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

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According to the given conditions, \( f(x) \) can be expressed as:
According to the given conditions, \( f(x) \) can be expressed as:
<cmath>f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.</cmath>
<cmath>f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.</cmath>
Combining the sign analysis from Step 1, we can rewrite this expression as:
Combining the sign analysis from Step 1, we can rewrite this expression as:
<cmath>f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.</cmath>
<cmath>f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.</cmath>
Furthermore, we can break down the expression into two parts:
Furthermore, we can break down the expression into two parts:
<cmath>f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.</cmath>
<cmath>f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.</cmath>
Defining \( f(x) = f_1(x) \cdot f_2(x) \), we have:
Defining \( f(x) = f_1(x) \cdot f_2(x) \), we have:
<cmath>f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.</cmath>
<cmath>f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.</cmath>
Step 3: Analysis of \( p_3 \) and \( q_3 \)
Step 3: Analysis of \( p_3 \) and \( q_3 \)


Revision as of 04:33, 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?

Solution 1

Step 1: Sign Chart Analysis

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 vertical asymptotes or holes of $f$ since $f(x)$ is undefined at $c$ and $d$. 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:

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

Step 2: Structure of $f(x)$

According to the given conditions, \( f(x) \) can be expressed as: \[f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.\] Combining the sign analysis from Step 1, we can rewrite this expression as: \[f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.\] Furthermore, we can break down the expression into two parts: \[f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.\] Defining \( f(x) = f_1(x) \cdot f_2(x) \), we have: \[f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.\] Step 3: Analysis of \( p_3 \) and \( q_3 \)

  • \( f_1(x) \) completely satisfies the sign chart for \( f(x) \). We can conclude that \( f_2(x) \) must be positive in every interval; otherwise, \( f_2(x) \) would be negative in interval \( p_3, q_3 \), which introduces extra sign changes. This forces the condition \( p_3 = q_3 \).Let \( p_3 = q_3 = t \).
  • Furthermore, \( t \) cannot lie within \( [a,b] \cup (c,d) \). If it did, it would contradict the given condition that the set \( \{x : f(x) \le 0\} \) consists of the closed interval \( [a, b] \) and the open interval \( (c, d) \).
  • Additionally, \( t \) cannot equal to \( a \) or \( b \). If it were, that point would be a hole (a point of discontinuity) in the graph of \( f(x) \), which contradicts the closed interval \( [a, b] \).

Thus, \( t \) can only equal to \( c \) or \( d \), or lie in an interval outside of \( [a,b] \cup (c,d) \).

Step 4: Ways to choose $a,b,c,d$ from $\{1,2,3,4,5\}$

Given $a < b < c < d$ and the condition that $\{x: f(x) \leq 0\} = [a,b] \cup (c,d)$, the number of ways to choose $a,b,c,d$ from $\{1,2,3,4,5\}$ is:

\[\binom{5}{4} = 5\]

Step 5: The possible value of $a$, $b$, $c$, $d$, and $t$ for the five cases are:

  • Case 1: $[1,2]\cup (3,4)$, $t$ can be 3, 4, or 5
  • Case 2: $[1,2]\cup (3,5)$, $t$ can be 3 or 5
  • Case 3: $[1,2]\cup (4,5)$, $t$ can be 3, 4, or 5.
  • Case 4: $[1,3]\cup (4,5)$, $t$ can be 4 or 5.
  • Case 5: $[2,3]\cup (4,5)$, $t$ can be 1, 4, or 5.

Summing up all possible cases, $3 + 2 + 3 + 2 + 3 = 13$ possible functions $f(x)$.

Thus the answer is $\boxed{E}$.


- Victor Zhang

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

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