The function f = {(5, 3), (5, 4), (6, 4), (8, 9)} is neither one-one nor onto.
The function is not one – one 8 does not have an image in the codomain N and we know that a function can only be one – one if every element in the set M has an image in the codomain N.
A function can be onto only if each element in the co-domain has a pre-image in the domain X. In the function f = {(5, 3), (5, 4), (6, 4), (8, 9)}, 10 in the co-domain N does not have a pre-image in the domain X.
f = {(5, 3), (6, 4), (7, 9), (8, 10)} is both one-one and onto.
f = {(5, 4), (5, 9), (6, 3), (7, 10), (8, 10)} and f = {(6, 4), (7, 3),(7, 9), (8, 10)} are many – one onto.