Question

$f\left(x\right)=2\sin x+\cos 2x$, $0\le x\le 2\pi$ is maximum at-

• A. $x=\frac{\pi }{2}$
• B. $x=\frac{3\pi }{2}$
• C. $x=\frac{\pi }{6}$
• D. no where

Answer 1

Answer: (C)

$x=\frac{\pi }{6}$

$f\left(x\right)=2\sin x+\cos 2x$, $0\le x\le 2\pi$
$f^{\prime }\left(x\right)=2\cos x-2\sin 2x$
$\Rightarrow$ $f^{\prime }\left(x\right)=0$

$\Rightarrow$ $2\cos x(1-2\sin x)=0$
$\Rightarrow$ $x=\frac{\pi }{6},\frac{\pi }{2}$
Now, $f^{\prime }\left(x\right)=-2\sin x-4\cos 2x$
$\Rightarrow$ $f^{\prime }\left(\frac{\pi }{6}\right)=-2.\frac{1}{2}-4.\frac{1}{2}=-3<0$
$\Rightarrow$ $f^{\prime }\left(\frac{\pi }{2}\right)=-2-4(-1)=2>0$
Hence, $f\left(x\right)$ is maximum at $x=\frac{\pi }{6}$