AQA A-Level Maths Example Responses: MA03 Unit P2 | AQA A-level 数学例题解析：MA03 卷二

📚 AQA A-Level Maths Example Responses: MA03 Unit P2 | AQA A-level 数学例题解析：MA03 卷二

This article presents a collection of worked examples that reflect the style of Exam Paper 2 in the AQA A-level Mathematics MA03 series. Each example is broken down into a model answer and examiner commentary, so you can see exactly what is needed to score full marks.

本文整理了 AQA A-level 数学 MA03 系列试卷二的典型例题，每道题均包含标准答案和考官点评，帮助您明确得分要点。

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1. Binomial Distribution | 二项分布

Question: A random variable X has distribution B(5, 0.6). Find P(X = 3).

题目：设随机变量 X 服从二项分布 B(5, 0.6)，求 P(X = 3)。

Using the binomial formula: P(X = r) = C(n, r) pʳ (1−p)ⁿ⁻ʳ. Here n = 5, p = 0.6, r = 3.

根据二项分布公式：P(X = r) = C(n, r) pʳ (1−p)ⁿ⁻ʳ。其中 n = 5，p = 0.6，r = 3。

P(X = 3) = C(5, 3) × 0.6³ × 0.4² = 10 × 0.216 × 0.16 = 0.3456

State the values of n, p, and r clearly. Show the substitution into the formula and the calculation of the binomial coefficient. Round only at the final step to avoid compounding errors.

务必清晰写出 n、p、r 的取值，显示代入公式和二项系数的计算过程。仅在最后一步进行四舍五入，避免累积误差。

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2. Hypothesis Testing | 假设检验

Question: A drug is effective in 30% of patients. A new drug is tested on 20 patients and 9 are cured. Test at the 5% significance level whether the new drug is more effective than the old one.

题目：某药物对 30% 的患者有效。新药在 20 名患者身上试验，其中 9 人被治愈。在 5% 的显著性水平下检验新药是否比旧药更有效。

Let p be the probability that a patient is cured. H₀: p = 0.3, H₁: p > 0.3. We use a one-tailed test. Under H₀, X ~ B(20, 0.3).

设 p 为患者被治愈的概率。H₀：p = 0.3，H₁：p > 0.3，采用单尾检验。在 H₀ 下，X ~ B(20, 0.3)。

P(X ≥ 9) = 1 − P(X ≤ 8) = 1 − 0.8867 = 0.1133

Since 0.1133 > 0.05, the result is not significant. There is insufficient evidence to conclude that the new drug is more effective. State the conclusion in context.

由于 0.1133 > 0.05，结果不显著，没有足够证据表明新药更有效。结论必须结合题目背景进行表述。

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3. Correlation and Regression | 相关与回归

Question: For 6 students, the number of hours studied x and exam score y gave Σx = 30, Σy = 210, Σx² = 190, Σxy = 1260. Find the regression line of y on x and interpret the product moment correlation coefficient r.

题目：对 6 名学生，学习时长 x 与考试成绩 y 的统计量为 Σx = 30，Σy = 210，Σx² = 190，Σxy = 1260。求 y 对 x 的回归直线，并解释积矩相关系数 r 的含义。

Compute Sxy = Σxy − ΣxΣy/n = 1260 − (30 × 210)/6 = 210. Sxx = Σx² − (Σx)²/n = 190 − 900/6 = 40.

计算 Sxy = Σxy − ΣxΣy/n = 1260 − (30 × 210)/6 = 210。Sxx = Σx² − (Σx)²/n = 190 − 900/6 = 40。

b = Sxy / Sxx = 210 / 40 = 5.25, a = ȳ − b x̄ = 35 − 5.25 × 5 = 8.75

Regression line: y = 8.75 + 5.25x. For correlation, r = Sxy / √(Sxx·Syy). A positive r close to 1 indicates strong positive linear correlation.

回归直线为 y = 8.75 + 5.25x。相关系数 r = Sxy / √(Sxx·Syy)。r 为正且接近 1 表明强正线性相关。

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4. Differentiation – Stationary Points | 微分 – 驻点

Question: Find the coordinates of the stationary points of y = x³ − 6x² + 9x, and determine their nature.

题目：求函数 y = x³ − 6x² + 9x 的驻点坐标，并判断其性质。

Differentiate: dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).

求导：dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3)。

Set dy/dx = 0, giving x = 1 and x = 3. Then y(1) = 4 and y(3) = 0. The second derivative is d²y/dx² = 6x − 12.

令 dy/dx = 0，得 x = 1 和 x = 3。则 y(1) = 4，y(3) = 0。二阶导数为 d²y/dx² = 6x − 12。

At x = 1, d²y/dx² = −6 < 0, so (1, 4) is a local maximum. At x = 3, d²y/dx² = 6 > 0, so (3, 0) is a local minimum.

在 x = 1 处，d²y/dx² = −6 < 0，因此 (1, 4) 为局部极大值点；在 x = 3 处，d²y/dx² = 6 > 0，因此 (3, 0) 为局部极小值点。

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5. Integration – Area Under a Curve | 积分 – 曲线下面积

Question: Find the area enclosed by the curve y = 2x − x² and the x-axis.

题目：求曲线 y = 2x − x² 与 x 轴所围成的面积。

First find the roots: 2x − x² = 0 ⇒ x(2 − x) = 0 ⇒ x = 0 or x = 2.

先求交点：2x − x² = 0 ⇒ x(2 − x) = 0 ⇒ x = 0 或 x = 2。

∫₀² (2x − x²) dx = [x² − x³/3]₀² = (4 − 8/3) − 0 = 4/3

The area is 4/3 square units. Remember to state the units and show the substitution of limits.

面积为 4/3 平方单位。记得写出单位并演示上下限的代入。

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6. Trigonometry – Solving Equations | 三角 – 解方程

Question: Solve 2 sin² θ − sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.

题目：在 0° ≤ θ ≤ 360° 范围内，解方程 2 sin² θ − sin θ − 1 = 0。

Let u = sin θ. Then 2u² − u − 1 = 0 ⇒ (2u + 1)(u − 1) = 0 ⇒ u = 1 or u = −1/2.

设 u = sin θ。则 2u² − u − 1 = 0 ⇒ (2u + 1)(u − 1) = 0 ⇒ u = 1 或 u = −1/2。

For sin θ = 1, θ = 90°. For sin θ = −1/2, θ = 210° or 330°. Therefore the solutions are 90°, 210°, 330°.

当 sin θ = 1 时，θ = 90°；当 sin θ = −1/2 时，θ = 210° 或 330°。因此解为 90°、210°、330°。

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7. Proof by Contradiction | 反证法

Question: Prove that √2 is irrational.

题目：证明 √2 是无理数。

Assume √2 is rational. Then √2 = a/b, where a, b are integers with no common factor and

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