A) 2
B) \[-3\]
C) \[-1\]
D) 1
Correct Answer: A
Solution :
Given that, \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}}\] \[\therefore \] \[f(a)=\frac{{{e}^{a}}}{1+{{e}^{a}}}\] ?.(i) And \[f(-a)=\frac{1}{1+{{e}^{a}}}\] ?..(ii) On adding Eqs. (i) and (ii), we get \[f(a)+f(-a)=1\] \[\Rightarrow \] \[f(a)=1-f(-a)\] Let \[f(-a)=t\] \[\Rightarrow \] \[f(a)=1-t\] Now, \[{{l}_{1}}=\int_{t}^{1-t}{xg}[x(1-x)]dx\] ...(iii) \[{{l}_{1}}=\int_{t}^{1-t}{(1-x)g}([x(1-x)]dx\] ...(iv) On adding Eqs. (iii) and (iv), we get \[2{{l}_{1}}=\int_{t}^{1-t}{g}[x(1-x)](1-x+x)dx\] \[\Rightarrow \] \[2{{l}_{1}}=\int_{t}^{1-t}{g[x(1-x)]}dx={{l}_{2}}\] \[\Rightarrow \] \[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{2}{1}=2\]You need to login to perform this action.
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