2025 AMC 12A Problems/Problem 25
Polynomials
and
each have degree
and leading coefficient
, and their roots are all elements of
. The function
has the property that there exist real numbers
such that the set of all real numbers
such that
consists of the closed interval
together with the open interval
. How many functions
are possible?
We are told that
, where
and
are monic cubics with roots among
, and that
for some
.
---
- Step 1. Sign analysis
Since
changes sign only at
, we must have
- zeros at
and
,
- vertical asymptotes (poles) at
and
.
Thus the sign chart is:
\[(+)\ a\ (−)\ b\ (+)\ c\ (−)\ d\ (+)\] (Error compiling LaTeX. Unknown error_msg)
so
is positive outside
and nonpositive on those intervals.
---
- Step 2. General form
Because both
and
are monic cubics,
for some
(possibly equal to one of the existing roots or poles).
This ensures that
and the sign pattern is controlled by
.
To avoid adding extra sign changes, the extra factor
must not change sign, so
must either be one of the poles (
or
) or lie outside the intervals
.
---
- Step 3. Counting possibilities
We must choose four distinct numbers
from
:
giving these possible intervals:
Now count possible
for each:
| Case |
| Possible
| Count |
|------|-------------------|---------------|--------|
| 1 |
|
| 3 |
| 2 |
|
| 2 |
| 3 |
|
| 3 |
| 4 |
|
| 2 |
| 5 |
|
| 3 |
Total possibilities:
---
possible functions
, corresponding to answer choice
.