If f(x)`=(e^(2))/(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`, where g is not identify function. Then the value of `I_(2)//I_(1)`, is

Options

A. -1
B. `1//2`
C. 2
D. 1

Answer

C

Explanation

We have, f(x)`=(e^(x))/(1+e^(x))`
`f(-x)=(e^(-x))/(1+e^(-x))=(1)/(1+e^(x))`
`:. f(x)+f(-x)=(e^(x))/(1+e^(x))+(1)/(1+e^(x))=1`
`rArr f(a)+f(-x)=1`
Now,
`I_(1)=overset(f(a))underset(f(-x))int (1-x)g{x(1-x)}dx " "`..........(i)
Using `overset(b)underset(a)int f(a+b-x)=overset(b)underset(a)intf(x)" and "f(-a)+f(a)=1` we get,
`I_(1)=overset(f(a))underset(f(-a))int(1-x)g{(1-x)(1-(1-x))}dx`
`rArr I_(1)=overset(f(a))underset(f(-a))int (1-x)g{x(1-x)}dx" "`...........(ii)
Adding (i) and (ii), we get
`2I_(1)=overset(f(a))underset(f(-a))int g{x(1-x)}dx`
`rArr 2I_(1)=I_(2)rArr(I_(2))/(I_(1))=2`