Question

The maximum value of $f\left(x\right)=2\sin x+\cos 2x,0\le x\le \frac{\pi }{2}$ occurs at $x$ is equal to

• A. $0$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

The correct option is B $\frac{\pi }{6}$

Given, $f\left(x\right)=2\sin x+\cos 2x$

On differentiating w.r.t $x$, we get

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

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

$2\cos x-4\sin x\cos x=0$

$\Rightarrow 2\cos x(1-2\sin x)=0$

$\Rightarrow \cos x=0,\sin x=\frac{1}{2}$

$\Rightarrow x=\frac{\pi }{2},x=\frac{\pi }{6}$

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

At $x=\frac{\pi }{2}$

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

$=2>0$., Minima

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

$f^{\prime \prime }\left(\frac{\pi }{6}\right)=-2\sin \left(\frac{\pi }{6}\right)-4\cos 2\left(\frac{\pi }{6}\right)$

$=-1-2=-3$, Maxima