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If f(x)=e^x/(1+e^x), I1=int(f(-a))^(f(a)...

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 `I_2/I_1`= (A) `-1` (B) `-3` (C) `2` (D) `1`

A

`-1`

B

`-2`

C

`2`

D

`1`

Text Solution

Verified by Experts

`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`
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