If $$f:R \to R$$, then $$f\left( x \right) = x\left| x \right|$$ will be A. many-one-ontoB. one-one-ontoC. many-one-intoD. one-one-into

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Vedantu Content Team · Accepted Answer

Hint: Here $$R$$ means a set of real numbers. The function $$f:R \to R$$ implies that the domain of the given function is $$R$$and corresponding range is also $$R$$. In simple words, if you put real numbers in the function you will get a set of real numbers. Plot the graph for the given function. So, use this concept to reach the solution of the problem.Complete step-by-step answer:Given, $$f:R \to R$$, $$f\left( x \right) = x\left| x \right|$$This function can be redefined as $$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}  { - {x^2},}&{x   {0,}&{x = 0} \   {{x^2}}&{x > 0} \end{array}} \right\}$$If we plot the graph for $$f\left( x \right)$$, it will be Here the function $$f:R \to R$$ is bijective since its graph meets every horizontal and vertical line exactly once.Since, the graph of $$f\left( x \right)$$ shows that it is a bijective (one-one-onto) function.Thus, the correct option is D. one-one-into.Note: Here a one-one function is a function of which the answers never repeat. If each element in the codomain ‘Y’ has at least one pre-image in the domain ‘X’ then the function is said to be onto. For a bijective function both one-one and onto functions must be satisfied.

If \[f:R \to R\], then \[f\left( x \right) = x\left| x \right|\] will be
A. many-one-onto
B. one-one-onto
C. many-one-into
D. one-one-into

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If \[f:R \to R\], then \[f\left( x \right) = x\left| x \right|\] will be A. many-one-ontoB. one-one-ontoC. many-one-intoD. one-one-into

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