A-Level OCR Further Mathematics: Unit Test Mock Paper Walkthrough | A-Level OCR 进阶数学：单元测试模拟卷解析

📚 A-Level OCR Further Mathematics: Unit Test Mock Paper Walkthrough | A-Level OCR 进阶数学：单元测试模拟卷解析

This article walks through a full unit test mock paper designed for OCR A-Level Further Mathematics. Covering topics such as complex numbers, matrices, hyperbolic functions, differential equations, polar coordinates, vectors, Maclaurin series, recurrence relations and mechanics, each question is explained in detail. Whether you are preparing for an end-of-unit test or the final exam, these step-by-step solutions will sharpen your problem-solving skills.

本文详细解析了一份为 OCR A-Level 进阶数学设计的单元测试模拟卷。内容涵盖复数、矩阵、双曲函数、微分方程、极坐标、向量、麦克劳林级数、递推关系及力学等核心主题。无论你是在准备单元测验还是大考，这些分步解答都能提升你的解题能力。

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1. Complex Numbers: Roots and Argand Diagram | 复数：求根与Argand图

Question 1: Solve z³ = 8i, giving answers in the form r e^(iθ) with r > 0, -π < θ ≤ π. Plot the roots on an Argand diagram.

问题1：解方程 z³ = 8i，答案表示为 r e^(iθ) 形式，其中 r > 0，-π < θ ≤ π。并在 Argand 图上标出根。

First, express 8i in polar form. Its modulus is 8 and its argument is π/2, so 8i = 8 e^(iπ/2).

首先，将 8i 写成极坐标形式。模长为 8，辐角为 π/2，因此 8i = 8 e^(iπ/2)。

Taking cube roots gives z = [8 e^(iπ/2)]^(1/3) = 2 e^(i(π/6 + 2kπ/3)), where k = 0, 1, 2.

开立方根可得 z = [8 e^(iπ/2)]^(1/3) = 2 e^(i(π/6 + 2kπ/3))，其中 k = 0, 1, 2。

The three roots are:

三个根分别为：

k = 0: z₁ = 2 e^(iπ/6)

k = 1: z₂ = 2 e^(i5π/6)

k = 2: z₃ = 2 e^(-iπ/2)

These roots lie on a circle of radius 2, separated by angles of 2π/3. On an Argand diagram, they form the vertices of an equilateral triangle with one vertex on the negative imaginary axis.

这些根位于半径为 2 的圆上，互隔 2π/3 的角度。在 Argand 图中，它们构成等边三角形的顶点，其中一个顶点在负虚轴上。

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2. Matrices: Eigenvalues and Eigenvectors | 矩阵：特征值与特征向量

Question 2: Given the matrix A =

2	1
-1	3

, find its eigenvalues and corresponding eigenvectors. Hence describe the nature of the invariant lines through the origin.

问题2：已知矩阵 A =

2	1
-1	3

，求其特征值和对应的特征向量。并由此说明过原点的不变线的性质。

The characteristic equation is det(A – λI) = 0. Compute:

特征方程为 det(A – λI) = 0。计算如下：

det(A – λI) = (2 – λ)(3 – λ) – (1)(-1) = λ² – 5λ + 7 = 0

Solving, λ = (5 ± √(25 – 28)) / 2 = (5 ± i√3) / 2. Thus the eigenvalues are complex conjugates λ₁ = (5 + i√3)/2 and λ₂ = (5 – i√3)/2.

解得 λ = (5 ± √(25 – 28)) / 2 = (5 ± i√3) / 2。因此特征值为共轭复数 λ₁ = (5 + i√3)/2，λ₂ = (5 – i√3)/2。

For λ₁, solve (A – λ₁I)v = 0. Choosing the first row gives a relationship: (2 – λ₁)x + y = 0 → y = (λ₁ – 2)x. Taking x = 1 yields an eigenvector v₁ =

1

λ₁ – 2

, which simplifies to a complex vector.

对于 λ₁，解 (A – λ₁I)v = 0。取第一行得关系式：(2 – λ₁)x + y = 0 → y = (λ₁ – 2)x。取 x = 1 得特征向量 v₁ =

1

λ₁ – 2

，这是一个复向量。

The conjugate eigenvalue λ₂ will have the conjugate eigenvector. Since the eigenvalues are non-real, there are no real invariant lines through the origin; the transformation consists of a rotation combined with an enlargement.

共轭特征值 λ₂ 具有共轭特征向量。由于特征值非实数，不存在过原点的实不变线；该变换由旋转和伸缩组合而成。

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3. Hyperbolic Functions: Exact Values and Logarithms | 双曲函数：精确值及对数式

Question 3: Given sinh x = 3/4, find the exact values of cosh x and tanh x. Hence express x in terms of natural logarithms.

问题3：已知 sinh x = 3/4，求 cosh x 和 tanh x 的精确值。并用自然对数表示 x。

Using the identity cosh² x – sinh² x = 1, we have cosh² x = 1 + (3/4)² = 1 + 9/16 = 25/16, so cosh x = 5/4 (positive root).

利用恒等式 cosh² x – sinh² x = 1，得 cosh² x = 1 + (3/4)² = 1 + 9/16 = 25/16，因此 cosh x = 5/4（取正根）。

Then tanh x = sinh x / cosh x = (3/4) / (5/4) = 3/5.

进而 tanh x = sinh x / cosh x = (3/4) / (5/4) = 3/5。

To find x, use the logarithmic form of arsinh: arsinh y = ln(y + √(y² + 1)). With y = 3/4, √(y² + 1) = √(9/16 + 1) = √(25/16) = 5/4.

为求 x，利用反双曲正弦的对数形式：arsinh y = ln(y + √(y² + 1))。代入 y = 3/4，得 √(y² + 1) = √(9/16 + 1) = √(25/16) = 5/4。

Thus x = ln(3/4 + 5/4) = ln(2).

因此 x = ln(3/4 + 5/4) = ln(2)。

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4. First-Order Differential Equation: Integrating Factor | 一阶微分方程：积分因子

Question 4: Solve dy/dx + 2y tan x = sin x, given that y(0) = 1.

问题4：解微分方程 dy/dx + 2y tan x = sin x，已知 y(0) = 1。

Identify the integrating factor I = e^(∫ P dx) with P(x) = 2 tan x. ∫ 2 tan x dx = -2 ln|cos x| = ln(sec² x). Hence I = e^(ln(sec² x)) = sec² x.

识别积分因子 I = e^(∫ P dx)，其中 P(x) = 2 tan x。∫ 2 tan x dx = -2 ln|cos x| = ln(sec² x)。因此 I = e^(ln(sec² x)) = sec² x。

Published by TutorHao | A-Level 进阶数学 Revision Series | aleveler.com

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