Question

If $f\left(x\right)=\frac{e^x}{1+e^x},A={\int }_{f(-a)}^{f\left(a\right)}xg\left\{x\right(1-x\left)\right\}dx$ and $B={\int }_{f(-a)}^{f\left(a\right)}g\left\{x\right(1-x\left)\right\}dx$, then $\frac{B}{A}$ is equal to

• A. -1
• B. -3
• C. 2
• D. 1

Answer 1

Answer: (C)

2

Let $f\left(x\right)=\frac{e^x}{1+e^x}$
$f\left(a\right)=\frac{e^a}{1+e^a}$
$f(-a)=\frac{e^{-a}}{1+e^{-a}}=\frac{1}{e^a+1}$
$\Rightarrow f\left(a\right)+f(-a)=1$
Let $b=f(-a)$
$\Rightarrow f\left(a\right)=1-b$
Now $A={\int }_b^{1-b}xg\left\{x\right(1-x\left)\right\}dx$

Using the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$, we get
$A={\int }_b^{1-b}(1-b+b-x)g\left\{\right(1-b+b-x\left)\right(1-1+b-b+x\left)\right\}dx$
$={\int }_b^{1-b}(1-x)g\left\{\right(1-x\left)x\right\}dx=B-A$
$\Rightarrow 2A=B\Rightarrow \frac{B}{A}=2.$