AMC12 2021 B

AMC12 2021 B · Q16

AMC12 2021 B · Q16. It mainly tests Polynomials, Rational expressions.

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$

设 $g(x)$ 是一个首项系数为 $1$ 的多项式，其三个根是 $f(x)=x^3+ax^2+bx+c$ 的三个根的倒数，其中 $1<a<b<c$。$g(1)$ 用 $a,b,c$ 表示是什么？

(A) \frac{1+a+b+c}c \frac{1+a+b+c}c

(B) 1+a+b+c 1+a+b+c

(D) \frac{a+b+c}{c^2} \frac{a+b+c}{c^2}

(E) \frac{1+a+b+c}{a+b+c} \frac{1+a+b+c}{a+b+c}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Note that $f(1/x)$ has the same roots as $g(x)$, if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] so we can see that \[g(x) = \frac{x^3}{c}f(1/x)\] Therefore \[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]

注意到 $f(1/x)$ 的根与 $g(x)$ 相同，如果乘以某个单项式使首项为 $x^3$，它们就相等。我们有 \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] 因此 \[g(x) = \frac{x^3}{c}f(1/x)\] 所以 \[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]

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