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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

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

    A)  \[0\]                    

    B)         \[\pi \]

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

    D)         \[2\pi \]

    E)  \[3\pi \]

    Correct Answer: B

    Solution :

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


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