Let `f: Z->Z` be given by `f(x)={x/2,\ if\ x\ i s\ e v e n,0,\ if\ x\ i s\ od d` . Then, f is (a) onto but not one-one (b) one-one but not onto (c) one-one and onto (d) neither one-one nor onto

Explanation

Correct option is A)

For Injectivity:

Let `x_1`​ and `x_2`​ be two elements in the domain (Z), such that,

`f(x_1​)=f(x_2​)`

CaseI:

Let both `x_1`​ and `x_2`​ be even.

Then, `f(x_1​)=f(x_2​)`

`⇒ 2x_1​​=2x_2​​`

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