Plz answer this

Q4.

f: N ® N is defined by

For n = 1, we have

For n = 2, we have

Thus, f (1) = f (2) for 1 ≠ 2.

So, f is not one-one as two distinct elements in the domain have the same image under function f.

Suppose n be an arbitrary element of N.

If n is odd natural number, then 2n –1 is also an odd natural number.

If n is even natural number, then 2n is also an even natural number.

Thus, for every n in the codomain of f, there exists its pre image in the domain N. Hence, the range of f is equal to the codomain of f. So, f is onto.

The function is not one-one but onto. Hence, f is not bijective.