$\Rightarrow \quad f^{\prime}(x)=4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x$

$\Rightarrow \quad f^{\prime}(x)=4 \sin x \cos x\left(\sin ^{2} x-\cos ^{2} x\right)\\\\ \Rightarrow \quad f^{\prime}(x)=2(2 \sin x \cos x)(-\cos 2 x)\ \ \ \ \quad\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right] \\\\ \Rightarrow \quad f^{\prime}(x)=-2(\sin 2 x)(\cos 2 x) \ \ \ \ \ \ \ \ [\because \sin 2 x=2 \sin x \cos x] \\\\ \Rightarrow \quad f^{\prime}(x)=-\sin 4 x$
We have

$f^{\prime}(c)=0 \\\\ \Rightarrow \quad-\sin 4 c=0 \\\\ \Rightarrow \quad \sin 4 c=0 \\\\ \Rightarrow \quad 4 c=0_{\text {or }} \pi \\\\ \Rightarrow \quad c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

Hence Rolle’s Theorem is verified.