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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    If \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}},\,\,\,\ {{I}_{1}}=\int_{f(-a)}^{f(a)}{xg\{x(1-x)\}dx}\], and \[{{I}_{2}}=\int_{f(-a)}^{f(a)}{g\{x(1-x))\}dx}\], then the value of \[\frac{{{I}_{2}}}{{{I}_{1}}}\] is [AIEEE 2004]

    A) 1    

    B) -3

    C) -1  

    D) 2

    Correct Answer: D

    Solution :

    • Given, \[f(x)=\frac{{{e}^{x}}}{x+1},{{I}_{1}}=\int_{f(-a)}^{f(a)}{xg\{x(1-x)\}dx}\]                   
    • and \[{{I}_{2}}=\int_{f(-a)}^{f(a)}{g\{x(1-x)\}dx}\]                   
    • \[f(a)=\frac{{{e}^{a}}}{1+{{e}^{a}}},f(-a)=\frac{{{e}^{-a}}}{1+{{e}^{-a}}}\]                   
    • \ \[f(a)+f(-a)=1\]                   
    • Now, \[2{{I}_{1}}=\int_{f(-a)}^{f(a)}{xg\{x(1-x)\}dx}\] + \[\int_{f(-a)}^{f(a)}{\{f(a)+f(-a)-x\}g\{(1-x)(x)}\}dx\]                   
    • Þ \[2{{I}_{1}}=\int_{f(-a)}^{f(a)}{g\{x(1-x)\}dx={{I}_{2}}}\], \[(\because f(a)+f(-a)=1)\]                   
    • \ \[2{{I}_{1}}={{I}_{2}}\]Þ \[\frac{{{I}_{2}}}{{{I}_{1}}}=2\].


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