2018/19 AHSMC Part 1 Problem 3

Problem 3 of 16

Let $f$ be a function such that $f\bigl(\dfrac{x+1}{3}\bigr) = x^2 + x + 3$ for any real number $x$ . The sum of all real numbers $a$ such that $f\bigl(\dfrac{a}{9}\bigr) = 10$ is

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If you set $y = \dfrac{x+1}{3}$ then $x = 3y - 1$ , hence $f(y) = (3y - 1)^2 + (3y - 1) + 3 = 9y^2 - 3y + 3$ . Now, $f\bigl(\dfrac{a}{9}\bigr) = 10$ is equivalent to $a^2 - 3a - 63 = 0$ . This equation has real roots having the sum $3$ . The answer is (C).

University of Alberta

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