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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • 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, \[a=\{\phi \}\]


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