Consider the real-valued function satisfying `2f(sinx)+f(cosx)=xdot` then the domain of `f(x)i sR` domain of `f(x)i s[-1,1]` range of `f(x)` is `[-(2pi)/3,pi/3]` range of `f(x)i sR`

Options

A. domain of `f(x)` is R
B. domain of `f(x)" is " [-1,1]`
C. range of `f(x) " is " [-(2pi)/(3),(pi)/(3)]`
D. range of `f(x)` is R

Answer

B, C

Explanation

Given `2f(sinx)+f(cosx)+x " (1)" `
Replacing x by `(pi)/(2)-x,` we get
`2f(cosx)+f(sinx)=(pi)/(2)-x " (2)" `
Eliminating `f(cosx)` from (1) and (2), we get
`3f(sinx)=3x-(pi)/(2)`
` or f(sinx)=x-(pi)/(2)`
` or f(x) ="sin"^(-1)x-(pi)/(6)`
`f(x)` has the domain [-1, 1].
Also, `sin^(-1)x in[-(pi)/(2),(pi)/(2)] or sin^(-1)x - (pi)/(6) in [-(2pi)/(3),(pi)/(3)].`