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

  • question_answer
    The maximum value of function \[{{x}^{3}}-12{{x}^{2}}+36x+\]17  in the interval [1, 10] is

    A)            17

    B)            177

    C)            77

    D)            None of these

    Correct Answer: B

    Solution :

               Let \[f(x)={{x}^{3}}-12{{x}^{2}}+36x+17\]                    \[\therefore f'(x)=3{{x}^{2}}-24x+36=0\] at \[x=2,\,6\]                    Again \[{f}''(x)=6x-24\] is \[-ve\]at \[x=2\]                    So that \[f(6)=17,\ \ f(2)=49\]                    At the end points \[=f(1)=42,\,\,f(10)=177\]                    So that \[f(x)\] has its maximum value as 177.


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