📚 Exam Report & Revision Guide: AQA International AS Mathematics 9660/MA01 Pure Mathematics Unit 1 June 2018 | 考试报告与复习指南：AQA 国际AS数学 9660/MA01 纯数学第1单元 2018年6月
This article provides a detailed analysis of the June 2018 examination for AQA International AS Mathematics 9660/MA01 Pure Mathematics Unit 1. We examine common student errors, identify recurring themes, and offer targeted revision strategies to help you maximise your score.
本文深入分析 AQA 国际AS数学 9660/MA01 纯数学第1单元 2018年6月考试。我们将检视学生常见错误、识别反复出现的考点，并提供针对性的复习策略，帮助你最大化得分。
1. Overview of the Examination | 考试概览
The Pure Mathematics Unit 1 paper assesses core algebraic, geometric and calculus skills. In the June 2018 session, candidates were required to answer all questions within the time limit, with marks distributed across algebraic manipulation, coordinate geometry, calculus and trigonometric reasoning.
纯数学第1单元试卷考查代数学、几何学与微积分的核心技能。在2018年6月考季中，考生须在规定时间内完成所有题目，分值分布在代数运算、坐标几何、微积分与三角推理等板块。
Key statistical observations from the examiner’s report indicated that candidates performed well on routine differentiation and quadratic-solving questions, but struggled with the following areas:
考官报告中的关键统计观察表明，考生在常规微分与二次方程求解题目中表现良好，但在以下方面存在困难：
- Translating worded problems into mathematical equations | 将文字问题转化为数学方程
- Manipulating surds and rationalising denominators | 处理根式与有理化分母
- Applying the chain rule and product rule correctly | 正确运用链式法则与乘法法则
- Understanding the conditions for a tangent versus a normal | 理解切线与法线的条件
- Integration by inspection and evaluating definite integrals | 凑微分与计算定积分
2. Algebraic Manipulation and Quadratics | 代数运算与二次函数
A significant proportion of the June 2018 paper involved solving quadratic equations and inequalities. The most common error observed was sign errors when applying the quadratic formula, particularly when the coefficient ‘a’ was negative.
2018年6月试卷中很大比例涉及求解二次方程与不等式。观察到最常见的错误是在套用二次公式时出现符号错误，尤其是当系数 ‘a’ 为负数时。
For a quadratic equation in the form ax² + bx + c = 0, the formula is:
x = (−b ± √(b² − 4ac)) / 2a
Candidates often forgot to substitute negative coefficients in brackets, leading to incorrect discriminant values. For example, solving 2x² − 5x − 3 = 0 requires careful substitution of b = −5:
考生常常忘记将负系数加括号代入，导致判别式计算错误。例如，解 2x² − 5x − 3 = 0 时需要仔细代入 b = −5：
x = (5 ± √(25 + 24)) / 4 = (5 ± 7) / 4
This yields x = 3 or x = −1/2. Always write out every step and use brackets around negative values when substituting.
由此得 x = 3 或 x = −1/2。代负值时务必写出每一步并加括号。
For inequalities involving quadratics, remember the three-step method:
对于涉及二次函数的不等式，记住三步法：
- Rearrange to make one side zero | 移项使一边为零
- Factorise or use the quadratic formula to find critical values | 因式分解或用二次公式求临界值
- Sketch the graph or use a sign table to determine the solution interval | 画草图或使用符号表确定解区间
3. Surds and Indices | 根式与指数
The examiner’s report highlighted that rationalising denominators was a topic where many marks were lost. Recall the technique for rationalising a denominator of the form a + √b:
考官报告指出，有理化分母是许多考生失分的主题。回忆 a + √b 形式分母的有理化技巧：
Multiply both numerator and denominator by the conjugate a − √b, because (a + √b)(a − √b) = a² − b.
将分子与分母同时乘以共轭式 a − √b，因为 (a + √b)(a − √b) = a² − b。
Example: Simplify 3 / (2 + √5).
示例： 化简 3 / (2 + √5)。
3 / (2 + √5) = 3(2 − √5) / [(2 + √5)(2 − √5)] = 3(2 − √5) / (4 − 5) = −3(2 − √5) = 3√5 − 6
Common errors in this topic:
该主题的常见错误：
- Forgetting to multiply the numerator as well as the denominator | 忘记分子也要乘以同一个因式
- Sign errors when expanding (a + √b)(a − √b) | 展开 (a + √b)(a − √b) 时出现符号错误
- Incorrect application of index laws such as (aᵐ)ⁿ = aᵐⁿ | 指数法则 (aᵐ)ⁿ = aᵐⁿ 应用错误
When simplifying expressions involving fractional indices, recall that a^(m/n) = (ⁿ√a)ᵐ. For example, 16^(3/4) = (⁴√16)³ = 2³ = 8.
化简分数指数表达式时，记住 a^(m/n) = (ⁿ√a)ᵐ。例如，16^(3/4) = (⁴√16)³ = 2³ = 8。
4. Coordinate Geometry: Straight Lines and Circles | 坐标几何：直线与圆
Question 4 of the paper involved finding the equation of a perpendicular bisector and determining whether a point lay on a circle. Many candidates who understood the concept of the perpendicular bisector still lost marks on algebraic accuracy.
试卷第4题涉及求垂直平分线方程并判断一点是否在圆上。许多理解垂直平分线概念的考生仍在代数准确性上失分。
Recall the key formulas for coordinate geometry:
回忆坐标几何的关键公式：
- Midpoint of (x₁, y₁) and (x₂, y₂): ((x₁ + x₂)/2, (y₁ + y₂)/2) | 中点坐标：((x₁ + x₂)/2, (y₁ + y₂)/2)
- Gradient: m = (y₂ − y₁) / (x₂ − x₁) | 斜率：m = (y₂ − y₁) / (x₂ − x₁)
- Perpendicular gradients: m₁ × m₂ = −1 | 垂直斜率满足 m₁ × m₂ = −1
- Equation of a line: y − y₁ = m(x − x₁) | 直线方程：y − y₁ = m(x − x₁)
- Equation of a circle, centre (a, b), radius r: (x − a)² + (y − b)² = r² | 圆方程：圆心 (a, b)，半径 r：(x − a)² + (y − b)² = r²
When finding the perpendicular bisector of a line segment AB, follow these steps:
求线段 AB 的垂直平分线时，按以下步骤：
- Find the midpoint M of AB | 求 AB 的中点 M
- Find the gradient of AB | 求 AB 的斜率
- Take the negative reciprocal to get the perpendicular gradient | 取负倒数得到垂直斜率
- Use y − y₁ = m(x − x₁) with M as the point | 以 M 为已知点代入 y − y₁ = m(x − x₁)
The examiner noted that common mistakes included using the original gradient instead of its negative reciprocal, and sign errors when completing the square for a circle equation. Completing the square: x² + 6x = (x + 3)² − 9. To find the centre and radius from x² + y² − 4x + 6y − 12 = 0, rewrite as (x − 2)² + (y + 3)² = 25, giving centre (2, −3) and radius 5.
考官指出，常见错误包括使用原斜率而非其负倒数，以及在将圆方程配方时出现符号错误。配方法：x² + 6x = (x + 3)² − 9。要从 x² + y² − 4x + 6y − 12 = 0 求圆心和半径，改写为 (x − 2)² + (y + 3)² = 25，得圆心 (2, −3)，半径 5。
5. Differentiation from First Principles | 从第一原理求导
The June 2018 paper included a question asking candidates to differentiate a simple polynomial from first principles. The definition of the derivative is:
2018年6月试卷包含一道从第一原理求简单多项式导数的题目。导数的定义是：
f′(x) = lim (h→0) [f(x + h) − f(x)] / h
A typical solution for f(x) = x² is:
f(x) = x² 的典型解法：
f′(x) = lim (h→0) [(x + h)² − x²] / h = lim (h→0) [x² + 2xh + h² − x²] / h = lim (h→0) (2x + h) = 2x
Common errors observed:
观察到的常见错误：
- Not writing the ‘lim (h→0)’ notation throughout | 没有全程写出 ‘lim (h→0)’ 记号
- Algebraic expansion errors with (x + h)² or (x + h)³ | 展开 (x + h)² 或 (x + h)³ 时出现代数错误
- Dividing only some terms by h instead of all terms | 只将部分而非所有项除以 h
When differentiating f(x) = x³ from first principles, expand (x + h)³ = x³ + 3x²h + 3xh² + h³. The terms containing h² and h³ vanish in the limit, leaving f′(x) = 3x².
从第一原理对 f(x) = x³ 求导时，展开 (x + h)³ = x³ + 3x²h + 3xh² + h³。含有 h² 和 h³ 的项在取极限时趋于零，得到 f′(x) = 3x²。
6. Rules of Differentiation | 微分法则
The product rule and chain rule were tested directly in this paper. The product rule states that if y = uv, then:
乘法法则与链式法则在本试卷中被直接考查。乘法法则指出，若 y = uv，则：
dy/dx = u(dv/dx) + v(du/dx)
The chain rule states that if y = f(g(x)), then:
链式法则指出，若 y = f(g(x))，则：
dy/dx = f′(g(x)) × g′(x)
When differentiating y = x² sin x, apply the product rule with u = x² and v = sin x:
对 y = x² sin x 求导时，以 u = x²，v = sin x 应用乘法法则：
- du/dx = 2x | du/dx = 2x
- dv/dx = cos x | dv/dx = cos x
- dy/dx = x² cos x + 2x sin x | dy/dx = x² cos x + 2x sin x
For the chain rule, consider y = (3x² + 1)⁵. Let u = 3x² + 1, then y = u⁵, dy/du = 5u⁴, du/dx = 6x. Therefore dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.
关于链式法则，考虑 y = (3x² + 1)⁵。令 u = 3x² + 1，则 y = u⁵，dy/du = 5u⁴，du/dx = 6x。因此 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。
The examiner’s report highlighted that when differentiating fractions of the form y = (2x + 1)/(x − 3), candidates who attempted to use the quotient rule often made sign errors. Remember the quotient rule:
考官报告强调，在求 y = (2x + 1)/(x − 3) 这类分式的导数时，尝试使用商法则的考生常犯符号错误。记住商法则：
dy/dx = [v(du/dx) − u(dv/dx)] / v²
Where u = 2x + 1 and v = x − 3. This gives dy/dx = [(x − 3)(2) − (2x + 1)(1)] / (x − 3)² = (2x − 6 − 2x − 1) / (x − 3)² = −7 / (x − 3)². Alternatively, you may find it easier to rewrite y = (2x + 1)(x − 3)⁻¹ and apply the product rule combined with the chain rule.
其中 u = 2x + 1，v = x − 3。由此得 dy/dx = [(x − 3)(2) − (2x + 1)(1)] / (x − 3)² = (2x − 6 − 2x − 1) / (x − 3)² = −7 / (x − 3)²。或者，将 y 改写为 y = (2x + 1)(x − 3)⁻¹ 并结合链式法则使用乘法法则，可能更简单。
7. Stationary Points and Applications | 驻点与应用
A key question in this paper required finding and classifying stationary points of a cubic function. The procedure is:
本试卷中一道关键题要求找到三次函数的驻点并分类。步骤为：
- Differentiate the function to find f′(x) | 对函数求导得到 f′(x)
- Set f′(x) = 0 and solve for x | 令 f′(x) = 0 并解出 x
- Calculate the corresponding y-values | 计算对应的 y 值
- Use the second derivative f″(x) to classify: f″(x) > 0 means minimum, f″(x) < 0 means maximum | 用二阶导数 f″(x) 分类：f″(x) > 0 为极小值，f″(x) < 0 为极大值
If f″(x) = 0, the test is inconclusive; use a sign table for f′(x) around the stationary point instead.
若 f″(x) = 0，该检验无定论；应改用 f′(x) 在驻点附近的符号表来判断。
For example, given f(x) = x³ − 3x² + 2:
例如，已知 f(x) = x³ − 3x² + 2：
- f′(x) = 3x² − 6x = 3x(x − 2) | f′(x) = 3x² − 6x = 3x(x − 2)
- Set f′(x) = 0, giving x = 0 or x = 2 | 令 f′(x) = 0，得 x = 0 或 x = 2
- f″(x) = 6x − 6 | f″(x) = 6x − 6
- At x = 0: f″(0) = −6 < 0, so (0, 2) is a maximum | 在 x = 0 处：f″(0) = −6 < 0，故 (0, 2) 为极大值点
- At x = 2: f″(2) = 6 > 0, so (2, −2) is a minimum | 在 x = 2 处：f″(2) = 6 > 0，故 (2, −2) 为极小值点
Common errors in this question type included miscalculating the y-coordinates and writing ‘maximum’ when f″(x) < 0 was correctly determined, indicating confusion about the sign convention.
此类题目的常见错误包括错算 y 坐标，以及虽然正确算出 f″(x) < 0 却写成 ‘极大值’，表明对符号约定的混淆。
8. Integration: Indefinite and Definite | 积分：不定积分与定积分
Integration questions appeared in the second half of the paper. The fundamental rule for integrating powers of x is:
积分题目出现在试卷后半部分。对 x 的幂次积分的基本法则是：
∫ xⁿ dx = x^(n+1) / (n+1) + C, for n ≠ −1
Candidates frequently omitted the constant of integration in indefinite integrals, which cost them marks. For definite integrals, the constant is not needed, but the evaluation must be shown clearly:
考生经常在不定积分中遗漏积分常数 C，因此丢分。对于定积分，不需要常数，但必须清楚展示计算过程：
∫ from 1 to 3 (2x + 1) dx = [x² + x] evaluated from x = 1 to x = 3 = (9 + 3) − (1 + 1) = 12 − 2 = 10
When integrating expressions that are not in standard form, algebraic manipulation may be required first. For example, before integrating (x² + 1)/√x, rewrite it as:
当被积表达式不是标准形式时，可能需先进行代数变形。例如，在积分 (x² + 1)/√x 前，先改写为：
(x² + 1)/√x = x^(3/2) + x^(−1/2)
Then integrate term by term: ∫ x^(3/2) dx = (2/5)x^(5/2) + C, and ∫ x^(−1/2) dx = 2x^(1/2) + C.
然后逐项积分：∫ x^(3/2) dx = (2/5)x^(5/2) + C，且 ∫ x^(−1/2) dx = 2x^(1/2) + C。
The examiner’s report noted that the most common integration error in this paper was incorrect manipulation of fractional indices, particularly the step from x^(−1/2) to the integrated form x^(1/2). Always check: when n = −1/2, n + 1 = 1/2, so the result is 2x^(1/2).
考官报告指出，本试卷中积分最常见的错误是分数指数运算不当，尤其是从 x^(−1/2) 到积分形式 x^(1/2) 的步骤。务必检查：当 n = −1/2 时，n + 1 = 1/2，所以结果是 2x^(1/2)。
9. Area Under a Curve | 曲线下的面积
Determining the area between a curve and the x-axis was examined. The area is given by the definite integral, but caution is required when the curve crosses the x-axis:
本试卷考查了求曲线与 x 轴之间面积的问题。面积由定积分给出，但当曲线穿过 x 轴时需要格外谨慎：
Area = ∫ from a to b f(x) dx, provided f(x) ≥ 0 on [a, b]
If the function takes negative values, the integral gives a negative result. To find the total enclosed area between a curve and the x-axis over an interval where the curve crosses the axis, split the integral at the roots. For example, for f(x) = x² − 4x + 3, which crosses the x-axis at x = 1 and x = 3:
若函数取负值，积分结果为负数。为求曲线与 x 轴在曲线穿过轴的区间内所围成的总面积，需在根处拆分积分。例如，f(x) = x² − 4x + 3 在 x = 1 与 x = 3 处穿过 x 轴：
Area = |∫ from 0 to 1 f(x) dx| + |∫ from 1 to 2 f(x) dx| (if both intervals are relevant). In practice, calculate each interval separately and take absolute values.
面积 = |∫ 从 0 到 1 f(x) dx| + |∫ 从 1 到 2 f(x) dx|（若两个区间都相关）。实际操作中，分别计算每个区间并取绝对值。
Exam technique tip: when finding the area between two curves, determine the points of intersection first, and then integrate the difference (upper curve minus lower curve) between those x-values. When finding the area between a line and a curve, the line may be above the curve within the interval of interest; draw a quick sketch to verify which is above.
考试技巧提示：求两曲线之间的面积时，先确定交点，然后对两曲线之差（上方曲线减去下方曲线）在相应 x 值之间积分。求直线与曲线之间的面积时，直线在相关区间内可能位于曲线上方；快速画草图以确认何者在上方。
10. Trigonometry: Identities and Equations | 三角学：恒等式与方程
The final questions of the paper involved trigonometric identities and solving equations. The fundamental identities that you must know:
试卷最后部分涉及三角恒等式与解三角方程。你必须掌握的基本恒等式：
- sin²θ + cos²θ = 1 | sin²θ + cos²θ = 1
- tanθ = sinθ / cosθ | tanθ = sinθ / cosθ
To solve an equation such as 3sinθ = 2cos²θ, use cos²θ = 1 − sin²θ to obtain a quadratic in sinθ:
解 3sinθ = 2cos²θ 这类方程时，利用 cos²θ = 1 − sin²θ 得到关于 sinθ 的二次方程：
3sinθ = 2(1 − sin²θ) → 2sin²θ + 3sinθ − 2 = 0
Factorise: (2sinθ − 1)(sinθ + 2) = 0. Since sinθ = −2 has no solution, we take sinθ = 1/2. Within the range 0 ≤ θ ≤ 360°, the solutions are θ = 30° and θ = 150°.
因式分解：(2sinθ − 1)(sinθ + 2) = 0。由于 sinθ = −2 无解，取 sinθ = 1/2。在 0 ≤ θ ≤ 360° 范围内，解为 θ = 30° 与 θ = 150°。
Common errors in this question included:
此题常见错误包括：
- Forgetting the second quadrant solution when solving sinθ = k | 解 sinθ = k 时遗漏第二象限解
- Incorrectly applying the identity sin²θ = 1 − cos²θ | 错误应用恒等式 sin²θ = 1 − cos²θ
- Not checking whether all solutions lie within the specified range | 未检查所有解是否在指定范围内
When asked to solve equations in radians, convert degree-based thinking: remember that π radians = 180°, so the range 0 ≤ θ ≤ 2π corresponds to 0 ≤ θ ≤ 360°.
当要求以弧度为单位解方程时，转换基于角度的思维：记住 π 弧度 = 180°，因此 0 ≤ θ ≤ 2π 对应 0 ≤ θ ≤ 360°。
11. Exponentials and Logarithms | 指数函数与对数
Although sometimes integrated into the calculus questions, exponential and logarithmic functions were a notable feature of the 2018 examination’s earlier sections. The key definitions and laws you must recall:
虽然有时融入微积分题目，但指数函数与对数函数是2018年试卷前部的一个重要特色。你必须回忆起关键定义与法则：
- y = eˣ is the exponential function; its derivative is itself | y = eˣ 是指数函数；其导数等于其自身
- y = ln x is the natural logarithm; it is the inverse of eˣ | y = ln x 是自然对数；它是 eˣ 的反函数
- ln(ab) = ln a + ln b | ln(ab) = ln a + ln b
- ln(a/b) = ln a − ln b | ln(a/b) = ln a − ln b
- ln(aᵏ) = k ln a | ln(aᵏ) = k ln a
For equations involving exponential functions, taking natural logarithms of both sides is often the most efficient strategy. To solve 2eˣ = 5:
对于涉及指数函数的方程，两边取自然对数通常是最有效的方法。求解 2eˣ = 5：
eˣ = 2.5 → x = ln 2.5
For equations of the form aᵏ = b, take logs of both sides: k ln a = ln b, so k = ln b / ln a.
对于 aᵏ = b 形式的方程，两边取对数：k ln a = ln b，故 k = ln b / ln a。
The examiner reported that candidates often confused ln x with log₁₀ x on calculators. In A-level mathematics at AQA, unless otherwise stated, ‘ln’ denotes the natural logarithm and ‘log’ generally defaults to base 10, but in pure mathematics contexts base e is standard. Always read the question carefully.
考官报告指出，考生常常混淆计算器上的 ln x 与 log₁₀ x。在 AQA A-level 数学中，除非另有说明，’ln’ 表示自然对数，’log’ 一般默认为以10为底，但在纯数学背景下以 e 为底是标准做法。务必仔细读题。
12. Examination Strategy and Common Pitfalls | 考试策略与常见陷阱
Based on the examiner’s report for June 2018, we recommend the following strategies to maximise your performance:
基于2018年6月的考官报告，我们推荐以下策略以最大化你的表现：
| Pitfall | 陷阱 | Solution | 对策 |
| Omitting the constant of integration | 遗漏积分常数 | Always write + C for indefinite integrals | 对不定积分始终写 + C |
| Sign errors in the quadratic formula | 二次公式符号错误 | Substitute negative values in brackets; check by substituting solutions back | 代负值时加括号；将解代回验证 |
| Using the wrong derivative rule | 使用错误的求导法则 | Identify the form first: product, quotient, or composite | 先认清代数形式：乘积、商或复合 |
| Forgetting the second solution in trig equations | 在三角方程中遗漏第二个解 | Sketch the sine/cosine graph or CAST diagram | 画出正弦/余弦图或 CAST 象限图 |
| Misreading the question range | 误读题目范围 | Highlight the range (degrees or radians) before solving | 先高亮范围（度或弧度）再求解 |
| Not laying out working clearly | 解题过程不清晰 | Show every step; method marks can be gained even with a slip | 展示每一步；即使有小失误也可能获得方法分 |
Time management is crucial. Aim to complete all questions, leaving approximately 10 minutes at the end for checking. When stuck on a part, move on and return if time permits.
时间管理至关重要。争取完成所有题目，在最后预留约10分钟检查。被某一部分卡住时，先跳过去，若时间允许再回来。
Finally, remember that examiners award method marks. Even if your final answer is incorrect, a clear and logical method can earn significant partial credit. Write neatly, define your variables, and annotate each step with reason.
最后，记住考官按方法给分。即使最终答案错误，清晰且逻辑严密的方法仍可获得大部分步骤分。书写工整，定义你的变量，并在每一步标注理由。
For further practice, revisit all past papers from 2018–2023, paying special attention to the topics covered in this guide. Consistent practice with timed conditions will build both speed and accuracy.
如需进一步练习，请重做 2018–2023 年的历年真题，特别注意本指南所涵盖的主题。在限时条件下持续练习将同时提升速度与准确度。
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