Question

The maximum value of $f\left(x\right)=2sinx+cos2x,0\le x\le \frac{\pi }{2}$ occurs at $x$ is equal to

• A. $0$
• B. $\pi /6$
• C. $\pi /2$
• D. None of these

Answer 1

Answer: (B)

Given, $f(x)=2\sin x+\cos 2x$
On differentiating w.r.t. $x$, we get $f^{\prime }(x)=2\cos x-2\sin 2x$
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, $f^{\prime \prime }(x)=-2\sin x-4\cos 2x$
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
$\text{ At }x=\frac{\pi }{6},$
$f^{\prime \prime }\left(\frac{\pi }{6}\right)=-2\sin \left(\frac{\pi }{6}\right)--4\cos 2\left(\frac{\pi }{6}\right)$
$=-1-2=-3,$ Maxima.