On the set of integers , define as
Then is
surjective but not injective
Explanation of correct option:
Finding :
Given
Thus every odd value of yields zero i.e. , for all .
So, it is a many-one function.
Hence, the function cannot be injective.
Let be any nonzero integer. Then such that .
For , such that , for all .
So, the function is surjective.
Hence, option (C) is the correct answer.