Question

一组数的调和平均数定义为这些数的倒数的算术平均数的倒数。例如， $4, 4$ 和 $5$ 的调和平均数为下面这个 $4050$ 次多项式所有实根的调和平均数是多少？ $\frac{1}{\frac{1}{3}\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{5}\right)} = \frac{30}{7}.$ $\begin{aligned} &\small \prod_{k=1}^{2025}(kx^2 - 4x - 3) \\ &= (x^2 - 4x - 3) \\ &\quad {}\cdot (2x^2 - 4x - 3) \\ &\quad {}\cdot (3x^2 - 4x - 3)\cdots \\ &\quad (2025x^2 - 4x - 3)? \end{aligned}$

The harmonic mean of a collection of numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers in the collection. For example, the harmonic mean of $4, 4,$ and $5$ is $\frac{1}{\frac{1}{3}\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{5}\right)} = \frac{30}{7}.$ What is the harmonic mean of all the real roots of the $4050$ th degree polynomial $\begin{aligned} &\small \prod_{k=1}^{2025}(kx^2 - 4x - 3) \\ &= (x^2 - 4x - 3) \\ &\quad {}\cdot (2x^2 - 4x - 3) \\ &\quad {}\cdot (3x^2 - 4x - 3)\cdots \\ &\quad (2025x^2 - 4x - 3)? \end{aligned}$

$-\dfrac{5}{3}$

$-\dfrac{3}{2}$

$-\dfrac{6}{5}$

$-\dfrac{5}{6}$

$-\dfrac{2}{3}$

答案：B

知识点：韦达定理二次方程调和平均数

难度评级：1630

解答：

每个因式 $kx^2 - 4x - 3$ 的判别式为 $16 + 12k \gt 0$ ，所以它有两个实根；总共有 $4050$ 个根。

对 $kx^2 - 4x - 3$ 的根，倒数之和为 $\dfrac{\text{sum}}{\text{product}} = \dfrac{4/k}{-3/k} = -\dfrac{4}{3}$ ，与 $k$ 无关。

对全部 $2025$ 个因式求和， $\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700$ 。调和平均数为 $\frac{4050}{-2700} = -\frac{3}{2}.$

因此，正确答案是 B。

Each factor $kx^2 - 4x - 3$ has discriminant $16 + 12k \gt 0,$ so it has two real roots; there are $4050$ roots in all.

For the roots of $kx^2 - 4x - 3,$ the sum of reciprocals is $\dfrac{\text{sum}}{\text{product}} = \dfrac{4/k}{-3/k} = -\dfrac{4}{3},$ independent of $k.$

Summing over all $2025$ factors, $\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700.$ The harmonic mean is $\frac{4050}{-2700} = -\frac{3}{2}.$

Thus, the correct answer is B.

2025 AMC 12A 第 12 题

答案：B

难度评级：1630

解答：

其他年份的第 12 题