If f: R rightarrow R,then the function f(x) = | vedclass

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(B) Given,$f: R ightarrow R$ defined by $f(x) = x|x|$.We can redefine the function as:$f(x) = \begin{cases} -x^2, & x < 0 \ 0, & x = 0 \ x^2, & x

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one-one-onto

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(B) Given,$f: R ightarrow R$ defined by $f(x) = x|x|$.We can redefine the function as:$f(x) = \begin{cases} -x^2, & x  0 \end{cases}$$1$. One-one check: Since $f(x)$ is a strictly increasing function (as $f'(x) = 2|x| \ge 0$ for all $x \in R$ and $f'(x) > 0$ for $x 
eq 0$),it is a one-one function.$2$. Onto check: As $x ightarrow \infty$,$f(x) ightarrow \infty$ and as $x ightarrow -\infty$,$f(x) ightarrow -\infty$. Since the range of the function is $(-\infty, \infty)$,which is equal to the codomain $R$,the function is onto.Therefore,the function is one-one and onto (bijective).

If $f: R \rightarrow R$,then the function $f(x) = x|x|$ is:

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