A) \[-\,1\]
B) \[-\,2\]
C) 2
D) 1
Correct Answer: C
Solution :
[c] \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}}\] \[\therefore f(a)=\frac{{{e}^{q}}}{1+{{e}^{a}}}\] And \[f(-a)=\frac{{{e}^{-a}}}{1+{{e}^{-a}}}=\frac{{{e}^{-a}}}{1+\frac{1}{{{e}^{a}}}}=\frac{1}{1+{{e}^{a}}}\] \[\therefore f(a)+f(-a)=\frac{{{e}^{a}}+1}{1+{{e}^{q}}}=1\] Let \[f(-a)=\alpha \] or \[f(a)=1-\alpha \] Now, \[{{I}_{1}}=\int\limits_{\alpha }^{1-\alpha }{xg(x(1-x))dx}\] \[=\int\limits_{\alpha }^{1-\alpha }{(1-x)g((1-x)(1-(1-x))dx}\] \[=\int\limits_{\alpha }^{1-\alpha }{(1-x)g(x(1-x)dx}\] \[\therefore 2{{I}_{1}}=\int\limits_{\alpha }^{1-\alpha }{g(x(1-x))dx}={{I}_{2}}\] Or \[\frac{{{I}_{2}}}{{{I}_{1}}}=2\]
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