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Let $f: R \rightarrow R$ be a function defined as $f(x)=x|x|$; for each $x \in R, R$ being the set of real numbers. Which one of the following is correct?
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The correct answer is:
$f$ is both one-one and onto
Given, $f(x)=x|x|$
If $\quad f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow x_{1}\left|x_{1}\right|=x_{2}\left|x_{2}\right|$
$\Rightarrow x_{1}=x_{2}$
$\therefore f(x)$ is one-one.
Also, range of $f(x)=$ co-domain of $f(x)$
$\therefore f(x)$ is onto.
Hence, $f(x)$ is both one-one and onto.
If $\quad f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow x_{1}\left|x_{1}\right|=x_{2}\left|x_{2}\right|$
$\Rightarrow x_{1}=x_{2}$
$\therefore f(x)$ is one-one.
Also, range of $f(x)=$ co-domain of $f(x)$
$\therefore f(x)$ is onto.
Hence, $f(x)$ is both one-one and onto.
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