Question

if $f\left(x\right)=\frac{e^x}{1+e^x},I_1={\int }_{f(-a)}^{f\left(a\right)}xg\left(x\right(1-x\left)\right)dx$, and $I_2={\int }_{f(-a)}^{f\left(a\right)}g\left(x\right(1-x\left)\right)dx$,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 $2$

$f\left(x\right)=\frac{e^x}{1+e^x}$
$\therefore f\left(a\right)=\frac{e^a}{1+e^a}$
and $f(-a)=\frac{e^{-a}}{1+e^{-a}}=\frac{e^{-a}}{1+\frac{1}{e^a}}=\frac{1}{1+e^a}$
$\Rightarrow$ $f\left(a\right)+f(-a)=\frac{e^a+1}{1+e^a}=1$
Let $f(-a)=\alpha$ or $f\left(a\right)=1-\alpha$
Now,$I_1={\int }_{\alpha }^{1-\alpha }xg\left(x\right(1-x\left)\right)dx$ ....(1)
$={\int }_{\alpha }^{1-\alpha }(1-x)g\left(\right(1-x\left)\right(1-(1-x)\left)\right)dx$
$I_1={\int }_{\alpha }^{1-\alpha }(1-x)g\left(x\right(1-x\left)\right)dx$ .....(2)
$\therefore 2I_1={\int }_{\alpha }^{1-\alpha }g\left(x\right(1-x\left)\right)dx=I_2$
$\Rightarrow \frac{I_2}{I_1}=2$