If \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}},\,\,\,\ {{I}_{1}}=\int_{f(-a)} – McqOptions

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1.	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
Answer» E.

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