Functions (XII)

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Question

For, $x \in R, x \neq 0, x \neq 1$ , let $f_{0} (x) = \frac{1}{1 - x}$ and $f_{n + 1} (x) = f_{0} (f_{n} (x)), n = 0, 1, 2,$ ….Then the value of $f_{100} (3) + f_{1} (\frac{2}{3}) + f_{2} (\frac{3}{2})$ is equal to :

Moderate

Solution

$f_{0} (x) = \frac{1}{1 - x} f_{1} (x) = f_{0} (\frac{1}{1 - x}) = \frac{x - 1}{x} f_{2} (x) = x; f_{100} (3) + f_{1} (\frac{2}{3}) + f_{2} (\frac{3}{2}) = \frac{2}{3} + \frac{\frac{2}{3} - 1}{\frac{2}{3}} + \frac{3}{2} = \frac{5}{3}$

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