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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 33

  • question_answer
    If the functions f(x) and g(x) are defined on \[R\to R\] such that \[f(x)=\left\{ \begin{matrix}    0,\,\,\,x\in rational  \\    x,\,\,\,\,x\in irrational\,  \\ \end{matrix} \right.\]\[g(x)=\left\{ \begin{matrix}    0,\,\,\,x\in \,\,\,irrational  \\    x,\,\,\,\,x\in \,\,rational\,  \\ \end{matrix} \right.\] then

    A) one-one and onto

    B) neither one-one nor onto

    C) one-one but not onto

    D) onto but not one-one

    Correct Answer: A

    Solution :

    We are given that \[f:R\to R\,\,such\,\,that\,\,f(x)=\left\{ \begin{matrix}    0,\,\,\,\,x\in rational  \\    x,\,\,\,\,\,\,x\,\,\,\in \,\,irrational  \\ \end{matrix} \right.\]i \[g:R\to R\,\,such\,\,that\,\,g(x)=\left\{ \begin{matrix}    0,\,\,\,\,x\in irrational  \\    x,\,\,\,\,\,\,x\,\,\,\in \,\,rational  \\ \end{matrix} \right.\] \[\therefore \,\,\,\,(f\,-g):R\to R\,\,such\,\,that\] \[(f\,-g)\,(x)\,\,=\,\,\left\{ \begin{matrix}    -x,\,\,if\,\,x\in rational  \\    x,\,\,if\,\,x\in irrational  \\ \end{matrix} \right.\] Since \[f\,-g:R\to R\] for any x there is only one value of \[\left( f(x)-g(x) \right)\] whether x is rational or irrational. Moreover as \[x\in R,\,\,f(x)-g\,(x)\] also belongs to R. Therefore, \[(f-g)\] is one-one onto.


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