AS Further Mathematics: Unit Test Paper | AS 进阶数学：单元测试卷

📚 AS Further Mathematics: Unit Test Paper | AS 进阶数学：单元测试卷

This comprehensive unit test paper is designed for AS Further Mathematics students. It covers core topics including complex numbers, matrices, hyperbolic functions, differential equations, Maclaurin series, vectors, and more. Each question is paired with a fully worked solution, providing an excellent revision resource to strengthen your understanding and exam technique.

这份全面的单元测试卷专为 AS 进阶数学学生设计。试卷涵盖复数、矩阵、双曲函数、微分方程、麦克劳林级数、向量等核心主题。每道题均配有详细解答，是一份绝佳的复习资料，有助于巩固理解并提高应试技巧。

1. Complex Numbers: Modulus-Argument Form & Roots | 复数：模-辐角形式与求根

Express the complex number z = (1 + i√3)² in modulus-argument form. Hence find all cube roots of z, giving your answers in exact Cartesian form.

将复数 z = (1 + i√3)² 表示为模-辐角形式。由此求出 z 的所有立方根，并以精确的直角坐标形式给出答案。

First write w = 1 + i√3. |w| = √(1² + (√3)²) = √4 = 2. arg(w) = tan⁻¹(√3/1) = π/3. Then z = w², so |z| = 2² = 4, arg(z) = 2 × π/3 = 2π/3. Thus z = 4(cos 2π/3 + i sin 2π/3) or 4e^(i2π/3). The cube roots of z satisfy |r| = ³√4, with arguments (2π/3 + 2kπ)/3 for k = 0,1,2. k=0: arg = 2π/9; k=1: arg = 8π/9; k=2: arg = 14π/9. Cartesian forms: ³√4 (cos 2π/9 + i sin 2π/9), ³√4 (cos 8π/9 + i sin 8π/9), ³√4 (cos 14π/9 + i sin 14π/9).

先令 w = 1 + i√3。|w| = √(1² + (√3)²) = √4 = 2，辐角 arg(w) = tan⁻¹(√3/1) = π/3。于是 z = w²，所以 |z| = 4，arg(z) = 2π/3。因此 z = 4(cos 2π/3 + i sin 2π/3) 或 4e^(i2π/3)。z 的立方根满足模为 ³√4，辐角分别为 (2π/3 + 2kπ)/3, k = 0,1,2。即 2π/9, 8π/9, 14π/9。直角坐标形式：³√4 (cos 2π/9 + i sin 2π/9)、³√4 (cos 8π/9 + i sin 8π/9)、³√4 (cos 14π/9 + i sin 14π/9)。

2. Matrices: Inverse and Solving Linear Systems | 矩阵：逆矩阵与求解线性方程组

Given matrix A = [[2, 1], [3, 4]], find A⁻¹. Hence solve the system 2x + y = 5, 3x + 4y = 6.

已知矩阵 A = [[2, 1], [3, 4]]，求 A⁻¹。并由此解方程组 2x + y = 5, 3x + 4y = 6。

det(A) = 2×4 – 1×3 = 5. A⁻¹ = (1/5) [[4, -1], [-3, 2]]. Write the system as A [x; y] = [5; 6]. Hence [x; y] = A⁻¹ [5; 6] = (1/5) [[4, -1], [-3, 2]] [5; 6] = (1/5) [4×5 -1×6; -3×5 +2×6] = (1/5) [20-6; -15+12] = (1/5) [14; -3] = [14/5; -3/5]. Thus x = 14/5, y = -3/5.

行列式 det(A) = 2×4 – 1×3 = 5。逆矩阵 A⁻¹ = (1/5) [[4, -1], [-3, 2]]。将方程组写为 A [x; y] = [5; 6]。因此 [x; y] = A⁻¹ [5; 6] = (1/5) [4×5 -1×6; -3×5 +2×6] = (1/5) [14; -3] = [14/5; -3/5]。于是 x = 14/5, y = -3/5。

3. Further Calculus: Derivative and Integral of Hyperbolic Functions | 进阶微积分：双曲函数的导数与积分

(a) Differentiate y = arctanh(x) with respect to x. (b) Evaluate ∫ sech²(3x) dx.

(a) 求 y = arctanh(x) 的导数。 (b) 计算 ∫ sech²(3x) dx。

(a) Let y = artanh(x). Then x = tanh(y), dx/dy = sech²(y) = 1 – tanh²(y) = 1 – x², for |x|<1. Hence dy/dx = 1/(1 - x²). (b) Recall d/dx [tanh(x)] = sech²(x). So ∫ sech²(3x) dx = (1/3) tanh(3x) + C.

(a) 设 y = artanh(x)，则 x = tanh(y)，dx/dy = sech²(y) = 1 – tanh²(y) = 1 – x² (|x|<1)。因此 dy/dx = 1/(1 - x²)。(b) 已知 d/dx [tanh(x)] = sech²(x)，故 ∫ sech²(3x) dx = (1/3) tanh(3x) + C。

4. Hyperbolic Identities and Equations | 双曲恒等式与方程

Prove that cosh²x – sinh²x = 1. Hence solve the equation sinh u = 2, giving your answer in logarithmic form.

证明 cosh²x – sinh²x = 1。并据此解方程 sinh u = 2，答案用对数形式表示。

By definition, cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2. Then cosh²x – sinh²x = [(eˣ + e⁻ˣ)² – (eˣ – e⁻ˣ)²]/4 = [e²ˣ + 2 + e⁻²ˣ – (e²ˣ – 2 + e⁻²ˣ)]/4 = (4)/4 = 1. For sinh u = 2: (eᵘ – e⁻ᵘ)/2 = 2 ⇒ eᵘ – e⁻ᵘ = 4. Multiply by eᵘ: e²ᵘ – 4eᵘ – 1 = 0. Solve quadratic in eᵘ: eᵘ = (4 ± √(16+4))/2 = 2 ± √5. Since eᵘ > 0, take eᵘ = 2 + √5. Thus u = ln(2 + √5).

根据定义，cosh x = (eˣ + e⁻ˣ)/2，sinh x = (eˣ – e⁻ˣ)/2。于是 cosh²x – sinh²x = [(eˣ+e⁻ˣ)² – (eˣ-e⁻ˣ)²]/4 = [4]/4 = 1。对于 sinh u = 2： (eᵘ – e⁻ᵘ)/2 = 2 ⇒ eᵘ – e⁻ᵘ = 4。两边乘 eᵘ 得 e²ᵘ – 4eᵘ – 1 = 0。解二次方程：eᵘ = 2 ± √5。由于 eᵘ > 0，取 eᵘ = 2 + √5，所以 u = ln(2 + √5)。

5. First-Order Linear Differential Equation | 一阶线性微分方程

Solve the differential equation dy/dx + 2y = eˣ, given that y(0) = 1.

求解微分方程 dy/dx + 2y = eˣ，其中 y(0) = 1。

The integrating factor I = e^(∫2 dx) = e²ˣ. Multiply both sides: e²ˣ dy/dx + 2e²ˣ y = e³ˣ. LHS becomes d/dx (y e²ˣ). Integrate: y e²ˣ = ∫ e³ˣ dx = (1/3)e³ˣ + C. Apply condition: y(0)=1 ⇒ 1·e⁰ = (1/3)e⁰ + C ⇒ 1 = 1/3 + C ⇒ C = 2/3. Hence y = (1/3)eˣ + (2/3)e⁻²ˣ.

积分因子 I = e^(∫2 dx) = e²ˣ。方程两边同乘：e²ˣ dy/dx + 2e²ˣ y = e³ˣ。左边化为 d/dx (y e²ˣ)。积分得 y e²ˣ = ∫ e³ˣ dx = (1/3)e³ˣ + C。代入初始条件：y(0)=1 ⇒ e⁰ = 1/3 + C ⇒ C = 2/3。因此 y = (1/3)eˣ + (2/3)e⁻²ˣ。

6. Second-Order Homogeneous ODE | 二阶齐次常微分方程

Find the particular solution of d²y/dx² + 4y = 0 satisfying y(0) = 1 and y'(0) = 0.

求方程 d²y/dx² + 4y = 0 满足 y(0) = 1、y'(0) = 0 的特解。

Auxiliary equation: m² + 4 = 0 ⇒ m = ±2i. General solution: y = A cos 2x + B sin 2x. Condition y(0)=1 ⇒ A = 1. Derivative y’ = -2A sin 2x + 2B cos 2x. y'(0)=0 ⇒ 2B = 0 ⇒ B=0. Particular solution: y = cos 2x.

辅助方程：m² + 4 = 0 ⇒ m = ±2i。通解为 y = A cos 2x + B sin 2x。由 y(0)=1 得 A = 1。求导得 y’ = -2A sin 2x + 2B cos 2x，代入 y'(0)=0 得 2B = 0 ⇒ B = 0。特解为 y = cos 2x。

7. Maclaurin Series Expansion | 麦克劳林级数展开

Determine the Maclaurin series for f(x) = ln(1+x) up to and including the term in x⁴.

求 f(x) = ln(1+x) 的麦克劳林级数，精确到含 x⁴ 项为止。

f'(x) = (1+x)⁻¹, f”(x) = -(1+x)⁻², f”'(x) = 2(1+x)⁻³, f⁽⁴⁾(x) = -6(1+x)⁻⁴, f⁽⁵⁾(x) = 24(1+x)⁻⁵. Evaluate at x=0: f(0)=0, f'(0)=1, f”(0)=-1, f”'(0)=2, f⁽⁴⁾(0)=-6. Maclaurin series: ln(1+x) = 0 + 1·x + (-1)·x²/2! + 2·x³/3! + (-6)·x⁴/4! + … = x – x²/2 + x³/3 – x⁴/4 + … . So up to x⁴: x – x²/2 + x³/3 – x⁴/4.

f'(x) = (1+x)⁻¹, f”(x) = -(1+x)⁻², f”'(x) = 2(1+x)⁻³, f⁽⁴⁾(x) = -6(1+x)⁻⁴。在 x=0 处求值：f(0)=0, f'(0)=1, f”(0)=-1, f”'(0)=2, f⁽⁴⁾(0)=-6。麦克劳林级数：ln(1+x) = x – x²/2! + 2x³/3! – 6x⁴/4! + … = x – x²/2 + x³/3 – x⁴/4 + … 。因此到 x⁴ 项为 x – x²/2 + x³/3 – x⁴/4。

8. Vector Product and Triangle Area | 向量积与三角形面积

Given vectors a = 2i + 3j – k and b = i – 2j + 4k, find a × b and the area of the triangle formed by a and b.

已知向量 a = 2i + 3j – k 与 b = i – 2j + 4k，求 a × b 以及由 a 和 b 构成的三角形的面积。

a × b = det|i j k; 2 3 -1; 1 -2 4| = i(3×4 – (-1)×(-2)) – j(2×4 – (-1)×1) + k(2×(-2) – 3×1) = i(12 – 2) – j(8 + 1) + k(-4 – 3) = 10i – 9j – 7k. Magnitude |a × b| = √(10² + (-9)² + (-7)²) = √(100+81+49) = √230. Area = ½ |a × b| = (√230)/2.

a × b = det|i j k; 2 3 -1; 1 -2 4| = i(12 – 2) – j(8 + 1) + k(-4 – 3) = 10i – 9j – 7k。模长 |a × b| = √(10²+9²+7²) = √230。三角形面积 = ½ |a × b| = (√230)/2。

9. De Moivre’s Theorem: Cosine of Multiple Angles | 棣莫弗定理：多倍角的余弦

Using de Moivre’s theorem, express cos 3θ in terms of powers of cos θ.

利用棣莫弗定理，将 cos 3θ 用 cos θ 的幂表示。

By de Moivre: (cos θ + i sin θ)³ = cos 3θ + i sin 3θ. Expand LHS: cos³θ + 3i cos²θ sin θ + 3i² cos θ sin²θ + i³ sin³θ = cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ. Equating real parts: cos 3θ = cos³θ – 3 cos θ sin²θ. Using sin²θ = 1 – cos²θ: cos 3θ = cos³θ – 3 cos θ (1 – cos²θ) = cos³θ – 3 cos θ + 3 cos³θ = 4 cos³θ – 3 cos θ.

由棣莫弗定理：(cos θ + i sin θ)³ = cos 3θ + i sin 3θ。左边展开：cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ。比较实部：cos 3θ = cos³θ – 3 cos θ sin²θ。利用 sin²θ = 1 – cos²θ，得 cos 3θ = cos³θ – 3 cos θ (1 – cos²θ) = 4 cos³θ – 3 cos θ。

10. Matrix Transformations: Reflection | 矩阵变换：反射

The transformation matrix M = [[0, 1], [1, 0]] represents a reflection in the plane. Determine the image of the point P(3, 2) under this transformation and identify the line of reflection.

变换矩阵 M = [[0, 1], [1, 0]] 表示平面内的一个反射。求点 P(3, 2) 在此变换下的像，并指出反射对称轴。

Apply matrix: [x’] = [[0,1],[1,0]] [3] = [0·3+1·2; 1·3+0·2] = [2; 3]. So image is (2, 3). The matrix swaps x and y coordinates, which corresponds to reflection in the line y = x.

应用矩阵：[x’; y’] = [[0,1],[1,0]] [3; 2] = [2; 3]。因此像点为 (2, 3)。该矩阵交换 x 与 y 坐标，对应关于直线 y = x 的反射。

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AS Further Mathematics: Unit Test Paper | AS 进阶数学：单元测试卷