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

Revision as of 04:17, 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: Possible value of $a$ , $b$ , $c$ , $d$ , 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: $\binom{5}{4} = 5$ \\ 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

@@ Line 7: / Line 7: @@
 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>c</imath> and <imath>d</imath> are 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 vertical asymptotes or holes of <imath>f</imath> since <imath>f(x)</imath> is undefined at <imath>c</imath> and <imath>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>.
 Thus the sign pattern is:
@@ Line 18: / Line 18: @@
 </imath>
-Step 2: Structure of <imath>f(x)</imath>
+Step 2: Structure of <imath>f(x)</imath>\\
 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>
@@ Line 32: / Line 32: @@
-Step 3: Analysis of \( p_3 \) and \( q_3 \)
+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, it would introduce an extra sign change. This forces the condition \( p_3 = q_3 \), otherwise  \( f_2(x) \) would be negative in interval \( p_3, q_3 \). Let \( p_3 = q_3 = t \).
+* \( f_1(x) \) 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 be 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 \( [a, b] \).
+* 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 be equal to \( c \) or \( d \), or lie in an interval outside of \( [a,b] \cup (c,d) \).
+Thus, \( t \) can only equal to \( c \) or \( d \), or lie in an interval outside of \( [a,b] \cup (c,d) \).\\
 Step 4: Possible value of <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>
+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 possible value of <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, and <imath>t</imath> for thefive cases are:
+Step 5: The possible value of <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, and <imath>t</imath> for 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.
+* Case 5: <imath>[2,3]\cup (4,5)</imath>, <imath>t</imath> can be 1, 4, or 5.\\
 Summing up all possible cases, <imath>3 + 2 + 3 + 2 + 3 = 13</imath> possible functions <imath>f(x)</imath>.

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

Revision as of 04:17, 7 November 2025

Solution 1

Video Solution 1 by OmegaLearn