The difference between the greatest and least value of the functions, `f(x)=cosx+(1)/(2)cos2x-(1)/(3)cos 3x` is

Options

A. `2//3`
B. `8//7`
C. `9//4`
D. `3//8`

Answer

C

Explanation

Clearly , f(x) is a periodic function with period `2 pi` So the difference betweenthe greatest and the least value of the function f(x) is the difference between the greatest and the least values on the interval `[0,2 pi]`
Let us find these on the interval `[0,2 pi]`
We have
`f'(x)=-(sinx+sin2x-sin3x)`
`rArrf'(x)=-(2sin""(3x)/2cos""x/2-2sin""(3x)/2cos""(3x)/2)`
`rArr f'(x) = -2sin""(3x)/2(cos""x/2-cos""(3x)/2)`
`therefore f'(x)=0rArrx=0,2 pi//3,pi and 2 pi`
Now
`f(0)=1+1/2-1/3=7/6,f((2 pi)/3)=(13)/(12), f(pi)=-1/6 and , f(2 pi)=7/6`
The largest and the smallest of these values are `7//6` and `-(13)//(12)` respectively.
Hence , the required difffernce `7//6-(-13//12)=9//4`