Answer: (b) = one - one
f (x) = x + 2 and f : A → B
∴ f (1) = 1 + 2 = 3, f (2) = 2 + 2 = 4
f (3) = 3 + 2 = 5, f (4) = 4 + 2 = 6
Thus each element in A has a unique image in B and no two elements in B have the same pre-image in A.
So f is one-one.
But not all the elements of B, i.e., 1 and 2 have a pre-image in A.
So, the function f is not onto