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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        If\[\int_{0}^{\pi }{xf(\sin x)}dx=A\int_{0}^{\pi /2}{f(\sin x)dx,}\]then A is equal to

    A)  0                                            

    B)  \[\pi \]

    C)  \[\frac{\pi }{4}\]                                             

    D)  \[2\pi \]

    Correct Answer: B

    Solution :

                    Let \[I=\int_{0}^{\pi }{x}f(\sin x)dx\]                      ....(i) \[I=\int_{0}^{\pi }{(\pi -}x)f[\sin (\pi -x)]dx\] \[I=\int_{0}^{\pi }{(\pi -}x)f(\sin x)dx\]                  ?..(ii) On adding Eq. (i) and Eq. (ii) \[2I=\int_{0}^{\pi }{(x+\pi -x)f(\sin x)dx}\] \[2I=\pi \int_{0}^{\pi }{f(\sin x)dx}\] \[2I=2\pi \int_{0}^{\pi /2}{f(\sin x)dx}\] \[I=\pi \int_{0}^{\pi /2}{f(\sin x)dx}\] \[\Rightarrow \] \[A\int_{0}^{\pi /2}{f(\sin x)}\,dx=\pi \int_{0}^{\pi /2}{f(\sin x)}\,dx\] \[\Rightarrow \]               \[A=\pi \]


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