2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Indefinite Integrals

JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Self Evaluation Test - Integrals

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

    A) 1

    B) \[-3\]

    C) \[-1\]

    D) \[2\]

    Correct Answer: D

    Solution :

    [d] \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}}\Rightarrow f(-x)=\frac{{{e}^{-x}}}{1+{{e}^{-x}}}\] \[=\frac{1}{{{e}^{x}}+1}\] \[\therefore f(x)+f(-x)=1\forall x\] Now \[{{I}_{1}}=\int\limits_{f(-a)}^{f(a)}{xg\{x(1-x)\}dx}\] \[=\int\limits_{f(-a)}^{f(a)}{(1-x)g\{x(1-x)\}dx}\] \[={{I}_{2}}-{{I}_{1}}\Rightarrow 2{{I}_{1}}={{I}_{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner