A function f(x) is given by f(x) = frac{5^{x} | vedclass

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(D) Given $f(x) = \frac{5^{x}}{5^{x} + \sqrt{5}}$.Note that $f(1-x) = \frac{5^{1-x}}{5^{1-x} + \sqrt{5}} = \frac{5/5^{x}}{5/5^{x} + \sqrt{5}} = \frac{

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$\frac{39}{2}$

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(D) Given $f(x) = \frac{5^{x}}{5^{x} + \sqrt{5}}$.Note that $f(1-x) = \frac{5^{1-x}}{5^{1-x} + \sqrt{5}} = \frac{5/5^{x}}{5/5^{x} + \sqrt{5}} = \frac{5}{5 + \sqrt{5} \cdot 5^{x}} = \frac{\sqrt{5}}{\sqrt{5} + 5^{x}}$.Thus,$f(x) + f(1-x) = \frac{5^{x} + \sqrt{5}}{5^{x} + \sqrt{5}} = 1$.The series has $39$ terms from $x = \frac{1}{20}$ to $x = \frac{39}{20}$.Pairing terms $f(x) + f(1-x) = 1$ is not directly applicable here as the sum is up to $x = \frac{39}{20}$.Let $S = \sum_{k=1}^{39} f\left(\frac{k}{20}\right)$.Since $f(x) + f(1-x) = 1$,we have $f\left(\frac{k}{20}\right) + f\left(1 - \frac{k}{20}\right) = 1$,which is $f\left(\frac{k}{20}\right) + f\left(\frac{20-k}{20}\right) = 1$.Summing from $k=1$ to $19$,we get $19$ pairs each summing to $1$,plus the middle term $f\left(\frac{20}{20}\right) = f(1) = \frac{5}{5+\sqrt{5}}$.Wait,the original problem implies $f(x) = \frac{5^x}{5^x + 5^{1/2}}$.Sum $= 19 + f(1) + \sum_{k=21}^{39} f\left(\frac{k}{20}\right) = 19 + f(1) + \sum_{j=1}^{19} f\left(1 + \frac{j}{20}\right)$.Using $f(x) + f(2-x) = 1$,the sum is $19 + f(1) = 19 + \frac{5}{5+\sqrt{5}} = 19 + \frac{\sqrt{5}}{\sqrt{5}+1} = \frac{19\sqrt{5}+19+\sqrt{5}}{\sqrt{5}+1} = \frac{20\sqrt{5}+19}{\sqrt{5}+1} \approx 19.5$.

$A$ function $f(x)$ is given by $f(x) = \frac{5^{x}}{5^{x} + \sqrt{5}}$. Then the sum of the series $f\left(\frac{1}{20}\right) + f\left(\frac{2}{20}\right) + f\left(\frac{3}{20}\right) + \ldots + f\left(\frac{39}{20}\right)$ is equal to ....... .

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