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

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

This article provides a complete worked solution walkthrough for a WJEC Year 13 Pure Mathematics (Unit 3) mock paper. Each section tackles a typical exam-style question, covering a broad range of A2 topics: partial fractions, functions, trigonometry, implicit differentiation, parametric equations, integration by parts, vectors, numerical methods, differential equations, and proof. Step-by-step reasoning is presented in both English and Chinese, highlighting common pitfalls and efficient methods to maximise marks in the real exam.

本文为 WJEC Year 13 纯数学（Unit 3）模拟卷提供全程逐题精讲。每节讲解一道典型考题，覆盖部分分式、函数、三角学、隐函数微分、参数方程、分部积分、向量、数值方法、微分方程与证明等 A2 核心主题。每个关键步骤均以中英双语展示，并指出常见失分点与高分技巧，帮助考生在真实考试中稳取高分。

1. Partial Fractions and Binomial Expansion | 第1题：部分分式与二项展开

Question: ) in partial fractions. Hence find the series expansion of f(x) up to and including the term in x³, stating the range of values of x for which the expansion is valid.

题目：将 f(x) = (5x − 1) / ((x + 1)(2x − 1)) 写成部分分式。由此求 f(x) 的级数展开式，写至 x³ 项，并说明展开有效的 x 取值范围。

n decomposition as A/(x+1) + B/(2x−1). Multiply both sides by (x+1)(2x−1) to obtain 5x − 1 ≡ A(2x−1) + B(x+1). By choosing x = 1/2, the term with A vanishes: 5(1/2) − 1 = B(1/2+1) ⇒ (3/2) = (3/2)B ⇒ B = 1. Choosing x = −1 gives A: 5(−1) − 1 = A(2(−1)−1) ⇒ −6 = A(−3) ⇒ A = 2. Thus f(x) = 2/(x+1) + 1/(2x−1).

设部分分式为 A/(x+1) + B/(2x−1)。方程两边同乘 (x+1)(2x−1) 得 5x − 1 ≡ A(2x−1) + B(x+1)。代入 x = 1/2 可消去 A 项：5(1/2) − 1 = B(1/2+1) ⇒ 3/2 = 3/2 B ⇒ B = 1。代入 x = −1 求 A：5(−1) − 1 = A(2(−1)−1) ⇒ −6 = −3A ⇒ A = 2。因此 f(x) = 2/(x+1) + 1/(2x−1)。

positive powers, write (2x−1)⁻¹ = −(1 − 2x)⁻¹. Expand using (1 + u)⁻¹ = 1 − u + u² − u³ + … : 2(1 + x)⁻¹ = 2[1 − x + x² − x³ + …] and −(1 − 2x)⁻¹ = −[1 + 2x + 4x² + 8x³ + …] = −1 − 2x − 4x² − 8x³ + … .

将每项转化为可二项展开的形式：2(x+1)⁻¹ 与 (2x−1)⁻¹ = −(1−2x)⁻¹。使用公式 (1+u)⁻¹ = 1 − u + u² − u³ + … ：2(1+x)⁻¹ = 2[1 − x + x² − x³ + …]； −(1−2x)⁻¹ = −[1 + 2x + 4x² + 8x³ + …] = −1 − 2x − 4x² − 8x³ + … 。

Adding gives f(x) = (2 − 1) + (−2 − 2)x + (2 − 4)x² + (−2 − 8)x³ + … = 1 − 4x − 2x² − 10x³ + … . The expansion of (1 + x)⁻¹ is valid for |x| < 1, and the expansion of (1 − 2x)⁻¹ is valid for |2x| < 1 ⇒ |x| < 1/2. The overall interval of validity is the more restrictive |x| < 1/2.

相加得 f(x) = (2 − 1) + (−2 − 2)x + (2 − 4)x² + (−2 − 8)x³ + … = 1 − 4x − 2x² − 10x³ + … 。(1+x)⁻¹ 的展开要求 |x| < 1，(1−2x)⁻¹ 要求 |x| < 1/2，最终有效区间取较严的 |x| < 1/2。

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2. Domain of Composite Functions | 第2题：复合函数的定义域

Question: For f(x) = ln(2x + 1), x > −1/2, and g(x) = eˣ − 3, find fg(x) and state its maximal domain.

题目：已知 f(x) = ln(2x + 1)，x > −1/2；g(x) = eˣ − 3。求 fg(x) 并写出其最大定义域。

Substitute g into f: fg(x) = f(eˣ − 3) = ln(2(eˣ − 3) + 1) = ln(2eˣ − 6 + 1) = ln(2eˣ − 5). For the natural logarithm to be defined, the argument must be strictly positive: 2eˣ − 5 > 0 ⇒ eˣ > 5/2 ⇒ x > ln(5/2). Therefore the maximal domain is (ln(5/2), ∞).

将 g 代入 f：fg(x) = f(eˣ − 3) = ln(2(eˣ − 3) + 1) = ln(2eˣ − 5)。自然对数要求真数大于零：2eˣ − 5 > 0 ⇒ eˣ > 5/2 ⇒ x > ln(5/2)。因此最大定义域为 (ln(5/2), ∞)。

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3. Trigonometric Equation Using R-Formula | 第3题：三角方程与R公式

Question: Solve 3 sin 2θ − 2 cos 2θ = 1 for 0 ≤ θ ≤ π, giving answers to 2 decimal places.

题目：解方程 3 sin 2θ − 2 cos 2θ = 1，其中 0 ≤ θ ≤ π，答案精确至两位小数。

Express the left side as R sin(2θ − α): R = √(3² + (−2)²) = √13, and tan α = (2)/(3) ⇒ α = arctan(2/3) ≈ 0.588 rad. The equation becomes √13 sin(2θ − 0.588) = 1 ⇒ sin(2θ − 0.588) = 1/√13.

将左边写成 R sin(2θ − α) 形式：R = √(3²+(−2)²)=√13，α = arctan(2/3) ≈ 0.588 rad。方程化为 √13 sin(2θ − 0.588) = 1 ⇒ sin(2θ − 0.588) = 1/√13。

Let φ = 2θ − 0.588. The principal value is arcsin(1/√13 0.588. The two relevant solutions are φ ≈ 0.281 and φ ≈ 2.861. Then 2θ = φ + 0.588 gives 2θ = 0.869 or 3.449, so θ = 0.4345 rad and 1.7245 rad. To 2 decimal places: θ ≈ 0.43 rad and 1.72 rad.

设 φ = 2θ − 0.588。主值 arcsin(1/√13) ≈ 0.281 rad。正弦在第二象限也为正，故 φ = π − 0.281 ≈ 2.861 rad。因 0 ≤ θ ≤ π，2θ 在 0 到 2π 间，φ 范围约为 −0.588 至 2π−0.588，两个符合条件的解为 φ ≈ 0.281 与 φ ≈ 2.861。得 2θ = 0.869 或 3.449，所以 θ ≈ 0.4345 rad 和 1.7245 rad。保留两位小数：θ ≈ 0.43 rad, 1.72 rad。

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4. Implicit Differentiation and Tangent Equation | 第4题：隐函数微分与切线方程

Question: 3xy − y² = 4. Find dy/dx, and hence the equation of the tangent to C at the point (2, 1).

题目：曲线 C 方程为 x² + 3xy − y² = 4。求 dy/dx，并由此写出 C 在点 (2, 1) 处的切线方程。

y/dx = 0. Group dy/dx terms: (3x − 2y) dy/dx + 2x + 3y = 0. Hence dy/dx = −(2x + 3y)/(3x − 2y).

逐项对 x 求导：d/dx(x²)=2x；d/dx(3xy)=3y+3x dy/dx（乘法法则）；d/dx(−y²)=−2y dy/dx。得 2x+3y+3x dy/dx−2y dy/dx=0。整理含 dy/dx 的项：(3x−2y)dy/dx + 2x+3y=0，故 dy/dx = −(2x+3y)/(3x−2y)。

) = −(4+3)/(6−2) = −7/4. Tangent line: y−1 = (−7/4)(x−2). Multiply by 4: 4y − 4 = −7x + 14 ⇒ 7x + 4y = 18.

在 (2,1) 处代入得斜率 dy/dx = −7/4。切线方程：y−1 = (−7/4)(x−2)。两边乘 4 整理得：7x + 4y = 18。

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5. Parametric Differentiation and Area | 第5题：参数方程微分与面积

Question: x and d²y/dx². Calculate the area bounded by the curve and the x-axis from t = 0 to t = 2.

题目：曲线由参数方程 x = t² + 1，y = t³ − 2t（t ≥ 0）给定。求 dy/dx 和 d²y/dx²。计算曲线与 x 轴之间从 t = 0 到 t = 2 围成的面积。

dt). Derivative of dy/dx: d/dt[(3t² − 2)/(2t)] = [6t·2t − (3t²−2)·2] / (4t²) = (12t² − 6t² + 4)/(4t²) = (6t² + 4)/(4t²) = (3t² + 2)/(2t²). Divide by dx/dt = 2t gives d²y/dx² = (3t²+2)/(2t² · 2t) = (3t²+2)/(4t³).

先求一阶导数：dx/dt = 2t，dy/dt = 3t² − 2，故 dy/dx = (3t² − 2)/(2t)。二阶导：d²y/dx² = d(dy/dx)/dt ÷ (dx/dt)。先对 dy/dx 求导：d/dt[(3t²−2)/(2t)] = (12t²−6t²+4)/4t² = (6t²+4)/4t² = (3t²+2)/(2t²)。再除以 dx/dt (2t) 得 d²y/dx² = (3t²+2)/(4t³)。

Area A = ∫ y dx. Since x = t²+1, dx = 2t dt. When t=0, x=1; t=2, x=5. So A = ∫_t=0² /3 ]₀² = 2[ (32/5 − 16/3) − 0 ] = 2[ (96/15 − 80/15) ] = 2 × (16/15) = 32/15 square units.

面积 A = ∫ y dx。由 x = t²+1 得 dx = 2t dt。积分限 t 从 0 到 2， −2t)·2t dt = 2∫₀² (t⁴−2t²) dt = 2 [ t⁵/5 − 2t³/3 ]₀² = 2 (32/5 − 16/3) = 2 (16/15) = 32/15 平方单位。

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6. Integration by Parts | 第6题：分部积分法

Question: Find the exact value of ∫₀¹ x² eˣ dx.

题目：求 ∫₀¹ x² eˣ dx 的精确值。

Use integration by parts twice. Let u = x², dv/dx = eˣ, then du/dx = 2x, v = eˣ. Formula: ∫ u dv = uv − ∫ v du. First application: ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx.

使用两次分部积分。设 u = x²，dv/dx = eˣ，则 du/dx = 2x，v = eˣ。公式 ∫ u dv = uv − ∫ v du。第一次：∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx。

For the remaining integral ∫ 2x eˣ dx, apply parts again: u₁ = 2x, dv/dx = eˣ, du₁/dx = 2, v = eˣ. ∫ 2x eˣ dx = 2x eˣ − ∫ 2 eˣ dx = 2x eˣ − 2eˣ + C. Combining: ∫ x² eˣ dx = x² eˣ − (2x eˣ − 2eˣ) + C = eˣ (x² − 2x + 2) + C.

对剩余积分 ∫ 2x eˣ dx 再次分部：设 u₁ = 2x，dv/dx = eˣ，则 ∫ 2x eˣ dx = 2x eˣ − ∫ 2 eˣ dx = 2x eˣ − 2eˣ + C。合并得 ∫ x² eˣ dx = eˣ (x² − 2x + 2) + C。

Now evaluate definite integral from 0 to 1: [eˣ 2x + 2)]₀¹ = e¹(1 − 2 + 2) − e⁰(0 − 0 + 2) = e·1 − 2 = e − 2. So the exact value is e − 2.

计算定积分：代入上限 1 得 e(1−2+2)=e；代入下限 0 得 e⁰·2=2。结果 = e − 2。

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7. Vectors: Shortest Distance from a Point to a Line | 第7题：向量：点到直线的最短距离

Question: : Find the shortest distance from point A(4, 3, −1) to the line L: r = (2i − j + k) + t(i + 2j − 2k).

题目：求点 A(4, 3, −1) 到直线 L: r = (2i − j + k) + t(i + 2j − 2k) 的最短距离。

distance, AB must be perpendicular to the direction vector d = i + 2j − 2k. So AB · d = 0

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