Question

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

• A. $1$
• B. $-3$
• C. $-1$
• D. $2$

Answer 1

The correct option is D $2$

$f\left(x\right)=\frac{e^x}{1+e^x}\Rightarrow f(-x)=\frac{e^{-x}}{1+e^{-x}}=\frac{1}{1+e^x}$

$\therefore f\left(x\right)+f(-x)=1-1)\forall x$

$I_1={\int }_{f(-a)}^{f\left(a\right)}xg\left\{x(1-x)\right\}dx={\int }_{f(-a)}^{f\left(a\right)}(1-x)g\left\{x(1-x)\right\}dx$ (Replacing $x$ by $1-x$ in $I_1$)

(Also, limit becomes $1-f(-a)$ to $1-f\left(a\right)$, use equation $1)$ to simplify limits)

$I_1=I_2-I_1\Rightarrow 2I_1=I_2$