if f:N→N is defined by f(n)=n−(−1)n, then
f is one-one but not onto
f is both one-one and onto
f is neither one-one nor onto
f is onto but not one-one
1→2, 2→13→4, 4→35→6, 6→5i.e., 2m−1 →2m and 2m→ 2m−1 So, f is one-one and onto.
f:R→R, f(x)=x|x| is