Let `f:[-(pi)/(3), (2pi)/(3)] to [0,4]` be a function defined as `f(x)=sqrt(3) sinx -cos x +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, C

Explanation

`f(x)=sqrt(3)sinx-cosx+2`
`=2(((sqrt(3))/(2)sinx-((1)/(2))cosx)+2`
`=2(cos((pi)/(6))sinx-sin((pi)/(6))cosx)+2`
`2sin(x-(pi)/(6))+2`
`x in[-(pi)/(3),(2pi)/(3)]impliesx-(pi)/(6)in[-(pi)/(3)-(pi)/(6),(2pi)/(3)-(pi)/(6)]`
i.e., `x-(pi)/(6)in[-(pi)/(2),(pi)/(2)]`
if `x-(pi)/(6)in[-(pi)/(2),(pi)/(2)]`
`impliessin(x-(pi)/(6))in[-1,1]`
`implies2sin(x-(pi)/(6))+2in[0.4]`
`impliesf(x)` is one-one & onto
`becausef^(-1)(x)` exists
`y=2sin(x-(pi)/(6))+2`
`(y-2)/(2)=sin(-(pi)/(6))`
`x=sin^(-1)((y-2)/(2))+(pi)/(6)`
i.e., `f^(-1)(x)=sin^(-1)((x-2)/(2))+(pi)/(6)`