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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    The maximum value of the function \[f(x)=3{{x}^{3}}-18{{x}^{2}}+27x-40\]on the set \[S=\{x\in R:{{x}^{2}}+30\le 11x\}\] is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[-122~\]                                  

    B)               \[-222\] 

    C)               222                              

    D)               122

    Correct Answer: D

    Solution :

    Here, \[{{x}^{2}}-11x+30\le 0\] \[\Rightarrow \]\[{{x}^{2}}-5x-6x+30\le 0\] \[\Rightarrow \]\[(x-5)(x-6)\le 0\] \[\Rightarrow \]\[5\le x\le 6\] \[\therefore \]\[S=\{x\in R,5\le x\le 6\}\] Now,\[f(x)=3{{x}^{3}}-18{{x}^{2}}+27x-40\] \[\therefore \]\[f'(x)=9{{x}^{2}}-36x+27=9(x-1)(x-3),\]which is positive in [5, 6]. So, f(x) is increasing in [5,6] Hence, maximum value of f(x) = f(6) =122


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