Q113 / 180 · KCET 2009 · Mathematics

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
  • Correct answer
    C
    surjective but not injective
  • D
    bijective

Solution

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.

Hence, the correct answer is option: 'C'.

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