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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    The value of integral \[\int_{0}^{2a}{\frac{f(x)}{f(x)+f(2a-x)}}dx,\]where\[f(x)\]is a continuous function, is

    A)  0                

    B)  1

    C)  \[a\]                

    D)  \[2a\]

    Correct Answer: C

    Solution :

     Let\[I=\int_{0}^{2a}{\frac{f(x)}{f(x)+f(2a-x)}}dx\]                ...(i) \[\Rightarrow \]\[I=\int_{0}^{2a}{\frac{f(2a-x)}{f(x)+f(2a-x)}}dx\] ?(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{2a}{1\,}dx\Rightarrow 2I[x]_{0}^{2a}\] \[\Rightarrow \] \[2I=2a\] \[\Rightarrow \] \[I=a\]


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