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

  • question_answer
    The maximum value of \[2{{x}^{3}}-24x+107\] in the interval       [?3, 3] is

    A)            75

    B)            89

    C)            125

    D)            139

    Correct Answer: D

    Solution :

               Let \[f(x)=2{{x}^{3}}-24x+107\]            At \[x=-3,\ f(-3)=2{{(-3)}^{3}}-24(-3)+107=125\]            At \[x=3,\ \ f(3)=2{{(3)}^{3}}-24(3)+107=89\]            For maxima or minima, \[{f}'\,(x)=6{{x}^{2}}-24=0\]            \[\Rightarrow x=2,\ \ -2\]            So at \[x=2,\ f(2)=2{{(2)}^{3}}-24(2)+107=75\]            at \[x=-2,\ \ f(-2)=2{{(-2)}^{3}}-24(-2)+107=139\]            Thus the maximum value of the given function in           [? 3, 3] is 139.


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