If f(x) = frac{2^{2x}}{2^{2x} + 2},x in R,the | vedclass

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(D) Given $f(x) = \frac{4^x}{4^x + 2}$.Consider $f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{4^{1-x}}{4^{1-x} + 2}$.$= \frac{4^x}{4^x + 2} + \frac{4/4

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$1011$

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(D) Given $f(x) = \frac{4^x}{4^x + 2}$.Consider $f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{4^{1-x}}{4^{1-x} + 2}$.$= \frac{4^x}{4^x + 2} + \frac{4/4^x}{4/4^x + 2} = \frac{4^x}{4^x + 2} + \frac{4}{4 + 2 \cdot 4^x} = \frac{4^x}{4^x + 2} + \frac{2}{2 + 4^x} = \frac{4^x + 2}{4^x + 2} = 1$.Thus,$f(x) + f(1-x) = 1$.The given sum is $S = \sum_{k=1}^{2022} f\left(\frac{k}{2023}ight)$.Since there are $2022$ terms,we can pair them as $f\left(\frac{k}{2023}ight) + f\left(1 - \frac{k}{2023}ight) = 1$.The number of such pairs is $\frac{2022}{2} = 1011$.Therefore,the sum is $1011 	imes 1 = 1011$.

If $f(x) = \frac{2^{2x}}{2^{2x} + 2}$,$x \in R$,then $f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + \dots + f\left(\frac{2022}{2023}\right)$ is equal to

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