Question

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

$I_1={\int }_{f(-a)}^{f\left(a\right)}x.g\left\{x(1-x)\right\}dx$ and $I_2={\int }_{f(-a)}^{f\left(a\right)}g\left\{x(1-x)\right\}dx$

then the value of $I_2/I_1$ is?

• A. $2$
• B. $-3$
• C. $-1$
• D. $1$

Answer 1

Answer: (A)

$2$

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

$=\frac{e^a}{1+e^a}+\frac{1}{1+e^a}$

$=1$

Now

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

Applying ${\int }_a^{b}f\left(x\right).dx={\int }_a^{b}f(a+b-x).dx$

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

Hence

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

$={\int }_{f(-a)}^{f\left(a\right)}\left(f\right(a)+f(-a)-x)g\left(\right(f\left(a\right)+f(-a)-x\left)\right)(1-(f\left(a\right)+f(-a)-x\left)\right)\left)\right).dx$

$={\int }_{f(-a)}^{f\left(a\right)}(1-x)g\left(\right(1-x\left)\right(x\left)\right).dx$

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

$=I_2-I_1$

Hence

$I_1=I_2-I_1$

$\Rightarrow 2I_1=I_2$

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