Question

If $f\left(x\right)=\sin x+2{\cos }^{2}x,\frac{\pi }{4}\le x\le \frac{3\pi }{4}$. Then, $f\left(x\right)$ attains its

• A. minimum at $x=\frac{\pi }{4}$
• B. maximum at $x=\frac{\pi }{2}$
• C. minimum at $\frac{\pi }{2}$
• D. maximum at $x={\sin }^{-1}\left(\frac{1}{4}\right)$

Answer 1

The correct option is C minimum at $\frac{\pi }{2}$

Given $f\left(x\right)=\sin x+2{\cos }^{2}x,x\in [\frac{\pi }{4},\frac{3\pi }{4}]$
$\therefore f^{\prime }\left(x\right)=\cos x-4\cos x\sin x$
and $f^{\prime \prime }\left(x\right)=-\sin x-4\cos 2x$
For maximum or minimum of $f\left(x\right)$
Put $f^{\prime }\left(x\right)=0$
$\Rightarrow \cos x-4\cos x.\sin x=0$
$\Rightarrow \cos x(1-4\sin x)=0$
$\Rightarrow \cos x=0=\cos \frac{\pi }{2};\sin x\ne \frac{1}{4}$
$\because x\in [\frac{\pi }{4},\frac{3\pi }{4}]\Rightarrow x=\frac{\pi }{2}$
Now $f^{\prime \prime }\left(\frac{\pi }{2}\right)=-\sin \frac{\pi }{2}-4\cos \pi$
$=-1+4=3>0$(min)
So, $f\left(x\right)$ is minimum at $x=\frac{\pi }{2}$
and its minimum value is
$f\left(\frac{\pi }{2}\right)=\sin \frac{\pi }{2}+2{\cos }^2\frac{\pi }{2}=1-2\times 0=1$