If f(x) = frac{e^x}{1+e^x},l_1 = int_{f(-a)}^ | vedclass

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(C) First,observe that $f(a) + f(-a) = \frac{e^a}{1+e^a} + \frac{e^{-a}}{1+e^{-a}} = \frac{e^a}{1+e^a} + \frac{1}{e^a+1} = \frac{e^a+1}{e^a+1} = 1$. L

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$2$

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(C) First,observe that $f(a) + f(-a) = \frac{e^a}{1+e^a} + \frac{e^{-a}}{1+e^{-a}} = \frac{e^a}{1+e^a} + \frac{1}{e^a+1} = \frac{e^a+1}{e^a+1} = 1$. Let $t = f(-a)$,then $f(a) = 1-t$. Now,$l_1 = \int_{t}^{1-t} x g(x(1-x)) dx$. Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A+B-x) dx$,we have: $l_1 = \int_{t}^{1-t} (t + (1-t) - x) g((t + (1-t) - x)(1 - (t + (1-t) - x))) dx$ $l_1 = \int_{t}^{1-t} (1-x) g((1-x)(1-(1-x))) dx = \int_{t}^{1-t} (1-x) g(x(1-x)) dx$. Adding the two expressions for $l_1$: $2l_1 = \int_{t}^{1-t} (x + 1 - x) g(x(1-x)) dx = \int_{t}^{1-t} g(x(1-x)) dx$. Since $l_2 = \int_{t}^{1-t} g(x(1-x)) dx$,we get $2l_1 = l_2$. Therefore,$\frac{l_2}{l_1} = 2$.

If $f(x) = \frac{e^x}{1+e^x}$,$l_1 = \int_{f(-a)}^{f(a)} x g(x(1-x)) dx$ and $l_2 = \int_{f(-a)}^{f(a)} g(x(1-x)) dx$,then the value of $\frac{l_2}{l_1}$ is

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