Question

If $f\left(x\right)=\frac{e^x}{1+e^x},I_1=\underset{f(-a)}{\overset{f\left(a\right)}{\int }}xg\left(x\right(1-x\left)\right)dx$ and $I_2=\underset{f(-a)}{\overset{f\left(a\right)}{\int }}g\left(x\right(1-x\left)\right)dx$, then the value of $\frac{I_2}{I_1}$ is

• A. $-1$
• B. $-2$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

$f\left(x\right)=\frac{e^x}{1+e^x}f\left(a\right)=\frac{e^a}{1+e^a}f(-a)=\frac{e^{-a}}{1+e^{-a}}=\frac{1}{1+e^a}$
So,
$f(-a)+f\left(a\right)=\frac{e^a+1}{1+e^a}=1$
Let
$f(-a)=\alpha ;f\left(a\right)=1-\alpha$
Now,
$I_1=\underset{\alpha }{\overset{1-\alpha }{\int }}xg\left(x\right(1-x\left)\right)dx\cdots \left(1\right)$
Using property,
$\Rightarrow I_1=\underset{\alpha }{\overset{1-\alpha }{\int }}(1-x)g\left(\right(1-x\left)\right(1-(1-x)\left)\right)dx\Rightarrow I_1=\underset{\alpha }{\overset{1-\alpha }{\int }}(1-x)g\left(x\right(1-x\left)\right)dx\cdots \left(2\right)$
Adding (1) and (2),m
$\Rightarrow 2I_1=\underset{\alpha }{\overset{1-\alpha }{\int }}g\left(x\right(1-x\left)\right)dx=I_2\Rightarrow \frac{I_2}{I_1}=2$