The difference between the greatest between the greatest and least value of the function `f(x)=sin2x-x" on"[-pi//2,pi//6]`, is

Options

A. `(sqrt3+sqrt2)/2`
B. `(sqrt3+sqrt2)/2+pi/6`
C. `(sqrt3)/2+pi/2`
D. `(sqrt3+sqrt2)/2-pi/3`

Answer

C

Explanation

We have
f'(x)=2cos2x-1
`therefore f'(x)=0`
`rArr2cos2x-1=0`
`rArr cos2x=1/2rArr2x=-pi//3,pi//3rArrx=-pi//6,pi//6`
Now
`f(-pi//2)=pi//2,f(pi//2)=-pi/2`
`f(-pi/6)=-sqrt3/2+pi/6 " and " f(pi/6)=sqrt3/2-pi/6`
`" Clearly " ,sqrt3/2-pi/6` is the greatest value of f(x) and its least value is `pi//2`.
Hence the required difference is `sqrt3/2-pi/6-(-pi/2)=sqrt3/2+pi/3`