Question

首项系数为 $1$ 、且系数为实数的二次多项式称为“无礼”的，如果方程 $p(p(x)) = 0$ 恰好有三个实数解。在所有无礼二次多项式中，有唯一一个多项式 $\tilde{p}(x)$ ，使得其根的和最大。 $\tilde{p}(1)$ 是多少？

A quadratic polynomial with real coefficients and leading coefficient $1$ is called disrespectful if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)?$

$\dfrac{5}{16}$

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$1$

$\dfrac{9}{8}$

答案：A

知识点：多项式二次方程最优化

难度评级：2380

解答：

设 $p$ 的根为 $r$ 和 $s.$ 那么 $p(p(x)) = 0$ 分解成 $p(x) = r$ 与 $p(x) = s,$ 两个判别式分别为 $(r - s)^2 + 4r$ 和 $(r - s)^2 + 4s.$ 恰有三个实根意味着一个判别式为 $0$ ，另一个为正。

取 $(r - s)^2 + 4s = 0$ ，并令 $u = r - s.$ 则 $s = -\tfrac{u^2}{4}$ ，另一个判别式为 $4u,$ 所以 $u>0.$ 此外， $r+s=-\tfrac{u^2}{2}+u,$ 它在 $u=1,$ 时达到最大值，此时 $r=\tfrac34$ 且 $s=-\tfrac14.$

所以 $\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right),$ 并且 $\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}.$

因此，正确答案是 A。

Let $p$ have roots $r$ and $s.$ Then $p(p(x)) = 0$ splits into $p(x) = r$ and $p(x) = s,$ with discriminants $(r - s)^2 + 4r$ and $(r - s)^2 + 4s.$ Exactly three real roots means one discriminant is $0$ and the other positive.

Take $(r - s)^2 + 4s = 0$ and set $u = r - s.$ Then $s = -\tfrac{u^2}{4}$ and the other discriminant is $4u,$ so $u>0.$ Also $r+s=-\tfrac{u^2}{2}+u,$ maximized at $u=1,$ giving $r=\tfrac34$ and $s=-\tfrac14.$

So $\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right),$ and $\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}.$

Thus, the correct answer is A.

2021 AMC 12A Fall 第 23 题

答案：A

难度评级：2380

解答：

其他年份的第 23 题