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Manipal Engineering Manipal Engineering Solved Paper-2012

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

    \[{{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}\,\,dx}\]     \[=\int_{1-k}^{k}{(1-x)f\{(1-x)}(1-(1-x))\}dx\]                 \[\left( \because \,\,\int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)dx}} \right)\]                 \[=\int_{1-k}^{k}{(1-x)f\{(1-x)x\}}\,\,dx\]                 \[=\int_{k-1}^{k}{f\{x(1-x)\}\,\,dx}\]                                                 \[-\int_{k-1}^{k}{xf\{x(1-x)\}dx}\] \[\Rightarrow \]               \[{{I}_{1}}={{I}_{2}}-{{I}_{1}}\]   \[\Rightarrow \]   \[2{{I}_{1}}={{I}_{2}}\] \[\Rightarrow \]               \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2}\]


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