AMC12 2024 B

AMC12 2024 B · Q17

AMC12 2024 B · Q17. It mainly tests Vieta / quadratic relationships (basic), Probability (basic).

Integers $a$ and $b$ are randomly chosen without replacement from the set of integers with absolute value not exceeding $10$. What is the probability that the polynomial $x^3 + ax^2 + bx + 6$ has $3$ distinct integer roots?

从绝对值不超过 $10$ 的整数集合中，不放回地随机选择整数 $a$ 和 $b$。多项式 $x^3 + ax^2 + bx + 6$ 具有 $3$ 个不同整数根的概率是多少？

(A) \frac{1}{240} \frac{1}{240}

(B) \frac{1}{221} \frac{1}{221}

(D) \frac{1}{84} \frac{1}{84}

(E) \frac{1}{63}. \frac{1}{63}.

Answer

Correct choice: (C)

正确答案：（C）

Solution

Since $-10 \le a,b \le 10$, there are 21 integers to choose from, and $P(21,2) = 21 \times 20 = 420$ equally likely ordered pairs $(a,b)$. Applying Vieta's formulas, $x_1 \cdot x_2 \cdot x_3 = -6$ $x_1 + x_2+ x_3 = -a$ $x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b$ Cases: (1) $(x_1,x_2,x_3) = (-1,1,6) , b = -1, a=-6$ valid (2) $(x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0$ valid (3) $(x_1,x_2,x_3) = (1,-2,3) , b = -5, a=-2$ valid (4) $(x_1,x_2,x_3) = (-1,2,3) , b = 1, a=-4$ valid (5) $(x_1,x_2,x_3) = (-1,-2,-3) , b = 11$ invalid the total event space is $21 \cdot (21- 1)$ (choice of select a times choice of selecting b given no-replacement) hence, our answer is $\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}$

由于 $-10 \le a,b \le 10$，共有 $21$ 个整数，选择有序对 $(a,b)$ 的方式有 $P(21,2) = 21 \times 20 = 420$ 种等可能的有序对。应用 Vieta 公式， $x_1 \cdot x_2 \cdot x_3 = -6$ $x_1 + x_2+ x_3 = -a$ $x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b$ 情况： (1) $(x_1,x_2,x_3) = (-1,1,6) , b = -1, a=-6$ 有效 (2) $(x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0$ 有效 (3) $(x_1,x_2,x_3) = (1,-2,3) , b = -5, a=-2$ 有效 (4) $(x_1,x_2,x_3) = (-1,2,3) , b = 1, a=-4$ 有效 (5) $(x_1,x_2,x_3) = (-1,-2,-3) , b = 11$ 无效总事件空间是 $21 \cdot (21- 1)$（选择 $a$ 的方式乘以给定不放回选择 $b$ 的方式）因此，答案是 $\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}$

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