A) \[x=\frac{\pi }{2}\]
B) \[x=\frac{\pi }{6}\]
C) \[x=\frac{\pi }{3}\]
D) \[x=\pi \]
Correct Answer: C
Solution :
\[y=\sin x(1+\cos x)\]\[=\sin x+\frac{1}{2}\sin 2x\] \ \[\frac{dy}{dx}=\cos x+\cos 2x\]and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\sin x-2\sin 2x\] On putting \[\frac{dy}{dx}=0\], \[\cos x+\cos 2x=0\] Þ \[\cos x=-\cos 2x=\cos (\pi -2x)\]Þ \[x=\pi -2x\] \ \[x=\frac{\pi }{3}\]; \ \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\pi /3}}=-\sin \left( \frac{1}{3}\pi \right)-2\sin \left( \frac{2}{3}\pi \right)\] = \[\frac{-\sqrt{3}}{2}-2.\frac{\sqrt{3}}{2}\] = \[\frac{-3\sqrt{3}}{2}\], which is negative. \ At \[x=\frac{\pi }{3}\] the function is maximum.
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