1987 AMC 12 第 24 题

先试着解答 1987 AMC 12 第 24 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1987 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

24.

有多少个多项式函数 ff 的次数 ≥1\ge1,且满足 f(x2)=[f(x)]2=f(f(x))? f(x^2)=[f(x)]^2=f(f(x))\text{?}
How many polynomial functions ff of degree ≥1\ge1 satisfy f(x2)=[f(x)]2=f(f(x))? f(x^2)=[f(x)]^2=f(f(x))?
00
11
22
有限多个,但多于 22 个
finitely many but more than 22
无限多个
infinitely many
答案:B
知识点:polynomial degreeleading coefficient函数方程
小提示:
比较这三个多项式的次数和首项系数
Compare the degrees and leading coefficients of the three polynomials
大提示:
证明 ff 是首一二次多项式后,在 f(x2)=[f(x)]2f(x^2)=[f(x)]^2 中比较各项系数
After proving ff is a monic quadratic, compare coefficients in f(x2)=[f(x)]2f(x^2)=[f(x)]^2
解答:
设 ff 的次数为 n≥1n\ge1,首项系数为 aa。三个表达式的次数分别为 2n,2n,n22n,2n,n^2,所以 n2=2nn^2=2n,从而 n=2n=2。比较 f(x2)=[f(x)]2f(x^2)=[f(x)]^2 两边的首项系数,得到 a=a2a=a^2,因此 a=1a=1。
令 f(x)=x2+bx+cf(x)=x^2+bx+c。那么 x4+bx2+c=(x2+bx+c)2。 x^4+bx^2+c=(x^2+bx+c)^2\text{。}三次项系数迫使 b=0b=0,随后二次项系数迫使 c=0c=0。因此唯一可能的函数是 f(x)=x2f(x)=x^2,而它确实使三个表达式都相等。所以恰有一个这样的函数。
因此,正确答案是 B。
Let ff have degree n≥1n\ge1 and leading coefficient a.a. The three expressions have degrees 2n,2n,n2,2n,2n,n^2, so n2=2nn^2=2n and therefore n=2.n=2. Comparing leading coefficients in f(x2)=[f(x)]2f(x^2)=[f(x)]^2 gives a=a2,a=a^2, hence a=1.a=1.
Write f(x)=x2+bx+c.f(x)=x^2+bx+c. Then x4+bx2+c=(x2+bx+c)2. x^4+bx^2+c=(x^2+bx+c)^2. The cubic coefficient forces b=0,b=0, and then the quadratic coefficient forces c=0.c=0. Thus the only candidate is f(x)=x2,f(x)=x^2, which indeed satisfies all three expressions. There is exactly one function.
Thus the correct answer is B.
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