Year 12 CAIE Further Mathematics: Case Study Practice | CAIE 12年级进阶数学：案例分析实战演练

📚 Year 12 CAIE Further Mathematics: Case Study Practice | CAIE 12年级进阶数学：案例分析实战演练

Welcome to this case study revision guide for CAIE AS Level Further Mathematics. This article walks you through carefully selected problems that mirror the style of Paper 1 (Further Pure Mathematics 1). Each example is dissected step by step, highlighting key techniques and common pitfalls. By working through these cases, you will develop a deeper understanding of complex numbers, matrices, series, induction, polar coordinates, and more.

欢迎阅读这本针对CAIE AS阶段进阶数学的案例分析复习指南。本文带领大家逐步解析精心挑选的题目，这些题目模拟了试卷一（纯数进阶1）的风格。每个例题都逐步拆解，强调关键技巧与常见错误。通过练习这些案例，您将更深入地理解复数、矩阵、级数、归纳法、极坐标等核心内容。

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1. Complex Numbers: Loci and Roots | 复数：轨迹与根

Problem: (a) Find the Cartesian equation of the locus of points z such that |z − 2i| = |z + 2|. (b) Determine all three cube roots of −8, giving your answers in exact Cartesian form.

题目：(a) 求满足 |z − 2i| = |z + 2| 的点 z 的轨迹的笛卡尔方程。(b) 求出 −8 的全部三个立方根，用精确的笛卡尔形式表示。

Substitute z = x + iy. Then |(x + iy) − 2i| = |x + i(y − 2)|, and |z + 2| = |(x + 2) + iy|.

设 z = x + iy，则 |(x + iy) − 2i| = |x + i(y − 2)|，而 |z + 2| = |(x + 2) + iy|。

moduli: √(x² + (y − 2)²) = √((x + 2)² + y²). Squaring both sides removes the square roots.

模相等：√(x² + (y − 2)²) = √((x + 2)² + y²)。两边平方消去根号。

x² + (y − 2)² = (x + 2)² + y² → x² + y² − 4y + 4 = x² + 4x + 4 + y².

展开得 x² + y² − 4y + 4 = x² + 4x + 4 + y²。

Cancelling x², y² and 4 gives −4y = 4x → y = −x. The locus is the straight line y = −x.

消去 x²、y² 和 4 得到 −4y = 4x → y = −x。轨迹为直线 y = −x。

). The cube roots are found via De Moivre’s theorem: z³ = 8(cos(π + 2kπ) + i sin(π + 2kπ)) ⇒ z = 2[cos((π + 2kπ)/3) + i sin((π + 2kπ)/3)], k = 0, 1, 2.

(b) 部分：将 −8 写成极坐标形式 8(cos π + i sin π)。由棣莫弗定理，z³ = 8(cos(π + 2kπ) + i sin(π + 2kπ)) ⇒ z = 2[cos((π + 2kπ)/3) + i sin((π + 2kπ)/3)]，k = 0, 1, 2。

k = 0: 2(cos(π/3) + i sin(π/3)) = 2(1/2 + i√3/2) = 1 + i√3.

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

k = 1: 2(cos π + i sin π) = 2(−1 + 0i) = −2.

k = 1 时：2(cos π + i sin π) = −2。

k = 2: 2(cos(5π/3) + i sin(5π/3)) = 2(1/2 − i√3/2) = 1 − i√3. The three cube roots are 1 + i√3, −2, 1 − i√3.

k = 2 时：2(cos(5π/3) + i sin(5π/3)) = 1 − i√3。三个立方根为 1 + i√3、−2、1 − i√3。

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2. Matrices: Inverses and Transformations | 矩阵：逆矩阵与线性变换

Problem: Let

2	1
−1	3

us equations 2x + y = 5, −x + 3y = 1. Describe the linear transformation represented by A.

题目：设矩阵 A =

2	1
−1	3

，求 A⁻¹ 并用它解方程组 2x + y = 5, −x + 3y = 1。描述 A 所表示的线性变换。

(1)(−1) = 6 + 1 = 7. The adjugate swaps the diagonal elements and changes signs of the off‑diagonal: adj(A) =

3	−1
1	2

.

对于2×2矩阵，A⁻¹ = (1/det(A)) × adj(A)。det(A) = 2×3 − 1×(−1) = 7。伴随矩阵交换主对角线元素并改变副对角线符号：adj(A) =

3	−1
1	2

。

Hence A⁻¹ = (1/7)

3	−1
1	2

=

3/7	−1/7
1/7	2/7

.

因此 A⁻¹ = (1/7) adj(A) =

3/7	−1/7
1/7	2/7

。

Write the system as Ax = b, where x = (x, y)ᵀ and b = (5, 1)ᵀ. Then x = A⁻¹b. Compute:

x

y

=

3/7	−1/7
1/7	2/7

5

1

=

(3/7)×5 + (−1/7)×1

(1/7)×5 + (2/7)×1

=

(15−1)/7

(5+2)/7

=

2

1

. So x = 2, y = 1.

将方程组写成 Ax = b，其中 x = (x, y)ᵀ，b = (5, 1)ᵀ。那么 x = A⁻¹b。计算得 x = 2, y = 1。

The transformation T: v ↦ Av elogram. The absolute value of det(A) = 7 gives the area scale factor. The columns (2, −1)ᵀ and (1, 3)ᵀ are images of the basis vectors. Describing fully: a shear combined with scaling.

变换 T: v ↦ Av 将单位正方形映成平行四边形。|det(A)| = 7 是面积缩放因子。列向量 (2, −1)ᵀ 和 (1, 3)ᵀ 是基向量的像。该变换可描述为剪切与缩放的复合。

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3. Summation of Series: Standard Results | 级数求和：标准公式应用

Problem: Show that Σᵣ₌₁ⁿ r(r + 1) = (n/3)(n + 1)(n + 2).

题目：证明 Σᵣ₌₁ⁿ r(r + 1) = (n/3)(n + 1)(n + 2)。

Write Σ r(r + 1) = Σ (r² + r) = Σ r² + Σ r, where both sums run from 1 to n.

将 Σ r(r + 1) 拆成 Σ (r² + r) = Σ r² + Σ r，求和指标均从1到 n。

Standard results: Σᵣ₌₁ⁿ r = n(n + 1)/2, Σᵣ₌₁ⁿ r² = n(n + 1)(2n + 1)/6.

Published by TutorHao | Year 12 进阶数学 Revision Series | aleveler.com

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