Fermat 2020 Q23

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Problem Statement

There are real numbers a and b for which the function f has the properties that \(f(x) = ax + b\) for all real numbers \(x\), and \(f(bx + a) = x\) for all real numbers \(x\). What is the value of \(a + b\)?

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Solution

$$a, b, x \in \mathbb{R}$$ $$f(x) = ax + b \;\;\; \forall \; x \in \mathbb{R}$$ $$f(bx + a) = x \;\;\; \forall \; x \in \mathbb{R}$$ $$\implies f(bx + a) = a(bx + a) + b = x \;\;\; \forall \; x \in \mathbb{R}$$ $$abx + a^2 + b - x = 0 \;\;\; \forall \; x \in \mathbb{R}$$ $$x(ab - 1) + a^2 + b = 0 \;\;\; \forall \; x \in \mathbb{R}$$ $$\text{When } x = 0 \implies 0(ab - 1) + a^2 + b = 0 \implies a^2 + b = 0$$ $$x(ab - 1) = 0 \;\;\; \forall \; x \in \mathbb{R}$$ $$\text{When } x \ne 0, ab - 1 = 0 \implies ab = 1 \implies a = \dfrac{1}{b}, b \ne 0$$ $$\text{Sub } a = \dfrac{1}{b} \text{ into } abx + a^2 + b - x = 0 \text{:}$$ $$(\dfrac{1}{b})(b)(x) + (\dfrac{1}{b})^2 + b - x = 0$$ $$\dfrac{1}{b^2} + b = 0$$ $$1 + b^3 = 0$$ $$b = -1$$ $$a = \dfrac{1}{b} = \dfrac{1}{-1} = -1$$ $$\therefore a + b = -2$$