A) \[\frac{3\sqrt{3}}{2}\]
B) \[\frac{3\sqrt{3}}{2}-2\]
C) \[\frac{3\sqrt{3}}{2}+2\]
D) None of these
Correct Answer: C
Solution :
[c] \[f(x)=2\sin x+\sin 2x\] \[f'(x)=2\cos x+2\cos 2x=2(\cos x+cos2x)\] \[\therefore \,\,\,\,\,\,f'(x)=0\Rightarrow 2co{{s}^{2}}x+cos\,x-1=0\] \[\cos x=\frac{-1\pm 3}{4}=-1,\frac{1}{2}\therefore \,\,\,\,x=\pi ,\,\,\frac{\pi }{3}\] Now, \[f(0)=0,f\left( \frac{3\pi }{2} \right)=-2\] \[f(\pi )=0,f\left( \frac{\pi }{3} \right)=2\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}\] \[\therefore \] Difference between greatest value and least value \[=\frac{3\sqrt{3}}{2}+2\]
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