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

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.	\[-\,2\]
C.	2
D.	1
Answer» D. 1

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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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