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 :
\[\operatorname{f}(x)=2\sin \,x\,\,+\,\,sin2\,x\] \[\operatorname{f}(x)=\,\,2cos\,x+2cos\,\,2x=2\left( cos\,\,x+cos\,\,2\,x \right)\] \[\therefore \,\,\,{f}'(x)=0\,\,\,\Rightarrow \,\,2\,\,{{\cos }^{2}}x\,\,+cos\,\,x-1\,\,=\,\,0\] \[\cos \,\,x=\,\,\frac{-1\pm 3}{4}\,\,=-1,\,\,\frac{1}{2};\,\,\,\therefore \,\,x=\pi ,\,\,\frac{\pi }{3}\] 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
You need to login to perform this action.
You will be redirected in
3 sec
