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

We have, $f(x) = {\sin ^4}x + {\cos ^4}x$ in $\left[ {0,\frac{\pi }{2}} \right]$ …….(i)

(i) $f(x)$ is continuous in $\left[ {0,\frac{\pi }{2}} \right]$

[since, ${\sin ^4}x$ and ${\cos ^4}x$ are continuous functions and we know that, if $g$ and $h$ be continuous functions, then $(g + h)$ is a continuous function.]

(ii) ${f^\prime }(x) = 4{(\sin x)^3} \cdot \cos x + 4{(\cos x)^3} \cdot ( - \sin x)$

$= 4{\sin ^3}x \cdot \cos x - 4\sin x \cdot {\cos ^3}x$
$= 4\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right)$ which exists in $\left( {0,\frac{\pi }{2}} \right)$ …….(ii)

Hence, $f(x)$ is differentiable in $\left( {0,\frac{\pi }{2}} \right)$.

(iii) Also, $f(0) = 0 + 1 = 1$ and $f\left( {\frac{\pi }{2}} \right) = 1 + 0 = 1$

$\Rightarrow$ $f(0) = f\left( {\frac{\pi }{2}} \right)$
Conditions of Rolle's theorem are satisfied.

Hence, there exists atleast one $c \in \left( {0,\frac{\pi }{2}} \right)$ such that ${f^\prime }(c) = 0$.

therefore,$4\sin c\cos c\left( {{{\sin }^2}c - {{\cos }^2}c} \right) = 0$

$\Rightarrow$ $4\sin c\cos c( - \cos 2c) = 0$

$\Rightarrow$ $- 2\sin 2c \cdot \cos 2c = 0$

$\Rightarrow$ $\quad - \sin 4c = 0$

$\Rightarrow$ $\sin 4c = 0$

$\Rightarrow$ $4c = \pi$

$\Rightarrow$ $C = \frac{\pi }{4}$

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

Hence, Rolle's theorem has been verified.