Year 13 AQA Further Maths Unit Test Mock Paper Walkthrough | AQA 进阶数学 Year 13 单元测试模拟卷解析

📚 Year 13 AQA Further Maths Unit Test Mock Paper Walkthrough | AQA 进阶数学 Year 13 单元测试模拟卷解析

This article presents a full walkthrough of a mock unit test designed for Year 13 AQA Further Mathematics. The paper covers key topics from the Core Pure 2 specification, including complex numbers, matrices, hyperbolic functions, polar coordinates, differential equations, Maclaurin series and improper integrals. Each question is broken down step by step with detailed reasoning, helping you master the techniques required for exam success.

本文详细解析了一份针对 Year 13 AQA 进阶数学设计的单元测试模拟卷。试卷涵盖了核心纯数 2 的主要考点，包括复数、矩阵、双曲函数、极坐标、微分方程、麦克劳林级数和反常积分。每道题都配有逐步推导和详细解释，帮助你掌握考试所需的关键方法。

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1. Complex Numbers & de Moivre's Theorem | 复数与棣莫弗定理

Question: Let z = 1 + i√3. (a) Express z in modulus-argument form. (b) Hence find z⁶. (c) Find all cube roots of 8i, giving your answers in both modulus-argument and Cartesian form.

问题：设 z = 1 + i√3。(a) 将 z 表示成模-辐角形式。(b) 由此计算 z⁶。(c) 求 8i 的所有立方根，并将答案同时用模-辐角形式和直角坐标形式给出。

First, find the modulus of z: r = √(1² + (√3)²) = √1 + 3 = 2.

首先求模：r = √(1² + (√3)²) = √4 = 2。

The argument θ satisfies tan θ = √3 / 1 = √3, and since z lies in the first quadrant, θ = π/3. Thus z = 2(cos(π/3) + i sin(π/3)).

辐角 θ 满足 tan θ = √3，因为 z 在第一象限，所以 θ = π/3。于是 z = 2(cos(π/3) + i sin(π/3))。

Using de Moivre' rem, z⁶ = 2⁶ (cos(6 × π/3) + i sin(6 × π/3)) = 64 (cos(2π) + i sin(2π)) = 64(1 + 0i) = 64.

利用棣莫弗定理，z⁶ = 2⁶ (cos(6π/3) + i sin(6π/3)) = 64 (cos 2π + i sin 2π) = 64。

it in modulus-argument form: 8i = 8(cos(π/2) + i sin(π/2)). The cube roots are given by 8^1/3 + i sin( (π/2 + 2kπ)/3 ) ) = 2 ( cos(π/6 + 2kπ/3) + i sin(π/6 + 2kπ/3) ) for k = 0, 1, 2.

对于 8i，先写成模-辐角形式：8i = 8(cos(π/2) + i sin(π/2))。立方根由公式 8^1/3 ( cos((π/2+2kπ)/3) + i sin((π/2+2kπ)/3) ) = 2(cos(π/6 + 2kπ/3) + i sin(π/6 + 2kπ/3)) 给出，k = 0, 1, 2。

When k = 0: root₁ = 2(cos(π/6) + i sin(π/6)) = 2(√3/2 + i/2) = √3 + i.

当 k = 0：根₁ = 2(cos π/6 + i sin π/6) = √3 + i。

When k = 1: root₂ = 2(cos(5π/6) + i sin(5π/6)) = 2(-√3/2 + i/2) = -√3 + i.

当 k = 1：根₂ = 2(cos 5π/6 + i sin 5π/6) = -√3 + i。

When k = 2: root₃ = 2(cos(3π/2) + i sin(3π/2)) = 2(0 - i) = -2i.

当 k = 2：根₃ = 2(cos 3π/2 + i sin 3π/2) = -2i。

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2. Matrices & Linear Equations | 矩阵与线性方程组

Question: Given matrix A =

2	-1
3	4

. (a) Find det A and A⁻¹. (b) Hence solve the simultaneous equations 2x – y = 5, 3x + 4y = 6.

问题：已知矩阵 A =

2	-1
3	4

。(a) 求 det A 和 A⁻¹。(b) 由此解方程组 2x – y = 5，3x + 4y = 6。

The determinant of A is: det A = (2)(4) – (–1)(3) = 8 + 3 = 11.

行列式 det A = 2×4 – (–1)×3 = 11。

ed cofactor matrix. The cofactors are: C₁₁ = 4, C₁₂ = -3, C₂₁ = 1, C₂₂ = 2. Hence adj(A) =

4	1
-3	2

.

伴随矩阵由代数余子式矩阵的转置构成。余子式：C₁₁ = 4，C₁₂ = -3，C₂₁ = 1，C₂₂ = 2，因此 adj(A) =

4	1
-3	2

。

The inverse is A⁻¹ = (1/det A) adj(A) = 1/11

4	1
-3	2

.

逆矩阵 A⁻¹ = (1/11) adj(A) = 1/11

4	1
-3	2

。

The system can be written as A

x

y

=

5

6

. Multiplying both sides by A⁻¹ gives the solution.

方程组可写为 A

x

y

=

5

6

。两边左乘 A⁻¹ 即得解。

Compute:

x

y

= 1/11

4	1
-3	2

5

6

= 1/11

4×5 + 1×6

-3×5 + 2×6

= 1/11

26

-3

.

计算：

x

y

= 1/11

26

-3

。

Therefore x = 26/11 and y = –3/11.

因此 x = 26/11，y = –3/11。

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3. Hyperbolic Functions Equation | 双曲函数方程

stion: Solve the equation 4 sinh x + 3 cosh x = 5, giving your answer in logarithmic form.

问题：解方程 4 sinh x + 3 cosh x = 5，答案用对数形式表示。

Recall the definitions: sinh x = (eˣ – e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2.

回忆定义：sinh x = (eˣ – e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2。

Substitute: 4 × (eˣ – e⁻ˣ)/2 + 3 × (eˣ + e⁻ˣ)/2 = 5 ⇒ 2(eˣ – e⁻ˣ) + 1.5(eˣ + e⁻ˣ) = 5 ⇒ 3.5eˣ – 0.5e⁻ˣ = 5.

代入得：2(eˣ – e⁻ˣ) + 1.5(eˣ + e⁻ˣ) = 5 ⇒ 3.5eˣ – 0.

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