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

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

This walkthrough unpacks a Year 13 AQA Further Mathematics unit test mock paper, concentrating on Further Pure 2 topics. Each question is analysed step by step, with model solutions that clarify reasoning, highlight common missteps, and strengthen exam technique. Use these solutions to review complex numbers, hyperbolic functions, matrix diagonalisation, polar area, second-order differential equations, integration by substitution, series, Maclaurin expansions, and roots of polynomial equations.

本篇解析详细拆解一份AQA 13年级进阶数学单元测试模拟卷，聚焦Further Pure 2核心内容。每道题均给出逐步推理与范例解法，旨在理清思路、点明常见错误、巩固应考技能。通过本文可系统复习复数、双曲函数、矩阵对角化、极坐标面积、二阶微分方程、换元积分法、级数求和、麦克劳林展开以及多项式求根。

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

Problem: Let z = 1 + i√3. Find |z| and arg z (in radians). Express z in the form re^iθ. Hence evaluate z⁶.

题目：设z = 1 + i√3。求|z|和辐角arg z（以弧度表示）。将z表示为re^iθ的形式，并由此计算z⁶。

First compute the modulus: |z| = √(1² + (√3)²) = √(1+3) = 2.

首先计算模：|z| = √(1² + (√3)²) = √(1+3) = 2。

The argument satisfies tan θ = (√3)/1 = √3. Since z lies in the first quadrant, θ = π/3.

辐角满足 tan θ = √3/1 = √3。因为z位于第一象限，故 θ = π/3。

Therefore, the exponential form is z = 2 e^iπ/3.

因此指数形式为 z = 2 e^iπ/3。

By de Moivre’s theorem, z⁶ = 2⁶ e^{i(6 × π/3)} = 64 e^i2π = 64(cos 2π + i sin 2π) = 64.

由棣莫弗定理，z⁶ = 2⁶ e^i(6×π/3) = 64 e^i2π = 64(cos 2π + i sin 2π) = 64。

Thus z⁶ = 64.

因此 z⁶ = 64。

2. Hyperbolic Equation in Logarithmic Form | 双曲方程的对数形式求解

Problem: Solve 4 sinh x + 3 cosh x = 2, giving the answer in logarithmic form.

题目：解方程 4 sinh x + 3 cosh x = 2，并给出对数形式的解。

Use the exponential definitions: sinh x = (e^x – e^–x)/2, cosh x = (e^x + e^–x)/2.

利用指数定义：sinh x = (e^x – e^–x)/2，cosh x = (e^x + e^–x)/2。

Substituting gives: 4 × ½(e^x – e^–x) + 3 × ½(e^x + e^–x) = 2 → 2e^x – 2e^–x + 1.5e^x + 1.5e^–x = 2 → 3.5e^x – 0.5e^–x = 2.

代入得：4×½(e^x – e^–x) + 3×½(e^x + e^–x) = 2 → 2e^x – 2e^–x + 1.5e^x + 1.5e^–x = 2 → 3.5e^x – 0.5e^–x = 2。

Multiply through by 2e^x: 7e^2x – 1 = 4e^x → 7e^2x – 4e^x – 1 = 0.

两边乘以2e^x：7e^2x – 1 = 4e^x → 7e^2x – 4e^x – 1 = 0。

Let y = e^x. Then 7y² – 4y – 1 = 0. Using the quadratic formula: y = [4 ± √(16 + 28)]/14 = [4 ± 2√11]/14 = [2 ± √11]/7.

令 y = e^x。则 7y² – 4y – 1 = 0。由求根公式：y = [4 ± √(16 + 28)]/14 = [4 ± 2√11]/14 = [2 ± √11]/7。

Since e^x > 0, reject the negative root (2 – √11 < 0). Hence e^x = (2 + √11)/7.

由于 e^x > 0，舍去负根（2 – √11 < 0）。故 e^x = (2 + √11)/7。

Therefore, x = ln((2 + √11)/7).

因此，x = ln

Published by TutorHao | Year 13 进阶数学 Revision Series | aleveler.com

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