Year 10 Edexcel Further Mathematics: Mock Unit Test Analysis | 十年级Edexcel进阶数学：单元模拟测试解析

📚 Year 10 Edexcel Further Mathematics: Mock Unit Test Analysis | 十年级Edexcel进阶数学：单元模拟测试解析

This article provides a detailed walkthrough of a mock unit test designed for Year 10 students following the Edexcel Further Mathematics (Further Pure) specification. Each question targets key skills from algebra, functions, trigonometry, series and vectors, with fully worked solutions that clarify common pitfalls and efficient methods.

本文为遵循Edexcel进阶数学（进阶纯数）教学大纲的十年级学生，提供了一套单元模拟测试卷的详尽解析。每道题目都针对代数、函数、三角、级数和向量中的关键技能，并配有完整的解题过程，旨在阐明常见错误和高效方法。

1. Question 1: Simplifying Algebraic Fractions | 第1题：化简代数分式

Question: Simplify fully (x² – 4)/(x² – x – 6) ÷ (x + 2)/(x – 3).

题目：完全化简 (x² – 4)/(x² – x – 6) ÷ (x + 2)/(x – 3)。

First, rewrite the division as multiplication by the reciprocal: (x² – 4)/(x² – x – 6) × (x – 3)/(x + 2). Factorise each polynomial where possible. The numerator x² – 4 is a difference of squares: (x – 2)(x + 2). The denominator x² – x – 6 factorises to (x – 3)(x + 2).

首先，将除法改写为乘以倒数：(x² – 4)/(x² – x – 6) × (x – 3)/(x + 2)。尽可能对每个多项式进行因式分解。分子 x² – 4 是平方差：(x – 2)(x + 2)。分母 x² – x – 6 可分解为 (x – 3)(x + 2)。

Substituting the factors gives: [(x – 2)(x + 2)] / [(x – 3)(x + 2)] × (x – cel the common factor (x + 2) from the first numerator and denominator, and also cancel (x – 3) appearing in both numerator and denominator after multiplication.

代入因式后得到：[(x – 2)(x + 2)] / [(x – 3)(x + 2)] × (x – 3)/(x + 2)。约去第一个分子分母中的公因式 (x + 2)，并约去相乘后分子分母中均出现的 (x – 3)。

Result: (x – 2)/(x + 2), where x ≠ –2, 3.

结果：(x – 2)/(x + 2)，其中 x ≠ –2, 3。

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2. Question 2: Solving a Quadratic Inequality | 第2题：解二次不等式

Question: Solve 2x² – 5x – 3 ≤ 0.

题目：解不等式 2x² – 5x – 3 ≤ 0。

Factorise the quadratic expression: 2x² – 5x – 3 = (2x + 1)(x – . The critical values are found by setting each factor equal to zero: 2x + 1 = 0 gives x = –½, and x – 3 = 0 gives x = 3.

将二次式进行因式分解：2x² – 5x – 3 = (2x + 1)(x – 3)。令每个因式等于零，求出临界值：2x + 1 = 0 得 x = –½，x – 3 = 0 得 x = 3。

Sketch a graph or use a sign table for (2x + 1)(x – 3). The parabola opens upwards (positive coefficient of x²). The product is less than or equal to zero between the critical values, including the endpoints.

画出草图或使用符号表分析 (2x + 1)(x – 3)。由于二次项系数为正，抛物线开口向上。乘积小于或等于零的范围在两临界值之间，包含端点。

Solution: –½ ≤ x ≤ 3.

解：–½ ≤ x ≤ 3。

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3. Question 3: Factor Theorem and Polynomial Division | 第3题：因式定理与多项式除法

Question: f(x) = 2x³ – 3x² – 3x + 2. Show that (x – 2) is a factor and hence factorise f(x) completely.

题目：f(x) = 2x³ – 3x² – 3x + 2。证明 (x – 2) 是一个因式，并由此将 f(x) 完全分解。

Apply the Factor Theorem: substitute x = 2 into f(x). f(2) = 2(2)³ – 3(2)² – 3(2) + 2 = 16 – 12 – 6 + 2 = 0. Since f(2) = 0, (x – 2) is indeed a factor.

应用因式定理：将 x = 2 代入 f(x)。f(2) = 2(2)³ – 3(2)² – 3(2) + 2 = 16 – 12 – 6 + 2 = 0。因为 f(2) = 0，所以 (x – 2) 是一个因式。

Divide f(x) by (x – 2) using algebraic long division or synthetic division. The quotient is 2x² + x – 1.

用代数长除法或综合除法，将 f(x) 除以 (x – 2)。商式为 2x² + x – 1。

Then factorise the quadratic quotient: 2x² + x – 1 = (2x – 1)(x + 1). Thus the complete factorisation is f(x) = (x – 2)(2x – 1)(x + 1).

再对二次商式进行因式分解：2x² + x – 1 = (2x – 1)(x + 1)。因此完全分解为 f(x) = (x – 2)(2x – 1)(x + 1)。

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4. Question 4: Exponential and Logarithmic Equations | 第4题：指数与对数方程

Question: Solve 3^2x = 5, giving your answer correct to 3 significant figures.

题目：解方程 3^2x = 5，答案精确至三位有效数字。

Take logarithms of both sides: log(3^2x) = log 5. Use the power rule to bring the exponent down: 2x log 3 = log 5.

两边取对数：log(3^2x) = log 5。利用幂法则将指数下移：2x log 3 = log 5。

calculator, log 5 ≈ 0.69897 and log 3 ≈ 0.47712. Thus x ≈ 0.69897 / (2 × 0.47712) ≈ 0.732.

移项求 x：x = log 5 / (2 log 3)。使用计算器，log 5 ≈ 0.69897，log 3 ≈ 0.47712。因此 x ≈ 0.69897 / (2 × 0.47712) ≈ 0.732。

x ≈ 0.732 (3 s.f.).

x ≈ 0.732（保留三位有效数字）。

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5. Question 5: Binomial Expansion | 第5题：二项式展开

Question: Find the first three terms in the expansion of (1 + 3x)⁵ in ascending powers of x. Use your expansion to estimate (1.03)⁵.

题目：按 x 的升幂展开 (1 + 3x)⁵ 的前三项。利用展开式估算 (1.03)⁵。

Apply the binomial theorem: (1 + 3x)⁵ = Σ (⁵Cᵣ) (1)^5–r (3x)ʳ. The first three terms correspond to r = 0, 1, 2.

应用二项式定理：(1 + 3x)⁵ = Σ (⁵Cᵣ) (1)^5–r (3x)ʳ。前三项对应 r = 0, 1, 2。

• r = 0: ⁵C₀ × 1⁵ × (3x)⁰ = 1

  r = 0: ⁵C₀ × 1⁵ × (3x)⁰ = 1

• r = 1: ⁵C₁ × 1⁴ × (3x)¹ = 5 × 3x = 15x

  r = 1: ⁵C₁ × 1⁴ × (3x)¹ = 5 × 3x = 15x

• r = 2: ⁵C₂ × 1³ × (3x)² = 10 × 9x² = 90x²

  r = 2: ⁵C₂ × 1³ × (3x)² = 10 × 9x² = 90x²

ubstitute into the first three terms: 1 + 15(0.01) + 90(0.01)² = 1 + 0.15 + 0.009 = 1.159.

因此展开式为：1 + 15x + 90x²。要估算 (1.03)⁵，注意到 1.03 = 1 + 3(0.01)，因此令 x = 0.01。代入前三项：1 + 15(0.01) + 90(0.01)² = 1 + 0.15 + 0.009 = 1.159。

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6. Question 6: Arithmetic Series | 第6题：等差级数

Question: The sum of the first 20 terms of an arithmetic series is 450, and the 10th term is 22. Find the first term a and the common difference d.

题目：一个等差级数的前 20 项之和为 450，第 10 项为 22。求首项 a 与公差 d。

Use the formulas: Sₙ = n/2 [2a + (n–1)d] and uₙ = a + (n– S₂₀ = 20/2 (2a + 19d) = 10(2a + 19d) = 450. This simplifies to 2a + 19d = 45. (Equation 1)

使用公式：Sₙ = n/2 [2a + (n–1)d] 以及 uₙ = a + (n–1)d。当 n = 20，S₂₀ = 20/2 (2a + 19d) = 10(2a + 19d) = 450。化简得 2a + 19d = 45。（方程1）

10th term: u₁₀ = a + 9d = 22. (Equation 2) Solve the system: subtract (2) doubled from (1)? Use elimination: multiply (2) by 2: 2a + 18d = 44. Subtract from (1): (2a + 19d) – (2a + 18d) = 45 – 44, giving d = 1. Substitute d = 1 into (2): a + 9 = 22, so a = 13.

第 10 项：u₁₀ = a + 9d = 22。（方程2）解方程组：将方程2乘以2：2a + 18d = 44。从方程1中减去该式：(2a + 19d) – (2a + 18d) = 45 – 44，得 d = 1。将 d = 1 代入方程2：a + 9 = 22，因此 a = 13。

a = 13, d = 1.

首项 a = 13，公差 d = 1。

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7. Question 7: Trigonometric Equation | 第7题：三角方程

Question: Solve 2sin²θ – sinθ – 1 = 0 for 0° ≤ θ ≤ 360°.

题目：解方程 2sin²θ – sinθ – 1 = 0，其中 0° ≤ θ ≤ 360°。

Treat as a quadratic in sinθ. Factorise: (2sinθ + 1)(sinθ – 1) = 0. Thus either 2sinθ + 1 = 0 or sinθ – 1 = 0.

将此视为关于 sinθ 的二次方程。因式分解：(2sinθ + 1)(sinθ – 1) = 0。因此要么 2sinθ + 1 = 0，要么 sinθ – 1 = 0。

Case 1: sinθ = – ince sin is negative in the third and fourth quadrants, θ = 180° + 30° = 210° and θ = 360° – 30° = 330°.

情况1：sinθ = –½。参考角为 30°。正弦在第三、四象限为负，故 θ = 180° + 30° = 210° 及 θ = 360° – 30° = 330°。

Case 2: sinθ = 1. This occurs at θ = 90°.

情况2：sinθ = 1。此情况出现在 θ = 90°。

Solutions: θ = 90°, 210°, 330°.

解：θ = 90°, 210°, 330°。

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8. Question 8: Vector Geometry | 第8题：向量几何

Question: Points A and B have coordinates (2, – tor AB, its magnitude, and the position vector of point C on AB such that AC : CB = 2 : 1.

题目：点 A 和 B 的坐标分别为 (2, –1) 和 (5, 3)。求向量 AB、其模长，以及线段 AB 上一点 C 的位置向量，使得 AC : CB = 2 : 1。

The vector AB = (5 – 2)i + (3 – (–1))j = 3i + 4j. In column form, AB = (3, 4).

向量 AB = (5 – 2)i + (3 – (–1))j = 3i + 4j。列向量形式为 AB = (3, 4)。

The magnitude |AB| = √3² + 4² = √9 + 16 = √25 = 5.

模长 |AB| = √3² + 4² = √9 + 16 = √25 = 5。

B) / (2+1) ) is not correct if AC:CB = 2:1. The correct weighting: C = ( (1×A + 2×B) / 3 )? Let' , then C is 2/3 of the way from A to B. Hence the position vector OC = OA + (2/3) AB = (2, –1) + (2/3)(3, 4) = (2 + 2, – a: C = ( (1·A + 2·B) / (2+1) ) gives ( (2 + 10, -1 + 6) / 3 ) = (12, 5)/3 = (4, 5/3). Yes.

要求内分点 C 使 AC:CB = 2:1，利用定比分点公式：C = ( (1·A + 2·B) / 3 )，注意权重与比例反向。所以 OC = ( (2 + 10)/3, (–1 + 6)/3 ) = (12/3, 5/3) = (4, 5/3)。或者用向量法：OC = OA + (2/3) AB = (2, –1) + (2/3)(3, 4) = (2 + 2, –1 + 8/3) = (4, 5/3)。

AB = (3, 4); |AB| = 5; C = (4, 5/3).

AB = (3, 4)；|AB| = 5；C = (4, 5/3)。

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