2025 AMC 12A Problems/Problem 25: Difference between revisions

Revision as of 02:27, 7 November 2025

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?

We are told that $f(x) = \frac{P(x)}{Q(x)}$ , where $P$ and $Q$ are monic cubics with roots among $\{1,2,3,4,5\}$ , and that $\{x : f(x) \le 0\} = [a,b] \cup (c,d)$ for some $a < b < c < d$ .

---

1. 1. Step 1. Sign analysis

Since $f(x)$ changes sign only at $a,b,c,d$ , we must have - zeros at $a$ and $b$ , - vertical asymptotes (poles) at $c$ and $d$ .

Thus the sign chart is:

\[(+)\ a\ (−)\ b\ (+)\ c\ (−)\ d\ (+)\] (Error compiling LaTeX. Unknown error_msg)

so $f(x)$ is positive outside $[a,b]\cup(c,d)$ and nonpositive on those intervals.

---

1. 1. Step 2. General form

Because both $P$ and $Q$ are monic cubics, $f(x) = \frac{(x-a)(x-b)(x-t)}{(x-c)(x-d)(x-t)}$ for some $t$ (possibly equal to one of the existing roots or poles). This ensures that $\deg P = \deg Q = 3$ and the sign pattern is controlled by $a,b,c,d$ .

To avoid adding extra sign changes, the extra factor $\frac{x-t}{x-t}$ must not change sign, so $t$ must either be one of the poles ( $c$ or $d$ ) or lie outside the intervals $[a,b]\cup(c,d)$ .

---

1. 1. Step 3. Counting possibilities

We must choose four distinct numbers $a<b<c<d$ from $\{1,2,3,4,5\}$ : $\binom{5}{4} = 5$ giving these possible intervals: $[1,2]\cup(3,4),\ [1,2]\cup(3,5),\ [1,2]\cup(4,5),\ [1,3]\cup(4,5),\ [2,3]\cup(4,5).$

Now count possible $t$ for each:

| Case | $[a,b]\cup(c,d)$ | Possible $t$ | Count | |------|-------------------|---------------|--------| | 1 | $[1,2]\cup(3,4)$ | $3,4,5$ | 3 | | 2 | $[1,2]\cup(3,5)$ | $3,5$ | 2 | | 3 | $[1,2]\cup(4,5)$ | $3,4,5$ | 3 | | 4 | $[1,3]\cup(4,5)$ | $4,5$ | 2 | | 5 | $[2,3]\cup(4,5)$ | $1,4,5$ | 3 |

Total possibilities: $3 + 2 + 3 + 2 + 3 = 13.$

---

$\boxed{13}$ possible functions $f(x)$ , corresponding to answer choice $\boxed{E}$ .

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

@@ Line 1: / Line 1: @@
 Polynomials <imath>P(x)</imath> and <imath>Q(x)</imath> each have degree <imath>3</imath> and leading coefficient <imath>1</imath>, and their roots are all elements of <imath>\{1,2,3,4,5\}</imath>. The function <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath> has the property that there exist real numbers <imath>a < b < c < d</imath> such that the set of all real numbers <imath>x</imath> such that <imath>f(x) \leq 0</imath> consists of the closed interval <imath>[a,b]</imath> together with the open interval <imath>(c,d)</imath>. How many functions <imath>f(x)</imath> are possible?
-==Solution 1 ==
+We are told that <imath>f(x) = \frac{P(x)}{Q(x)}</imath>, where <imath>P</imath> and <imath>Q</imath> are monic cubics with roots among <imath>\{1,2,3,4,5\}</imath>, and that
-We begin by analyzing the sign chart. From the condition <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce the following:
-- <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath>, since we transition from <imath>f>0</imath> to <imath>f\le0</imath> at <imath>a</imath>, and from <imath>f\le0</imath> to <imath>f>0</imath> at <imath>b</imath>.
-- <imath>c</imath> and <imath>d</imath> are poles or holes of <imath>f</imath>, since on <imath>(c,d)</imath> we have <imath>f\le0</imath>, while immediately outside that interval <imath>f>0</imath>.
-Thus, the sign pattern of <imath>f(x)</imath> is
 <cmath>
-\begin{array}{cccccccccc}
+\{x : f(x) \le 0\} = [a,b] \cup (c,d)
- & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\
-f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & +
-\end{array}
 </cmath>
+for some <imath>a < b < c < d</imath>.
 ---
-**Step 2. Structure of <imath>f(x)</imath>**
+### Step 1. Sign analysis
+Since <imath>f(x)</imath> changes sign only at <imath>a,b,c,d</imath>, we must have
+- zeros at <imath>a</imath> and <imath>b</imath>,
+- vertical asymptotes (poles) at <imath>c</imath> and <imath>d</imath>.
-According to the given conditions, we can express
+Thus the sign chart is:
-<cmath>
-f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.
-</cmath>
-Combining this with the sign analysis, we can write
-<cmath>
-f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.
-</cmath>
-We can separate this into two factors:
-<cmath>
-f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.
-</cmath>
-Define
 <cmath>
-f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \qquad f_2(x) = \frac{x-p_3}{x-q_3}.
+(+)\ a\ (−)\ b\ (+)\ c\ (−)\ d\ (+)
 </cmath>
+so <imath>f(x)</imath> is positive outside <imath>[a,b]\cup(c,d)</imath> and nonpositive on those intervals.
 ---
-**Step 3. Determining <imath>p_3</imath> and <imath>q_3</imath>**
+### Step 2. General form
+Because both <imath>P</imath> and <imath>Q</imath> are monic cubics,
-By analyzing the sign changes of <imath>f_1(x)</imath>, we see that it already matches the required sign pattern for <imath>f(x)</imath>. Therefore, <imath>f_2(x)</imath> must remain positive on every interval; otherwise, it would introduce extra sign changes.
-This means <imath>p_3 = q_3</imath>, since if <imath>p_3 \ne q_3</imath>, the factor <imath>\frac{x-p_3}{x-q_3}</imath> would change sign between <imath>p_3</imath> and <imath>q_3</imath>.
-Let <imath>p_3 = q_3 = t</imath>.
-Furthermore:
-- <imath>t</imath> cannot lie within <imath>[a,b] \cup (c,d)</imath>, since that would alter the set <imath>\{x : f(x) \le 0\}</imath>.
-- <imath>t</imath> cannot be equal to <imath>a</imath> or <imath>b</imath>, since that would make those endpoints holes (discontinuities), contradicting that <imath>[a,b]</imath> is a *closed* interval.
-Hence, <imath>t</imath> must either be equal to <imath>c</imath> or <imath>d</imath>, or lie outside <imath>[a,b] \cup (c,d)</imath>.
----
-**Step 4. Possible values of <imath>a,b,c,d,t</imath>**
-Given <imath>a<b<c<d</imath> and <imath>\{x : f(x) \le 0\} = [a,b] \cup (c,d)</imath>, we choose four distinct numbers from <imath>\{1,2,3,4,5\}</imath> for <imath>a,b,c,d</imath>:
 <cmath>
-\binom{5}{4} = 5.
+f(x) = \frac{(x-a)(x-b)(x-t)}{(x-c)(x-d)(x-t)}
-</cmath>
-The five possible cases are:
-<cmath>
-[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).
 </cmath>
+for some <imath>t</imath> (possibly equal to one of the existing roots or poles).
+This ensures that <imath>\deg P = \deg Q = 3</imath> and the sign pattern is controlled by <imath>a,b,c,d</imath>.
+To avoid adding extra sign changes, the extra factor <imath>\frac{x-t}{x-t}</imath> must not change sign, so <imath>t</imath> must either be one of the poles (<imath>c</imath> or <imath>d</imath>) or lie outside the intervals <imath>[a,b]\cup(c,d)</imath>.
 ---
-**Case 1:** <imath>[1,2]\cup(3,4)</imath>
+### Step 3. Counting possibilities
-<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+We must choose four distinct numbers <imath>a<b<c<d</imath> from <imath>\{1,2,3,4,5\}</imath>:
 <cmath>
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}, \quad
+\binom{5}{4} = 5
-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)}.
 </cmath>
+giving these possible intervals:
-**Case 2:** <imath>[1,2]\cup(3,5)</imath>
-<imath>t</imath> can be <imath>3</imath> or <imath>5</imath>, giving 2 functions:
 <cmath>
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}, \quad
+[1,2]\cup(3,4),\ [1,2]\cup(3,5),\ [1,2]\cup(4,5),\ [1,3]\cup(4,5),\ [2,3]\cup(4,5).
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}.
 </cmath>
-**Case 3:** <imath>[1,2]\cup(4,5)</imath>
+Now count possible <imath>t</imath> for each:
-<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
-<cmath>
-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)}.
-</cmath>
-**Case 4:** <imath>[1,3]\cup(4,5)</imath>
+| Case | <imath>[a,b]\cup(c,d)</imath> | Possible <imath>t</imath> | Count |
-<imath>t</imath> can be <imath>4</imath> or <imath>5</imath>, giving 2 functions:
+|------|-------------------|---------------|--------|
-<cmath>
+| 1 | <imath>[1,2]\cup(3,4)</imath> | <imath>3,4,5</imath> | 3 |
-f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
+| 2 | <imath>[1,2]\cup(3,5)</imath> | <imath>3,5</imath> | 2 |
-f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
+| 3 | <imath>[1,2]\cup(4,5)</imath> | <imath>3,4,5</imath> | 3 |
-</cmath>
+| 4 | <imath>[1,3]\cup(4,5)</imath> | <imath>4,5</imath> | 2 |
+| 5 | <imath>[2,3]\cup(4,5)</imath> | <imath>1,4,5</imath> | 3 |
-**Case 5:** <imath>[2,3]\cup(4,5)</imath>
+Total possibilities:
-<imath>t</imath> can be <imath>1</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
 <cmath>
-f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}, \quad
++ 2 + 3 + 2 + 3 = 13.
-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)}.
 </cmath>
 ---
-**Step 5. Final Answer**
+<imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, corresponding to answer choice <imath>\boxed{E}</imath>.
-Summing all possible cases:
-<cmath>
-+ 2 + 3 + 2 + 3 = 13.
-</cmath>
-Thus, there are <imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, corresponding to choice <imath>\boxed{E}</imath>.
--Victor Zhang
 ==Video Solution 1 by OmegaLearn==
 https://youtu.be/SPbTyq3Dz_0

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2025 AMC 12A Problems/Problem 25: Difference between revisions

Revision as of 02:27, 7 November 2025

Video Solution 1 by OmegaLearn