A) Maximum
B) Minimum
C) Neither maximum nor minimum
D) None of these
Correct Answer: A
Solution :
Let \[f(x)=\sin x(1+\cos x)\] Þ \[{f}'(x)=\cos 2x+\cos x\] and \[{f}''(x)=-2\sin 2x-\sin x=-(2\sin 2x+\sin x)\] For maximum or minimum value of \[f(x)\], \[f'(x)=0\] \[\cos 2x+\cos x=0\]Þ \[\cos x=-\cos 2x\] Þ \[\cos x=\cos (\pi \pm 2x)\] \[\therefore x=\pi \pm 2x\] 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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