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Question
f(x) = \[\sin + \sqrt{3} \cos x\] is maximum when x = ___________ .
Options
\[\frac{\pi}{3}\]
\[\frac{\pi}{4}\]
\[\frac{\pi}{6}\]
0
Solution
\[\frac{\pi}{6}\]
\[\text { Given }: f\left( x \right) = \sin x + \sqrt{3} \cos x\]
\[ \Rightarrow f'\left( x \right) = \cos x - \sqrt{3} \sin x\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \cos x - \sqrt{3} \sin x = 0\]
\[ \Rightarrow \cos x = \sqrt{3} \sin x\]
\[ \Rightarrow \tan x = \frac{1}{\sqrt{3}}\]
\[ \Rightarrow x = \frac{\pi}{6}\]
\[\text { Now,} \]
\[f''\left( x \right) = - \sin x - \sqrt{3} \cos x\]
\[ \Rightarrow \Rightarrow f''\left( \frac{\pi}{2} \right) = - \sin\frac{\pi}{2} - \sqrt{3} \cos\frac{\pi}{2}\frac{- 1}{2} - \frac{3}{2} = - 2 < 0\]
\[\text { So,} x = \frac{\pi}{2}\text { is a local maxima }. \]
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