$f(x)=1+2 \sin x+3 \cos ^{2} x, 0 \leq x \leq \frac{2 \pi}{3}$ is
(a) Minimum at $x=\pi / 2$
(b) Maximum at $x=\sin ^{-1}(1 / \sqrt{3})$
(c) Minimum at $x=\pi / 6$
(d) Maximum at $\sin ^{-1}(1 / 6)$
(a) Minimum at $x=\frac{\pi}{2}$
Given : $f(x)=1+2 \sin x+3 \cos ^{2} x$
$\Rightarrow f^{\prime}(x)=2 \cos x-6 \cos x \sin x$
$\Rightarrow f^{\prime}(x)=2 \cos x(1-3 \sin x)$
For a local maxima or a local minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow 2 \cos x(1-3 \sin x)=0$
$\Rightarrow 2 \cos x=0$ or $(1-3 \sin \mathrm{x})=0$
$\Rightarrow \cos x=0$ or $\sin x=\frac{1}{3}$
$\Rightarrow x=\frac{\pi}{2}$ or $x=\sin ^{-1}\left(\frac{1}{3}\right)$
Now,
$f^{\prime \prime}(x)=-2 \sin x-6 \cos 2 x$
$\Rightarrow f^{\prime \prime}\left(\frac{\pi}{2}\right)=-2 \sin \frac{\pi}{2}-6 \cos \left(2 \times \frac{\pi}{2}\right)=-2+6=4>0$
So, $x=\frac{\pi}{2}$ is a local minima.
Also,
$f^{\prime \prime}\left(\sin ^{-1}\left(\frac{1}{3}\right)\right)=-2 \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)-6 \cos \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)=\frac{-2}{3}-6 \times \frac{2 \sqrt{2}}{3}=-\left(\frac{2}{3}+4 \sqrt{2}\right)<0$
So, $x=\sin ^{-1}\left(\frac{1}{3}\right)$ is a local maxima.