2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

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

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

    B)  \[[4,\infty )\]

    C)  \[\{4\}\]                             

    D)  \[\phi \]

    Correct Answer: D

    Solution :

     f(x)=x2-4x+a always attains its minimum value.  So its range must be closed. So, a= \[\{\phi \}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner