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.
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