2025 AMC 12A 第 25 题

Question

多项式

P(x)

和

Q(x)

都是

3

次，首项系数都是

1

，并且它们的根都属于

\{1, 2, 3, 4, 5\}

。函数

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

具有如下性质：存在实数

a \lt b \lt c \lt d

，使得所有满足

f(x) \le 0

的实数

x

的集合由闭区间

[a, b]

与开区间

(c, d)

组成。可能有多少个函数

f(x)

？

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?

7

9

11

12

13

答案：E

知识点：函数多项式分类讨论

难度评级：2540

小提示：

闭区间的端点必须是

P

的零点，开区间的端点必须是

Q

的零点

The closed endpoints must be zeros of

P,

and the open endpoints must be zeros of

Q

大提示：

这四个端点因子固定后，分子和分母剩下的因子必须相消；要数的是不同的函数，而不是多项式对

After those four endpoint factors are fixed, the remaining numerator and denominator factors must cancel; count distinct functions, not polynomial pairs

解答：

本题已作废：按题面作答，正确答案不在选项中。下面给出计数。

对于

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

，闭区间端点

a, b

必须是

P

的零点，且在这些点

Q\ne0

，而

c,d

必须是极点。因此符号的变化必须与下面这个函数一致：

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

P

中未使用的第三个因子必须与

Q

中未使用的第三个因子相同；否则会多出零点、极点或符号变化。

如果这个公共因子是

a,b,c,d

之外的第五个值，它不能在任一区间内部形成空点。在五个数从小到大的排列中，它只能位于

a

之前、

b

与

c

之间，或

d

之后，因此给出

3

个函数。

另一种可能是公共因子等于

c

或

d

。对

a,b,c,d

的每一种选择（共

\binom54=5

种），这两个多项式对化简为同一个公式

g

，且定义域相同（都排除

c,d

），所以只定义一个函数。这再给出

5

个函数，按题面字面理解总数为

3+5=8

。

如果在后五种情形中把两个多项式对分别计数，就会得到

3+2\cdot5=13

，即临时答案 (E)，但这没有回答题目所问的函数个数。因此印刷的选项都不正确。

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

or

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.

2025 AMC 12A 第 25 题

答案：E

难度评级：2540

小提示：

大提示：

解答：

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