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. $2$
• B. $-3$
• C. $-1$
• D. $1$

Answer 1

Answer: (A)

$2$

Given, $f\left(x\right)=\frac{e^x}{1+e^x}$
Thus $f\left(a\right)=\frac{e^a}{1+e^a}$
and $f(-a)=\frac{e^{-a}}{1+e^{-a}}=\frac{1}{e^a+1}$
Therefore, $f\left(a\right)+f(-a)=\frac{e^a}{1+e^a}+\frac{1}{e^a+1}=1$
$\Rightarrow f\left(a\right)=1-f(-a)$
Let $f(-a)=t\Rightarrow f\left(a\right)=1-t$
Now, $I_1={\int }_{f(-a)}^{f\left(a\right)}xg\left\{x(1-x)\right\}dx$
$I_1={\int }_t^{1-t}xg\left\{x(1-x)\right\}dx$ ...(i)
$I_1={\int }_t^{1-t}(1-x)g\left\{x(1-x)\right\}dx$ ...(ii)
On adding equations (i) and (ii), we get
$2I_1={\int }_t^{1-t}g\left\{x(1-x)\right\}(x+1-x)dx$
$={\int }_t^{1-t}g\left\{x(1-x)\right\}dx=I_2$
$\Rightarrow 2I_1=I_2$
$\Rightarrow \frac{I_2}{I_1}=2$