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 $\frac{I_2}{I_1}$ equals

• A. $-1$
• B. $-3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

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

$\therefore$ $\frac{I_2}{I_1}=\frac{2}{1}$