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If $f: Z \rightarrow N$ is defined by
$f(n)=\left\{\begin{array}{cll}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0, \text { then } f \text { is } \\ -2 n-1, & \text { if } & n < 0\end{array}\right.$
Options:
$f(n)=\left\{\begin{array}{cll}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0, \text { then } f \text { is } \\ -2 n-1, & \text { if } & n < 0\end{array}\right.$
Solution:
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Verified Answer
The correct answer is:
onto but not one-one
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.
$f(n)= \begin{cases}2 n, & \text { if } n>0 \\ 1, & \text { if } n=0 \\ -2 n-1, & \text { if } n < 0\end{cases}$
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.
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