2025 AMC 12A Problem 25

Problem 25 of 25HarderAlgebra

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) = \dfrac{P(x)}{Q(x)}

has the property that there exist real numbers

a \lt b \lt c \lt d

such that the set of all real numbers

x

such that

f(x) \le 0

consists of the closed interval

[a, b]

together with the open interval

(c, d).

How many functions

f(x)

are possible?

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Solution

This problem was voided: as written, its answer is not among the choices. Here is the count. For

\{f \le 0\} = [a, b] \cup (c, d),

the endpoints

a, b

of the closed interval must be zeros of

P

at which

Q\ne0,

while

c,d

must be poles. The required sign pattern is therefore that of

g(x)=\frac{(x-a)(x-b)}{(x-c)(x-d)}.

The unused third factor of

P

must match the unused third factor of

Q;

otherwise there would be an extra zero, pole, or sign change. If the common factor is the fifth value not among

a,b,c,d,

it cannot make a hole inside either interval. In the ordered list of five values it may therefore occur before

a,

between

b

and

c,

or after

d,

giving

3

functions. Alternatively, the common factor can equal

c

d.

For each of the

\binom54=5

choices of

a,b,c,d,

these two polynomial pairs simplify to the same formula

g

and have the same domain (both omit

c,d

), so they define only one function. This gives

5

more functions, for a literal total of

3+5=8.

Counting the two polynomial pairs separately in each of the last five cases gives

3+2\cdot5=13,

the provisional answer (E), but that does not answer the stated question about functions. Thus none of the printed choices is correct.