If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in the domain .

Options

A. `[0, (pi)/(2)]`
B. `[-(pi)/(4),(pi)/(4)]`
C. `[-(pi)/(2),(pi)/(2)]`
D. `[0,pi]`

Answer

B

Explanation

`f(x)=sinx+cosx, g(x)=x^(2)-1`
`implies g(f(x))=(sinx+cosx)^(2)-1=sin2x`
Clearly g(f(x)) is invertible in `-(pi)/(2) le 2x le (pi)/(2)`
` " " [ :' sin theta " is invertible when " -pi//2 le theta le pi//2]`
`implies -(pi)/(4) le x le (pi)/(4)`