The interval in which function given by $f(x)=sin^4x+cos^4x, x\in \left[0, \frac{\pi}{2}\right]$ is decreasing is :

$\left[0, \frac{\pi}{4}\right]$

The correct answer is Option (4) → $\left[0, \frac{\pi}{4}\right]$

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

$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$

$=1-2\sin^2x\cos^2x$

$=1-\frac{1}{2}(\sin 2x)^2$

$f'(x)=-\frac{1}{2}×2\sin 2x×\cos 2x×2$

$=-2\sin 2x\cos 2x$

$=-\sin 4x$

The function $f(x)$ is decreasing where $f'(x)<0$.

$⇒-\sin 4x<0$

$⇒\sin 4x>0$

$⇒4x∈[0,\pi]$

$⇒x∈\left[0, \frac{\pi}{4}\right]$