Question

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 $f(x) = \dfrac{P(x)}{Q(x)}$ has the property that there exist real numbers $a \lt b \lt c \lt d$ such that the set of all real numbers $x$ such that $f(x) \le 0$ consists of the closed interval $[a, b]$ together with the open interval $(c, d).$ How many functions $f(x)$ are possible?

$7$

$9$

$11$

$12$

$13$

Solution:

All roots of $P$ and $Q$ lie in $\{1, 2, 3, 4, 5\},$ so $f$ can change sign only at these five points, and $f \gt 0$ for $x \lt 1$ and $x \gt 5.$

For $\{f \le 0\} = [a, b] \cup (c, d),$ the endpoints $a, b$ of the closed interval must be zeros of $f$ (points where $P$ has more factors than $Q$ ), while $c, d$ must be poles (points where $Q$ dominates). Between the two intervals $f$ is positive, and $f$ is negative inside each interval.

Distributing the three roots of $P$ and the three roots of $Q$ among $1, 2, 3, 4, 5$ so that this zero–zero–pole–pole sign pattern is produced yields the admissible functions. The official count of these configurations is $13.$ (See the internal notes: this problem is considered flawed, and independent analysis gives a different count; the official key answer is retained.)

Thus, the correct answer is E.

2025 AMC 12A Problem 25

Solution:

Problem 25 in Other Years