Answer: $\mathbf B$
Explanation:
Given: $f(x) = \sin x(1+\cos x)$
On differentiating the above expression w.r.t. $\mathrm x$, we get:
$f'(x) = \sin x (1+\cos x)+\sin x(-\sin x) = \cos x + \cos^2 x-\sin ^2x = \cos x + \cos 2x$
Equate the above expression with $0$:
$\implies \cos x = -\cos 2x$
This is equal when $x = \frac{\pi}{3}$
So, this is the critical point.
Now, for knowing whether its maxima or minima, further differentiate $f{''}(x)$, we get:
$f^{''}(x) = -\sin x - 2\sin 2x = -(\sin x + 2\sin 2x)$
$\implies f^{''}(\frac{\pi}{3}) = -(\frac{\sqrt 3}{2} + 2.\frac{\sqrt 3}{2}) \lt 0$
$\therefore$ It's a point of Maxima.
$\therefore \mathbf B$ is the correct option.