Question

If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in the domain .
A. `[0,(pi)/(2)]`
B. `[-(pi)/(4),(pi)/(4)]`
C. `[-(pi)/(2),(pi)/(2)]`
D. `[0,pi]`

Answer 1

Correct Answer - B
Be definition of composition of function,
`g(f(x))=(sinx+cosx)^(2)-1,` is invertible
(i.e. bijective)
We, know, sin x is bijective, only when `x in [-(pi)/(2),(pi)/(2)]`.
Thus, g{f(x)} is bijective, if `-(pi)/(2) le 2x le (pi)/(2)`
`rArr -(pi)/(4) le x le (pi)/(4)`