If , then is equal to [2025]

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Solution

Q.
If $f (x) = \frac{2^{x}}{2^{x} + \sqrt{2}}, x \in R$ , then $\sum_{k = 1}^{81} f (\frac{k}{82})$ is equal to [2025]

1 $\frac{81}{2}$

2 82

3 41

4 $81 \sqrt{2}$

Ans.
(1)

We have, $f (x) = \frac{2^{x}}{2^{x} + \sqrt{2}}, x \in R$

$\Rightarrow f (x) + f (1 - x) = \frac{2^{x}}{2^{x} + \sqrt{2}} + \frac{2^{1 - x}}{2^{1 - x} + \sqrt{2}}$

$= \frac{2^{x}}{2^{x} + \sqrt{2}} + \frac{2}{2 + \sqrt{2} \cdot 2^{x}} = 1$

Now, $S = \sum_{k = 1}^{81} f (\frac{k}{82}) = f (\frac{1}{82}) + f (\frac{2}{82}) + . . . + f (\frac{81}{82})$

$\Rightarrow S = f (\frac{1}{82}) + . . . + f (1 - \frac{2}{82}) + f (1 - \frac{1}{82})$

$= f (\frac{1}{82}) + f (1 - \frac{1}{82}) + f (\frac{2}{82}) + f (1 - \frac{2}{82}) + . . . + upto 40 + f (\frac{41}{82})$

$= \underset{40 times}{\underset{⏟}{(1 + 1 + 1 + . . . + 1)}} + \frac{2^{1 / 2}}{2^{1 / 2} + \sqrt{2}} = 40 + \frac{1}{2} = \frac{81}{2}$ .

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