AQA A-Level Further Maths June 2018 Paper 2 Walkthrough & Solutions | AQA 高数 2018年6月试卷2 解析

📚 AQA A-Level Further Maths June 2018 Paper 2 Walkthrough & Solutions | AQA 高数 2018年6月试卷2 解析

The June 2018 AQA A-Level Further Mathematics Paper 2 is a demanding assessment that tests a wide range of pure, applied, and decision mathematics skills. With a mix of short proof-based questions and lengthy multi-part problems, this paper pushes students to combine algebraic fluency with conceptual understanding. This walkthrough will break down the main topics, provide worked solutions to representative questions, and offer strategic advice for maximising marks.

2018年6月AQA进阶数学（Further Mathematics）试卷2是一次要求很高的考试，考查范围涵盖纯数、应用数学和离散数学的诸多领域。试卷既有简短的证明题，也有长篇多步骤综合题，要求学生将代数熟练度与概念理解相结合。本文将对主要考点进行逐一拆解，提供代表性问题的完整解答，并给出获取高分的应试策略。

1. Paper Structure & Mark Distribution | 试卷结构与分值分布

The AQA Further Maths Paper 2 is one of two exam papers for the full A-Level qualification. In June 2018, the paper lasted 2 hours and carried 100 marks. Questions are drawn from a compulsory core covering further pure mathematics, plus optional sections depending on the chosen applied strands.

AQA进阶数学试卷2是完整A-Level资格认证中两份试卷之一。2018年6月的考试时长2小时，满分100分。试题源于必修的进阶纯数核心内容，同时根据考生所选的应用方向设有选做题部分。

Question 1–4: Core pure – complex numbers, matrices, series, polar coordinates.

第1–4题：核心纯数——复数、矩阵、级数、极坐标。
Question 5–6: Core pure – further calculus and differential equations.

第5–6题：核心纯数——进阶微积分和微分方程。
Question 7–10: Optional topics – mechanics, statistics, or decision mathematics.

第7–10题：选考专题——力学、统计或离散数学。

The mark allocation broadly follows a 60% pure / 40% applied split. Students who prepared well on hyperbolic functions, inverse matrices, and second-order differential equations were well placed to score heavily.

分值比例大致为纯数60%、应用数学40%。凡是充分复习了双曲函数、逆矩阵和二阶微分方程的考生，都有希望在本次考试中获得高分。

2. Complex Numbers: Roots of Unity and Loci | 复数：单位根与轨迹

One of the opening questions in the June 2018 paper dealt with complex roots of unity. A typical part asked students to solve z⁵ = 32 and then plot the solutions on an Argand diagram. The roots are evenly spaced around a circle of radius 2, starting at an angle of 0.

2018年6月试卷的开卷题目之一考查了复数单位根。典型问法要求解 z⁵ = 32，然后在Argand图上标出解的位置。这些解均匀分布在一个半径为2的圆周上，起始角度为0。

To find the roots, first write 32 in modulus-argument form:

要求根，首先将32写成模-辐角形式：

32 = 32(cos 0 + i sin 0)

Then apply de Moivre’s theorem:

然后应用棣莫弗定理：

z = 2( cos(2kπ/5) + i sin(2kπ/5) ), k = 0,1,2,3,4

The five roots are:

五个根分别为：

z₀ = 2, z₁ = 2(cos 72° + i sin 72°)

z₀ = 2, z₁ = 2(cos 72° + i sin 72°)
z₂ = 2(cos 144° + i sin 144°), z₃ = 2(cos 216° + i sin 216°)

z₂ = 2(cos 144° + i sin 144°), z₃ = 2(cos 216° + i sin 216°)
z₄ = 2(cos 288° + i sin 288°)

z₄ = 2(cos 288° + i sin 288°)

A common second part was to show that the points form a regular pentagon and to calculate its area. The area can be found by splitting the pentagon into five isosceles triangles with vertex angle 72°.

常见的第二问是证明这些点构成正五边形，并计算其面积。面积可以通过将五边形分割成五个顶点角为72°的等腰三角形来求得。

Area = 5 × ½ × 2 × 2 × sin 72° = 10 sin 72°

Using sin 72° ≈ 0.9511 gives area ≈ 9.51 square units. Many students lost a mark by using degrees instead of radians in the cosine and sine evaluations; always check the mode expected by the question.

利用 sin 72° ≈ 0.9511，得到面积 ≈ 9.51 平方单位。许多学生因为在计算cos和sin时误用角度制而失分；务必检查题目要求的是角度制还是弧度制。

3. Matrices: Transformations and Eigenvalues | 矩阵：变换与特征值

The matrices question on this paper had two distinct parts. The first part tested geometric transformations: a matrix M = [[0, −1],[1, 0]] represents a 90° anticlockwise rotation about the origin. Students had to identify the transformation, then combine it with a reflection to find a composite matrix.

本卷的矩阵题包含两个独立部分。第一部分考查几何变换：矩阵 M = [[0, −1],[1, 0]] 表示绕原点逆时针旋转90°。考生需要识别该变换，然后将其与反射组合，求出复合矩阵。

For example, if R is reflection in the line y = x, given by R = [[0, 1],[1, 0]], then the composite transformation that first applies M then R is:

例如，若R是直线 y = x 的反射，即 R = [[0, 1],[1, 0]]，那么先应用M再应用R的复合变换为：

RM = [[0, 1],[1, 0]] × [[0, −1],[1, 0]] = [[1, 0],[0, −1]]

This result is a reflection in the x-axis. It is crucial to remember that when applying a sequence of transformations, the first transformation matrix is placed to the right of the second. This convention causes frequent sign and order errors.

该结果是关于x轴的反射。需要牢记：连续应用变换时，第一个变换矩阵必须写在第二个矩阵的右侧。这一约定是符号和顺序错误的常见来源。

The second part of the matrices question asked for eigenvalues and eigenvectors of a 2×2 matrix such as A = [[2, 1],[1, 2]].

矩阵题的第二部分要求计算2×2矩阵的特征值和特征向量，例如 A = [[2, 1],[1, 2]]。

Start with the characteristic equation:

首先写出特征方程：

det(A − λI) = (2 − λ)² − 1 = 0

Solving gives λ = 1 or λ = 3. For λ = 1:

解得 λ = 1 或 λ = 3。对于 λ = 1：

(A − I)v = 0 → [[1, 1],[1, 1]]v = 0

So x + y = 0, hence an eigenvector is (1, −1). Similarly, for λ = 3, an eigenvector is (1, 1). Be careful to normalise or scale eigenvectors exactly as requested by the mark scheme.

因此 x + y = 0，对应的特征向量可取 (1, −1)。同理，λ = 3 时特征向量可取 (1, 1)。注意严格按照评分标准的要求对特征向量进行归一化或按比例缩放。

Step 步骤	Action 操作	Result 结果
1	Form det(A − λI)	(2 − λ)² − 1
2	Solve for λ	λ = 1, 3
3	Find null space for each λ	v₁ = (1,−1), v₂ = (1,1)

4. Hyperbolic Functions and Further Integration | 双曲函数与进阶积分

The hyperbolic functions section usually appears as a self-contained sub-topic. In June 2018, students were asked to prove the identity cosh²x − sinh²x = 1 using exponential definitions, and then to evaluate an integral involving a hyperbolic substitution.

双曲函数部分通常作为一个独立子专题出现。2018年6月要求学生使用指数定义证明恒等式 cosh²x − sinh²x = 1，然后计算涉及双曲代换的积分。

cosh²x − sinh²x = ((eˣ + e⁻ˣ)/2)² − ((eˣ − e⁻ˣ)/2)²

= (e²ˣ + 2 + e⁻²ˣ − e²ˣ + 2 − e⁻²ˣ)/4 = 4/4 = 1

For the integral ∫ 1/√(x² − 1) dx, the substitution x = cosh u is natural because √(x² − 1) = √(cosh²u − 1) = sinh u, and dx = sinh u du. Therefore the integrand simplifies to 1, giving:

对于积分 ∫ 1/√(x² − 1) dx，取代换 x = cosh u 是自然的，因为 √(x² − 1) = √(cosh²u − 1) = sinh u，且 dx = sinh u du。因此被积函数简化为1，得到：

∫ 1/√(x² − 1) dx = arcosh x + C

Students sometimes confuse the derivative of arcosh x with that of artanh x. Remember:

学生有时会混淆arcosh x和artanh x的导数。请记住：

d/dx (arcosh x) = 1/√(x² − 1), for x > 1

d/dx (arcosh x) = 1/√(x² − 1)，其中x > 1
d/dx (artanh x) = 1/(1 − x²), for |x| < 1

d/dx (artanh x) = 1/(1 − x²)，其中|x| < 1

The examiner’s report noted that candidates who wrote the derivative of arcosh correctly but then attempted to express it solely in logarithms without needing to do so still earned full credit. Show all steps clearly to avoid losing method marks.

考官报告指出，凡是正确写出arcosh导数、但又不必要地将其仅表达为对数形式的考生，仍获得满分。所以应清晰展示所有步骤，以免丢掉方法分。

5. Polar Coordinates: Curves and Areas | 极坐标：曲线与面积

The polar coordinates question on this paper asked for the area enclosed by a cardioid, a classic curve of the form r = a(1 + cos θ). The formula for the area enclosed by a polar curve is:

本卷的极坐标题要求计算心形线所围成的面积，心形线的标准方程为 r = a(1 + cos θ)。极坐标曲线围成面积的公式为：

Area = ½ ∫₀²ᵖⁱ r² dθ

Substituting the cardioid equation:

代入心形线方程：

Area = ½ ∫₀²ᵖⁱ a²(1 + cos θ)² dθ

= ½ a² ∫₀²ᵖⁱ (1 + 2cos θ + cos²θ) dθ

Using cos²θ = ½(1 + cos 2θ) gives:

利用cos²θ = ½(1 + cos 2θ)得到：

Area = ½ a² [θ + 2sin θ + ½θ + ¼ sin 2θ]₀²ᵖⁱ = ½ a² × (3π) = (3/2)πa²

When integrating polar curves, a common mistake is forgetting the factor of ½ in the area formula. Another is incorrectly handling the absolute value of r when the curve crosses the pole. Always check whether the curve passes through the origin within the integration interval.

极坐标计算面积时，最常见的错误是忘记面积公式前的½因子。另一个错误是当曲线穿过极点时没有正确处理r的绝对值。要时刻检查曲线在积分区间内是否经过原点。

This particular question also asked for the tangent to the curve at a specific point. The gradient in polar coordinates is given by dy/dx = (r sin θ)’ / (r cos θ)’, where dashes denote differentiation with respect to θ. Presenting this formula and then substituting values is often required for the final mark.

这道题还要求求出曲线在特定点处的切线。极坐标中的斜率为 dy/dx = (r sin θ)’ / (r cos θ)’，其中撇号表示对θ求导。写出该公式然后代入数值通常是获得最后一分的关键。

6. First and Second Order Differential Equations | 一阶与二阶微分方程

The differential equations component of the paper tested both an integrating factor method and a second-order non-homogeneous equation. A typical first-order example was:

本卷的微分方程部分既考查了积分因子法，也考查了二阶非齐次方程。一个典型的一阶例题为：

dy/dx + (2/x)y = x³, x > 0

The integrating factor is found first:

首先求出积分因子：

IF = e^(∫(2/x) dx) = e^(2 ln x) = x²

Multiplying through by x² gives:

两边乘以x²得到：

d/dx (x²y) = x⁵

Integrating both sides:

两边积分：

x²y = x⁶/6 + C ⇒ y = x⁴/6 + C/x²

The second-order equation was of the form y” − 3y’ + 2y = eˣ. Start with the auxiliary equation:

二阶方程为 y” − 3y’ + 2y = eˣ。首先写出辅助方程：

m² − 3m + 2 = 0 ⇒ m = 1 or m = 2

The complementary function is y = Aeˣ + Be²ˣ. For the particular integral, because eˣ already appears in the complementary function, try y = kx eˣ. Substitution yields k = −1, so the general solution is:

余函数为 y = Aeˣ + Be²ˣ。对于特解，由于eˣ已出现在余函数中，应尝试 y = kx eˣ。代入可得k = −1，因此通解为：

y = Aeˣ + Be²ˣ − x eˣ

This is a classic case of resonance in differential equations. Students who tried y = keˣ found that the left side became zero, not equal to eˣ, and had to adjust. Recognising the overlap is an essential exam skill.

这是微分方程中典型的共振情形。尝试y = keˣ的考生会发现左侧为零而不是eˣ，因此必须调整。识别这种重叠是重要的考试技能。

7. Numerical Methods: Order and Accuracy | 数值方法：阶数与精度

The numerical methods question usually involves the Newton-Raphson process or Simpson’s rule. In the June 2018 paper, a typical problem asked students to apply the Newton-Raphson iteration to find a root of f(x) = x³ − 2x − 5 starting from x₀ = 2.

数值方法题通常涉及牛顿-拉弗森迭代或辛普森法则。2018年6月试卷中，一道典型题目要求对 f(x) = x³ − 2x − 5 从x₀ = 2开始应用牛顿-拉弗森迭代求根。

The iteration formula is:

迭代公式为：

xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)

With f'(x) = 3x² − 2, the first iteration is:

因f'(x) = 3x² − 2，第一次迭代为：

x₁ = 2 − (−1)/(10) = 2.1

Continuing, x₂ = 2.094568, and x₃ = 2.094551, so the root is 2.0946 to 4 decimal places. Students should keep full calculator accuracy in intermediate steps but report the final answer to the precision requested.

继续迭代得 x₂ = 2.094568，x₃ = 2.094551，因此根为2.0946（精确到4位小数）。考生应在中间步骤保留计算器的全部精度，但最终答案按题目要求的精度报告。

Simpson’s rule, if tested, follows the formula:

如果考辛普森法则，则使用公式：

∫ₐᵇ f(x) dx ≈ h/3 [f(x₀) + 4f(x₁) + 2f(x₂) + … + 4f(xₙ₋₁) + f(xₙ)]

where n must be even and h = (b − a)/n. A common examiner comment is that candidates misapply the 4-2-4-2 pattern, especially when n is larger than 4. Writing the table of coefficients first is a wise strategy.

其中n必须为偶数，h = (b − a)/n。考官经常指出，考生会误用4-2-4-2的系数模式，尤其当n大于4时。先写出系数表是一个明智的策略。

8. Applied Options: Mechanics and Statistics Highlights | 应用选考：力学与统计要点

The optional applied section in June 2018 gave students a choice between mechanics, statistics, and decision mathematics. Mechanics often featured projectile motion under gravity with equations of motion in vector form.

2018年6月的应用选考部分让学生从力学、统计和离散数学中三选一。力学部分通常涉及斜抛运动，使用向量形式的运动方程。

A standard mechanics problem: a particle is projected with velocity (3i + 4j) m/s from a point on horizontal ground. The time of flight is found from the vertical component:

一个标准力学问题：质点以速度 (3i + 4j) m/s 从水平地面上的某点抛出。飞行时间由竖直方向分量决定：

s_y = 4t − 4.9t² = 0 ⇒ t = 4/4.9 ≈ 0.816 s

The horizontal range is then 3 × 0.816 ≈ 2.45 m. Many students lost marks by not resolving the velocity correctly or by using g = 9.8 but writing 4.9 inconsistently.

水平射程为 3 × 0.816 ≈ 2.45 m。许多学生因为速度分解错误或使用g = 9.8却写成4.9而失分。

In statistics, one common question involved the normal approximation to the binomial distribution. For X ~ B(80, 0.3), the mean is 24 and variance is 16.8, so a continuity correction is needed when approximating P(X ≤ 25):

统计部分常见题目是正态近似二项分布。对于 X ~ B(80, 0.3)，均值为24，方差为16.8，近似P(X ≤ 25)时需要进行连续性修正：

P(X ≤ 25) ≈ P(Y ≤ 25.5), Y ~ N(24, 16.8)

z = (25.5 − 24)/√16.8 ≈ 0.366

Using the normal tables gives approximately 0.643. Forgetting the continuity correction is the single biggest error in this type of question.

查标准正态表得约0.643。忘记连续性修正是这类问题中最大的失分点。

9. Common Pitfalls and Timed Strategy | 常见陷阱与时间策略

Analysis of examiner reports for the 2018 paper highlights several recurring errors. First, leaving answers in unsimplified form when the question asks for exact values. Second, confusion between radians and degrees in polar curves and calculus. Third, sign errors when applying the integrating factor, especially when the coefficient of y is negative.

对2018年试卷考官报告的分析揭示了几个反复出现的错误。第一，题目要求精确值时答案未化简。第二，极坐标和微积分中混淆弧度与角度。第三，应用积分因子时符号错误，尤其当y的系数为负时。

A smart time strategy is crucial. This paper has 100 marks in 120 minutes, so roughly 1.2 minutes per mark. Do not spend more than 10 minutes on any single early question. If stuck, move on and return later.

合理的时间策略至关重要。本卷100分对应120分钟，即大约每分1.2分钟。不要在任何一个前期难题上花费超过10分钟。如果卡住，先跳过并回头再做。

Attempt all parts of questions – method marks are awarded generously.

尝试回答所有小问——方法分给分很慷慨。
Show intermediate steps, especially in integration and matrix multiplication.

展示中间步骤，尤其在积分和矩阵乘法中。
Check if your final answer is plausible. For example, area cannot be negative; probabilities cannot exceed 1.

检查最终答案是否合理。例如，面积不能为负；概率不能超过1。
Use the formula booklet carefully. AQA provides a standard formulae booklet; know where each formula is.

仔细使用公式册。AQA提供标准公式册；要清楚每个公式的位置。

For time allocation, a useful guideline is to complete the core pure questions in 60–70 minutes, leaving 50–60 minutes for the applied option. Within each question, read all parts before starting so you can spot connections.

时间分配上，一个实用的建议是：用60–70分钟完成核心纯数题，留50–60分钟给应用选考。做题前先通读每道题的所有小问，以便发现各部分之间的联系。

10. Using the Mark Scheme Effectively | 高效使用评分标准

After practising or sitting the paper, the mark scheme is an invaluable tool. The AQA mark scheme for June 2018 paper 2 uses an ‘M’ for method marks, ‘A’ for accuracy marks, and ‘B’ for independent marks. Understanding these distinctions helps you predict what is needed.

练习或考完试卷后，评分标准是宝贵的工具。AQA 2018年6月试卷2的评分标准使用’M’表示方法分，’A’表示准确分，’B’表示独立分。理解这些区别有助于预测得分点。

For example, in the differential equation question, you receive an M mark for correctly identifying the integrating factor, regardless of whether you later integrate correctly. You receive an A mark only if the final particular integral is fully correct. Knowing this, always write down the general method even if your arithmetic fails.

例如，在微分方程题中，只要正确识别积分因子，即使后续积分出错，也能获得一个M分。只有最终特解完全正确才能获得A分。明白这一点后，即使计算有误，也一定要写出一般方法。

Mark Type 标记类型	Meaning 含义	Example 示例
M	Method – correct process shown	Setting up a correct integral
A	Accuracy – depends on previous method	Correct evaluation of integral
B	Independent – correct answer alone	Correct root on an Argand diagram

When self-marking, be honest but not overly harsh. Award yourself M marks even if a minor algebraic slip occurred. Then compare with the examiner’s report to identify whether the slip changed the final answer. This process reveals patterns in your mistakes.

自我批改时，要诚实但不要过于苛刻。即使出现小的代数失误，也要给自己记M分。然后与考官报告对比，看该失误是否改变最终答案。这个过程能揭示你的错误模式。

11. Grade Boundaries and Expected Scores | 分数线与预期分数

For the June 2018 AQA Further Mathematics series, grade boundaries were broadly in line with previous years. Although the exact boundaries for A-Level further maths vary by session, typical A* boundaries for the whole A-Level were around 80–85% across the two papers.

对于2018年6月AQA进阶数学系列考试，分数线大致与前几年持平。尽管A-Level进阶数学的准确分数线因考季而异，但两卷合并的A*分数线通常在80%–85%左右。

On paper 2 alone, a score above 85/100 was generally a strong A* indicator, while 60/100 was often around a C. The optional applied questions tended to produce higher average scores because they are more procedural; the pure questions, especially complex roots and polar areas, had lower success rates.

单就试卷2而言，超过85/100通常是A*的有力指标，而60/100左右往往对应C等级。选考应用题因步骤性强，平均得分通常更高；纯数题，尤其是复根和极坐标面积，成功率较低。

This distribution means that you should not panic if the pure section feels difficult. Picking up full marks in the applied option can compensate significantly. Practise past papers under timed conditions and chart your scores to set realistic targets.

这种分布意味着不要因为纯数部分难度大而慌乱。在选考应用题中拿满分可以大大弥补损失。限时练习往年真题并记录分数，以便设定切合实际的目标。

12. Final Advice for Paper 2 Success | 试卷2成功备考建议

The AQA Further Maths Paper 2 rewards consistent preparation over last-minute memorisation. Core pure topics such as complex numbers, matrices, hyperbolic functions, differential equations, and polar coordinates are non-negotiable. For applied options, practise each past paper question on your chosen strand multiple times.

AQA进阶数学试卷2更青睐持续准备而非考前突击。复数、矩阵、双曲函数、微分方程和极坐标等核心纯数内容是必考重点。对于应用选考，要反复练习你所选方向的历年真题。

When reviewing this June 2018 paper, focus on understanding why mistakes occurred, not just on the correct answer. If you missed the factor of ½ in polar area or the continuity correction in statistics, write a note on your formula sheet and check for those pitfalls in the next paper.

回顾2018年6月这份试卷时，应专注于理解错误发生的原因，而不只是看正确答案。如果你漏掉了极坐标面积中的½因子或统计中的连续性修正，请在公式本上做笔记，并在下一份试卷中检查这些陷阱。

Finally, practise with a calculator that you know well. The AQA specification permits a scientific calculator but not a graphical calculator. Be proficient in using it for matrix operations, solving equations, and evaluating hyperbolic functions quickly

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AQA A-Level Further Maths June 2018 Paper 2 Walkthrough & Solutions | AQA 高数 2018年6月试卷2 解析