A) \[\frac{{{3}^{3/2}}}{4}\]
B) \[\frac{{{3}^{5/3}}}{4}\]
C) \[\frac{3}{2}\]
D) \[\frac{{{3}^{7/5}}}{4}\]
Correct Answer: A
Solution :
Given, \[f(x)=sin\text{ }x(1+cos\text{ }x)\] \[\Rightarrow \] \[f(x)=\sin x+\frac{1}{2}\sin 2x\] On differentiating w.r.t.\[x,\]we get \[f'(x)=\cos x+\cos 2x\] Put \[f'(x)=0,\] \[cos\text{ }x+cos\text{ }2x=0\] \[\Rightarrow \] \[2\cos \left( \frac{3x}{2} \right)\cos \left( \frac{x}{2} \right)=0\] \[\Rightarrow \] \[\cos \frac{3x}{2}=0,\cos \frac{x}{2}=0\] \[\Rightarrow \] \[x=\frac{\pi }{3},x=\pi \] Now, \[f'\,'(x)=-\sin x-2\sin 2x\] At \[x=\frac{\pi }{3},f'\,'(x)=-\sin \frac{\pi }{3}-2\sin \frac{2\pi }{3}\] \[=-\frac{\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}<0,\]maxima \[\therefore \]Maximum value, \[f\left( \frac{\pi }{3} \right)=\sin \left( \frac{\pi }{3} \right)\left( 1+\cos \frac{\pi }{3} \right)\] \[=\frac{\sqrt{3}}{2}\left( 1+\frac{1}{2} \right)=\frac{3\sqrt{3}}{4}=\frac{{{3}^{3/2}}}{4}\]
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