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JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

  • question_answer
    The function \[f:[2,\,\,\infty )\to (0,\,\,\infty )\] defined by\[f(x)={{x}^{2}}-4x+a\], then the set of values of \['a'\] for which\[f(x)\] becomes onto is

    A) \[(4,\,\,\infty )\]                              

    B) \[[4,\,\,\infty )\]

    C) \[\{4\}\]                                              

    D) \[\phi \]

    Correct Answer: D

    Solution :

     \[f(x)={{x}^{2}}-4x+a\] always attains its minimum value. So its range must be closed. So,\[a=\left\{ \phi  \right\}\]


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