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(A) Consider the property $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2}$.$= \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x + 2} = \frac{4^x}{

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$19$

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(A) Consider the property $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2}$.$= \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x + 2} = \frac{4^x}{4^x+2} + \frac{4}{4 + 2 \cdot 4^x} = \frac{4^x}{4^x+2} + \frac{2}{2 + 4^x} = \frac{4^x+2}{4^x+2} = 1$.Let $S = f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + \dots + f\left(\frac{39}{40}\right)$.Pairing terms $f\left(\frac{k}{40}\right) + f\left(1 - \frac{k}{40}\right) = f\left(\frac{k}{40}\right) + f\left(\frac{40-k}{40}\right) = 1$.There are $39$ terms in the sum. The middle term is $f\left(\frac{20}{40}\right) = f\left(\frac{1}{2}\right)$.There are $19$ pairs that sum to $1$,plus the middle term $f\left(\frac{1}{2}\right)$.So,$S = 19 + f\left(\frac{1}{2}\right)$.Therefore,$S - f\left(\frac{1}{2}\right) = 19 + f\left(\frac{1}{2}\right) - f\left(\frac{1}{2}\right) = 19$.

Let the function $f:[0,1] \rightarrow \mathbb{R}$ be defined by $f(x) = \frac{4^x}{4^x+2}$. Then the value of $f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \dots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right)$ is:

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