Question

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

• A. $2$
• B. $-3$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $2$

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

$\therefore f\left(a\right)=\frac{e^a}{1+e^a}$ and $f(-a)=\frac{e^{-a}}{1+e^{-a}}=\frac{1}{1+e^a}$

$\Rightarrow f\left(a\right)+f(-a)=1\Rightarrow f\left(a\right)=1-f(-a)f(-a)=t\Rightarrow f\left(a\right)=1-t$

Now $I_1={\int }_t^{1-t}xg\left[x(1-x)\right]dx$ ...(1)

$I_1={\int }_t^{1-t}(1-x)g\left[x(1-x)\right]dx$ ...(2)

Adding (1) and (2)

$2I_1={\int }_t^{1-t}g\left[x(1-x)\right](1-x+x)dx={\int }_t^{1-t}g\left[x(1-x)\right]dx=I_2$

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