Applications Of Derivatives Question 64
Question: The difference between greatest and least value of $ f(x)=2\sin x+\sin 2x,x\in [ 0,\frac{3\pi }{2} ] $ is-
Options:
A) $ \frac{3\sqrt{3}}{2} $
B) $ \frac{3\sqrt{3}}{2}-2 $
C) $ \frac{3\sqrt{3}}{2}+2 $
D) None of these
Show Answer
Answer:
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 2cos^{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( \frac{3\pi }{2} )=-2 $
$ f(\pi )=0,f( \frac{\pi }{3} )=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 $
BETA
