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Q.4 Show that the Modulus Function f : R → R, given by $f (x) = | x |$ , is neither one-one nor onto, where $| x |$ is $x,$ if $x$ is positive or 0 and $| x |$ is $- x$ , if $x$ is negative.

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$f : R \rightarrow R$

$f (x) = | x |$

$f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0$

One- one:

For $-1,1 \in R$ then $f (-1) = | -1 |= 1$

$f (1) = | 1 |= 1$

$-1\neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-2 \in R$ ,

We know $f (x) = | x |$ is always positive there is no x in R such that $f (x) = | x |=-2$

$\therefore$ f is not onto i.e. not surjective.

Hence, $f (x) = | x |$ , is neither one-one nor onto.

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seema garhwal

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