Question

If $f\left(x\right)=\frac{e^x}{1+e^x},I_1={\int }_{f(-a)}^{f\left(a\right)}xg\left\{x\right(1-x\left)\right\}dxand{I}_2={\int }_{f(-a)}^{f\left(a\right)}g\left\{x\right(1-x\left)\right\}dx,$ then the value of $\frac{I_2}{I_1}is$

• A. 2
• B. 1
• C. -1
• D. -3

Answer 1

The correct option is A 2

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