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 $\frac{I_2}{I_1}$ is

• A. 2
• B. -3
• C. -1
• D. 1

Answer 1

The correct option is A 2

$f\left(x\right)=\frac{e^x}{1+e^x}\Rightarrow f(-a)+f\left(a\right)=\frac{e^a}{1+e^a}+\frac{e^{-a}}{1+e^{-a}}=1$
Let $I_1={\int }_{f(-a)}^{f\left(a\right)}xg\left(x(1-x)\right)dx$.....(1)
Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I_1={\int }_{f(-a)}^{f\left(a\right)}(f(-a)+f\left(a\right)-x)g\left((f(-a)+f\left(a\right)-x)(1-(f(-a)+f\left(a\right)-x))\right)dx$
$={\int }_{f(-a)}^{f\left(a\right)}(1-x)g\left((1-x)x\right)dx$ ...(2)
Adding (1) and (2), we get
$2I_1={\int }_{f(-a)}^{f\left(a\right)}g\left((1-x)x\right)dx=I_2\Rightarrow \frac{I_2}{I_1}=2$