Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

← Older editNewer edit →
Tiankaizhang (talk | contribs)
editor
42 edits
Tiankaizhang (talk | contribs)
editor
42 edits
Line 3: Line 3:
==Solution 1 ==
==Solution 1 ==
From the structure <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 structure <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce:
\begin{itemize}
* <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>).
    \item <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 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>).
    \item <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 \le 0</imath>, at <imath>c^+</imath> and <imath>d^-</imath> the sign is negative, and immediately outside the interval <imath>f > 0</imath>).
\end{itemize}


Thus the sign pattern is:\\
Thus the sign pattern is:


<imath>\textstyle
<imath>\textstyle

Revision as of 02:49, 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

From the structure $f(x) \le 0$ 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 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:

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

- Victor Zhang

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2025_AMC_12A_Problems/Problem_25&oldid=266786"