Art of Problem Solving

2025 AMC 12A Problems/Problem 25

Polynomials $P(x)$ and $Q(x)$ each have degree $3$ and leading coefficient $1$, and their roots are all elements of $\{1,2,3,4,5\}$. The function $f(x) = \tfrac{P(x)}{Q(x)}$ has the property that there exist real numbers $a < b < c < d$ such that the set of all real numbers $x$ such that $f(x) \leq 0$ consists of the closed interval $[a,b]$ together with the open interval $(c,d)$. How many functions $f(x)$ are possible?

Solution 1

From the given $f(x) \le 0$ on $[a,b] \cup (c,d)$ and $f(x) > 0$ elsewhere, we deduce:

  • $a$ and $b$ are zeros of $f$ (since we transition from $f > 0$ to $f \le 0$ at $a$ and from $f \le 0$ to $f > 0$ at $b$).
  • $c$ and $d$ are poles (vertical asymptotes or holes) of $f$ since on $(c,d)$ we have $f \le 0$, at $c^+$ and $d^-$ the sign is negative, and immediately outside the interval $f > 0$.

Thus the sign pattern is:

$\textstyle \begin{array}{cccccccccc}  & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\ f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & + \end{array}$

- Victor Zhang

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0