AMC12 2025 A

None of the answer choices on the official test (which asked for the number of possible functions $f(x)$) were correct, but choice E would be correct if this problem asked for the number of pairs of functions $(P(x), Q(x))$. Let $R(x) = \frac{P(x)}{Q(x)}$. Since $R(x) \leq 0$ on $[a, b]$ but not for values slightly less than $a$ or slightly more than $b$, $P(x) = 0$ at $x = a$ and $x = b$. Therefore, $P(x) = (x-a)(x-b)(x-r)$ for some $r \in \{1, 2, 3, 4, 5\}$. Since $R(x) \leq 0$ on $(c, d)$ but not at $x = c$ or $x = d$, $R(x)$ is not continuous at $x = c$ or $x = d$. Therefore, $R(x)$ must be undefined at $x = c$ and $x = d$, that is, $Q(x) = 0$ at $x = c$ and $x = d$. So $Q(x) = (x-c)(x-d)(x-s)$ for some $s \in \{1, 2, 3, 4, 5\}$. Therefore, $R(x) = \frac{(x-a)(x-b)(x-r)}{(x-c)(x-d)(x-s)}$. Notice that $\frac{(x-a)(x-b)}{(x-c)(x-d)} \leq 0$ only on $[a, b]$ and $(c, d)$. Therefore, $\frac{x-r}{x-s}$ must be nonnegative for all $x \notin \{a, b, c, d, r, s\}$. This only happens if $r = s$. Thus $R(x) = \frac{(x-a)(x-b)(x-r)}{(x-c)(x-d)(x-r)}$, which is the same as $\frac{(x-a)(x-b)}{(x-c)(x-d)}$ except that it is undefined at $x = r$. Thus $R(x)$ satisfies the desired property as long as $r \notin [a, b] \cup (c, d)$. Note that each quintuple $(a, b, c, d, r)$ defines a unique pair of functions $(P(x), Q(x))$. If $(a, b, c, d) = (1, 2, 3, 4)$, $r$ can be $3$, $4$, or $5$. If $(a, b, c, d) = (1, 2, 3, 5)$, $r$ can be $3$ or $5$. If $(a, b, c, d) = (1, 2, 4, 5)$, $r$ can be $3$, $4$, or $5$. If $(a, b, c, d) = (1, 3, 4, 5)$, $r$ can be $4$ or $5$. If $(a, b, c, d) = (2, 3, 4, 5)$, $r$ can be $1$, $4$, or $5$. Therefore, there are $\boxed{(\textbf{E})\ 13}$ possible pairs of functions $(P(x), Q(x))$.