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If the function $f: R \rightarrow R$ is defined by $f(x)=x|x|$, then .
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Verified Answer
The correct answer is:
$f$ is both one-one and onto
Given, $f(x)=x|x|=\left\{\begin{array}{l}x(x), x \geq 0 \\ x(-x), x < 0\end{array}\right.$
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2, x \geq 0 \\ -x^2, x < 0\end{array}\right.$, Also, $f: R \rightarrow R$
According to the graph,
domain of $f(x)=$ Range of $f(x)=R$
$\therefore$ Range $=$ Codomain $=R$
$\Rightarrow f(x)$ is an onto function.
If a line parallel to $X$-axis is drawn,
it will intersect at only one point of $f(x)$.
Hence, $f(x)$ is one-one also.
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2, x \geq 0 \\ -x^2, x < 0\end{array}\right.$, Also, $f: R \rightarrow R$
According to the graph,
domain of $f(x)=$ Range of $f(x)=R$
$\therefore$ Range $=$ Codomain $=R$
$\Rightarrow f(x)$ is an onto function.
If a line parallel to $X$-axis is drawn,
it will intersect at only one point of $f(x)$.
Hence, $f(x)$ is one-one also.
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