If f: Z rightarrow Z is defined by f(x)=begin | vedclass

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(A) Given,$f: Z ightarrow Z$ defined as $f(x)=\begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cas

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onto but not one-to-one

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(A) Given,$f: Z ightarrow Z$ defined as $f(x)=\begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cases}$.For one-to-one check: Consider $x_1 = 1$ and $x_2 = 3$. Both are odd,so $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 
eq 3$,the function is not one-to-one.For onto check: The range of $f$ consists of all integers. For any integer $y \in Z$,we can find an $x \in Z$ such that $f(x) = y$. If $y = 0$,$f(1) = 0$. If $y 
eq 0$,let $x = 2y$,which is even,then $f(2y) = \frac{2y}{2} = y$. Since the range equals the codomain $(Z)$,the function is onto.Therefore,$f$ is onto but not one-to-one.

If $f: Z \rightarrow Z$ is defined by $f(x)=\begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$,then $f$ is

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