If f:R→R given by f(x)=ax+sinx+a, then f is one-one and onto for all
a∈R
a∈R~[-1,1]
a∈R~{0}
a∈R~{-1}
For a≠0, the range of f is R. f is differentiable function so f is one-one and only if f is monotonic.
f′(x)=a+cosxIf a>1, f'(x)>0 i.e. f is increasingIf a<-1, f'(x)<0 i.e. f is decreasing
Thus f is monotonic if a∈R~[-1,1].