#### Question

f(x) = 1+2 sin x+3 cos^{2}x, `0<=x<=(2pi)/3` is ________________ .

##### Options

Minimum at x =\[\frac{\pi}{2}\]

Maximum at x = sin \[- 1\] ( \[\frac{1}{\sqrt{3}}\])

Minimum at x = \[\frac{\pi}{6}\]

Maximum at `sin^-1(1/6)`

#### Solution

\[\text { Minimum at } x = \frac{\pi}{2}\]

\[\text { Given }: f\left( x \right) = 1 + 2 \sin x + 3 \cos^2 x\]

\[ \Rightarrow f'\left( x \right) = 2 \cos x - 6 \cos x \sin x\]

\[ \Rightarrow f'\left( x \right) = 2 \cos x\left( 1 - 3 \sin x \right)\]

\[\text { For a local maxima or a local minima, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 2 \cos x\left( 1 - 3 \sin x \right) = 0\]

\[ \Rightarrow 2 \cos x = 0 or \left( 1 - 3 \sin x \right) = 0\]

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

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

\[\text { Now,} \]

\[f''\left( x \right) = - 2 \sin x - 6 \cos 2x\]

\[ \Rightarrow f''\left( \frac{\pi}{2} \right) = - 2 \sin \frac{\pi}{2} - 6 \cos \left( 2 \times \frac{\pi}{2} \right) = - 2 + 6 = 4 > 0\]

\[\text { So, x } = \frac{\pi}{2} \text { is a local minima }.\]

\[\text { Also }, \]

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

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