Question

$f\left(x\right)=1+2\sin x+2{\cos }^{2}x$, $0\le x\le \pi /2$ is maximum at-

• A. $x=\pi /2$
• B. $x=\pi /6$
• C. $x=\pi /3$
• D. No where

Answer 1

The correct option is B $x=\pi /6$

$f=1+2\sin x+2{\cos }^{2}x$

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

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

$f=2\cos x-2\sin 2x$

$f\left(x\right)=0\Rightarrow \cos x-\sin 2x=0$

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

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

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

$f\left(\frac{\pi }{2}\right)=-2+4=2$

$\Rightarrow$ min at

$x=\dfrac{\pi}{2}

$f\left(\frac{\pi }{6}\right)=-2\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)$

$=-2\Rightarrow$ max at

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

local min at $x=\frac{\pi }{2}$

local max at $x=\frac{\pi }{6}$

look at critical point and end points

$f\left(0\right)=3,f\left(\frac{\pi }{2}\right)=3$

$f\left(\frac{\pi }{6}\right)=3.5$

$\therefore$ Global more at $x=\frac{\pi }{6}$