📚 GCSE Eduqas Further Mathematics: Unit Test Mock Paper Walkthrough | GCSE Eduqas 进阶数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a GCSE Eduqas Further Mathematics unit test mock paper. The paper covers key topics from algebra, functions, coordinate geometry, and introductory calculus, as typically assessed in the Level 2 Certificate in Further Mathematics. By working through these questions, students can consolidate their understanding of factor theorem, binomial expansion, circle equations, differentiation, integration, and matrix algebra. Each solution is explained step by step, with reasoning given in both English and Chinese to support bilingual learners.
本文详细解析了一套 GCSE Eduqas 进阶数学单元测试模拟卷。该试卷涵盖了代数、函数、坐标几何以及微积分初步等核心内容，与二级进阶数学证书的考试范围一致。通过逐题讲解，学生可以巩固因式定理、二项式展开、圆的方程、微分、积分和矩阵代数的知识。每道题的解答均以中英双语逐步呈现，帮助双语学习者全面理解。
1. Question 1: Solving a Quadratic Inequality | 第1题：解二次不等式
Question: Solve the inequality x² – 5x + 6 > 0.
题目：解不等式 x² – 5x + 6 > 0。
1: Factorise the quadratic expression. Look for two numbers that multiply to +6 and add to –5; these are –2 and –3. Thus, x² – 5x + 6 = (x – 2)(x – 3).
步骤1：对二次式进行因式分解。找到两个数，乘积为 +6，和为 –5，这两个数是 –2 和 –3。因此，x² – 5x + 6 = (x – 2)(x – 3)。
Step 2: Set up the inequality (x – 2)(x – = 3. These roots divide the number line into three intervals: x < 2, 2 < x < 3, and x > 3.
步骤2：建立不等式 (x – 2)(x – 3) > 0。临界值为 x = 2 和 x = 3。这两个根将数轴划分为三个区间：x < 2、2 < x < 3 和 x > 3。
Step 3: Test each interval. For x < 2 (e.g., x = 0), (–)(–) = +, so positive. For 2 < x < 3 (e.g., x = 2.5), (+)(–) = – , positive. The product must be greater than 0, so we require strictly positive intervals.
步骤3：检验每个区间。对于 x < 2（如 x = 0），(–)(–) = +，为正。对于 2 < x < 3（如 x = 2.5），(+)(–) = –，为负。对于 x > 3（如 x = 4），(+)(+) = +，为正。我们要求乘积大于 0，因此取严格为正的区间。
Step 4: Write the solution set. {x : x < 2} ∪ {x : x > 3}, or in interval notation, (–∞, 2) ∪ (3, ∞).
步骤4：写出解集。{x : x < 2} ∪ {x : x > 3}，或用区间表示法 (–∞, 2) ∪ (3, ∞)。
Answer: x < 2 or x > 3.
答案：x < 2 或 x > 3。
2. Question 2: Factor Theorem and Polynomial Division | 第2题：因式定理与多项式除法
Question: f(x) = 2x³ – 3x² – 3x + 2. Given that (x – a product of a linear factor and a quadratic factor, and hence factorise f(x) completely.
题目： f(x) = 2x³ – 3x² – 3x + 2。已知 (x – 2) 是一个因式，将 f(x) 表示为一次因式与二次因式的乘积，并由此对 f(x) 进行彻底因式分解。
Step 1: Verify the factor using the Factor Theorem: f(2) = 2(8) – 3(4) – 3(2) + 2 = 16 – 12 – 6 + 2 = 0, confirming (x – 2) is indeed a factor.
步骤1：利用因式定理验证：f(2) = 2(8) – 3(4) – 3(2) + 2 = 16 – 12 – 6 + 2 = 0，确认 (x – 2) 是一个因式。
Step 2: Perform polynomial division of f(x) by (x – 2). Using algebraic long division or synthetic division, divide 2x³ – 3x² – 3x + 2 by (x – 2). Synthetic division with root 2: coefficients 2, –3, –3, 2. Bring down 2; multiply 2×2=4, add to –3 gives 1; multiply 1×2=2, add to –3 gives –1; multiply –1×2=–2, add to 2 gives 0 (remainder). The quotient is 2x² + x – 1.
步骤2：用 (x – 2) 去除 f(x)。使用代数长除法或综合除法，综合除法以根 2 操作：系数为 2, –3, –3, 2。移下 2；2×2=4，加到 –3 得 1；1×2=2，加到 –3 得 –1；–1×2=–2，加到 2 得 0（余数）。商式为 2x² + x – 1。
Step 3: Factorise the quadratic quotient 2x² + x – 1. Calculate product a×c = 2×(–1) = –2. Two numbers that multiply to –2 and add to 1 are 2 and –1. Split the middle term: 2x² + 2x – x – 1. Factor by grouping: 2x(x + 1) – 1(x + 1) = (2x – 1)(x + 1).
步骤3：对二次商式 2x² + x – 1 进行因式分解。计算 a×c = 2×(–1) = –2。两个数乘积为 –2 且和为 1 的是 2 和 –1。拆分中间项：2x² + 2x – x – 1。分组分解：2x(x + 1) – 1(x + 1) = (2x – 1)(x + 1)。
Step 4: Combine all factors: f(x) = (x – 2)(2x – 1)(x + 1).
步骤4：合并所有因式：f(x) = (x – 2)(2x – 1)(x + 1)。
Answer: f(x) = (x – 2)(2x – 1)(x + 1).
答案：f(x) = (x – 2)(2x – 1)(x + 1)。
3. Question 3: Binomial Expansion | 第3题：二项式展开
Question: Expand (2 + 3x)⁴ in ascending powers of x up to and including the term in x³. Simplify each coefficient.
题目：按 x 的升幂展开 (2 + 3x)⁴，直至包含 x³ 项。化简各系数。
Step 1: Use the binomial theorem: (a + b)ⁿ = Σ (ⁿCᵣ) aⁿ⁻ʳ bʳ, from r = 0 to n. Here a = 2, b = 3x, n = 4.
步骤1：使用二项式定理：(a + b)ⁿ = Σ (ⁿCᵣ) aⁿ⁻ʳ bʳ，r 从 0 到 n。此处 a = 2，b = 3x，n = 4。
Step 2: Compute the general term for r = 0, 1, 2, 3. Term Tᵣ₊₁ = ⁴Cᵣ (2)⁴⁻ʳ (3x)ʳ.
步骤2：计算 r = 0, 1, 2, 3 时的通项。Tᵣ₊₁ = ⁴Cᵣ (2)⁴⁻ʳ (3x)ʳ。
r = 0: ⁴C₀ (2)⁴ (3x)⁰ = 1 × 16 × 1 = 16.
r = 1: ⁴C₁ (2)³ (3x)¹ = 4 × 8 × 3x = 96x.
r = 2: ⁴C₂ (2)² (3x)² = 6 × 4 × 9x² = 216x².
r = 3: ⁴C₃ (2)¹ (3x)³ = 4 × 2 × 27x³ = 216x³.
Step 3: Write the expansion up to x³: (2 + 3x)⁴ = 16 + 96x + 216x² + 216x³ + ...
步骤3：写出截至 x³ 的展开式：(2 + 3x)⁴ = 16 + 96x + 216x² + 216x³ + ...
Answer: 16 + 96x + 216x² + 216x³ (higher-order terms omitted).
答案：16 + 96x + 216x² + 216x³（更高次项省略）。
4. Question 4: Straight Line – Perpendicular Bisector | 第4题：直线 – 垂直平分线
Question: Points A(–1, 2) and B(5, – line segment AB. Give your answer in the form ax + by + c = 0, where a, b, c are integers.
题目：已知点 A(–1, 2) 和 B(5, –4)。求线段 AB 的垂直平分线的方程。答案写成 ax + by + c = 0 的形式，其中 a, b, c 为整数。
Step 1: Find the midpoint M of AB. Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) = ((–1 + 5)/2, (2 + (–4))/2) = (4/2, –2/2) = (2, –1).
步骤1：求线段 AB 的中点 M。中点 = ((–1+5)/2, (2–4)/2) = (4/2, –2/2) = (2, –1)。
Step 2: Calculate the gradient of AB. mAB = (y₂ – y₁) / (x₂ – x₁) = (–4 – 2) / (5 – (–1)) = –6 / 6 = –1.
步骤2：计算 AB 的斜率。mAB = (–4 – 2) / (5 – (–1)) = –6/6 = –1。
Step 3: The gradient of the perpendicular bisector is the negative reciprocal. Since mAB = –1, the perpendicular gradient mperp = 1.
步骤3：垂直平分线的斜率是原斜率的负倒数。因为 mAB = –1，所以垂直斜率 mperp = 1。
Step 4: Use point-slope form with midpoint M(2, –1) and slope 1: y – (–1) = 1(x – 2) → y + 1 = x – 2.
步骤4：使用点斜式，代入中点 M(2, –1) 和斜率 1：y + 1 = x – 2。
Step 5: Rearrange into the form ax + by + c = 0. y + 1 – x + 2 = 0 → –x + y + 3 = 0. Multiply by –1 to get x – y – 3 = 0.
步骤5：整理为 ax + by + c = 0 的形式。y + 1 – x + 2 = 0 → –x + y + 3 = 0。乘以 –1 得 x – y – 3 = 0。
Answer: x – y – 3 = 0.
答案：x – y – 3 = 0。
5. Question 5: Equation of a Circle | 第5题：圆的方程
Question: The points P(1, 7) and Q(9, – re the endpoints of a diameter of a circle. Find the equation of the circle in the form (x – a)² + (y – b)² = r².
题目：点 P(1, 7) 和 Q(9, –1) 是一个圆的一条直径的两个端点。求该圆的方程，写成 (x – a)² + (y – b)² = r² 的形式。
Step 1: The centre C is the midpoint of PQ. Centre = ((1 + 9)/2, (7 + (–1))/2) = (10/2, 6/2) = (5, 3).
步骤1：圆心 C 是 PQ 的中点。圆心 = ((1+9)/2, (7–1)/2) = (10/2, 6/2) = (5, 3)。
: The radius r is half the length of diameter PQ, or distance from C to P. Compute CP²: (5 – 1)² + (3 – 7)² = 4² + (–4)² = 16 + 16 = 32. Thus r² = 32.
步骤2：半径 r 是直径 PQ 长度的一半，或计算圆心 C 到 P 的距离。CP² = (5 – 1)² + (3 – 7)² = 4² + (–4)² = 16 + 16 = 32。因此 r² = 32。
Step 3: Substitute into the standard form: (x – 5)² + (y – 3)² = 32.
步骤3：代入标准式：(x – 5)² + (y – 3)² = 32。
Answer: (x – 5)² + (y – 3)² = 32.
答案：(x – 5)² + (y – 3)² = 32。
6. Question 6: Differentiation – Finding Tangents | 第6题：微分 – 求切线
Question: A curve has equation y = x³ – 4x + 1. Find the equation of the tangent to the curve at the point where x = 2.
题目：已知曲线方程为 y = x³ – 4x + 1。求该曲线在 x = 2 处的切线方程。
Step 1: Differentiate to find the gradient function. dy/dx = 3x² – 4.
步骤1：求导得到斜率函数。dy/dx = 3x² – 4。
Step 2: Evaluate the gradient at x = 2. m = 3(2)² – 4 = 3(4) – 4 = 12 – 4 = 8.
步骤2：计算 x = 2 处的斜率。m = 3(2)² – 4 = 12 – 4 = 8。
Step 3: Find the y-coordinate of the point of tangency. y = (2)³ – 4(2) + 1 = 8 – 8 + 1 = 1. The point is (2, 1).
步骤3：求切点的纵坐标。y = 2³ – 4×2 + 1 = 8 – 8 + 1 = 1。切点为 (2, 1)。
Step 4: Write the tangent equation using point-slope form: y – 1 = 8(x – 2).
步骤4：用点斜式写出切线方程：y – 1 = 8(x – 2)。
Step 5: Simplify to slope-intercept form: y = 8x – 16 + 1 → y = 8x – 15. Alternatively, in general form: 8x – y – 15 = 0.
步骤5：化简为斜截式：y = 8x – 15。或一般式：8x – y – 15 = 0。
Answer: y = 8x – 15 (or 8x – y – 15 = 0).
答案：y = 8x – 15（或 8x – y – 15 = 0）。
7. Question 7: Integration – Area under a Curve | 第7题：积分 – 曲线下方面积
Question: Find the area bounded by the curve y = 3x² + 2, the x-axis, and the lines x = 1 and x = 3.
题目：求由曲线 y = 3x² + 2、x 轴及直线 x = 1 和 x = 3 所围成的面积。
Step 1: Set up the definite integral. Area = ∫₁³ (3x² + 2) dx.
步骤1：建立定积分。面积 = ∫₁³ (3x² + 2) dx。
ate term by term. ∫ 3x² dx = x³, and ∫ 2 dx = 2x. So the antiderivative is F(x) = x³ + 2x.
步骤2：逐项积分。∫ 3x² dx = x³，∫ 2 dx = 2x。因此原函数为 F(x) = x³ + 2x。
te using limits. F(3) = 3³ + 2(3) = 27 + 6 = 33. F(1) = 1³ + 2(1) = 1 + 2 = 3. Area = F(3) – F(1) = 33 – 3 = 30.
步骤3：利用上下限求值。F(3) = 27 + 6 = 33。F(1) = 1 + 2 = 3。面积 = 33 – 3 = 30。
Step 4: Include units if applicable; the answer is 30 square units.
步骤4：根据题意带单位；答案为 30 平方单位。
Answer: 30.
答案：30。
8. Question 8: Matrices – Multiplication and Inverse | 第8题：矩阵 – 乘法与逆矩阵
Question: Given matrices A = [2, 1; 4, 3] and B = [3, –1; – nverse of matrix A, and use it to solve the system of equations: 2x + y = 7, 4x + 3y = 17.
题目：已知矩阵 A = [2, 1; 4, 3] 和 B = [3, –1; –2, 5]，(a) 求乘积 AB。(b) 求矩阵 A 的逆矩阵，并利用它解方程组：2x + y = 7, 4x + 3y = 17。
Part (a) - Multiplication: Matrix AB is computed as follows (first row dot columns of B).
部分 (a) – 乘法：计算矩阵 AB 如下（第一行与 B 的各列点乘）。
AB[1,1] = 2×3 + 1×(–2) = 6 – 2 = 4.
AB[1,1] = 2×3 + 1×(–2) = 6 – 2 = 4。
AB[1,2] = 2×(–1) + 1×5 = –2 + 5 = 3.
AB[1,2] = 2×(–1) + 1×5 = –2 + 5 = 3。
AB[2,1] = 4×3 + 3×(–2) = 12 – 6 = 6.
AB[2,1] = 4×3 + 3×(–2) = 12 – 6 = 6。
AB[2,2] = 4×(–1) + 3×5 = –4 + 15 = 11.
AB[2,2] = 4×(–1) + 3×5 = –4 + 15 = 11。
Thus, AB = [4, 3; 6, 11].
因此，AB = [4, 3; 6, 11]。
Part (b) - Inverse of A: For a 2×2 matrix A = [a, b; c, d], the inverse is (1/(ad – bc)) [d, –b; –c, a]. Here a=2, b=1, c=4, d=3. Determinant Δ = ad – bc = 2×3 – 1×4 = 6 – 4 = 2. Since Δ ≠ 0, inverse exists. A⁻¹ = (1/2) [3, –1; –4, 2] = [3/2, –1/2; –2, 1].
部分 (b) – A 的逆矩阵：对于 2×2 矩阵 A = [a, b; c, d]，逆矩阵为 (1/(ad – bc)) [d, –b; –c, a]。此处 a=2, b=1, c=4, d=3。行列式 Δ = 2×3 – 1×4 = 6 – 4 = 2。因 Δ ≠ 0，逆矩阵存在。A⁻¹ = (1/2) [3, –1; –4, 2] = [3/2, –1/2; –2, 1]。
Solve the system: Write as AX = C, where X = [x; y] and C = [7; 17]. Then X = A⁻¹C.
解方程组：写成 AX = C，其中 X = [x; y]，C = [7; 17]。则 X = A⁻¹C。
Compute: x = (3/2)×7 + (–1/2)×17 = 21/2 – 17/2 = 4/2 = 2. y = (–2)×7 + 1×17 = –14 + 17 = 3.
计算：x = (3/2)×7 + (–1/2)×17 = 21/2 – 17/2 = 4/2 = 2。y = (–2)×7 + 1×17 = –14 + 17 = 3。
Solution: x = 2, y = 3.
解为：x = 2, y = 3。
Answer: (a) AB = [4, 3; 6, 11]; (b) A⁻¹ = [3/2, –1
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