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) $\frac{\pi}{6}$
Given: $f(x)=\sin x+\sqrt{3} \cos x$
$\Rightarrow f^{\prime}(x)=\cos x-\sqrt{3} \sin x$
For a local maxima or a local minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow \cos x-\sqrt{3} \sin x=0$
$\Rightarrow \cos x=\sqrt{3} \sin x$
$\Rightarrow \tan x=\frac{1}{\sqrt{3}}$
$\Rightarrow x=\frac{\pi}{6}$
Now,
$f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x$
$\Rightarrow \Rightarrow f^{\prime \prime}\left(\frac{\pi}{2}\right)=-\sin \frac{\pi}{2}-\sqrt{3} \cos \frac{\pi}{2} \frac{-1}{2}-\frac{3}{2}=-2<0$
So, $x=\frac{\pi}{2}$ is a local maxima.