`Iff(x)=(e^x)/(1+e^x),I_1=int_(f(-a))^(f(a))xg(x(1-x)dx ,a n d` `I_2=int_(f(-a))^(f(a))g(x(1-x))dx ,t h e nt h ev a l u eof(I_2)/(I_1)i s` `-1` (b) `-2` (c) 2 (d) 1

Options

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

Answer

C

Explanation

`f(x)=(e^(x))/(1+e^(x))`
`:.f(a)=(e^(a))/(1+e^(a))`
and `f(-a)=(e^(-a))/(1+e^(-a))=(e^(-a))/(1+1/(e^(a)))=1/(1+e^(a))`
`:. f(a)+f(-a)=(e^(a)+1)/(1+e^(a))=1`
Let `f(-a)=alpha` or `f(a)=1-alpha`
Now `I_(1)=int_(alpha)^(1-alpha)xg(x(1-x))dx`
`=int_(alpha)^(1-alpha)(1-x)g((1-x)(1-(1-x))dx`
`=int_(alpha)^(1-alpha) (1-x)g(x(1-x))dx`
`:. 2I_(1)=int_(alpha)^(1-alpha) g(x(1-x))dx=I_(2)` or `(I_(2))/(I_(1))=2`