Correct Answer - B
For any `x, y in N`
`f(x)=f(y) Rightarrow 4x+3=4y+3 Rightarrow x=y`
`therefore` f is one-one
Clearly, `Y=(y in N: y=4x+3"for some "x in N)="Range (f)"`
`therefore` f:N-Y is onto.
Thus, `f:N to Y` is a bijection and hence invertible.
Let g be the inverse of f. Then.
`fog (y)=y"for all "y inY`
`Rightarrow f(g(y))=yRightarrow 4g(y)+3 =yRightarrow g(y)=(y-3)/(4)`