MCQ
The maximum value of $\sin x\,\,(1 + \cos x)$ will be at the
- A$x = {\pi \over 2}$
- B$x = {\pi \over 6}$
- ✓$x = {\pi \over 3}$
- D$x = \pi $
$\therefore$ $\frac{{dy}}{{dx}} = \cos x + \cos 2x$ and $\frac{{{d^2}y}}{{d{x^2}}} = - \sin x - 2\sin 2x$
On putting $\frac{{dy}}{{dx}} = 0$, $\cos x + \cos 2x = 0$
==> $\cos x = - \cos 2x = \cos (\pi - 2x)$==> $x = \pi - 2x$
$\therefore$ $x = \frac{\pi }{3}$;
$\therefore$ ${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{x = \pi /3}} = - \sin \left( {\frac{1}{3}\pi } \right) - 2\sin \left( {\frac{2}{3}\pi } \right)$
$= \frac{{ - \sqrt 3 }}{2} - 2.\frac{{\sqrt 3 }}{2} = \frac{{ - 3\sqrt 3 }}{2}$, which is negative.
$\therefore$ At $x = \frac{\pi }{3}$ the function is maximum.
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