Question

If $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$. Then $f$ is an increasing function in the interval

• A. $[\frac{5\pi }{8},\frac{3\pi }{4}]$
• B. $[\frac{\pi }{2},\frac{5\pi }{8}]$
• C. $[\frac{\pi }{4},\frac{\pi }{2}]$
• D. $[0,\frac{\pi }{4}]$

Answer 1

The correct option is C $[\frac{\pi }{4},\frac{\pi }{2}]$

$f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$

Diferentiating $f\left(x\right)$ w.r.t. $x$, we have

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

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

$f^{\prime }\left(x\right)=4\sin x\cos x({\sin }^{2}x-{\cos }^{2}x)$

$f^{\prime }\left(x\right)=2\sin 2x(-\cos 2x)$

$f^{\prime }\left(x\right)=-\sin 4x$

For $f\left(x\right)$ to be an increasing function,

$f^{\prime }\left(x\right)>0$

$\Rightarrow -\sin 4x>0$

$\Rightarrow \sin 4x<0$

$\Rightarrow \pi <4x<2\pi$

$\Rightarrow \frac{\pi }{4}<x<\frac{\pi }{2}$

Hence $f\left(x\right)$ will be increasing for all $x\in [\frac{\pi }{4},\frac{\pi }{2}]$.