2025 AMC 12A Problems/Problem 25

We begin by analyzing the sign chart. From the condition $f(x) \le 0$ on $[a,b] \cup (c,d)$ and $f(x) > 0$ elsewhere, we deduce the following:

- $a$ and $b$ are zeros of $f$ , since we transition from $f>0$ to $f\le0$ at $a$ , and from $f\le0$ to $f>0$ at $b$ . - $c$ and $d$ are poles or holes of $f$ , since on $(c,d)$ we have $f\le0$ , while immediately outside that interval $f>0$ .

Thus, the sign pattern of $f(x)$ is $\begin{array}{cccccccccc} & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\ f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & + \end{array}$

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- Step 2. Structure of $f(x)$ **

According to the given conditions, we can express $f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.$ Combining this with the sign analysis, we can write $f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.$ We can separate this into two factors: $f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.$ Define $f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \qquad f_2(x) = \frac{x-p_3}{x-q_3}.$

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- Step 3. Determining $p_3$ and $q_3$ **

By analyzing the sign changes of $f_1(x)$ , we see that it already matches the required sign pattern for $f(x)$ . Therefore, $f_2(x)$ must remain positive on every interval; otherwise, it would introduce extra sign changes.

This means $p_3 = q_3$ , since if $p_3 \ne q_3$ , the factor $\frac{x-p_3}{x-q_3}$ would change sign between $p_3$ and $q_3$ . Let $p_3 = q_3 = t$ .

Furthermore: - $t$ cannot lie within $[a,b] \cup (c,d)$ , since that would alter the set $\{x : f(x) \le 0\}$ . - $t$ cannot be equal to $a$ or $b$ , since that would make those endpoints holes (discontinuities), contradicting that $[a,b]$ is a *closed* interval.

Hence, $t$ must either be equal to $c$ or $d$ , or lie outside $[a,b] \cup (c,d)$ .

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- Step 4. Possible values of $a,b,c,d,t$ **

Given $a<b<c<d$ and $\{x : f(x) \le 0\} = [a,b] \cup (c,d)$ , we choose four distinct numbers from $\{1,2,3,4,5\}$ for $a,b,c,d$ : $\binom{5}{4} = 5.$ The five possible cases are: $[1,2]\cup(3,4), \quad [1,2]\cup(3,5), \quad [1,2]\cup(4,5), \quad [1,3]\cup(4,5), \quad [2,3]\cup(4,5).$

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- Case 1:** $[1,2]\cup(3,4)$

$t$ can be $3$ , $4$ , or $5$ , giving 3 functions: $f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}, \quad f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}, \quad f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}.$

- Case 2:** $[1,2]\cup(3,5)$

$t$ can be $3$ or $5$ , giving 2 functions: $f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}, \quad f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}.$

- Case 3:** $[1,2]\cup(4,5)$

$t$ can be $3$ , $4$ , or $5$ , giving 3 functions: $f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}, \quad f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}, \quad f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}.$

- Case 4:** $[1,3]\cup(4,5)$

$t$ can be $4$ or $5$ , giving 2 functions: $f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.$

- Case 5:** $[2,3]\cup(4,5)$

$t$ can be $1$ , $4$ , or $5$ , giving 3 functions: $f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}, \quad f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.$

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- Step 5. Final Answer**

Summing all possible cases: $3 + 2 + 3 + 2 + 3 = 13.$ Thus, there are $\boxed{13}$ possible functions $f(x)$ , corresponding to choice $\boxed{E}$ .

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2025 AMC 12A Problems/Problem 25