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JEE Main & Advanced Mathematics Definite Integration Question Bank Summation of series by Definite Integration, Gamma function, Leibnitz's rule

  • question_answer
    If \[F(x)=\frac{1}{{{x}^{2}}}\int_{4}^{x}{(4{{t}^{2}}-2{F}'(t))\,dt,}\] then \[{F}'(4)\] equals

    A)                 32          

    B)                 \[\frac{32}{3}\]

    C)                 \[\frac{32}{9}\]

    D)                 None of these

    Correct Answer: C

    Solution :

                       We have \[F(x)=\frac{1}{{{x}^{2}}}\int_{4}^{x}{(4{{t}^{2}}-2F'(t))dt}\]                    \[\therefore \,\,F'(x)=\frac{1}{{{x}^{2}}}\left( 4{{x}^{2}}-2F'(x) \right)-\frac{2}{{{x}^{3}}}\int_{4}^{x}{(4{{t}^{2}}-2F'(t))dt}\]                                 Þ \[F'(4)=\frac{1}{16}[64-2F'(4)]-0\Rightarrow F'(4)=\frac{32}{9}\].


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