A mapping `f: N -> N` where N is set of natural numbers is destined as `f(n) = n^2 , n=odd and 2n + 1, n = even` for `n in N` then `f` is

Options

A. surjective but not injective
B. injective but not surjective
C. bijective
D. Neither injective nor surjective

Answer

D

Explanation

Since, `f(n)={{:(n^(2)",",,"if n is odd."),(2n+1",",,"if n is even."):}`
`:." "f(1)=1^(2)=1,f(2)=2(2)+1=5`
`f(3)=3^(2)=9,f(4)=2(4)+1=9`
`because" "f(3)=f(4)`
So, f is not injective.
Also, f is not surjective as some element of N is not the image of any element of N.