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The function $f(x)=\sin x(1+ \cos x)$ which is defined for all real values of $x$

- has a maximum at $x= \pi /3$
- has a maximum at $x= \pi$
- has a minimum at $x= \pi /3$
- has neither a maximum nor a minimum at $x=\pi/3$

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__$\underline{\textbf{Answer:}\Rightarrow\;\mathbf A}$__

$\underline{\textbf{Explanation:}\Rightarrow }$

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.**

$\therefore \mathbf A$ is the correct option.