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If $\quad f: Z \rightarrow Z \quad$ is $\quad$ defined
$f(x)=\left\{\begin{array}{lll}\frac{x}{2}, & \text { if } & x \text { is even } \\ 0, & \text { if } & x \text { is odd }\end{array}\right.$, then $f$ is
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$f(x)=\left\{\begin{array}{lll}\frac{x}{2}, & \text { if } & x \text { is even } \\ 0, & \text { if } & x \text { is odd }\end{array}\right.$, then $f$ is
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onto but not one-to-one
Given, $f: Z \rightarrow Z, f(x)=\left\{\begin{array}{l}\frac{x}{2}, \text { if } x \text { is even } \\ 0, \text { if } x \text { is odd }\end{array}\right.$ when $x$ is odd, $f(1)=f(3)=0$ $\Rightarrow f(x)$ is not one-one function. when $x$ is even i.e., $x=0, \pm 2, \pm 4, \pm 6, \ldots \infty$ $\Rightarrow \frac{x}{2}=0, \pm 1, \pm 2, \pm 3, \ldots \infty$
Hence, range of $f$ is $Z$. So, it is onto.
Hence, $f$ is not one-one but it is onto.
Hence, range of $f$ is $Z$. So, it is onto.
Hence, $f$ is not one-one but it is onto.
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