A) \[3\]
B) \[\frac{3\sqrt{3}}{4}\]
C) \[4\]
D) \[3\sqrt{3}\]
Correct Answer: B
Solution :
Let \[f(x)=\sin x(1+\cos x)\] On differentiating w.r.t. x, we get \[f'(x)=\cos x(1+\cos x)+\sin x(-\sin x)\] \[=\cos x+{{\cos }^{2}}x-{{\sin }^{2}}x\] \[=\cos x+\cos 2x\] For maxima or minima, put\[f'(x)=0\] \[\Rightarrow \] \[\cos x=-\cos 2x\] \[\Rightarrow \] \[x=\pi -2x\] \[\Rightarrow \] \[x=\frac{\pi }{3}\] \[\therefore \] \[f''(x)=-\sin x-2\sin 2x\] At \[x=\frac{\pi }{3}\], \[f''(x)=-\sin \frac{\pi }{3}-2\sin \frac{2\pi }{3}=-\frac{3\sqrt{3}}{2}<0\], maximum \[\therefore \]The maximum value of the function at\[x=\frac{\pi }{3}\]
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