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Question 17

Given $$f(x) = \dfrac{2025x}{x+1}$$ where $$x \neq -1$$. Then the value of $$x$$ for which $$f(f(x)) = (2025)^2$$ is ------.


Correct Answer: -1/2025

Solution

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$$f(x) = \dfrac{2025x}{x+1}$$

$$f(f(x)) = \dfrac{2025\left(\dfrac{2025x}{x+1}\right)}{\dfrac{2025x}{x+1}+1} = \dfrac{2025^2x}{2025x+x+1} = \dfrac{2025^2x}{2026x+1} $$

Given

$$ 2025^2 = \dfrac{2025^2x}{2026x+1} \implies 2026x+1 = x \ \text{or} \ x = \dfrac{-1}{2025}$$

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