If the solve the problem

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.

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