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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
    The greatest value of the function \[F(x)=\int_{1}^{x}{\,\,|t|\,dt}\] on the interval \[\left[ -\frac{1}{2},\,\,\frac{1}{2} \right]\] is given by                                         [IIT Screening]

    A)                 \[\frac{3}{8}\]   

    B)                 \[-\frac{1}{2}\]

    C)                 \[-\frac{3}{8}\] 

    D)                 \[\frac{2}{5}\]

    Correct Answer: C

    Solution :

                       \[F'(x)=|x|>0\forall x\in \left[ -\frac{1}{2},\frac{1}{2} \right]\]            Hence the function is increasing on \[\left[ -\frac{1}{2},\frac{1}{2} \right]\] and therefore \[F(x)\]has maxima at the right end point of \[\left[ -\frac{1}{2},\frac{1}{2} \right]\].                                 Þ \[\text{Max}\,\,F(x)=F\left( \frac{1}{2} \right)=\int_{1}^{1/2}{\,\,\,\,|t|\,}dt=-\frac{3}{8}\].


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