The stationary point of $f(x)=sin^4x+cos^4x, x \in \left(0, \frac{\pi}{2}\right)$ is given by :

$x=\frac{\pi}{4}$

The correct answer is Option (1) → $x=\frac{\pi}{4}$

$f(x)=\sin^4x+\cos^4x$

so $y'=4\sin^3x\cos x-4\cos^3x\sin x$

$=4\sin x\cos x(\sin^2x-\cos^2x)=0$

at stationary point

$⇒\sin x=\cos x$ as $x∈\left(0, \frac{\pi}{2}\right)$

$⇒x=\frac{\pi}{4}$