IGCSE CAIE Further Pure Mathematics: Unit Test Mock Paper Walkthrough | IGCSE CAIE 进阶数学：单元测试模拟卷解析

📚 IGCSE CAIE Further Pure Mathematics: Unit Test Mock Paper Walkthrough | IGCSE CAIE 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 IGCSE CAIE Further Pure Mathematics syllabus. The mock paper covers key topics including functions, polynomials, logarithms, trigonometry, calculus, vectors, series, and combinatorics. Each question is broken down step by step to demonstrate effective problem-solving techniques and common pitfalls to avoid.

本文针对IGCSE CAIE进阶纯数学大纲，详细解析一份单元测试模拟卷。该模拟卷涵盖函数、多项式、对数、三角、微积分、向量、数列及排列组合等核心主题。每道题均逐步拆解，演示高效的解题技巧及常见易错点。

---

1. Mock Paper Overview | 模拟卷概览

The mock exam consists of 8 questions designed to test understanding of fundamental concepts in Further Pure Mathematics. It mirrors the style and difficulty of CAIE papers, with a mix of routine computations and application problems. Each question carries marks that reflect typical exam weighting. We will now examine each question in turn, with complete solutions and commentary.

本模拟卷包含8道题目，旨在考查学生对进阶纯数基础概念的理解。试卷风格与难度贴近CAIE真题，包含常规计算题与应用题。每题分值参考了实际考试权重。现在我们将逐一讲解每道题，并给出完整解答与评注。

---

2. Question 1: Functions and Composite Functions | 第1题：函数与复合函数

) = 2x − 3 and g(x) = x² + 1. Find (a) fg(x), (b) gf(x), (c) f⁻¹(x), and state its domain.

设 f(x) = 2x − 3，g(x) = x² + 1。求 (a) fg(x)，(b) gf(x)，(c) f⁻¹(x) 及其定义域。

To find fg(x), substitute g(x) into f: f(g(x)) = f(x² + 1) = 2(x² + 1) − 3.

为求 fg(x)，将 g(x) 代入 f：f(g(x)) = f(x² + 1) = 2(x² + 1) − 3。

fg(x) = 2x² + 2 − 3 = 2x² − 1

For gf(x), substitute f(x) into g: g(f(x)) = (2x − 3)² + 1.

求 gf(x) 时，将 f(x) 代入 g：g(f(x)) = (2x − 3)² + 1。

gf(x) = 4x² − 12x + 9 + 1 = 4x² − 12x + 10

nction f⁻¹(x), let y = 2x − 3, then solve for x: x = (y + 3) / 2, so f⁻¹(x) = (x + 3) / 2.

求反函数时，令 y = 2x − 3，解出 x：x = (y + 3) / 2，因此 f⁻¹(x) = (x + 3) / 2。

The domain of f⁻¹ is all real numbers because the original linear function has range all real numbers. No restriction is needed.

f⁻¹ 的定义域为全体实数，因为原线性函数的值域是任意实数，无需限制。

---

广告 / Sponsored

3. Question 2: Polynomials and Remainder Theorem | 第2题：多项式与余式定理

factor using the factor theorem. (b) Hence factorise p(x) completely. (c) Solve p(x) = 0.

已知 p(x) = 2x³ − 5x² + x + 2。(a) 用因式定理证明 (x − 2) 是因式。(b) 并完全分解 p(x)。(c) 解方程 p(x) = 0。

2(2)³ − 5(2)² + 2 + 2 = 16 − 20 + 2 + 2 = 0, thus by factor theorem, (x − 2) is a factor.

计算 p(2)：p(2) = 2(8) − 5(4) + 2 + 2 = 16 − 20 + 2 + 2 = 0，由因式定理可知 (x − 2) 为因式。

ng long division or synthetic division to find the quotient: p(x) ÷ (x − 2) = 2x² − x − 1.

用长除法或综合除法除以 (x − 2)，得商式 2x² − x − 1。

Factorise the quadratic: 2x² − x − 1 = (2x + 1)(x − 1).

分解二次式：2x² − x − 1 = (2x + 1)(x − 1)。

p(x) = (x − 2)(2x + 1)(x − 1)

Set each factor to zero: x − 2 = 0 ⇒ x = 2; 2x + 1 = 0 ⇒ x = −½; x − 1 = 0 ⇒ x = 1. The solution set is {2, −½, 1}.

令各因式为零：x = 2，x = −½，x = 1。解集为 {2, −½, 1}。

---

4. Question 3: Logarithmic and Exponential Equations | 第3题：对数与指数方程

Solve the equation 2^x+1 = 3^x, giving your answer in the form ln a / ln b.

解方程 2^x+1 = 3^x，答案以 ln a / ln b 的形式表示。

Take natural logarithms on both sides: ln(2^x+1) = ln(3^x).

两边取自然对数：ln(2^x+1) = ln(3^x)。

Apply the power rule for logarithms: (x+1) ln 2 = x ln 3.

运用对数的幂法则：(x+1) ln 2 = x ln 3。

Expand and collect terms in x: x ln 2 + ln 2 = x ln 3 → x ln 2 − x ln 3 = −ln 2.

展开并整理含 x 的项：x ln 2 + ln 2 = x ln 3 → x ln 2 − x ln 3 = −ln 2。

Factor out x: x (ln 2 − ln 3) = −ln 2.

提取 x：x (ln 2 − ln 3) = −ln 2。

x = −ln 2 / (ln 2 − ln 3) = ln 2 / (ln 3 − ln 2)

law of logs, ln 3 − ln 2 = ln(3/2), so the solution can be written as x = ln 2 / ln(3/2).

利用对数商法则，ln 3 − ln 2 = ln(3/2)，因此解可写作 x = ln 2 / ln(3/2)。

---

5. Question 4: Trigonometric Equations and Identities | 第4题：三角方程与恒等式

Solve sin 2x = cos x for 0° ≤ x ≤ 360°.

在 0° ≤ x ≤ 360° 范围内解方程 sin 2x = cos x。

double-angle identity sin 2x = 2 sin x cos x. The equation becomes 2 sin x cos x = cos x.

运用倍角公式 sin 2x = 2 sin x cos x，方程变为 2 sin x cos x = cos x。

l terms to one side: 2 sin x cos x − cos x = 0, and factor cos x: cos x (2 sin x − 1) = 0.

移项并提取公因式：2 sin x cos x − cos x = 0，即 cos x (2 sin x − 1) = 0。

Thus either cos x = 0 or sin x = ½.

因此有 cos x = 0 或 sin x = ½。

Solve cos x = 0 within the given interval: x =

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

更多咨询请联系16621398022（同微信）