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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Maxima and Minima

  • question_answer
    The function \[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+4\] is maximum at                                                                          [Karnataka CET 2001]

    A)            \[x=2\]

    B)            \[x=4\]

    C)            \[x=0\]

    D)            \[x=3\]

    Correct Answer: A

    Solution :

               \[f(x)=2{{x}^{3}}-15{{x}^{2}}+36x+4\]Þ \[{f}'(x)=6{{x}^{2}}-30x+36\]  ?..(i)            We know that for its maximum value \[{f}'(x)=0.\] \[6{{x}^{2}}-30x+36=0\] Þ \[(x-2)(x-3)=0\] Þ \[x=2,\,3.\]            Again differentiating equation (i), we get \[{f}''(x)=12x-30\]            Þ \[{f}''(2)=24-30=-6<0\].            Therefore \[f(x)\] is maximum at \[x=2.\]


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