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If     f(x)=\frac{e^{x}}{1+e^{x}},\; I_{1}=\int_{f(-a)}^{f(a)}xg\left \{x(1-x) \right \}{}dx    and   I_{2}=\int_{f(-a)}^{f(a)}g\left \{x(1-x) \right \}{}dx,    then the value of  \frac{I_{2}}{I_{1}}    is

  • Option 1)

    –1

  • Option 2)

    –3

  • Option 3)

    2

  • Option 4)

    1

 

Answers (1)

best_answer

As learnt in concept

Properties of Definite integration -

\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx

When \int_{0}^{b}f\left ( x \right )dx= \int_{0}^{b}f\left ( b-x \right )dx

 

- wherein

Put the \left ( a+b-x \right ) at the place of x in f\left ( x \right )

 

 Given f(x) =\frac{e^{x}}{1+e^{x}}

f(a)=\frac{e^{a}}{1+e^{a}} and f(-a)=\frac{e^{-a}}{1+e^{-a}}, Hence f(a)+f(-a)=1

I_{1}=\int_{f(-a)}^{f(a)}\:xg\left \{ x(1-x) \right \}dx

I_{1}=\int_{f(-a)}^{f(a)}(1-x)g\left \{ (1-x)x \right \}dx

2I_{1}=\int_{f(-a)}^{f(a)}g\left \{ x(1-x) \right \}dx=I_{2}

Thus, \frac{I_{1}}{I_{2}}=\frac{1}{2}\Rightarrow\frac{I_{2}}{I_{1}}=2


Option 1)

–1

This option is incorrect

Option 2)

–3

This option is incorrect

Option 3)

2

This option is correct

Option 4)

1

This option is incorrect

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