Let `f:[-pi/3,(2pi)/3] rarr [0,4]` be a function defined as `f(x)=sqrt(3)sinx-cosx+2.` Then `f^(-1)(x)` is given by
`(a)` `sin^(-1)((x-2)/2)-pi/6`
`(b)` `sin^(-1)((x-2)/2)+pi/6`
`(c)` `(2pi)/3+cos^(-1)((x-2)/2)`
`(d)` none of these

Options

A. `sin^(-1),((x-2)/(2))-(pi)/(6)`
B. `sin^(-1)((x-2)/(2))+(pi)/(6)`
C. `sin^(-1)((x+2)/(2))-(pi)/(6)`
D. `(2pi)/(3)+cos^(-1)((x-2)/(3))`

Answer

B

Explanation

`f (x) =sqrt(3)sin x-cos x+2=2sin (x-(pi)/(6))+2`
Since f (x) is one -one and onto, f is invertible.
`implies f^(-1)(x)=sin^(-1)((x)/(2)-1)+(pi)/(6). |(pi)/(2)-1|le 1`
Because for all `x [0,4]`
Also using `sin^(-1) alpha+cos^(-1) alpha=(pi)/(2)`
`f^(-1)(x) =(pi)/(2) -cos^(-1) ((x-2)/(2))+(pi)/(6)`
`(2pi)/(3)-cos^(-1)((x-2)/(3))`