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

  • A
    $f$ is one-one but not onto
  • B
    $f$ is onto but not one-one
  • C
    $f$ is both one-one and onto
  • D
    $f$ is neither one-one nor onto

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Show that the function $f: R_* \rightarrow R_*$ defined by $f(x) = \frac{1}{x}$ is one-one and onto,where $R_*$ is the set of all non-zero real numbers. Is the result true if the domain $R_*$ is replaced by $N$ with the co-domain remaining the same as $R_*$?

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$I$. $f$ is one-one
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