KCET · Maths · Functions

On the set of integers \(Z\), define \(f: Z \rightarrow Z\) as \(f(n)=\left\{\begin{array}{ll}\frac{\mathrm{n}}{2}, & \mathrm{n} \text { is even } \\ 0, & \mathrm{n} \text { is odd }\end{array}\right.\) then \(f\) is

A injective but not surjective
B neither injective nor surjective
C surjective but not injective
D bijective

Verified Solution

Answer & Solution

Correct Answer

(C) surjective but not injective

Step-by-step Solution

Detailed explanation

Given, \(f(n)= \begin{cases}\frac{n}{2}, & n \text { is even } \\ 0, & n \text { is odd }\end{cases}\)
Here, we see that for every odd values of \(z\), it will give zero. It means that it is a many one function.
For every even values of \(z\), we will get a set of integers \((-\infty, \infty)\). So, it is onto. Hence, it is surjective but not injective.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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