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

📚 IGCSE CIE Further Maths: Mock Test Paper Walkthrough | IGCSE 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 IGCSE CIE Further Mathematics syllabus. By examining typical questions on polynomials, matrices, complex numbers, calculus, vectors and inequalities, you can reinforce key concepts and exam techniques.

本文为 IGCSE CIE 进阶数学大纲的单元测试模拟卷提供详细解析。通过讲解多项式、矩阵、复数、微积分、向量和不等式等典型题目，帮助你巩固核心概念和应试技巧。

1. Polynomial Roots and Arithmetic Progression | 多项式求根与等差数列

The cubic equation x³ – 9x² + 26x – 24 = 0 has three roots that form an arithmetic progression. Find all roots and verify the common difference.

三次方程 x³ – 9x² + 26x – 24 = 0 的三个根构成等差数列。求出所有根并验证公差。

Use the Rational Root Theorem to test possible integer roots (±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24). Substituting x = 2 gives 8 – 36 + 52 – 24 = 0, so (x – 2) is a factor.

利用有理根定理检验可能的整数根（±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24）。代入 x = 2 得 8 – 36 + 52 – 24 = 0，因此 (x – 2) 是一个因式。

Perform polynomial division or synthetic division: (x³ – 9x² + 26x – 24) ÷ (x – 2) = x² – 7x + 12.

进行多项式除法或综合除法：(x³ – 9x² + 26x – 24) ÷ (x – 2) = x² – 7x + 12。

x³ – 9x² + 26x – 24 = (x – 2)(x² – 7x + 12) = (x – 2)(x – 3)(x – 4)

The roots are 2, 3 and 4. They indeed form an arithmetic progression with common difference d = 1.

根为 2、3 和 4。它们确实构成等差数列，公差 d = 1。

2. Matrix Inverse and Linear Equations | 矩阵的逆与线性方程组

Given matrix A =

2	1
5	3

, find A⁻¹ and hence solve the system Ax = b where b =

已知矩阵 A 如上，求 A⁻¹ 并由此解方程组 Ax = b，其中 b 如上。

Compute the determinant: det(A) = 2×3 – 1×5 = 6 – 5 = 1. Since det(A) ≠ 0, the inverse exists.

计算行列式：det(A) = 2×3 – 1×5 = 6 – 5 = 1。由于 det(A) ≠ 0，逆矩阵存在。

The adjugate matrix is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements: adj(A) =

3	-1
-5	2

伴随矩阵通过交换主对角线元素并改变副对角线元素符号得到：adj(A) 如上。

Thus A⁻¹ = (1/det(A)) adj(A) =

3	-1
-5	2

因此 A⁻¹ = (1/det(A)) adj(A) 如上所示。

Now solve for x: x = A⁻¹ b =

3	-1
-5	2

3×4 + (-1)×11

-5×4 + 2×11

12 – 11

-20 + 22

现在解 x：x = A⁻¹ b，乘积计算得 x = 1, y = 2。

3. Complex Numbers in Polar Form | 复数的极坐标形式

Express z = -√3 + i in the form r(cos θ + i sin θ) with -π < θ ≤ π. Hence evaluate z³ and give the answer in Cartesian form.

将 z = -√3 + i 表示为 r(cos θ + i sin θ) 形式，-π < θ ≤ π。由此计算 z³ 并以直角坐标形式给出答案。

First, find modulus r = √((-√3)² + 1²) = √(3+1) = 2.

首先，求模 r = √((-√3)² + 1²) = √(3+1) = 2。

The argument θ satisfies tan θ = 1/(-√3) = -1/√3. Since the point (-√3, 1) lies in the second quadrant, θ = π – π/6 = 5π/6.

辐角 θ 满足 tan θ = 1/(-√3) = -1/√3。由于点 (-√3, 1) 位于第二象限，θ = π – π/6 = 5π/6。

Thus z = 2 (cos(5π/6) + i sin(5π/6)).

因此 z = 2 (cos(5π/6) + i sin(5π/6))。

Using De Moivre’s theorem: z³ = 2³ (cos(3 × 5π/6) + i sin(3 × 5π/6)) = 8 (cos(5π/2) + i sin(5π/2)).

利用棣莫弗定理：z³ = 2³ (cos(3 × 5π/6) + i sin(3 × 5π/6)) = 8 (cos(5π/2) + i sin(5π/2))。

Since 5π/2 = 2π + π/2, cos(5π/2) = cos(π/2) = 0 and sin(5π/2) = sin(π/2) = 1. Therefore z³ = 8(0 + i·1) = 8i.

因为 5π/2 = 2π + π/2，cos(5π/2) = cos(π/2) = 0，sin(5π/2) = sin(π/2) = 1。所以 z³ = 8i。

4. Integration by Substitution | 换元积分法

Evaluate ∫ from 3 to 7 of (2x+1)√(x-3) dx, using the substitution u = x – 3.

利用代换 u = x – 3 计算定积分 ∫₃⁷ (2x+1)√(x-3) dx。

Let u = x – 3, then du = dx and x = u + 3. The limits change: when x = 3, u = 0; when x = 7, u = 4.

令 u = x – 3，则 du = dx 且 x = u + 3。积分限改变：当 x = 3 时 u = 0；当 x = 7 时 u = 4。

The integrand becomes (2(u+3)+1)√u = (2u + 7)√u. The integral I = ∫₀⁴ (2u + 7)u^(1/2) du = ∫₀⁴ (2u^(3/2) + 7u^(1/2)) du.

被积函数变为 (2(u+3)+1)√u = (2u + 7)√u。积分 I = ∫₀⁴ (2u^(3/2) + 7u^(1/2)) du。

I = [ (2/(5/2)) u^(5/2) + (7/(3/2)) u^(3/2) ]₀⁴ = [ (4/5) u^(5/2) + (14/3) u^(3/2) ]₀⁴

Evaluate at u = 4: u^(5/2) = (√4)⁵ = 2⁵ = 32; u^(3/2) = 2³ = 8.

在 u = 4 处计算：u^(5/2) = 32，u^(3/2) = 8。

So I = (4/5)×32 + (14/3)×8 = 128/5 + 112/3 = (384 + 560)/15 = 944/15.

因此 I = 128/5 + 112/3 = (384+560)/15 = 944/15。

5. Area Enclosed by a Polar Curve | 极坐标曲线围成面积

Find the total area enclosed by the cardioid r = 2 + cos θ.

求心形线 r = 2 + cos θ 所围成的总面积。

The area in polar coordinates is given by A = ½ ∫₀²π r² dθ. Here r² = (2 + cos θ)² = 4 + 4 cos θ + cos² θ.

极坐标下的面积公式为 A = ½ ∫₀²π r² dθ。此处 r² = 4 + 4 cos θ + cos² θ。

Use the identity cos² θ = (1 + cos 2θ)/2. Then r² = 4 + 4 cos θ + ½ + ½ cos 2θ = 9/2 + 4 cos θ + ½ cos 2θ.

利用恒等式 cos² θ = (1 + cos 2θ)/2，得 r² = 9/2 + 4 cos θ + ½ cos 2θ。

A = ½ ∫₀²π (9/2 + 4 cos θ + ½ cos 2θ) dθ = ½ [ (9/2)θ + 4 sin θ + (1/4) sin 2θ ]₀²π

At θ = 2π, sin 2π = 0 and sin 4π = 0. At θ = 0, all sine terms are zero. Thus A = ½ × (9/2 × 2π) = ½ × 9π = 9π/2.

在 θ = 2π 处正弦项均为零。因此 A = ½ × (9π) = 9π/2。

6. Vector Lines: Intersection and Perpendicularity | 向量直线：交点与垂直

Two lines are given by r₁ = (1,2,3) + λ(1,-1,2) and r₂ = (4,1,0) + μ(2,1,-1). Find their intersection point and show that the direction vectors are perpendicular.

给定两直线 r₁ 和 r₂，求它们的交点并证明方向向量互相垂直。

For intersection, we set r₁ = r₂: (1+λ, 2-λ, 3+2λ) = (4+2μ, 1+μ, -μ).

求交点时令 r₁ = r₂，得到三个分量方程。

From the third coordinate: 3 + 2λ = -μ ⇒ μ = -3 – 2λ. Substitute into the first: 1 + λ = 4 + 2(-3 – 2λ) ⇒ 1 + λ = 4 – 6

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