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

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

1.	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.	\[2\]
Answer» E.

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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

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