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Show that the Modulus Function $f: R \rightarrow R$ given by $f(x)$ $=|\mathbf{x}|$, is neither one-one nor onto, where $|\mathbf{x}|$ is $\mathbf{x}$, if $\mathbf{x}$ is positive or 0 and $|x|$ is $-x$, if $x$ is negative.
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$f: R \rightarrow R$ given by $f(x)=|x|$
(a) $\mathrm{f}(-1)=|-1|=1, \mathrm{f}(1)=|1|=1$
$\Rightarrow-1$ and 1 have the same image
$\therefore \mathrm{f}$ is not one-one
(b) No negative value belonging to codomain of f has any pre-image in its domain
$\therefore \mathrm{f}$ is not onto. Hence, $\mathrm{f}$ is neither one-one nor onto.
(a) $\mathrm{f}(-1)=|-1|=1, \mathrm{f}(1)=|1|=1$
$\Rightarrow-1$ and 1 have the same image
$\therefore \mathrm{f}$ is not one-one
(b) No negative value belonging to codomain of f has any pre-image in its domain
$\therefore \mathrm{f}$ is not onto. Hence, $\mathrm{f}$ is neither one-one nor onto.
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