If $f(x)=\frac{e^x}{1+e^x}, I_1=\int\limits_{f(-a)}^{f(a)} x g\{x(1-x)\} d x$ and $I_2=\int\limits_{f(-a)}^{f(a)} g\{x(1-x)\} d x$, then the value of $\frac{I_2}{I_1}$, is
Answer & explanation
Correct answer: option 4
We have,
$f(x)=\frac{e^x}{1+e^x}$
$\Rightarrow f(a)+f(-a)=\frac{e^a}{1+e^a}+\frac{e^{-a}}{1+e^{-a}}$
$\Rightarrow f(a)+f(-a)=\frac{e^a}{1+e^a}+\frac{1}{1+e^a}=1$
Using $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$, we have
$I_1 =\int\limits_{f(-a)}^{f(a)}(1-x) g\{(1-x) x\} d x$ $[∵ f(a)+f(-a)=1]$
$\Rightarrow I_1 =\int\limits_{f(-a)}^{f(a)} g\{(1-x) x\} d x-I_1$
$\Rightarrow \quad 2 =I_2-I_1 \Rightarrow 2 I_1=I_2 \Rightarrow \frac{I_2}{I_1}=2$