If the solve the problem


$f(x)=1+2 \sin x+3 \cos ^{2} x, 0 \leq x \leq \frac{2 \pi}{3}$ 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


$\Rightarrow 2 \cos x(1-3 \sin x)=0$

$\Rightarrow 2 \cos x=0$ or $(1-3 \sin \mathrm{x})=0$

$\Rightarrow \cos x=0$ or $\sin x=\frac{1}{3}$

$\Rightarrow x=\frac{\pi}{2}$ or $x=\sin ^{-1}\left(\frac{1}{3}\right)$


$f^{\prime \prime}(x)=-2 \sin x-6 \cos 2 x$

$\Rightarrow f^{\prime \prime}\left(\frac{\pi}{2}\right)=-2 \sin \frac{\pi}{2}-6 \cos \left(2 \times \frac{\pi}{2}\right)=-2+6=4>0$

So, $x=\frac{\pi}{2}$ is a local minima.


$f^{\prime \prime}\left(\sin ^{-1}\left(\frac{1}{3}\right)\right)=-2 \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)-6 \cos \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)=\frac{-2}{3}-6 \times \frac{2 \sqrt{2}}{3}=-\left(\frac{2}{3}+4 \sqrt{2}\right)<0$

So, $x=\sin ^{-1}\left(\frac{1}{3}\right)$ is a local maxima.

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