Q41 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

If $f: Z \rightarrow N$ is defined by $f(n)=\left\{\begin{array}{cll}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0, \text { then } f \text { is } \\ -2 n-1, & \text { if } & n < 0\end{array}\right.$

  • A
    one-one but not onto
  • Correct answer
    B
    onto but not one-one
  • C
    both one-one and onto
  • D
    neither one-one nor onto

Solution

We have, $f(n)= \begin{cases}2 n, & \text { if } n>0 \\ 1, & \text { if } n=0 \\ -2 n-1, & \text { if } n < 0\end{cases}$ When, $n>0$ $f(n)=2,4,6,8 \ldots .$ When, $n < 0$ $f(n)=1,3,5,7 \ldots . .$ $\therefore$ Range of $f(n)$ is $\{1,2,3,4, \ldots .$.$\} .$ $\therefore \quad$ Codomain $=$ Range Hence, $f(n)$ is onto. Here, $f(0)-f(-1)=1$ $\therefore f$ is not one one.

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

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