#### Question

f(x) = \[\sin + \sqrt{3} \cos x\] is maximum when x =

(a) \[\frac{\pi}{3}\]

(b) \[\frac{\pi}{4}\]

(c) \[\frac{\pi}{6}\]

(d) 0

#### Solution

\[(c) \frac{\pi}{6}\]

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

\[ \Rightarrow f'\left( x \right) = \cos x - \sqrt{3} \sin x\]

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

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

\[ \Rightarrow \cos x - \sqrt{3} \sin x = 0\]

\[ \Rightarrow \cos x = \sqrt{3} \sin x\]

\[ \Rightarrow \tan x = \frac{1}{\sqrt{3}}\]

\[ \Rightarrow x = \frac{\pi}{6}\]

\[\text { Now,} \]

\[f''\left( x \right) = - \sin x - \sqrt{3} \cos x\]

\[ \Rightarrow \Rightarrow f''\left( \frac{\pi}{2} \right) = - \sin\frac{\pi}{2} - \sqrt{3} \cos\frac{\pi}{2}\frac{- 1}{2} - \frac{3}{2} = - 2 < 0\]

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