Let f: N → N be defined as \(f(n)=\left\{\begin{matrix}\frac{(n+1)}{2}&if& n\, is\, odd \\ \frac{n}{2}&if &n\,is\,even\end{matrix}\right.\)

Then f is

onto

Here f(3) = 2, f(4) = 2. Hence f is not one-to-one. Also

f(1) = 1, f(2) = 1, f(3) = 2, f(4) = 2, f(5) = 3, f(6) = 3, ......

f(2n - 1) = n and f(2n) = n ⇒ Range of f = N.

Hence (C) is correct answer.