If f ( x ) = 2 2 x 2 2 x + 2 , x R , then f ( 1 2023 ) + f ( 2 2023 ) + . . . . + f ( 2022 2023 ) is equal to [2023]

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Solution

Q.
If $f (x) = \frac{2^{2 x}}{2^{2 x} + 2}, x \in R,$ then

$f (\frac{1}{2023}) + f (\frac{2}{2023}) + . . . . + f (\frac{2022}{2023})$ is equal to [2023]

1 1011

2 2010

3 1010

4 2011

Ans.
(1)

Given, $f (x) = \frac{2^{2 x}}{2^{2 x} + 2} = \frac{4^{x}}{4^{x} + 2}$

$f (1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2}$

Now, $f (x) + f (1 - x) = \frac{4^{x}}{4^{x} + 2} + \frac{4^{(1 - x)}}{4^{(1 - x)} + 2}$

$f (x) + f (1 - x) = \frac{4^{x}}{4^{x} + 2} + \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{4^{x}}{4^{x} + 2} + \frac{4}{4 + 2 (4^{x})}$

$= \frac{4^{x}}{4^{x} + 2} + \frac{2}{2 + 4^{x}} = 1$

$f (\frac{1}{2023}) + f (\frac{2}{2023}) + \dots + f (\frac{2022}{2023})$

$= f (\frac{1}{2023}) + f (\frac{2}{2023}) + f (\frac{3}{2023}) + \dots + f (1 - \frac{3}{2023}) + f (1 - \frac{2}{2023}) + f (1 - \frac{1}{2023})$

$= 1 + 1 + 1 + \dots + 1 (1011 times) = 1011$

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