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. $\pi/ 6$
• C. $\pi/ 2$
• D. None of these

Answer 1

Answer: (B)

Given, $f(x)=2 \sin x+\cos 2 x$
On differentiating w.r.t. $x$, we get $f^{\prime}(x)=2 \cos x-2 \sin 2 x$
Put $f^{\prime}(x)=0$,
$2 \cos x-4 \sin x \cos x=0$
$\Rightarrow 2 \cos x(1-2 \sin x)=0$
$\Rightarrow \cos x=0, \sin x=\frac{1}{2}$
$\Rightarrow x=\frac{\pi}{2}, x=\frac{\pi}{6}$
Now, $f^{\prime \prime}(x)=-2 \sin x-4 \cos 2 x$
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
$ \text { 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.