Year 13 CIE Maths: Unit Test Mock Paper Walkthrough | Year 13 CIE 数学：单元测试模拟卷解析

📚 Year 13 CIE Maths: Unit Test Mock Paper Walkthrough | Year 13 CIE 数学：单元测试模拟卷解析

This article goes through a typical Year 13 CIE Mathematics mock paper, section by section. It focuses on key topics from Pure Mathematics 3 (P3) and gives you step‑by‑step solutions, common mistakes to avoid, and revision tips for each question type.

本文带你逐题精析一份典型的 Year 13 CIE 数学单元测试模拟卷。内容紧扣纯数学 3（P3）的核心考点，每一类题型都配有详细分步解答、易错提醒和复习建议，帮助你在单元测试中稳定发挥。

1. Modulus Equations and Inequalities | 模方程与不等式

The exam often begins with a modulus problem. A common question is: Solve |x – 2| = 3x + 1.

考试常以模问题开场。常见题型如：解方程 |x – 2| = 3x + 1。

|x – 2| = 3x + 1

First, consider the case x – 2 ≥ 0, i.e. x ≥ 2. Then |x – 2| = x – 2, giving x – 2 = 3x + 1 → –2x = 3 → x = –1.5. But this fails the condition x ≥ 2, so it is not valid.

首先考虑 x – 2 ≥ 0，即 x ≥ 2。这时 |x – 2| = x – 2，方程化为 x – 2 = 3x + 1 → –2x = 3 → x = –1.5。但这个解不满足 x ≥ 2，因此舍去。

Next, for x – 2 < 0, i.e. x < 2, we have |x – 2| = –(x – 2) = –x + 2. The equation becomes –x + 2 = 3x + 1 → –4x = –1 → x = 0.25. This satisfies x < 2, so it is the valid solution.

接着考虑 x – 2 < 0，即 x < 2。这时模为 –x + 2，方程化为 –x + 2 = 3x + 1 → –4x = –1 → x = 0.25。它满足 x < 2，因此是有效解。

Always check the domain condition after solving each branch – many students forget and keep the extraneous root.

务必在解出每个分支后检验定义域条件——很多考生忘记这一步而保留了增根。

2. Exponential and Logarithmic Functions | 指数与对数函数

A typical P3 question asks you to express y in terms of x after applying logarithms, e.g. 2e²ˣ = 5eˣ + 3.

典型的 P3 题目要求取对数后把 y 表示成 x 的函数，例如解 2e²ˣ = 5eˣ + 3。

2e²ˣ – 5eˣ – 3 = 0

Let u = eˣ. Then 2u² – 5u – 3 = 0 → (2u + 1)(u – 3) = 0, so u = –½ or u = 3.

设 u = eˣ，则方程化为 2u² – 5u – 3 = 0 → (2u + 1)(u – 3) = 0，解得 u = –½ 或 u = 3。

Since eˣ > 0 for all real x, discard u = –½. Hence eˣ = 3 → x = ln 3.

因为对所有实数 x 均有 eˣ > 0，舍去 u = –½。因此 eˣ = 3 → x = ln 3。

Re‑check by substituting back: 2e²ˡⁿ³ = 2·9 = 18 and 5eˡⁿ³ + 3 = 15 + 3 = 18, so the solution is correct.

代回检验：2e²ˡⁿ³ = 2·9 = 18，5eˡⁿ³ + 3 = 15 + 3 = 18，解正确。

3. Trigonometric Equations and Identities | 三角方程与恒等式

Often you need to solve an equation like 2 sin² θ – cos θ = 1 for 0° ≤ θ ≤ 360°.

常需要解如 2 sin² θ – cos θ = 1 的三角方程，自变量范围 0° ≤ θ ≤ 360°。

2 sin² θ – cos θ = 1

Use sin² θ = 1 – cos² θ. Substituting: 2(1 – cos² θ) – cos θ – 1 = 0 → 2 – 2 cos² θ – cos θ – 1 = 0 → –2 cos² θ – cos θ + 1 = 0.

用 sin² θ = 1 – cos² θ 代换得：2(1 – cos² θ) – cos θ – 1 = 0 → 2 – 2 cos² θ – cos θ – 1 = 0 → –2 cos² θ – cos θ + 1 = 0。

Multiply by –1: 2 cos² θ + cos θ – 1 = 0. Factorise: (2 cos θ – 1)(cos θ + 1) = 0 → cos θ = ½ or cos θ = –1.

两边乘 –1：2 cos² θ + cos θ – 1 = 0。因式分解：(2 cos θ – 1)(cos θ + 1) = 0 → cos θ = ½ 或 cos θ = –1。

For cos θ = ½, θ = 60°, 300°. For cos θ = –1, θ = 180°. All within the given interval.

cos θ = ½ 给出 θ = 60°, 300°；cos θ = –1 给出 θ = 180°。所有解均在给定区间内。

4. Differentiation: Chain, Product and Quotient Rules | 微分：链式、乘积与商法则

Differentiate y = ln(cos 2x). This tests the chain rule within a logarithmic function.

对 y = ln(cos 2x) 求导，考查含对数函数的链式法则。

y = ln(cos 2x)

Let u = cos 2x, so y = ln u. Then dy/dx = (1/u) · du/dx. Now du/dx = –sin 2x · 2 = –2 sin 2x.

设 u = cos 2x，则 y = ln u。dy/dx = (1/u) · du/dx。而 du/dx = –sin 2x · 2 = –2 sin 2x。

Hence dy/dx = (1/(cos 2x)) · (–2 sin 2x) = –2 tan 2x. Simpler final form: dy/dx = –2 tan 2x.

因此 dy/dx = –2 sin 2x / cos 2x = –2 tan 2x。最终可以写为 dy/dx = –2 tan 2x。

Always simplify your derivative – it makes later work (like finding stationary points) much easier.

务必化简导数表达式，这对后续求驻点等操作会简便很多。

5. Parametric and Implicit Differentiation | 参数与隐函数微分

A question gives x = t² + t, y = t³ – 3t. Find dy/dx in terms of t, then the equation of the tangent at t = 1.

题目给出参数方程 x = t² + t, y = t³ – 3t，要求用 t 表示 dy/dx，并求 t = 1 处的切线方程。

dx/dt = 2t + 1, dy/dt = 3t² – 3. Then dy/dx = (dy/dt) / (dx/dt) = (3t² – 3) / (2t + 1).

dx/dt = 2t + 1，dy/dt = 3t² – 3。于是 dy/dx = (3t² – 3) / (2t + 1)。

At t = 1, slope m = (3 – 3)/(2+1) = 0. The point is (1²+1, 1³–3) = (2, –2). Tangent is horizontal: y = –2.

t = 1 时，斜率 m = 0。对应点为 (2, –2)。切线水平，方程为 y = –2。

If slope had been non‑zero, use y – y₁ = m(x – x₁). For parametrics, always find t first for a given point.

若斜率不为零，用点斜式。参数方程中，通常先由给定点反求 t 值。

6. Integration by Substitution and Parts | 换元积分与分部积分

Evaluate ∫ x eˣ² dx. Recognise the inner function and use substitution.

计算 ∫ x eˣ² dx。识别内函数后用换元法。

Let u = x² → du = 2x dx → x dx = ½ du.

Then ∫ x eˣ² dx = ∫ eᵁ · ½ du = ½ eᵁ + C = ½ eˣ² + C.

则 ∫ x eˣ² dx = ½ ∫ eᵁ du = ½ eᵁ + C = ½ eˣ² + C。

For integration by parts, a standard P3 example is ∫ x ln x dx. Let u = ln x, dv = x dx. Then du = (1/x) dx, v = ½ x².

分部积分的标准例题是 ∫ x ln x dx。设 u = ln x，dv = x dx，则 du = (1/x) dx，v = ½ x²。

∫ x ln x dx = ½ x² ln x – ∫ ½ x² · (1/x) dx = ½ x² ln x – ½ ∫ x dx = ½ x² ln x – ¼ x² + C.

Always decide the substitution by looking for a function and its derivative; for parts, follow LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).

选择换元时看是否函数和它的导数同时出现；分部积分按 LIATE 顺序优先选 u。

7. Numerical Solution of Equations | 方程的数值解

Show that x³ – 5x + 1 = 0 has a root between 2 and 3, then use the Newton‑Raphson method with x₀ = 2 to find it to 3 d.p.

证明方程 x³ – 5x + 1 = 0 在 2 和 3 之间有一个根，并用 Newton‑Raphson 方法以 x₀ = 2 起始求根，精确至三位小数。

Let f(x) = x³ – 5x + 1. f(2) = –1, f(3) = 13. Sign change confirms a root in (2,3). f'(x) = 3x² – 5.

设 f(x) = x³ – 5x + 1。f(2) = –1，f(3) = 13，符号改变确认 (2,3) 内有根。导数为 f'(x) = 3x² – 5。

The iteration is xₙ₊₁ = xₙ – f(xₙ)/f'(xₙ). Start x₀ = 2: x₁ = 2 – (–1)/(12‑5) = 2 + 1/7 ≈ 2.142857.

迭代公式为 xₙ₊₁ = xₙ – f(xₙ)/f'(xₙ)。以 x₀ = 2 开始：x₁ = 2 – (–1)/(7) = 2.142857。

Continue: x₂ ≈ 2.1284, x₃ ≈ 2.12842. The root is 2.128 (3 d.p.). Always show full working for at least two iterations in the exam.

继续迭代得 x₂ ≈ 2.1284，x₃ ≈ 2.12842。根为 2.128（3 位小数）。考试中至少展示完整的两步迭代过程。

8. Vectors: Dot Product and Intersections | 向量：点积与相交问题

Find the acute angle between two lines with direction vectors a = i + 2j – k and b = 2i – j + 3k.

求两直线方向向量 a = i + 2j – k 与 b = 2i – j + 3k 的锐夹角。

cos θ = |a·b| / (|a||b|)

a·b = 1·2 + 2·(–1) + (–1)·3 = 2 – 2 – 3 = –3. |a| = √(1+4+1)=√6, |b| = √(4+1+9)=√14.

点积：a·b = –3。模长 |a| = √6，|b| = √14。

cos θ = |–3| / (√6·√14) = 3 / √84 = 3 / (2√21) → θ = cos⁻¹(3/(2√21)) ≈ 72.2° (acute angle).

cos θ = 3 / √84 = 3/(2√21)，∴ θ ≈ 72.2°（锐角）。

For intersection of two lines, set parametric equations equal and solve for the parameters. If consistent, find the point; otherwise lines are skew.

直线相交问题需设参数方程相等，解出参数，若一致则存在交点，否则为异面直线。

9. Complex Numbers: Modulus and Argument | 复数：模与幅角

Express z = –√3 + i in modulus‑argument form. Then find z⁴ and its principal argument.

将 z = –√3 + i 表为模‑幅角形式，并求 z⁴ 及其主幅角。

|z| = √((–√3)² + 1²) = √(3+1) = 2.

Argument θ = π – tan⁻¹(1/√3) = π – π/6 = 5π/6. So z = 2(cos(5π/6) + i sin(5π/6)).

幅角：点(–√3, 1)在第二象限，θ = π – tan⁻¹(1/√3) = 5π/6。因此 z = 2 cis(5π/6)。

Using De Moivre: z⁴ = 2⁴ cis(4 × 5π/6) = 16 cis(10π/3) = 16 cis(10π/3 – 2π) = 16 cis(4π/3). The principal argument is –2π/3 (or 4π/3, but usually given in (–π, π] so –2π/3).

用棣莫弗定理：z⁴ = 16 cis(10π/3) = 16 cis(4π/3)（减去 2π），主幅角为 –2π/3（习惯取 (–π, π]）。

Always ensure the argument is stated in the correct interval, and remember to adjust for the quadrant.

务必保证幅角落在规定区间内，并注意根据象限修正结果。

10. Differential Equations and Applications | 微分方程及其应用

Solve the differential equation dy/dx = 2xy² given that y(0) = 1. Find y in terms of x.

解微分方程 dy/dx = 2xy²，初始条件 y(0) = 1。求 y 关于 x 的表达式。

Separate variables: (1/y²) dy = 2x dx.

Integrate both sides: ∫ y⁻² dy = ∫ 2x dx → –y⁻¹ = x² + C. So –1/y = x² + C.

两边积分：∫ y⁻² dy = ∫ 2x dx → –1/y = x² + C。

Use y = 1 when x = 0: –1/1 = 0 + C → C = –1. Thus –1/y = x² – 1 → 1/y = 1 – x² → y = 1/(1 – x²).

代入初始条件：–1 = C → –1/y = x² – 1 → y = 1/(1 – x²)。有效范围 |x| < 1 以确保分母不为零且 y 为正。

In modelling contexts, always state the domain for which the solution is valid – this often earns a mark.

在建模情境中，始终指明解的有效定义域，这在评分中通常占有一分。

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Year 13 CIE Maths: Unit Test Mock Paper Walkthrough | Year 13 CIE 数学：单元测试模拟卷解析