A-Level Cambridge Further Mathematics: Unit Test Mock Exam Analysis | A-Level Cambridge 进阶数学：单元测试模拟卷解析

📚 A-Level Cambridge Further Mathematics: Unit Test Mock Exam Analysis | A-Level Cambridge 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 Cambridge A-Level Further Mathematics. Each section presents a classic problem type, step-by-step reasoning in English, and the equivalent Chinese explanation to strengthen bilingual understanding. Key techniques from complex numbers, matrices, series, proof by induction, hyperbolic functions, polar coordinates, differential equations, and vectors are covered. Use these model solutions to identify common pitfalls and sharpen your exam technique.

本文详细讲解了一套剑桥 A-Level 进阶数学单元模拟试卷。每节选取经典题型，用英文展示分步推理，并配以相应的中文解释，以加深双语理解。内容涵盖复数、矩阵、级数、归纳法证明、双曲函数、极坐标、微分方程和向量等核心技巧。通过这些范例解答，你可以发现常见错误，提升应试能力。

1. Complex Numbers – Solving Equations | 复数方程求解

complex numbers z such that z³ = 8i. Express your answers in exact Cartesian form a + bi.

题目：求所有满足 z³ = 8i 的复数 z。将答案写成精确的代数形式 a + bi。

d side in polar form: 8i has modulus 8 and argument π/2, so 8i = 8[cos(π/2) + i sin(π/2)].

将右边转化为极坐标形式：8i 的模为 8，辐角为 π/2，因此 8i = 8[cos(π/2) + i sin(π/2)]。

θ) + i sin(3θ)]. Equate modulus: r³ = 8 → r = 2. Equate arguments: 3θ = π/2 + 2kπ, where k ∈ ℤ. Hence θ = π/6 + 2kπ/3.

设 z = r(cos θ + i sin θ)。则 z³ = r³[cos(3θ) + i sin(3θ)]。比较模：r³ = 8 → r = 2。比较辐角：3θ = π/2 + 2kπ，k ∈ ℤ。因此 θ = π/6 + 2kπ/3。

= 0, 1, 2 we obtain the three distinct roots. k=0: θ = π/6 → z = 2(cos π/6 + i sin π/6) = √3 + i. k=1: θ = 5π/6 → z = 2(cos 5π/6 + i sin 5π/6) = -√3 i. k=2: θ = 3π/2 → z = 2(cos 3π/2 + i sin 3π/2) = -2i. Always verify by cubing each root.

取 k = 0, 1, 2 得到三个不同的根。k=0: θ = π/6 → z = √3 + i。k=1: θ = 5π/6 → z = -√3 + i。k=2: θ = 3π/2 → z = -2i。务必通过三次方运算验证每个根。

---

2. Matrices – Determinants and Inverses | 矩阵行列式与逆矩阵

= [[2, -1], [3, 4]], find det(A) and A⁻¹. Hence solve the system 2x - y = 5, 3x + 4y = 6.

题目：已知矩阵 A = [[2, -1], [3, 4]]，求 det(A) 和 A⁻¹。并由此解方程组 2x - y = 5, 3x + 4y = 6。

formula gives A⁻¹ = (1/det(A)) [[4, 1], [-3, 2]]. So A⁻¹ = [[4/11, 1/11], [-3/11, 2/11]].

det(A) = 2×4 - (-1)×3 = 8 + 3 = 11。由于行列式非零，逆矩阵存在。利用伴随矩阵公式得 A⁻¹ = (1/11) [[4, 1], [-3, 2]] = [[4/11, 1/11], [-3/11, 2/11]]。

Write the system as A [x, y]ᵀ = [5, 6]ᵀ. Multiply both sides by A⁻¹: [x, y]ᵀ = A⁻¹ [5, 6]ᵀ = (1/11)[4×5 + 1×6, -3×5 + 2×6]ᵀ = (1/11)[26, -3]ᵀ. Thus x = 26/11, y = -3/11.

将方程组写成矩阵形式 A [x, y]ᵀ = [5, 6]ᵀ。两边同时左乘 A⁻¹：[x, y]ᵀ = A⁻¹ [5, 6]ᵀ = (1/11)[26, -3]ᵀ。因此 x = 26/11, y = -3/11。

A⁻¹ = (1/11) [[4, 1], [-3, 2]]

---

3. Roots of Polynomials | 多项式根的关系

quation x³ - px² + qx - r = 0 has roots α, β, γ. Express α² + β² + γ² in terms of p, q, r.

题目：三次方程 x³ - px² + qx - r = 0 的根为 α, β, γ。用 p, q, r 表示 α² + β² + γ²。

Use Vieta' e want Σα² = α²+β²+γ². Notice that (Σα)² = Σα² + 2Σαβ. Hence Σα² = (Σα)² - 2Σαβ = p² - 2q.

利用韦达定理：Σα = α+β+γ = p，Σαβ = αβ+βγ+γα = q，αβγ = r。我们需要求 Σα²。注意到 (Σα)² = Σα² + 2Σαβ，因此 Σα² = p² - 2q。

ith roots 1,2,3. Then p=6, q=11, Σα²=1+4+9=14, and p²-2q=36-22=14. The relationship holds.

该结果与 r 无关。快速验证：以简单三次方程 x³ - 6x² + 11x - 6 = 0 为例，根为 1,2,3。则 p=6, q=11，Σα²=14，而 p²-2q=14，关系成立。

---

4. Summation of Series | 级数求和

Problem: Find an expression for Σ_r=1ⁿ r(r+2). Simplify your answer fully.

题目：求 Σ_r=1ⁿ r(r+2) 的表达式，并将答案完全化简。

Σ (r² + 2r) = Σ r² + 2 Σ r. Use standard results Σ r = n(n+1)/2 and Σ r² = n(n+1)(2n+1)/6.

展开求和：Σ r(r+2) = Σ (r² + 2r) = Σ r² + 2 Σ r。运用已知公式 Σ r = n(n+1)/2，Σ r² = n(n+1)(2n+1)/6。

)/6 + n(n+1). Factor n(n+1): n(n+1)[(2n+1)/6 + 1] = n(n+1)[(2n+1 + 6)/6] = n(n+1)(2n+7)/6.

代入：Σ r² + 2 Σ r = n(n+1)(2n+1)/6 + n(n+1)。提取公因式 n(n+1)：n(n+1)[(2n+1)/6 + 1] = n(n+1)(2n+7)/6。

Σ_r=1ⁿ r(r+2) = n(n+1)(2n+7)/6

---

5. Proof by Induction | 归纳法证明

Problem: Prove by induction that for all positive integers n, Σ_r=1ⁿ (3r - 2) = (n/2)(3n - 1).

题目：用数学归纳法证明对所有正整数 n，有 Σ_r=1ⁿ (3r - 2) = (n/2)(3n - 1)。

case n=1: LHS = 3×1 - 2 = 1. RHS = (1/2)(3×1 - 1) = 1. Both sides equal, so true for n=1.

奠基：当 n=1 时，左边 = 1，右边 = (1/2)(3-1)=1，两边相等，命题成立。

Inductive step: Assume true for n = k, i.e. Σ_r=1^k (3r - 2) = (k/2)(3k - 1). For n = k+1, LHS = Σ_r=1^k merator: (k+1)(3k+2)/2 = (k+1)/2 × (3(k+1) - 1). This is exactly the RHS for n = k+1. Thus the statement holds for k+1 whenever it holds for k.

归纳递推：假设 n=k 时命题成立，即 Σ_r=1^k (3r - 2) = (k/2)(3k - 1)。则 n=k+1 时，左边 = 前 k 项和 + [3(k+1)-2] = (k/2)(3k-1) + (3k+1)。通分合并得 [3k²+5k+2]/2 = (k+1)(3k+2)/2 = (k+1)/2 × (3(k+1)-1)，这正是 n=k+1 时的右边。因此由归纳法原理，命题对所有正整数 n 成立。

---

广告 / Sponsored

6. Hyperbolic Functions – Solving Equations | 双曲函数方程求解

Problem: Solve cosh x - 3 sinh x = 2, giving your answer in logarithmic form.

题目：解方程 cosh x - 3 sinh x = 2，答案用对数形式表示。

Use definitions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ - e⁻ˣ)/2. Substitute: (eˣ + e⁻ˣ)/2 - 3(eˣ - e⁻ˣ)/2 = 2. Multiply by 2: eˣ + e⁻ˣ - 3eˣ + 3e⁻ˣ = 4. Collect terms: -2eˣ + 4e⁻ˣ = 4. Divide by 2: -eˣ + 2e⁻ˣ = 2.

利用定义 cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ - e⁻ˣ)/2。代入方程：(eˣ+e⁻ˣ)/2 - 3(eˣ-e⁻ˣ)/2 = 2。两边乘以 2 得 eˣ+e⁻ˣ-3eˣ+3e⁻ˣ = 4，整理得 -2eˣ+4e⁻ˣ = 4。除以 2：-eˣ + 2e⁻ˣ = 2。

Multiply both sides by eˣ to obtain a quadratic in eˣ: -e²ˣ + 2 = 2eˣ → e²ˣ + 2eˣ - 2 = 0. Let y = eˣ (>0). Then y² + 2y - 2 = 0. Quadratic formula: y = [-2 ± √4 + 8]/2 = [-2 ± √12]/2 = -1 ± √3. Since y>0, take y = √3 - 1. Hence eˣ = √3 - 1, so x = ln(√3 - 1).

两边同乘 eˣ 得到关于 eˣ 的二次式：-e²ˣ + 2 = 2eˣ → e²ˣ + 2eˣ - 2 = 0。令 y = eˣ (>0)，则 y²+2y-2=0。解得 y = -1 ± √3，取正根 y = √3 - 1。因此 x = ln(√3 - 1)。

Check that √3 - 1 > 0, so the logarithm is defined. Always verify that the solution satisfies the original equation.

验证 √3 - 1 > 0，因此对数有效。务必代回原方程检验。

---

7. Polar Coordinates – Area Enclosed | 极坐标围成面积

urve C has polar equation r = 2 + sin θ for 0 ≤ θ ≤ 2π. Find the total area enclosed by C.

题目：曲线 C 的极坐标方程为 r = 2 + sin θ，0 ≤ θ ≤ 2π。求 C 所围成的总面积。

The area enclosed by a polar curve r = f(θ) is A = (1/2)∫_α^β Because the curve is traced exactly once as θ goes from 0 to 2π, the total area is (1/2)∫₀^2π (2 + sin θ)² dθ.

极坐标曲线围成面积公式为 A = (1/2)∫_α^β r² dθ。曲线在 θ 从 0 到 2π 时描出完整一周，因此总面积 A = (1/2)∫₀^2π (2 + sin θ)² dθ。

in θ)² = 4 + 4 sin θ + sin²θ. Use sin²θ = (1 - cos 2θ)/2. Then the integral becomes (1/2)∫₀^2π [4 + 4 sin θ + 1/2 - (1/2) cos 2θ] dθ = (1/2)∫₀^2π (9/2 + 4 sin θ - 1/2 cos 2θ) dθ.

展开：(2 + sin θ)² = 4 + 4 sin θ + sin²θ。利用降幂公式 sin²θ = (1 - cos 2θ)/2。积分式变为 (1/2)∫₀^2π (9/2 + 4 sin θ - 1/2 cos 2θ) dθ。

tegrals give zero. Hence A = (1/2) × (9/2) × (2π) = (9π)/2. The area is 4.5π square units.

逐项积分：∫ sin θ 和 ∫ cos 2θ 在一个完整周期内的积分均为零。因此 A = (1/2) × (9/2) × 2π = 9π/2 平方单位。

---

8. First Order Differential Equations | 一阶微分方程

Problem: Solve the differential equation dy/dx + 2xy = x, given y(0) = 1.

题目：解微分方程 dy/dx + 2xy = x，并满足初始条件 y(0) = 1。

first-order: dy/dx + P(x)y = Q(x) with P(x)=2x, Q(x)=x. The integrating factor is μ(x) = e^{∫ 2x dx} = e^x². Multiply through: e^x² dy/dx + 2x e^x² y = x e^x². The left side is d/dx [y e^x²].

这是一阶线性微分方程，P(x)=2x，Q(x)=x。积分因子 μ(x) = e^{∫ 2x dx} = e^x²。方程两边乘以积分因子：e^x² dy/dx + 2x e^x² y = x e^x²，左边恰为 y e^x² 的导数。

Integrate both sides: y e^x² = ∫ x e^x² dx. Substitute u = x², du = 2x dx, thus ∫ x e^x² dx = (1/2) e^x² + C. So y e^x² = (1/2) e^x² + C → y = 1/2 + C e^-x².

两边积分：y e^x² = ∫ x e^x² dx。令 u = x²，则 ∫ x e^x² dx = (1/2) e^x² + C。于是 y = 1/2 + C e^-x²。

Apply y(0)=1: 1 = 1/2 + C → C = 1/2. Thus the particular solution is y = 1/2 + (1/2) e^-x² = (1/2)(1 + e^-x²).

代入初始条件 y(0)=1 得 C=1/2。因此特解为 y = (1/2)(1 + e^-x²)。

---

9. Second Order Differential Equations | 二阶微分方程

Problem: Solve d²y/dx² - 5 dy/dx + 6y = e^2x, given y(0)=0, y'(0)=1.

题目：解二阶微分方程 d²y/dx² - 5 dy/dx + 6y = e^2x，初始条件 y(0)=0, y'(0)=1。

Auxiliary equation: m² - 5m + 6 = 0 → m = 2, 3. The complementary function is y_c = A e^2x + B e^3x. Since the RHS e^2x duplicates a root, the particular integral takes the form y_p = C x e^2x.

辅助方程 m² - 5m + 6 = 0，根为 m=2, 3。齐次解为 y_c = A e^2x + B e^3x。右端项 e^2x 与特征根重复，故设特解形式为 y_p = C x e^2x。

Differentiate y_p: y_p' = C e^2x (1 + 2x), y_p'' = C e^2x (4 + 4x). Substitute into the DE: C e^2x[(4+4x) -5(1+2x) + 6x] = C e^2x[4+4x -5 -10x +6x] = C e^2x[-1] = e^2x. Hence C = -1. So y_p = -x e^2x.

对特解求导代入原方程，比较系数得 C = -1。因此 y_p = -x e^2x。通解为 y = A e^2x + B e^3x - x e^2x。

General solution: y = A e^2x + B e^3x - x e^2x. Apply y(0)=0: A + B = 0. y'(x) = 2A e^2x + 3B e^3x - e^2x - 2x e^2x. y' 3B -1 = 1 → 2A+3B=2. Solving A+B=0 and 2A+3B=2 gives A = -2, B = 2. Final answer: y = -2 e^2x + 2 e^3x - x e^2x.

利用初始条件求出 A = -2, B = 2。最终解为 y = -2 e^2x + 2 e^3x - x e^2x。

---

10. Vectors – Lines and Planes | 向量直线与平面

section point of the line r = (i + 2j - k) + t(2i - j + 3k) and the plane 3x - y + 2z = 5.

题目：求直线 r = (i + 2j - k) + t(2i - j + 3k) 与平面 3x - y + 2z = 5 的交点。

- t, z = -1 + 3t. Substitute into the plane equation: 3(1+2t) - (2 - t) + 2(-1 + 3t) = 5.

由直线的参数方程得 x = 1+2t, y = 2-t, z = -1+3t。代入平面方程：3(1+2t) - (2 - t) + 2(-1+3t) = 5。

-2 + t -2 + 6t = 5. Simplify: (3-2-2) + (6t + t + 6t) = -1 + 13t = 5 → 13t = 6 → t = 6/13.

展开并化简：3 + 6t -2 + t -2 + 6t = 5 → -1 + 13t = 5 → 13t = 6 → t = 6/13。

= 2 - 6/13 = 20/13, z = -1 + 18/13 = 5/13. The intersection point is (25/13, 20/13, 5/13).

回代求坐标：x = 25/13, y = 20/13, z = 5/13。交点坐标为 (25/13, 20/13, 5/13)。

Always check that the point satisfies both the line (by verifying the parameter gives consistent coordinates) and the plane equation.

务必验证该点既在直线上（参数值一致）又满足平面方程，确保解答无误。

---

Published by TutorHao | Further Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）