If f(x)`(e^(x))/(1+e^(x)),I_(1)=int_(f(-a))^(f(a)) xg{x(1-x)}dx` and `I_(2)=int_(f(-a))^(f(a)) g{x(1-x)}dx`, then the value of `(I_(2))/(I_(1))` is

Options

A. 1
B. -3
C. -1
D. 2

Answer

D

Explanation

We have,
`f(x)=(e^(x))/(1+e^(x))`
`rArr f(a)+f(-a)=(e^(a))/(1+e^(a))+(e^(-a))/(1+e^(-a))`
`rArr f(a)+f(-a)=(e^(a))/(1+e^(a))+(1)/(1+e^(a))=1`
Using `overset(b)underset(a)int f(x)dx=overset(b)underset(a)int f(a+b-x)dx`, we have
`I_(1)=overset(f(a))underset(f(a-))int (1-x)g{(1-x)x}dx[:'f(a)+f(-a)=1]`
`rArr I_(1)=overset(f(a))underset(f(-a))int g{(1-x)x}dx-I_(1)`
`rArr 2=I_(2)-I_(1) rArr 2I_(1)=I_(2)rArr (I_(2))/(I_(1))=2`