Let `f:[-pi/3,(2pi)/3]vec[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. `(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)`