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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009

  • question_answer
        Let\[f\]be a positive function. Let\[{{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}}dx,\]\[{{I}_{2}}=\int_{1-k}^{k}{f\{x(1-x)\}}dx\]where\[2k-1>0\]. Then, \[\frac{{{I}_{1}}}{{{I}_{2}}}\]is

    A)  2                                            

    B)  \[k\]

    C)  \[\frac{1}{2}\]                                  

    D)  1

    Correct Answer: C

    Solution :

                    \[{{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}}dx\] \[=\int_{1-k}^{k}{(1-x)f[(1-x)\{1-(1-x)\}]}dx\]                                                 [Put\[x=1-x\]] \[=\int_{1-k}^{k}{(1-x)f\{x(1-x)\}}dx\] \[=\int_{1-k}^{k}{f\{x(1-x)\}}dx-\int_{1-k}^{k}{xf\{x(1-x)\}}dx\] \[={{I}_{2}}-{{I}_{1}}\] \[\therefore \]  \[2{{I}_{1}}={{I}_{2}}\] \[\Rightarrow \]               \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2}\]


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