JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Rolle's theorem Lagrange's mean value theorem

  • question_answer
    If the function \[f(x)={{x}^{3}}-6{{x}^{2}}+ax+b\] satisfies Rolle?s theorem in the interval \[[1,\,3]\] and \[f'\left( \frac{2\sqrt{3}+1}{\sqrt{3}} \right)=0\], then                                                               [MP PET 2002]

    A)            \[a=-11\]

    B)            \[a=-6\]

    C)            \[a=6\]

    D)            \[a=11\]

    Correct Answer: D

    Solution :

               \[f(x)={{x}^{3}}-6{{x}^{2}}+ax+b\]                    Þ  \[{f}'(x)=3{{x}^{2}}-12x+a\]            Þ  \[{f}'(c)=0\]  Þ  \[{f}'\left( 2+\frac{1}{\sqrt{3}} \right)=0\]            Þ  \[3{{\left( 2+\frac{1}{\sqrt{3}} \right)}^{2}}-12\left( 2+\frac{1}{\sqrt{3}} \right)+a=0\]            Þ  \[3\left( 4+\frac{1}{3}+\frac{4}{\sqrt{3}} \right)-12\left( 2+\frac{1}{\sqrt{3}} \right)+a=0\]            \[12+1+4\sqrt{3}-24-4\sqrt{3}+a=0\] Þ \[a=11\].

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