Question

If $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x$ in $(0,\frac{\pi }{2})$, then which of the following options is INCORRECT ?

• A. $f$ is strictly increasing in $(\frac{\pi }{4},\frac{\pi }{2})$
• B. $f$ is strictly decreasing in $(0,\frac{\pi }{4})$
• C. $x=\frac{\pi }{4}$ is a point of local minima
• D. $x=\frac{\pi }{4}$ is a point of local maxima

Answer 1

The correct option is D $x=\frac{\pi }{4}$ is a point of local maxima

$f\left(x\right)={\sin }^{4}x+{\cos }^{4}x,x\in (0,\frac{\pi }{2})$
$f^{\prime }\left(x\right)=4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x$
$=-4\sin x\cos x({\cos }^{2}x-{\sin }^{2}x)$
$=-2\sin 2x\cdot \cos 2x$
$=-\sin 4x$

Now, $f^{\prime }\left(x\right)=0$
$\Rightarrow x=\frac{\pi }{4}$

When $0<x<\frac{\pi }{4}$
$\Rightarrow 0<4x<\pi$
Since, $\sin y$ is positive in $(0,\pi )$,
$\sin 4x>0$
$\Rightarrow -\sin 4x<0$
$\Rightarrow f^{\prime }\left(x\right)<0$
$\therefore f$ is strictly decreasing in $(0,\frac{\pi }{4})$

When $\frac{\pi }{4}<x<\frac{\pi }{2}$
$\Rightarrow \pi <4x<2\pi$
Since, $\sin y$ is negative in $(\pi ,2\pi )$,
$\sin 4x<0$
$\Rightarrow -\sin 4x>0$
$\Rightarrow f^{\prime }\left(x\right)>0$
$\therefore f$ is strictly increasing in $(\frac{\pi }{4},\frac{\pi }{2})$

For $x<\frac{\pi }{4},f^{\prime }\left(x\right)<0$
For $x>\frac{\pi }{4},f^{\prime }\left(x\right)>0$
$\Rightarrow x=\frac{\pi }{4}$ is local minima.