Question

Let $f\left(x\right)=sinx+2co{s}^{2}x,\frac{\pi }{4}\le x\le \frac{3\pi }{4}.$ Then f attains its

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

Answer 1

Answer: (C)

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