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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    If \[f(x)={{x}^{2}}+1\], then \[{{f}^{-1}}(17)\] and \[{{f}^{-1}}(-3)\]will be [UPSEAT 2003]

    A)                    4, 1

    B)            4, 0

    C)                    3, 2

    D)            None of these

    Correct Answer: D

    Solution :

               Let\[y={{x}^{2}}+1\] Þ  \[x=\pm \sqrt{y-1}\]            Þ \[{{f}^{-1}}(y)=\pm \sqrt{y-1}\] Þ \[{{f}^{-1}}(x)=\pm \sqrt{x-1}\]            Þ \[{{f}^{-1}}(17)=\pm \sqrt{17-1}=\pm 4\]            and \[{{f}^{-1}}(-3)=\pm \sqrt{-3-1}=\pm \sqrt{-4}\], which is not possible.


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