Question

Find the value of x for which $f\left(x\right)=1+2\sin x+2{\cos }^{2}x,0\le x\le \pi /2.$ is maximum.

• A. $x=\pi /6.$
• B. $x=\pi /2.$
• C. $x=\pi /4.$
• D. $x=\pi /3.$

Answer 1

The correct option is A $x=\pi /6.$

$f\left(x\right)=1+2\sin x+2{\cos }^{2}x$
$f^{\prime }\left(x\right)=2\cos x-4\cos .x\sin x$
For maximum or minimum,
$f^{\prime }\left(x\right)=0$
$\Rightarrow 2\cos x-4\cos .x\sin x=0$
$\cos x=0$ or $\sin x=\frac{1}{2}$
The solution in the interval $0\le x\le \pi /2$ is $\frac{\pi }{2}$ and $\frac{\pi }{6}$ respectively.
Now $f^{\prime \prime }\left(x\right)=-2\sin x-4\cos 2x$
$\therefore f^{\prime \prime }\left(x\right)>0$ at $x=\frac{\pi }{2}$
and $f^{\prime \prime }\left(x\right)<0$ at $x=\frac{\pi }{6}$
Hence f(x) is minimum at $x=\frac{\pi }{2}$ and maximum at $x=\frac{\pi }{6}$