Let . Then the value of is equal to [2025]

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Solution

Q.
Let $f (x) = \frac{2^{x + 2} + 16}{2^{2 x + 1} + 2^{x + 4} + 32}$ . Then the value of $8 (f (\frac{1}{15}) + f (\frac{2}{15}) + . . . + f (\frac{59}{15}))$ is equal to [2025]

1 118

2 102

3 92

4 108

Ans.
(1)

$f (x) = \frac{2^{x + 2} + 16}{2^{2 x + 1} + 2^{x + 4} + 32}$

$= \frac{2^{2} (2^{x} + 4)}{2 [{(2^{x})}^{2} + 8 \times 2^{x} + 16]} = \frac{2 (2^{x} + 4)}{{(2^{x} + 4)}^{2}} = \frac{2}{2^{x} + 4}$

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

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

$∴ f (\frac{1}{15}) + f (\frac{59}{15}) = f (\frac{2}{15}) + f (\frac{58}{15})$

$= f (\frac{3}{15}) + f (\frac{57}{15}) = . . . = f (\frac{29}{15}) + f (\frac{31}{15}) = \frac{1}{2}$

$f (\frac{30}{15}) = f (2) = \frac{2}{4 + 4} = \frac{1}{4}$

$8 [f (\frac{1}{15}) + f (\frac{2}{15}) + . . . + f (\frac{59}{15})]$

$= 8 [\frac{1}{2} + \frac{1}{2} + . . . upto (29 terms) + \frac{1}{4}] = 8 (\frac{29}{2} + \frac{1}{4}) = 118$

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