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

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==Solution 1 ==
==Solution 1 ==
Step 1: Sign Chart Analysis
From the given <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce:
From the given <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce:
* <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath> (since we transition from <imath>f > 0</imath> to <imath>f \le 0</imath> at <imath>a</imath> and from <imath>f \le 0</imath> to <imath>f > 0</imath> at <imath>b</imath>).
* <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath> since we transition from <imath>f > 0</imath> to <imath>f \le 0</imath> at <imath>a</imath> and from <imath>f \le 0</imath> to <imath>f > 0</imath> at <imath>b</imath>.
* <imath>c</imath> and <imath>d</imath> are poles (vertical asymptotes or holes) of <imath>f</imath> since on <imath>(c,d)</imath> we have <imath>f \le 0</imath>, at <imath>c^+</imath> and <imath>d^-</imath> the sign is negative, and immediately outside the interval <imath>f > 0</imath>.
* <imath>c</imath> and <imath>d</imath> are poles (vertical asymptotes or holes) of <imath>f</imath> since on <imath>(c,d)</imath> we have <imath>f \le 0</imath>, at <imath>c^+</imath> and <imath>d^-</imath> the sign is negative, and immediately outside the interval <imath>f > 0</imath>.


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\end{array}
\end{array}
</imath>
</imath>
Step 2: Structure of <imath>f(x)</imath>
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: \( p_3 \) and \( q_3 \)
* \( f_1(x) \) completely satisfies the sign chart for \( f(x) \). We can conclude that \( f_2(x) \) must be positive on every interval; otherwise, it would introduce an extra sign change. This forces the condition \( p_3 = q_3 \), otherwise it would be negative on certain intervals. Let \( p_3 = q_3 = t \).
* Furthermore, this means <imath>f(t)</imath> is undefined, forcing a hole at <imath>x=t</imath>. Thus, 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 \( (c, d) \).
Thus, we have cannot have <imath>t = a</imath> or <imath>b</imath>, but we can have <imath>t = c</imath> or <imath>d</imath> or any number outside \( [a,b] \cup (c,d) \).
Step 4: <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, and <imath>t</imath>
Given <imath>a < b < c < d</imath> and the condition that <imath>\{x: f(x) \leq 0\} = [a,b] \cup (c,d)</imath>, the number of ways to choose <imath>a,b,c,d</imath> from <imath>\{1,2,3,4,5\}</imath> is:<cmath>\binom{5}{4} = 5 </cmath>
The five cases are:
    <imath>[1,2]\cup (3,4)</imath>; <imath>[1,2]\cup (3,5)</imath>; <imath>[1,2]\cup (4,5)</imath>; <imath>[1,3]\cup (4,5)</imath>; <imath>[2,3]\cup (4,5)</imath>
* Case 1: <imath>[1,2]\cup (3,4)</imath>, <imath>t</imath> can be 3, 4, or 5
* Case 2: <imath>[1,2]\cup (3,5)</imath>, <imath>t</imath> can be 3 or 5
* Case 3: <imath>[1,2]\cup (4,5)</imath>, <imath>t</imath> can be 3, 4, or 5.
* Case 4: <imath>[1,3]\cup (4,5)</imath>, <imath>t</imath> can be 4 or 5.
* Case 5: <imath>[2,3]\cup (4,5)</imath>, <imath>t</imath> can be 1, 4, or 5.
Summing up all possible cases from Step 4 gives us <imath>3 + 2 + 3 + 2 + 3 = 13</imath> possible functions <imath>f(x)</imath>.
Thus the answer is <imath>\boxed{E}</imath>.


- Victor Zhang
- Victor Zhang

Revision as of 03:01, 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 poles (vertical asymptotes or holes) of $f$ since on $(c,d)$ 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: \( p_3 \) and \( q_3 \)

  • \( f_1(x) \) completely satisfies the sign chart for \( f(x) \). We can conclude that \( f_2(x) \) must be positive on every interval; otherwise, it would introduce an extra sign change. This forces the condition \( p_3 = q_3 \), otherwise it would be negative on certain intervals. Let \( p_3 = q_3 = t \).
  • Furthermore, this means $f(t)$ is undefined, forcing a hole at $x=t$. Thus, 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 \( (c, d) \).

Thus, we have cannot have $t = a$ or $b$, but we can have $t = c$ or $d$ or any number outside \( [a,b] \cup (c,d) \).

Step 4: $a$, $b$, $c$, $d$, and $t$ 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:\[\binom{5}{4} = 5\] The five cases are: 

   $[1,2]\cup (3,4)$; $[1,2]\cup (3,5)$; $[1,2]\cup (4,5)$; $[1,3]\cup (4,5)$; $[2,3]\cup (4,5)$
  • 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 from Step 4 gives us $3 + 2 + 3 + 2 + 3 = 13$ 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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