Q50 / 160 · WBJEE 2009 · Mathematics

A Mapping from IN to IN is defined as follows : $$ \begin{aligned} & \mathrm{f}: \mathrm{IN} \rightarrow \mathrm{IN} \\ & \mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2, \mathrm{n} \in \mathrm{IN} \end{aligned} $$ (IN is the set of natural numbers). Then

  • A
    $\mathrm{f}$ is not one-to-one
  • B
    $\mathrm{f}$ is onto
  • C
    $\mathrm{f}$ is both one-to-one and onto
  • Correct answer
    D
    $\mathrm{f}$ is one-to-one but not onto

Solution

$$ \begin{aligned} & \text { Hints: } \mathrm{f}: \mathrm{IN} \rightarrow \mathrm{IN} ; \mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2 \\ & \left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2 \\ & \Rightarrow\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1+\mathrm{n}_2+10\right)=0 \\ & \Rightarrow \mathrm{n}_1=\mathrm{n}_2 \rightarrow \text { one-to-one } \end{aligned} $$ There does not exist $n \in$ IN such that $(n+5)^2=1$ Hence $\mathrm{f}$ is not onto

Hence, the correct answer is option: 'D'.

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