AS AQA International Pure Maths MA02: Example Responses & Model Solutions | AS AQA 国际纯数学 MA02：示例作答与评分要点

📚 AS AQA International Pure Maths MA02: Example Responses & Model Solutions | AS AQA 国际纯数学 MA02：示例作答与评分要点

In the AQA International AS Pure Mathematics MA02 paper, short-answer and structured questions test algebra, coordinate geometry, calculus, trigonometry and mathematical reasoning. This article gives six exam-style questions with model responses and examiner comments, so you can see exactly how method marks and accuracy marks are earned.

在 AQA 国际 AS 纯数学 MA02 试卷中，简答题与结构化题目考查代数、坐标几何、微积分、三角函数和数学推理。本文提供六道考试风格题目，并配以示范作答和考官点评，帮助你清楚地看到方法分和精确分是如何获得的。

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1. How to Use Example Responses | 如何使用示范作答

Before reading the model answers, write your own solution under timed conditions. Then compare each line of your work with the example response and label every step with the mark it would earn. This turns a model answer into a diagnostic tool rather than a text to be read passively.

在阅读示范答案之前，请先在限时条件下写下自己的解答。然后逐行与示范作答比较，并给每一步标出它能拿到的分数。这样示范答案就变成了诊断工具，而不是被动阅读的材料。

AQA International AS Pure Mathematics mark schemes award M marks for method, A marks for accuracy, and B marks for statements such as a correct factorisation. A candidate who reaches the right answer with no visible method often scores fewer marks than a candidate who shows correct reasoning and slips at the final arithmetic step.

AQA 国际 AS 纯数学评分方案中，M 分给方法，A 分给精确结果，B 分给诸如正确因式分解这样的独立结论。一个直接写出正确答案但没有任何过程的考生，往往比一个思路正确、只在最后一步算错的考生得分更低。

After marking, write a short error log with three columns: the topic, the exact mistake, and the fix. Return to the same questions after 48 hours and after two weeks. Spaced repetition of your own mistakes is far more efficient than re-reading notes.

批改之后，请写一个简短错误日志，包含三列：知识点、具体错误、改进方法。48 小时后和两周后再回到同一批题目重做。对自己错误进行间隔重复，比反复阅读笔记高效得多。

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2. Mark Scheme Conventions You Must Recognise | 必须看懂的评分符号

AQA uses short abbreviations in mark schemes, and knowing them prevents you from throwing away marks. The most common ones are M (method), A (accuracy), B (independent statement), ft (follow through), cao (correct answer only), oe (or equivalent) and isw (ignore subsequent working).

AQA 评分方案使用简写符号，看懂它们能避免白白丢分。最常见的包括 M（方法分）、A（精确分）、B（独立结论分）、ft（后续追分）、cao（只接受正确答案）、oe（或等价形式）以及 isw（忽略后续过程）。

Code | 符号	Meaning | 含义	How to secure it | 如何拿到
M1	Method mark | 方法分	Write the correct formula, substitution or differentiating step | 写出正确公式、代入或求导步骤
A1	Accuracy mark | 精确分	Give the exact simplified answer, not a rounded decimal | 给出精确化简结果，不用四舍五入小数
B1	Independent mark | 独立结论分	State a correct result such as a factorisation or a condition | 写出正确结果，如因式分解或条件
ft	Follow through | 后续追分	Continue correctly from your own earlier answer | 在自己前面答案基础上正确继续

The single most common error in structured questions is skipping the substitution line. If a mark scheme allocates M1 for substituting into the discriminant or into the cosine rule, that mark is lost even when the final answer is correct.

结构化题目中最常见的错误是省略代入那一行。如果评分方案把 M1 给到"代入判别式"或"代入余弦定理"这一步，那么即使最终答案正确，这一分也会丢掉。

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3. Question 1 - Indices, Surds and Rationalising | 第 1 题：指数、根式与分母有理化

Question. (a) Evaluate 27^2/3 + 16^−1/2. (b) Simplify √50 + √18 − √8. (c) Express 6 ÷ (√5 − 1) in the form k(√5 + 1), where k is a rational number.

题目。(a) 计算 27^2/3 + 16^−1/2。(b) 化简 √50 + √18 − √8。(c) 将 6 ÷ (√5 − 1) 表示为 k(√5 + 1) 的形式，其中 k 为有理数。

Model response (a). Write 27^2/3 = (27^1/3)² = 3² = 9, and 16^−1/2 = 1 ÷ 16^1/2 = 1 ÷ 4. Adding gives 9 + 1/4.

示范作答 (a)。写出 27^2/3 = (27^1/3)² = 3² = 9，以及 16^−1/2 = 1 ÷ 16^1/2 = 1 ÷ 4。相加得 9 + 1/4。

27^2/3 + 16^−1/2 = 9 + 1 ÷ 4 = 37 ÷ 4 = 9.25

Model response (b). Factor out square numbers: √50 = √25 × 2 = 5√2, √18 = √9 × 2 = 3√2, √8 = √4 × 2 = 2√2. Then collect like surds.

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示范作答 (b)。提取平方因子：√50 = √25 × 2 = 5√2，√18 = √9 × 2 = 3√2，√8 = √4 × 2 = 2√2。然后合并同类根式。

√50 + √18 − √8 = 5√2 + 3√2 − 2√2 = 6√2

Model response (c). Multiply numerator and denominator by the conjugate √5 + 1, then use the difference of two squares on the denominator.

示范作答 (c)。分子分母同乘共轭式 √5 + 1，分母使用平方差公式。

6 ÷ (√5 − 1) = 6(√5 + 1) ÷ ((√5)² − 1²) = 6(√5 + 1) ÷ 4 = (3 ÷ 2)(√5 + 1)

So k = 3 ÷ 2. Examiner comment: B1 for each of the three surd simplifications in part (b), and M1 for multiplying by the conjugate in part (c). The final A1 requires the simplified form (3 ÷ 2)(√5 + 1); leaving 6(√5 + 1) ÷ 4 unsimplified is a classic dropped mark.

所以 k = 3 ÷ 2。考官点评：(b) 中三个根式化简各得 B1，(c) 中乘以共轭式得 M1。最后的 A1 要求化简为 (3 ÷ 2)(√5 + 1) 的形式；把 6(√5 + 1) ÷ 4 原样留着不化简，是典型的丢分点。

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4. Question 2 - Discriminants, Inequalities and Hidden Quadratics | 第 2 题：判别式、不等式与隐藏二次方程

Question. 9 = 0 has equal roots. Find the possible values of k. (b) Solve 2x² + 3x ≥ 5. (c) Solve 3^2x − 10(3^x) + 9 = 0.

题目。(a) 方程 x² + (k + 2)x + 9 = 0 有两个相等实根，求 k 的可能值。(b) 解不等式 2x² + 3x ≥ 5。(c) 解方程 3^2x − 10(3^x) + 9 = 0。

Model response (a). Equal roots mean the discriminant is zero, so b² − 4ac = 0 with a = 1, b = k + 2, c = 9.

示范作答 (a)。两根相等意味着判别式为零，即 b² − 4ac = 0，其中 a = 1，b = k + 2，c = 9。

(k + 2)² − 4 × 1 × 9 = 0 → (k + 2)² = 36 → k + 2 = ±6 → k = 4 or k = −8

Model response (b). nd x = 1. The parabola opens upwards, so the expression is non-negative outside the roots.

示范作答 (b)。移项得 2x² + 3x − 5 ≥ 0，因式分解为 (2x + 5)(x − 1) ≥ 0，得到临界值 x = −5 ÷ 2 与 x = 1。抛物线开口向上，因此表达式在两根之外取非负值。

x ≤ −5 ÷ 2 or x ≥ 1

Model response (c). This is a quadratic in disguise. Let u = 3^x, so u² = 3^2x ation becomes u² − 10u + 9 = 0, that is (u − 1)(u − 9) = 0, giving u = 1 or u = 9. Since 3⁰ = 1 and 3² = 9, the solutions are x = 0 and x = 2.

示范作答 (c)。这是"伪装的二次方程"。令 u = 3^x，则 u² = 3^2x，方程化为 u² − 10u + 9 = 0，即 (u − 1)(u − 9) = 0，得 u = 1 或 u = 9。由于 3⁰ = 1 且 3² = 9，所以解为 x = 0 和 x = 2。

Examiner comment: 0, A1 for both values. Do not divide by 2 or ignore the negative root. In part (b), a sketch or a sign line protects the direction of the inequality; simply writing

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