AS CIE Further Maths: Unit Test Mock Paper Walkthrough | AS CIE 进阶数学：单元测试模拟卷解析

📚 AS CIE Further Maths: Unit Test Mock Paper Walkthrough | AS CIE 进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for the AS CIE Further Mathematics syllabus. We will cover key topics including complex numbers, matrices, hyperbolic functions, differential equations, vectors, and proof by induction. Each question is explained step by step to help you understand the methods and common pitfalls.

本文详细解析了一份针对 AS CIE 进阶数学教学大纲的单元测试模拟卷。我们将涵盖复数、矩阵、双曲函数、微分方程、向量以及归纳法证明等核心专题。每道题都给出逐步解析，帮助你掌握解题方法和常见误区。

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1. Paper Structure and Assessment Overview | 试卷结构与考核概览

The mock paper follows the structure of the CIE Further Pure Mathematics 1 unit test: a two-hour paper worth 75 marks, divided into short routine questions and longer multi-part questions. Roughly half the marks are awarded for method rather than the final answer, so a clear written line of reasoning is essential even when your calculator gives a numerical result instantly.

这份模拟卷沿用 CIE 进阶纯数学 1 单元测试的结构：两小时、满分 75 分，分为短小的常规题和较长的多小问题。约有一半的分值给的是方法分而不是最终答案分，所以即使计算器能瞬间给出数值结果，清晰的书面推理过程依然不可或缺。

Questions on complex numbers and matrices usually open the paper because they are quick to set up, while differential equations and proof by induction typically appear near the end and carry higher tariffs. Hyperbolic functions and vectors are often embedded inside longer questions rather than being tested in isolation.

复数和矩阵题通常放在卷首，因为它们的建模过程很快；而微分方程和归纳法证明一般出现在卷末，分值更高。双曲函数和向量往往被嵌入较长的题目中考查，而不是单独成题。

• Section A short questions: 短小题，通常 3-5 分。 / Section A short questions: usually worth 3-5 marks.
• Section B extended questions: 长题，多为 8-12 分，含 "show that" 与 "hence" 提示。 / Section B extended questions: usually 8-12 marks, often with "show that" and "hence" prompts.
• Calculator rules: 全程允许使用图形计算器，但必须写出关键中间步骤。 / Calculator rules: a graphical calculator is permitted throughout, but key intermediate steps must still be written down.

Total marks = 75, Time allowed = 2 hours, Marks per minute ≈ 0.625

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2. Complex Numbers: De Moivre's Theorem | 复数：德莫弗定理

Question: Express (1 + √3 i)⁵ in the form a + bi, where a and b are real numbers.

题目：将 (1 + √3 i)⁵ 表示为 a + bi 的形式，其中 a 和 b 为实数。

Step 1: Find the modulus of the complex number. |z| = √(1² + (√3)²) = √1 + 3 = 2.

步骤 1：求复数的模。|z| = √(1² + (√3)²) = √1 + 3 = 2。

|z| = 2

Step 2: Determine the argument. arg(z) = arctan(√3 / 1) = π/3, and because the point lies in the first quadrant no adjustment is needed.

步骤 2：确定辐角。arg(z) = arctan(√3 / 1) = π/3，由于该点位于第一象限，无需再作调整。

arg(z) = π/3

(π/3) + i sin(π/3)). Always state the modulus and argument separately before applying the theorem, because that is where most method marks are earned.

步骤 3：将 z 写成极坐标形式：z = 2(cos(π/3) + i sin(π/3))。在应用定理前一定要分别写出模和辐角，因为方法分大多来自这一步。

Step 4: Apply De Moivre' 3) + i sin(5π/3)). Since 5π/3 lies in the fourth quadrant, cos(5π/3) = ½ and sin(5π/3) = −√3/2.

步骤 4：应用德莫弗定理：z⁵ = 2⁵ [cos(5 · π/3) + i sin(5 · π/3)] = 32(cos(5π/3) + i sin(5π/3))。由于 5π/3 位于第四象限，cos(5π/3) = ½，sin(5π/3) = −√3/2。

z⁵ = 32(½ − (√3/2) i) = 16 − 16√3 i

The most common error is forgetting to raise the modulus to the same power as the angle is multiplied by. A second frequent slip is writing the argument in degrees while the rest of the calculation uses radians; keep everything in radians unless the question states otherwise.

最常见的错误是忘记把模也同时乘方，而只把角度乘以 5。第二个高频失误是角度用度数、其他计算用弧度；除非题目另有说明，请全程使用弧度。

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

= 1, the n roots are equally spaced on the unit circle with arguments 2kπ/n for k = 0, 1, …, n − 1, and their sum is always zero for n ≥ 2.

单位根问题几乎总是在求出根之后要求求和或求积。若 zⁿ = 1，则 n 个根均匀分布在单位圆上，辐角为 2kπ/n（k = 0, 1, …, n − 1）；当 n ≥ 2 时，它们的和恒为 0。

For the cube roots of unity, the roots are 1, e^2πi/3 and e^4πi/3. Setting ω = e^2πi/3 ³ = 1, which are the two identities that unlock most algebraic manipulation in this topic.

对于三次单位根，三个根为 1、e^2πi/3 与 e^4πi/3。记 ω = e^2πi/3，则得到 1 + ω + ω² = 0 以及 ω³ = 1，这两个恒等式是解决本专题大多数代数变形题的钥匙。

1 + ω + ω² = 0, ω³ = 1

Loci in the Argand diagram follow two standard shapes. An equation of the form |z − (a + bi)| bi)) = θ describes a half-line starting at (a, b) at angle θ, with the endpoint excluded.

阿冈图上的轨迹只有两种标准形状。形如 |z − (a + bi)| = r 的方程表示以 (a, b) 为圆心、r 为半径的圆；而 arg(z − (a + bi)) = θ 表示从 (a, b) 出发、倾角为 θ 的半直线，且起点不包含在内。

• Circle locus: 圆轨迹，写成 |z − z₀| = r，并注明圆心与半径。 / Circle locus: write it as |z − z₀| = r and state the centre and radius.
• Half-line locus: 半直线轨迹，注意方向与不取端点。 / Half-line locus: watch the direction and the excluded endpoint.
• Perpendicular bisector: 若 |z − a| = |z − b|，轨迹是 a、b 两点连线的垂直平分线。 / Perpendicular bisector: if |z − a| = |z − b|, the locus is the perpendicular bisector of the segment joining a and b.

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4. Matrices: Determinants, Inverses and Simultaneous Equations | 矩阵：行列式、逆矩阵与联立方程

Question: Given A = [[3, 2], [1, 4]], find A⁻¹ and use it to solve 3x + 2y = 7, x + 4y = 9.

题目：已知 A = [[3, 2], [1, 4]]，求 A⁻¹，并用它解 3x + 2y = 7、x + 4y = 9。

) = 12 − 2 = 10. Since det A ≠ 0, the inverse exists and the system has a unique solution.

对于 2 × 2 矩阵，行列式为 ad − bc。此处 det A = (3)(4) − (2)(1) = 12 − 2 = 10。因为 det A ≠ 0，逆矩阵存在，方程组有唯一解。

det A = 10

The inverse is formed by swapping the leading diagonal entries, changing the sign of the other two, and dividing by the determinant.

求逆矩阵的方法是：交换主对角线元素，改变另外两个元素的符号，再整体除以行列式。

A⁻¹ = (1/10) [[4, −2], [−1, 3]]

Write the system as AX = B, so X = A⁻¹B. Then X = (1/10) [[4, −2], [−1, 3]] [7, 9]ᵀ = (1/10) [28 − 18, −7 + 27]ᵀ = (1/10) [10, 20]ᵀ = [1, 2]ᵀ.

把方程组写成 AX = B，于是 X = A⁻¹B。则 X = (1/10) [[4, −2], [−1, 3]] [7, 9]ᵀ = (1/10) [28 − 18, −7 + 27]ᵀ = (1/10) [10, 20]ᵀ = [1, 2]ᵀ。

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x = 1, y = 2

t an inverse; instead state that the system is either inconsistent or has infinitely many solutions, and use row operations to decide which case applies.

务必代回检验：3(1) + 2(2) = 7，1 + 4(2) = 9，解得到验证。若 det A = 0，不要试图求逆；应指出方程组无解或有无穷多解，并用行变换判断具体属于哪种情形。

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5. Matrices: Linear Transformations and Invariant Lines | 矩阵：线性变换与不变直线

sformation of the plane. The absolute value of its determinant gives the area scale factor, and the sign of the determinant tells you whether the orientation has been reversed.

2 × 2 矩阵表示平面上的一个线性变换。其行列式的绝对值给出面积伸缩因子，行列式的符号则说明方向是否被反转。

ry region has its area multiplied by 6, and the orientation is preserved because the determinant is positive. Points on the x-axis are mapped to (2x, 0), so the x-axis is an invariant line; points on the y-axis are mapped to (0, 3y), so the y-axis is also invariant.

对于 M = [[2, 0], [0, 3]]，det M = 6，因此任意区域的面积都放大 6 倍；由于行列式为正，方向保持不变。x 轴上的点被映到 (2x, 0)，所以 x 轴是不变直线；y 轴上的点被映到 (0, 3y)，所以 y 轴也是不变直线。

earranging gives (M − λI)v = 0, which has non-trivial solutions only when det(M − λI) = 0.

要求过原点的不变直线，需要解 Mv = λv，其中 λ 是特征值。整理得 (M − λI)v = 0；只有当 det(M − λI) = 0 时才有非零解。

det(M − λI) = 0

• Rotation by θ: 旋转 θ，矩阵 [[cos θ, −sin θ], [sin θ, cos θ]]，行列式为 1。 / Rotation by θ: matrix [[cos θ, −sin θ], [sin θ, cos θ]], determinant 1.
• Reflection in the x-axis: 关于 x 轴反射，矩阵 [[1, 0], [0, −1]]，行列式为 −1。 / Reflection in the x-axis: matrix [[1, 0], [0, −1]], determinant −1.
• Enlargement by factor k: 以 k 为比例放大，矩阵 [[k, 0], [0, k]]，行列式为 k²。 / Enlargement by factor k: matrix [[k, 0], [0, k]], determinant k².

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6. Hyperbolic Functions: Definitions and Identities | 双曲函数：定义与恒等式

hem much easier to handle than they first appear. The two core definitions are cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ − e⁻ˣ)/2, with tanh x = sinh x / cosh x.

双曲函数由指数函数定义，因此比它乍看之下容易处理得多。两个核心定义是 cosh x = (eˣ + e⁻ˣ)/2 与 sinh x = (eˣ − e⁻ˣ)/2，并有 tanh x = sinh x / cosh x。

ows by direct expansion. Note the sign: it is a minus, not a plus as in the trigonometric version, and this single difference explains most of the confusion students experience.

由这些定义直接展开即可得到基本恒等式 cosh²x − sinh²x = 1。注意符号：这里是减号，而不是三角版本中的加号；正是这一点差异造成了学生大多数混淆。

cosh²x − sinh²x = 1

h(x + y) = sinh x cosh y + cosh x sinh y, and cosh(x + y) = cosh x cosh y + sinh x sinh y.

加法公式与三角函数的对应公式形式相同，只是双曲恒等式中的加号取代了三角公式中的减号。于是 sinh(x + y) = sinh x cosh y + cosh x sinh y，cosh(x + y) = cosh x cosh y + sinh x sinh y。

• cosh 0 = 1, sinh 0 = 0: 基本取值。 / cosh 0 = 1 and sinh 0 = 0: basic values.
• cosh is even, sinh is odd: cosh(−x) = cosh x, sinh(−x) = −sinh x。 / cosh is even and sinh is odd: cosh(−x) = cosh x, sinh(−x) = −sinh x.
• Range: cosh x ≥ 1 for all real x, while sinh x takes every real value。 / Range: cosh x ≥ 1 for all real x, while sinh x takes every real value.

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7. Hyperbolic Functions: Inverse Forms and Calculus | 双曲函数：反函数形式与微积分

verse hyperbolic functions have logarithmic forms that CIE examiners frequently ask for in "show that" questions. The standard results are arsinh x = ln(x + √x² + 1) for all real x, and arcosh x = ln(x + √x² − 1) for x ≥ 1.

反双曲函数具有对数形式，CIE 考官经常在 "show that" 类题目中考查。标准结果为：对任意实数 x，arsinh x = ln(x + √x² + 1)；对 x ≥ 1，arcosh x = ln(x + √x² − 1)。

To prove the arcosh result, let y = arcosh x, so cosh y = x. Then (eʸ + e⁻ʸ)/2 = x, which gives e²ʸ − 2x eʸ + 1 = 0. Solving this quadratic in eʸ and taking the positive root yields eʸ = x + √x² − 1.

证明 arcosh 的结果时，令 y = arcosh x，则 cosh y = x。于是 (eʸ + e⁻ʸ)/2 = x，得 e²ʸ − 2x eʸ + 1 = 0。解这个关于 eʸ 的二次方程并取正根，得 eʸ = x + √x² − 1。

For example, solving cosh x = 2 gives x = ±arcosh 2 = ±ln(2 + √3 h is even, so a positive right-hand side always produces two solutions unless the question restricts the domain.

例如，解 cosh x = 2 得 x = ±arcosh 2 = ±ln(2 + √3)，数值约为 ±1.3169。记住 cosh 是偶函数，所以只要右端为正，就总有两个解，除非题目限定了定义域。

x = ±ln(2 + √3) ≈ ±1.3169

sinh x, and d/dx tanh x = sech²x, with the corresponding integrals obtained by reversing these results.

求导与积分可直接由指数定义推出。标准导数为 d/dx sinh x = cosh x、d/dx cosh x = sinh x、d/dx tanh x = sech²x；对应的积分结果即将这些式子反向使用。

• ∫ sinh x dx = cosh x + C: 由导数结果反向得到。 / ∫ sinh x dx = cosh x + C: obtained by reversing the derivative result.
• ∫ cosh x dx = sinh x + C: 记忆时注意符号不发生改变。 / ∫ cosh x dx = sinh x + C: note that no sign change occurs.
• ∫ 1/√x² + 1 dx = arsinh x + C: 标准反双曲积分形式。 / ∫ 1/√x² + 1 dx = arsinh x + C: the standard inverse hyperbolic integral.

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8. Differential Equations: First Order Methods | 微分方程：一阶方法

First order differential equations in Further Maths fall into two families: separable equations and linear equations solved with an integrating factor. Identifying the family correctly is the single most important step, because the wrong method wastes several minutes with no marks to show for it.

进阶数学中的一阶微分方程分为两大类：可分离变量的方程，以及用积分因子求解的线性方程。正确判断类型是最关键的一步，因为方法选错会浪费好几分钟却拿不到任何分数。

Question: Solve dy/dx + y tan x = sec x for 0 ≤ x < π/2.

题目：求解 dy/dx + y tan x = sec x，其中 0 ≤ x < π/2。

This is linear with P(x) = tan x, so the integrating factor is e^{∫tan x dx} = e^{ln sec x} = sec x. Multiplying through gives d/dx (y sec x) = sec²x.

这是线性方程，P(x) = tan x，因此积分因子为 e^{∫tan x dx} = e^{ln sec x} = sec x。两边相乘得 d/dx (y sec x) = sec²x。

Integrating factor = sec x

as y = 0 when x = 0 is given, then C = 0 and the particular solution is simply y = sin x.

两边积分得 y sec x = tan x + C，因此 y = sin x + C cos x。若给出初始条件如 x = 0 时 y = 0，则 C = 0，特解就是 y = sin x。

= xy with y = 1 when x = 0, separate the variables to obtain ∫(1/y) dy = ∫x dx, giving ln|y| = x²/2 + C. Applying the condition yields C = 0, so y = e^x²/2.

对于可分离方程，例如 dy/dx = xy 且 x = 0 时 y = 1，分离变量得 ∫(1/y) dy = ∫x dx，即 ln|y| = x²/2 + C。代入条件得 C = 0，故 y = e^x²/2。

n, or omitting the constant of integration before applying the initial condition. Write the constant explicitly at the first integration step and substitute the condition before rearranging.

常见错误是除以 y 时未检查 y = 0 是否为解，或者忘记在代入初始条件前写出积分常数。请在第一次积分时就明确写出常数，并在整理表达式之前先代入条件。

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9. Differential Equations: Second Order Linear Equations | 微分方程：二阶线性方程

A second order linear equation with constant coefficients is solved in two independent stages: the complementary function from the auxiliary equation, and a particular

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