AS Eduqas Further Maths: Case Study Walkthroughs | AS Eduqas 进阶数学：案例分析实战演练

📚 AS Eduqas Further Maths: Case Study Walkthroughs | AS Eduqas 进阶数学：案例分析实战演练

In AS Eduqas Further Mathematics, students encounter a range of advanced pure topics including complex numbers, matrices, hyperbolic functions, series, and vectors. To excel, it's crucial to apply theoretical knowledge to problem-solving. This article presents detailed case study walkthroughs, breaking down step-by-step solutions to typical exam-style questions. Each case study is explained in both English and Chinese to reinforce understanding.

在AS Eduqas进阶数学中，学生需要掌握一系列高级纯数主题，包括复数、矩阵、双曲函数、级数和向量。要取得优异成绩，将理论知识应用于解题至关重要。本文提供详细的案例分析演练，逐步解析典型考题的解题过程。每个案例均用中英双语解释以加深理解。

---

1. Complex Roots: Solving z⁴ = -16 | 复数根：求解 z⁴ = -16

Find all roots of the equation z⁴ = -16, giving answers in the form re^iθ where r > 0 and -π < θ ≤ π.

求方程 z⁴ = -16 的所有根，结果表示为 re^iθ 形式，其中 r > 0 且 -π < θ ≤ π。

: modulus is 16, argument is π because -16 lies on the negative real axis. Hence -16 = 16e^iπ.

将 -16 表示为极坐标形式：模为 16，辐角为 π，因为 -16 位于负实轴上。因此 -16 = 16e^iπ。

Let z = re^iθ. Then z⁴ = r⁴e^i4θ = 16e^{i(π + 2kπ)} teger k. Equate moduli: r⁴ = 16 → r = 2. Equate arguments: 4θ = π + 2kπ → θ = (π + 2kπ)/4.

设 z = re^iθ，则 z⁴ = r⁴e^i4θ = 16e^{i(π + 2kπ)}（k 为整数）。比较模长：r⁴ = 16 → r = 2。比较辐角：4θ = π + 2kπ → θ = (π + 2kπ)/4。

k = 0, 1, 2, 3 to generate four distinct roots, then adjust angles into the range (-π, π].

代入 k = 0, 1, 2, 3 得到四个不同的根，然后将角度调整到区间 (-π, π]。

k=0: θ = π/4 → z₀ = 2e^iπ/4
k=1: θ = 3π/4 → z₁ = 2e^i3π/4
k=2: θ = 5π/4, which exceeds π, so subtract 2π: -3π/4 → z₂ = 2e^-i3π/4
k=3: θ = 7π/4, equivalent to -π/4 → z₃ = 2e^-iπ/4

k=0：θ = π/4 → z₀ = 2e^iπ/4
k=1：θ = 3π/4 → z₁ = 2e^i3π/4
k=2：θ = 5π/4，超出范围，减去2π得 -3π/4 → z₂ = 2e^-i3π/4
k=3：θ = 7π/4，等价于 -π/4 → z₃ = 2e^-iπ/4

The four roots are therefore 2e^iπ/4, 2e^i3π/4, 2e^-i3π/4, 2e^-iπ/4. In Cartesian form they are √2 + i√2, -√2 + i√2, -√2 - i√2, √2 - i√2.

因此四个根为 2e^iπ/4, 2e^i3π/4, 2e^-i3π/4, 2e^-iπ/4。用笛卡尔坐标表示为 √2 + i√2, -√2 + i√2, -√2 - i√2, √2 - i√2。

---

广告 / Sponsored

2. Matrix Characteristic Equation and Inverse | 矩阵的特征方程与逆矩阵

Given matrix A =

2	1
-1	3

, find A² and determine constants p and q such that A² + pA + qI = 0. Hence obtain A⁻¹.

已知矩阵 A =

2	1
-1	3

，求 A²，并确定常数 p 和 q 使得 A² + pA + qI = 0。由此求出 A⁻¹。

First compute A² = A × A:

首先计算 A² = A × A：

A² =

2	1
-1	3

×

2	1
-1	3

=

2×2 + 1×(-1)	2×1 + 1×3
(-1)×2 + 3×(-1)	(-1)×1 + 3×3

=

3	5
-5	8

Now set up the equation A² + pA + qI = 0. Substituting A², pA and qI gives:

现在建立方程 A² + pA + qI = 0。代入 A²、pA 和 qI 得：

3	5
-5	8

+

2p	p
-p	3p

+

q	0
0	q

=

0	0
0	0

nto first equation: 3 - 10 + q = 0 → q = 7. The second equation also holds: 8 -15 + 7 = 0.

比较对应元素：(1,1)：3 + 2p + q = 0；(1,2)：5 + p = 0 → p = -5；(2,1)：-5 - p = 0 → 确认 p = -5；(2,2)：8 + 3p + q = 0。将 p = -5 代入第一个方程：3 - 10 + q = 0 → q = 7。第二个方程也成立：8 -15 + 7 = 0。

A(A - 5I) = -7I → A [-(1/7)(A - 5I)] = I. Therefore A⁻¹ = -(1/7)(A - 5I) = (1/7)(5I - A).

因此 p = -5，q = 7。矩阵满足 A² - 5A + 7I = 0。变形得：A² - 5A = -7I → A(A - 5I) = -7I → A [-(1/7)(A - 5I)] = I。所以 A⁻¹ = -(1/7)(A - 5I) = (1/7)(5I - A)。

Compute 5I - A =

5-2	0-1
0-(-1)	5-3

=

3	-1
1	2

. Hence A⁻¹ = (1/7)

3	-1
1	2

.

计算 5I - A =

5-2	0-1
0-(-1)	5-3

=

3	-1
1	2

。因此 A⁻¹ = (1/7)

3	-1
1	2

。

---

3. Hyperbolic Identity and Equation | 双曲恒等式与方程

Prove that cosh²x - sinh²x = 1. Hence solve the equation 2cosh²x + 3sinh x = 5.

证明 cosh²x - sinh²x = 1。由此求解方程 2cosh²x + 3sinh x = 5。

Recall definitions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ - e⁻ˣ)/2. Then cosh²x - sinh²x = [(eˣ+e⁻ˣ)/2]² - [(eˣ-e⁻ˣ)/2]² = (e²ˣ+2+e⁻²ˣ)/4 - (e²ˣ-2+e⁻²ˣ)/4 = (4)/4 = 1.

回忆定义：cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ - e⁻ˣ)/2。那么 cosh²x - sinh²x = [(eˣ+e⁻ˣ)/2]² - [(eˣ-e⁻ˣ)/2]² = (e²ˣ+2+e⁻²ˣ)/4 - (e²ˣ-2+e⁻²ˣ)/4 = 4/4 = 1。

: 2(1 + sinh²x) + 3sinh x = 5 → 2sinh²x + 3sinh x + 2 - 5 = 0 → 2sinh²x + 3sinh x - 3 = 0.

现在使用恒等式将 cosh²x 改写为 1 + sinh²x。代入方程：2(1 + sinh²x) + 3sinh x = 5 → 2sinh²x + 3sinh x + 2 - 5 = 0 → 2sinh²x + 3sinh x - 3 = 0。

inh x: 2u² + 3u - 3 = 0. The discriminant: Δ = 3² - 4×2×(-3) = 9 + 24 = 33. Thus u = [-3 ± √33] / 4.

将其视为关于 u = sinh x 的二次方程：2u² + 3u - 3 = 0。判别式：Δ = 3² - 4×2×(-3) = 9 + 24 = 33。因此 u = [-3 ± √33] / 4。

Sinh x takes all real values; check which solutions are valid. √33 o u₁ = (-3+5.744)/4 = 0.686, u₂ = (-3-5.744)/4 = -2.186. Both are real, so we accept both.

Sinh x 取所有实数；检查哪些解有效。√33 ≈ 5.744，所以 u₁ = (-3+5.744)/4 = 0.686，u₂ = (-3-5.744)/4 = -2.186。两者均为实数，故均接受。

Hence sinh x = 0.686 or sinh x = -2.186. Solve using x = arsinh(u) = ln(u + √u²+1). For u=0.686: x₁ = ln(0.686 + √0.686²+1) = ln(0.686 + √1.4706) = ln(0.686+1.2127) = ln(1.8987) ≈ 0.641. For u=-2.186: x₂ = ln(-2.186 + √5.778+1) = ln(-2.186 + √6.778) = ln(-2.186+2.603) = ln(0.417) ≈ -0.874. These are the real solutions.

因此 sinh x = 0.686 或 sinh x = -2.186。使用 x = arsinh(u) = ln(u + √u²+1) 求解。对于 u=0.686：x₁ = ln(0.686 + √0.686²+1) = ln(0.686 + √1.4706) = ln(0.686+1.2127) = ln(1.8987) ≈ 0.641。对于 u=-2.186：x₂ = ln(-2.186 + √5.778+1) = ln(-2.186 + √6.778) = ln(-2.186+2.603) = ln(0.417) ≈ -0.874。这些是实数解。

---

4. Summation of Series Using Standard Results | 利用标准结果对级数求和

Evaluate Σ(r=1 to n) (r² + 4r - 5).

计算 Σ（r=1 到 n）(r² + 4r - 5)。

Use standard formulas: Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6, and Σ1 = n. Split the sum:

使用标准公式

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

更多咨询请联系16621398022（同微信）