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

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(B) Given $f(x) = \frac{2^x}{2^x + \sqrt{2}}$.Consider $f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2/2^x}{2/2^x + \sqrt{2}} = \frac{2}{2 + \s

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$\frac{81}{2}$

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(B) Given $f(x) = \frac{2^x}{2^x + \sqrt{2}}$.Consider $f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2/2^x}{2/2^x + \sqrt{2}} = \frac{2}{2 + \sqrt{2} \cdot 2^x} = \frac{\sqrt{2}}{\sqrt{2} + 2^x}$.Thus,$f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{2^x + \sqrt{2}} = \frac{2^x + \sqrt{2}}{2^x + \sqrt{2}} = 1$.We need to evaluate $S = \sum_{k=1}^{81} f\left(\frac{k}{82}ight) = f\left(\frac{1}{82}ight) + f\left(\frac{2}{82}ight) + \dots + f\left(\frac{81}{82}ight)$.Pairing terms $f\left(\frac{k}{82}ight) + f\left(1 - \frac{k}{82}ight) = f\left(\frac{k}{82}ight) + f\left(\frac{82-k}{82}ight) = 1$.There are $40$ such pairs (for $k=1$ to $40$) and the middle term $f\left(\frac{41}{82}ight) = f\left(\frac{1}{2}ight)$.$f\left(\frac{1}{2}ight) = \frac{2^{1/2}}{2^{1/2} + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} + \sqrt{2}} = \frac{1}{2}$.So,$S = 40 	imes 1 + \frac{1}{2} = \frac{81}{2}$.

If $f(x) = \frac{2^x}{2^x + \sqrt{2}}$,$x \in R$,then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to :

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