Prove that $f(x) = \sin x + \sqrt 3 \cos x$ has maximum value at $x = \frac{\pi }{6}$.

We have, $f(x) = \sin x + \sqrt 3 \cos x$

Therefore,${f^\prime }(x) = \cos x + \sqrt 3 ( - \sin x)$
$= \cos x - \sqrt 3 \sin x$

For ${f^\prime }(x) = 0,$ $\cos x = \sqrt 3 \sin x$

$\Rightarrow$ $\tan x = \frac{1}{{\sqrt 3 }} = \tan \frac{\pi }{6}$

$\Rightarrow$ $x = \frac{\pi }{6}$

Again, differentiating ${f^\prime }(x)$, we get
${f^{\prime \prime }}(x) = - \sin x - \sqrt 3 \cos x$

At $x = \frac{\pi }{6},$ ${f^{\prime \prime }}(x) = - \sin \frac{\pi }{6} - \sqrt 3 \cos \frac{\pi }{6}$

$= - \frac{1}{2} - \sqrt 3 \cdot \frac{{\sqrt 3 }}{2}$

$= - \frac{1}{2} - \frac{3}{2} = - 2 < 0$

Hence, at $x = \frac{\pi }{6},f(x)$ has maximum value at $\frac{\pi }{6}$ is the point of local maxima.

Long Answer Questions (L.A.)