Year 13 CAIE Mathematics: Unit Test Mock Paper Solutions | Year 13 CAIE 数学：单元测试模拟卷解析

📚 Year 13 CAIE Mathematics: Unit Test Mock Paper Solutions | Year 13 CAIE 数学：单元测试模拟卷解析

This article provides detailed, step-by-step solutions to a mock unit test designed for Year 13 CAIE Mathematics. The paper focuses on the Pure Mathematics 3 syllabus and covers key techniques such as partial fractions, differential equations, complex numbers, vectors, iterative methods, integration skills, implicit differentiation, parametric equations and volumes of revolution. Every solution is presented in a clear bilingual format, highlighting essential working steps and exam strategies to help you strengthen your problem-solving accuracy and speed.

本文为 Year 13 CAIE 数学单元测试模拟卷提供了详细的逐步解析。试卷聚焦于纯数学 3 大纲，涵盖了分式分解、微分方程、复数、向量、迭代法、积分技巧、隐函数求导、参数方程以及旋转体体积等核心技巧。每道题的解答都以清晰的双语形式呈现，突出了关键的解题步骤与应试策略，帮助你提高解题的准确性和速度。

1. Partial Fractions and Integration | 分式分解与积分

Problem: Express (2x+1)/((x‑1)(x+2)) in partial fractions and hence evaluate ∫ (2x+1)/((x‑1)(x+2)) dx.

问题：将 (2x+1)/((x‑1)(x+2)) 表示为部分分式，并由此计算积分 ∫ (2x+1)/((x‑1)(x+2)) dx。

Step 1 – Set up the identity: (2x+1)/((x‑1)(x+2)) ≡ A/(x‑1) + B/(x+2). Multiply through by (x‑1)(x+2) to obtain 2x+1 ≡ A(x+2) + B(x‑1).

步骤1 – 建立恒等式：(2x+1)/((x‑1)(x+2)) ≡ A/(x‑1) + B/(x+2)。两边同乘 (x‑1)(x+2) 得到 2x+1 ≡ A(x+2) + B(x‑1)。

Step 2 – Choose strategic values to find A and B. Let x = 1: 2(1)+1 = A(1+2) → 3 = 3A ⇒ A = 1. Let x = –2: 2(–2)+1 = B(‑2‑1) → –3 = –3B ⇒ B = 1.

步骤2 – 选取特殊值求 A 和 B。令 x = 1：2(1)+1 = A(1+2) → 3 = 3A ⇒ A = 1。令 x = –2：2(–2)+1 = B(‑2‑1) → –3 = –3B ⇒ B = 1。

Step 3 – Rewrite the integrand: (2x+1)/((x‑1)(x+2)) = 1/(x‑1) + 1/(x+2). Integrate term by term: ∫ 1/(x‑1) dx = ln|x‑1|, ∫ 1/(x+2) dx = ln|x+2|. Combine logs: final result = ln| (x‑1)(x+2) | + C.

步骤3 – 重写被积函数：(2x+1)/((x‑1)(x+2)) = 1/(x‑1) + 1/(x+2)。逐项积分：∫ 1/(x‑1) dx = ln|x‑1|，∫ 1/(x+2) dx = ln|x+2|。合并对数得最终结果：ln| (x‑1)(x+2) | + C。

2. First Order Differential Equation | 一阶微分方程

Problem: Solve the differential equation dy/dx = y² sin x, given that y = 1 when x = 0. Give your answer in the form y = f(x).

问题：求解微分方程 dy/dx = y² sin x，已知当 x = 0 时 y = 1。将答案表示为 y = f(x) 的形式。

Step 1 – The equation is separable. Rearrange as 1/y² dy = sin x dx, valid for y ≠ 0. Integrate both sides: ∫ y⁻² dy = ∫ sin x dx.

步骤1 – 方程为可分离变量型。整理为 1/y² dy = sin x dx（需 y ≠ 0）。两边积分：∫ y⁻² dy = ∫ sin x dx。

Step 2 – Perform the integration: –1/y = –cos x + C, which simplifies to 1/y = cos x – C. (Note sign carefully: –1/y = –cos x + C ⇒ 1/y = cos x – C.)

步骤2 – 计算积分：–1/y = –cos x + C，化简得 1/y = cos x – C。（注意符号：–1/y = –cos x + C ⇒ 1/y = cos x – C。）

Step 3 – Apply the initial condition: when x = 0, y = 1 ⇒ 1/1 = cos 0 – C ⇒ 1 = 1 – C ⇒ C = 0. Therefore 1/y = cos x, so y = 1/cos x = sec x.

步骤3 – 代入初始条件：x = 0 时 y = 1 ⇒ 1/1 = cos 0 – C ⇒ 1 = 1 – C ⇒ C = 0。因此 1/y = cos x，于是 y = 1/cos x = sec x。

3. Complex Numbers – Modulus, Argument and Powers | 复数 – 模、辐角及幂

Problem: Given z = √3 – i, find the modulus and argument of z in the interval (–π, π]. Hence write z in polar form and calculate z⁶.

问题：已知 z = √3 – i，求 z 的模和辐角（区间 (–π, π]）。由此写出 z 的极形式，并计算 z⁶。

Step 1 – Modulus: |z| = √( (√3)² + (–1)² ) = √(3 + 1) = 2. Argument θ = arctan( –1/√3 ). Since the complex number lies in the fourth quadrant (real positive, imaginary negative), θ = –π/6.

步骤1 – 模：|z| = √( (√3)² + (–1)² ) = √(3 + 1) = 2。辐角 θ = arctan( –1/√3 )。由于复数位于第四象限（实部正，虚部负），θ = –π/6。

Step 2 – Polar form: z = 2( cos(–π/6) + i sin(–π/6) ) or equivalently 2( cos(π/6) – i sin(π/6) ).

步骤2 – 极形式：z = 2( cos(–π/6) + i sin(–π/6) )，也可写作 2( cos(π/6) – i sin(π/6) )。

Step 3 – Use De Moivre’s theorem: z⁶ = 2⁶ ( cos(6 × –π/6) + i sin(6 × –π/6) ) = 64 ( cos(–π) + i sin(–π) ) = 64( –1 + 0i ) = –64.

步骤3 – 利用棣莫弗定理：z⁶ = 2⁶ ( cos(6 × –π/6) + i sin(6 × –π/6) ) = 64 ( cos(–π) + i sin(–π) ) = 64( –1 + 0i ) = –64。

4. Vectors – Angle and Area of a Triangle | 向量 – 夹角与三角形面积

Problem: The points A, B and C have position vectors a = i + 2j – k, b = 3i – j + 2k and c = –i + j – 2k. Find the angle ABC and the area of triangle ABC.

问题：点 A、B、C 的位置向量分别为 a = i + 2j – k，b = 3i – j + 2k，c = –i + j – 2k。求角 ABC 及三角形 ABC 的面积。

Step 1 – Determine the vectors along the sides: AB = b – a = (3–1)i + (–1–2)j + (2–(–1))k = 2i – 3j + 3k. BC = c – b = (–1–3)i + (1–(–1))j + (–2–2)k = –4i + 2j – 4k.

步骤1 – 求出边向量：AB = b – a = (3–1)i + (–1–2)j + (2–(–1))k = 2i – 3j + 3k。BC = c – b = (–1–3)i + (1–(–1))j + (–2–2)k = –4i + 2j – 4k。

Step 2 – Use the dot product to find angle ABC (the angle between BA and BC, but BA = –AB = –2i + 3j – 3k; however the angle between vectors AB and BC is supplementary to angle ABC. Better to use vectors directed from B: BA = a – b = –2i + 3j – 3k, BC as above. Compute BA · BC = (–2)(–4) + (3)(2) + (–3)(–4) = 8 + 6 + 12 = 26. Magnitudes: |BA| = √(4+9+9)=√22, |BC| = √(16+4+16)=√36=6.

步骤2 – 使用点积求角 ABC（B 处的角，应使用从 B 出发的向量）。BA = a – b = –2i + 3j – 3k，BC = –4i + 2j – 4k。计算 BA · BC = (–2)(–4) + (3)(2) + (–3)(–4) = 8 + 6 + 12 = 26。模长：|BA| = √(4+9+9)=√22，|BC| = √(16+4+16)=√36=6。

Step 3 – cos(∠ABC) = (BA·BC) / (|BA||BC|) = 26 / (6√22) = 13 / (3√22). Thus ∠ABC = arccos( 13/(3√22) ). For the area, compute the magnitude of the cross product BA × BC. BA × BC = | i j k; –2 3 –3; –4 2 –4 | = i( (3)(–4) – (–3)(2) ) – j( (–2)(–4) – (–3)(–4) ) + k( (–2)(2) – (3)(–4) ) = i(–12+6) – j(8–12) + k(–4+12) = –6i + 4j + 8k. Magnitude = √(36+16+64) = √116 = 2√29. Area = ½ × 2√29 = √29 square units.

步骤3 – cos(∠ABC) = (BA·BC) / (|BA||BC|) = 26 / (6√22) = 13 / (3√22)。因此 ∠ABC = arccos( 13/(3√22) )。对于面积，计算向量积 BA × BC 的模。BA × BC = | i j k; –2 3 –3; –4 2 –4 | = i( (3)(–4) – (–3)(2) ) – j( (–2)(–4) – (–3)(–4) ) + k( (–2)(2) – (3)(–4) ) = i(–12+6) – j(8–12) + k(–4+12) = –6i + 4j + 8k。模 = √(36+16+64) = √116 = 2√29。面积 = ½ × 2√29 = √29 平方单位。

5. Newton‑Raphson Method | 牛顿‑拉夫森法

Problem: The equation x³ – 3x + 1 = 0 has a root between 0 and 1. Using the Newton‑Raphson formula xₙ₊₁ = xₙ – f(xₙ)/f ‘(xₙ) with initial approximation x₀ = 0.5, find the root correct to three decimal places.

问题：方程 x³ – 3x + 1 = 0 在 0 与 1 之间有一个根。使用牛顿‑拉夫森公式 xₙ₊₁ = xₙ – f(xₙ)/f ‘(xₙ)，并以 x₀ = 0.5 作为初始近似，求根的精确到三位小数的值。

Step 1 – Define f(x) = x³ – 3x + 1, then f ‘(x) = 3x² – 3. The iteration formula becomes xₙ₊₁ = xₙ – (xₙ³ – 3xₙ + 1)/(3xₙ² – 3).

步骤1 – 定义 f(x) = x³ – 3x + 1，则 f ‘(x) = 3x² – 3。迭代公式为 xₙ₊₁ = xₙ – (xₙ³ – 3xₙ + 1)/(3xₙ² – 3)。

Step 2 – First iteration (n = 0): f(0.5) = 0.125 – 1.5 + 1 = –0.375. f ‘(0.5) = 3(0.25) – 3 = 0.75 – 3 = –2.25. So x₁ = 0.5 – (–0.375)/(–2.25) = 0.5 – 0.166666… = 0.333333.

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Year 13 CAIE Mathematics: Unit Test Mock Paper Solutions | Year 13 CAIE 数学：单元测试模拟卷解析