Let `f:[-pi/3, 2pi/3] rarr [0,4]` be a function defined as `f(x)=sqrt3 sinx-cosx+2`. Then `f^-1(x)` is given by

Options

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

Answer

B

Explanation

`y=f(x)=sqrt(3) sin x - cosx +2 = 2 sin(x-(pi)/(6))+2 " (1)" `
Since f(x) is one-one and onto, f is invertible.
From (1), ` sin(x-(pi)/(6))=(y-2)/(2)`
`or x="sin"^(-1)(y-2)/(2)+(pi)/(6)`
`or f^(-1)(x)= sin^(-1)((x-2)/(2))+(pi)/(6)`