📚 Year 10 Edexcel Further Maths: Unit Test Mock Paper Walkthrough | 爱德思十年级进阶数学：单元测试模拟卷解析
This walkthrough breaks down a full Unit Test mock paper for Year 10 Edexcel Further Mathematics. Each question is explained with clear steps, guiding you through the key concepts and common pitfalls. Use this to self‑assess, strengthen your understanding, and boost exam confidence.
本解析详细拆解了一套针对爱德思十年级进阶数学的单元测试模拟卷。每道题目都配有清晰的步骤讲解，帮助你掌握核心概念并回避常见错误。你可以用来自我评估、夯实理解，并在考前建立信心。
1. Question 1 – Factor Theorem and Cubic Roots | 第1题 – 因式定理与三次方程求根
Given f(x) = x³ – eorem to find one linear factor and hence factorise f(x) completely. Then state all roots.
已知 f(x) = x³ – 4x² + x + 6，利用因式定理找出一个一次因式，进而将 f(x) 完全分解，并写出所有根。
Step 1: Test simple values like x = –1, 1, 2, –2, 3, –3. f(–1) = –1 – 4 –1 + 6 = 0, so (x + 1) is a factor.
第一步：检验简单数值，如 x = –1, 1, 2, –2, 3, –3。f(–1) = –1 – 4 –1 + 6 = 0，因此 (x + 1) 是一个因式。
Step 2: Divide f(x) by (x + 1) using algebraic long division or synthetic division. The quotient is x² – 5x + 6.
第二步：用长除法或综合除法将 f(x) 除以 (x + 1)。商式为 x² – 5x + 6。
Step 3: Factorise the quadratic: x² – 5x + 6 = (x – 2)(x – 3). Thus f(x) = (x + 1)(x – 2)(x – 3).
第三步：对二次式进行因式分解：x² – 5x + 6 = (x – 2)(x – 3)。因此 f(x) = (x + 1)(x – 2)(x – 3)。
Step 4: Roots are the solutions to f(x) = 0, so x = –1, x = 2, x = 3.
第四步：根即为方程 f(x) = 0 的解，故 x = –1，x = 2，x = 3。
2. Question 2 – Binomial Expansion | 第2题 – 二项式展开
Expand (1 + 2x)⁵ up to and including the term in x³. Write coefficients in simplest form.
展开 (1 + 2x)⁵，写出包含 x³ 项在内的各次项，系数化为最简形式。
Use the binomial theorem: (1 + u)ⁿ = 1 + nC₁ u + nC₂ u² + nC₃ u³ + … where u = 2x.
使用二项式定理：(1 + u)ⁿ = 1 + nC₁ u + nC₂ u² + nC₃ u³ + …，此处 u = 2x。
Coefficients: 5C₁ = 5, 5C₂ = 10, 5C₃ = 10. Then u = 2x, u² = 4x², u³ = 8x³.
组合数：5C₁ = 5，5C₂ = 10，5C₃ = 10。代入 u = 2x，得 u² = 4x²，u³ = 8x³。
0(4x²) + 10(8x³) = 1 + 10x + 40x² + 80x³. The expansion up to x³ is 1 + 10x + 40x² + 80x³.
各项：1 + 5(2x) + 10(4x²) + 10(8x³) = 1 + 10x + 40x² + 80x³。包含 x³ 项的展开式为 1 + 10x + 40x² + 80x³。
Common error: forgetting to raise the 2 inside u to the same power. Always apply the power to both the constant and the x.
常见错误：忘记将 u 中的数字 2 也乘方。务必对常数和 x 同时施加幂次。
3. Question 3 – Composite Functions | 第3题 – 复合函数
Given f(x) = 2x² – 3x + 1, find f(x + 1) in its simplest form.
已知 f(x) = 2x² – 3x + 1，求 f(x + 1) 的最简表达式。
Replace every x in f(x) with (x + 1): f(x + 1) = 2(x + 1)² – 3(x + 1) + 1.
将 f(x) 中的所有 x 替换为 (x + 1)：f(x + 1) = 2(x + 1)² – 3(x + 1) + 1。
Expand: (x + 1)² = x² + 2x + 1, so 2(x² + 2x + 1) = 2x² + 4x + 2. Then –3(x + 1) = –3x – 3.
展开：(x + 1)² = x² + 2x + 1，故 2(x² + 2x + 1) = 2x² + 4x + 2。接着 –3(x + 1) = –3x – 3。
Combine: 2x² + 4x + 2 – 3x – 3 + 1 = 2x² + (4x – 3x) + (2 – 3 + 1) = 2x² + x + 0 = 2x² + x.
合并同类项：2x² + 4x + 2 – 3x – 3 + 1 = 2x² + (4x – 3x) + (2 – 3 + 1) = 2x² + x + 0 = 2x² + x。
Always simplify completely. Leaving brackets unexpanded loses marks.
一定要化简到底。保留括号不展开会导致失分。
4. Question 4 – Finding a Tangent Equation | 第4题 – 求切线方程
Find the equation of the tangent to the curve y = x³ – 3x² + 2 at the point where x = 2.
求曲线 y = x³ – 3x² + 2 在 x = 2 处的切线方程。
Step 1: Find y-coordinate: y(2) = 8 – 3(4) + 2 = 8 – 12 + 2 = –2. Point is (2, –2).
第一步：求 y 坐标：y(2) = 8 – 3(4) + 2 = 8 – 12 + 2 = –2，点为 (2, –2)。
Step 2: Differentiate: dy/dx = 3x² – 6x. Gradient at x = 2: m = 3(4) – 6(2) = 12 – 12 = 0.
第二步：求导：dy/dx = 3x² – 6x。在 x = 2 处的斜率：m = 3(4) – 6(2) = 12 – 12 = 0。
Step 3: Tangent equation with m = 0 is y – y₁ = 0(x – x₁) → y = –2.
第三步：斜率 m = 0 的切线方程为 y – y₁ = 0(x – x₁) → y = –2。
curve has a stationary point at x = 2, which is confirmed by substituting back into dy/dx.
这是一条水平切线。该曲线在 x = 2 处有一个驻点，可通过代入导数表达式确认。
5. Question 5 – Integration by Reverse Chain Rule | 第5题 – 逆链式法则积分
Find ∫ (2x + 1)³ dx.
求 ∫ (2x + 1)³ dx。
Recognise the inner function u = 2x + 1, du/dx = 2, so dx = du/2.
识别内层函数 u = 2x + 1，du/dx = 2，故 dx = du/2。
Substitute: ∫ u³ (du/2) = ½ ∫ u³ du = ½ × (u⁴/4) = u⁴ / 8 + C.
换元：∫ u³ (du/2) = ½ ∫ u³ du = ½ × (u⁴/4) = u⁴ / 8 + C。
Replace u: (2x + 1)⁴ / 8 + C. Always include the constant of integration.
回代 u： (2x + 1)⁴ / 8 + C。切勿遗漏积分常数。
2x²+6x+1 and integrate term by term to check: 2x⁴+4x³+3x²+x+C. Factorising confirms that both forms are equivalent (factor out 1/8 and 8, it matches (2x+1)⁴/8).
另解：先展开 (2x+1)³ = 8x³+12x²+6x+1，再逐项积分得 2x⁴+4x³+3x²+x+C。将结果提取 1/8 并配出 8，可验证与原答案等价。
6. Question 6 – Matrix Inverse and Simultaneous Equations | 第6题 – 逆矩阵与联立方程组
Let A = [[2,1],[3,4]]. Find A⁻¹ and use it to solve the system: 2x + y = 5, 3x + 4y = 6.
设 A = [[2,1],[3,4]]。求 A⁻¹ 并利用它解方程组：2x + y = 5，3x + 4y = 6。
Determinant: det(A) = 2×4 – 1×3 = 8 – 3 = 5. Since det ≠ 0, A⁻¹ exists.
行列式：det(A) = 2×4 – 1×3 = 8 – 3 = 5。因行列式非零，故 A⁻¹ 存在。
A⁻¹ = (1/5) × [[4, –1],[–3, 2]] = [[0.8, –0.2],[–0.6, 0.4]] or leave as fractions [[4/5, –1/5],[–3/5, 2/5]].
A⁻¹ = (1/5) × [[4, –1],[–3, 2]] = [[0.8, –0.2],[–0.6, 0.4]]，亦可用分数形式 [[4/5, –1/5],[–3/5, 2/5]]。
Write system as A × [x; y] = [5; 6]. Multiply both sides by A⁻¹: [x; y] = A⁻¹ × [5; 6].
将方程组写作 A × [x; y] = [5; 6]。两边同乘 A⁻¹：[x; y] = A⁻¹ × [5; 6]。
x = (4/5)×5 + (–1/5)×6 = 4 – 1.2 = 2.8 = 14/5. y = (–3/5)×5 + (2/5)×6 = –3 + 2.4 = –0.6 = –3/5.
x = (4/5)×5 + (–1/5)×6 = 4 – 1.2 = 2.8 = 14/5。y = (–3/5)×5 + (2/5)×6 = –3 + 2.4 = –0.6 = –3/5。
Always check by substituting back into original equations.
务必将解代回原方程组进行验证。
7. Question 7 – Vectors and Distance in 3D | 第7题 – 三维向量及距离
Points A(1, 2, 3) and B(4, 6, 8). Find vector AB and the distance |AB|.
已知点 A(1, 2, 3) 和 B(4, 6, 8)。求向量 AB 以及距离 |AB|。
Vector AB = (4–1, 6–2, 8–3) = (3, 4, 5). This can be written as 3i + 4j + 5k.
向量 AB = (4–1, 6–2, 8–3) = (3, 4, 5)，可记作 3i + 4j + 5k。
Distance |AB| = √3² + 4² + 5² = √9 + 16 + 25 = √50 = 5√2 (approx 7.07).
距离 |AB| = √3² + 4² + 5² = √9 + 16 + 25 = √50 = 5√2（约等于 7.07）。
Simplifying surds is expected – √50 = √25×2 = 5√2.
要求将根式化简——√50 = √25×2 = 5√2。
8. Question 8 – Arithmetic Sequences and Series | 第8题 – 等差数列与求和
The 5th term of an arithmetic sequence is 20 and the 10th term is 45. Find the first term a and common difference d. Hence calculate the sum of the first 20 terms.
一个等差数列的第5项为20，第10项为45。求首项 a 和公差 d，并由此计算前20项的和。
nth term formula: uₙ = a + (n–1)d. For n=5: a + 4d = 20. For n=10: a + 9d = 45.
通项公式：uₙ = a + (n–1)d。代入 n=5：a + 4d = 20；代入 n=10：a + 9d = 45。
Subtract first equation from second: (a+9d) – (a+4d) = 45 – 20 → 5d = 25 → d = 5.
将第二个方程减去第一个：(a+9d) – (a+4d) = 45 – 20 → 5d = 25 → d = 5。
Substitute d=5 into a+4(5)=20 → a+20=20 → a=0.
将 d=5 代入 a+4(5)=20 → a+20=20 → a=0。
Sum of first 20 terms: Sₙ = n/2 [2a + (n–1)d] → S₂₀ = 20/2 [0 + 19×5] = 10 × 95 = 950.
前20项和公式：Sₙ = n/2 [2a + (n–1)d] → S₂₀ = 20/2 [0 + 19×5] = 10 × 95 = 950。
Even with a=0, the sequence is still arithmetic (0, 5, 10, 15, …). Always show the full working.
即使首项为0，该数列仍是等差数列（0, 5, 10, 15, …）。展示完整解题过程至关重要。
9. Question 9 – Parametric Differentiation | 第9题 – 参数方程求导
A curve is defined by x = 2t² + 1, y = 4t³ – t. Find dy/dx in terms of t, and hence find the gradient at t = 2.
某曲线由参数方程 x = 2t² + 1，y = 4t³ – t 定义。求 dy/dx 关于 t 的表达式，并据此求出 t = 2 处的梯度。
Differentiate separately: dx/dt = 4t, dy/dt = 12t² – 1.
分别求导：dx/dt = 4t，dy/dt = 12t² – 1。
By chain rule, dy/dx = (dy/dt) / (dx/dt) = (12t² – 1) / (4t). Simplify if possible.
根据链式法则，dy/dx = (dy/dt) / (dx/dt) = (12t² – 1) / (4t)。尽可能化简。
At t = 2: dy/dx = (12×4 – 1) / (4×2) = (48 – 1) / 8 = 47 / 8 = 5.875. The gradient is 47/8.
代入 t = 2：dy/dx = (12×4 – 1) / (4×2) = (48 – 1) / 8 = 47 / 8 = 5.875。梯度为 47/8。
Leaving answer as an exact fraction is preferred in further maths unless asked otherwise.
在进阶数学中，除非另有要求，答案保留为精确分数更佳。
10. Common Mistakes and Top Tips | 常见错误与高分技巧
After marking hundreds of mocks, we see the same slip‑ups: forgetting the constant of integration, algebraic sign errors when subtracting equations, incorrect factorisation after synthetic division, and mishandling negative signs in binomial expansions. Always double‑check sign changes and constants.
在批阅了数百份模拟卷后，我们发现同样的失分点反复出现：遗漏积分常数、联立方程相减时符号错误、综合除法后的因式分解出错，以及二项式展开中负号处理不当。务必反复检查符号变化和常数项。
Top tips: (1) Show every step – method marks often save grades. (2) On calculus questions, write “+ C” divisors efficiently. (4) For matrices, verify by multiplying A × A⁻¹ to see if you get I.
高分建议：(1) 写出每一步——方法分经常能挽救等级。(2) 在积分题中，完成积分后立刻写上“+ C”。(3) 利用因式定理高效检验除数。(4) 对于矩阵，用 A × A⁻¹ 验算是否得到单位矩阵 I。
Time management: spend 1 minute per mark. If stuck, move on and return later – the paper rewards accuracy across all topics.
时间管理：每题按分值分配一分钟。如果卡住，先跳过，稍后回来再答——试卷看重的是各知识点的整体正确率。
Practise past papers under timed conditions. This walkthrough mirrors the style of Edexcel further maths unit tests. Consistent practice leads to fluency and speed.
在计时的条件下练习往年真题。本解析与爱德思进阶数学单元测试的风格一致。持续练习会带来解题的流畅度和速度。
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