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JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    If\[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}},{{I}_{1}}=\int_{f(-a)}^{f(a)}{x}g\{x(1-x)\}dx\]and\[{{I}_{2}}=\int_{f(-a)}^{f(a)}{g\{x(1-x)\}dx},\] then the value of\[\frac{{{I}_{2}}}{{{I}_{1}}}\]is

    A) 2      

    B)                        \[-3\]                    

    C)        \[-1\]                    

    D)        1

    Correct Answer: A

    Solution :

    Given that, \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}}\] \[\therefore \]  \[f(a)=\frac{{{e}^{a}}}{1+{{e}^{a}}}\]                      ?.(i) And        \[f(-a)=\frac{1}{1+{{e}^{a}}}\]                    ?..(ii) On adding Eqs. (i) and (ii), we get \[f(a)+f(-a)=1\] \[\Rightarrow \]               \[f(a)=1-f(-a)\] Let          \[f(-a)=t\] \[\Rightarrow \]               \[f(a)=1-t\]         Now,     \[{{l}_{1}}=\int_{t}^{1-t}{xg}[x(1-x)]dx\]                            ...(iii) \[{{l}_{1}}=\int_{t}^{1-t}{(1-x)g}([x(1-x)]dx\]         ...(iv) On adding Eqs. (iii) and (iv), we get \[2{{l}_{1}}=\int_{t}^{1-t}{g}[x(1-x)](1-x+x)dx\] \[\Rightarrow \] \[2{{l}_{1}}=\int_{t}^{1-t}{g[x(1-x)]}dx={{l}_{2}}\] \[\Rightarrow \]               \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{2}{1}=2\]


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