MCQ
The function $\sin x(1 + \cos x)$ at $x = {\pi \over 3}$, is
- ✓Maximum
- BMinimum
- CNeither maximum nor minimum
- DNone of these
==> $f'(x) = \cos 2x + \cos x$
and $f''(x) = - 2\sin 2x - \sin x = - (2\sin 2x + \sin x)$
For maximum or minimum value of $f(x)$, $f'(x) = 0$
$\cos 2x + \cos x = 0$==> $\cos x = - \cos 2x$
==> $\cos x = \cos (\pi \pm 2x)$
$\therefore x = \pi \pm 2x$ or $x = \frac{\pi }{3},\,\, - \pi $
Now $f''\,\left( {\frac{\pi }{3}} \right) = - 2\sin \frac{{2\pi }}{3} - \sin \frac{\pi }{3} $
$= - 2\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{2} = - \frac{{3\sqrt 3 }}{2} = - ve$
Hence $f(x)$ is maximum at $x = \frac{\pi }{3}$.
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$( S 1) P \left( A ^{\prime} \cup B \right)=\frac{5}{6}$
$( S 2) P \left( A ^{\prime} \cap B ^{\prime}\right)=\frac{1}{18}$. Then.