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 $\frac{I_2}{I_1}$ [AIEEE 2004]

• A. 1
• B. -3
• C. -1
• D. 2

Answer 1

The correct option is D 2

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