Question
The difference between greatest and least value of $$f\left( x \right) = 2\,\sin \,x + \sin \,2x,\,x\, \in \left[ {0,\,\frac{{3\pi }}{2}} \right]{\text{ is :}}$$
A.
$$\frac{{3\sqrt 3 }}{2}$$
B.
$$\frac{{3\sqrt 3 }}{2} - 2$$
C.
$$\frac{{3\sqrt 3 }}{2} + 2$$
D.
none of these
Answer :
$$\frac{{3\sqrt 3 }}{2} + 2$$
Solution :
$$\eqalign{
& f\left( x \right) = 2\,\sin \,x + \sin \,2x \cr
& f'\left( x \right) = 2\,\cos \,x + 2\,\cos \,2x = 2\left( {\cos \,x + \cos \,2x} \right) \cr
& \therefore \,f'\left( x \right) = 0 \Rightarrow 2\,{\cos ^2}x + \cos \,x - 1 = 0 \cr
& \cos \,x = \frac{{ - 1 \pm 3}}{4} = - 1,\,\frac{1}{2} \cr
& \therefore \,x = \pi ,\,\frac{\pi }{3} \cr
& {\text{Now, }}f\left( 0 \right) = 0,\,f\left( {\frac{{3\pi }}{2}} \right) = - 2 \cr
& f\left( \pi \right) = 0,\,f\left( {\frac{\pi }{3}} \right) = 2\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2} \cr} $$
$$\therefore $$ Difference between greatest value and least value $$ = \frac{{3\sqrt 3 }}{2} + 2$$