📚 Year 12 Edexcel Further Mathematics: Unit Test Mock Paper Analysis | Year 12 Edexcel 进阶数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a typical Year 12 Edexcel Further Mathematics unit test mock paper, covering Core Pure 1 topics such as complex numbers, proof by induction, matrices, series, polynomial roots, vectors, and volumes of revolution. Each section breaks down common question types, key methods, and examiner tips to help you excel.
本文详细解析一份典型的 Year 12 Edexcel 进阶数学单元测试模拟卷，涵盖核心纯数 1 的主题，包括复数、归纳法证明、矩阵、级数、多项式根、向量和旋转体体积。每节拆解常见题型、关键方法和考官提示，助你取得优异成绩。
1. Complex Numbers: Solving Quadratic Equations and Loci | 复数：解二次方程与轨迹
5 + i) = 0. Represent the solutions on an Argand diagram and describe the locus defined by |z - (2 - i)| = 3.
题目：解方程 z² - (3 + 2i)z + (5 + i) = 0。在阿尔冈图上表示其根，并描述满足 |z - (2 - i)| = 3 的轨迹。
² = 9 + 12i - 4 = 5 + 12i. Then 4(5+i) = 20 + 4i. So Δ = (5 + 12i) - (20 + 4i) = -15 + 8i.
使用二次公式求根，z = [ (3+2i) ± √Δ ] / 2，其中判别式 Δ = (3+2i)² - 4(5+i)。展开：(3+2i)² = 9 + 12i + 4i² = 5 + 12i，4(5+i) = 20 + 4i，故 Δ = -15 + 8i。
To find √-15+8i ² = -15 and 2ab = 8, so ab = 4. Solving yields a=1, b=4 or a=-1, b=-4. Hence √Δ = ±(1+4i).
为求 √-15+8i，设 (a+bi)² = -15+8i，得 a² - b² = -15，2ab = 8 即 ab = 4。解得 a=1, b=4 或 a=-1, b=-4。因此 √Δ = ±(1+4i)。
)/2 = 2+3i and z₂ = (2-2i)/2 = 1-i. Plot (2,3) and (1,-1) on the Argand diagram. The locus |z - (2-i)| = 3 is a circle centred at (2,-1) with radius 3.
于是 z = [ (3+2i) ± (1+4i) ] / 2。两根为 z₁ = 2+3i 和 z₂ = 1-i，在阿尔冈图上标出 (2,3) 与 (1,-1)。轨迹 |z - (2 - i)| = 3 是以 (2,-1) 为圆心、半径为 3 的圆。
Common loci to remember: |z - a| = r is a circle; |z - a| = |z - b| is the perpendicular bisector; arg(z - a) = θ is a half-line.
常见轨迹：|z - a| = r 为圆；|z - a| = |z - b| 为垂直平分线；arg(z - a) = θ 为射线。
- our solutions by substituting back into the original equation and be careful with i² = -1.
- 考官提示：务必代入原方程检验解，并谨慎处理 i² = -1。
2. Proof by Induction: Summation of Series | 归纳法证明：级数求和
Question: Prove by induction that for all n ∈ ℕ, ∑r=1n r(r+1) = (1/3)n(n+1)(n+2).
题目：用数学归纳法证明对所有正整数 n，∑r=1n r(r+1) = (1/3)n(n+1)(n+2)。
Base case n=1: LHS = 1×2 = 2. RHS = (1/3)×1×2×3 = 2. True.
基础情况 n=1：左边 = 1×2 = 2，右边 = (1/3)×1×2×3 = 2，成立。
Inductive step: assume true for n=k, so ∑r=1k r(r+1) = (1/3)k(k+1)(k+2). For n=k+1, LHS = ∑r=1k r(r+1) + (k+1)(k+2). Substitute the assumption:
归纳步骤：假设 n=k 时成立，即 ∑r=1k r(r+1) = (1/3)k(k+1)(k+2)。对于 n=k+1，左边 = ∑r=1k r(r+1) + (k+1)(k+2)。代入假设：
s the formula for n=k+1. Hence true for n=k+1. By induction, the statement holds for all n ∈ ℕ.
左边 )k(k+1)(k+2) + (k+1)(k+2) = (k+1)(k+2)(k/3+1) = (k+1)(k+2)((k+3)/3) = (1/3)(k+1)(k+2)(k+3)，与 n=k+1 时公式一致。因此由归纳法，命题对所有正整数成立。
- Always state the three parts clearly: basis, assumption, and inductive conclusion. Factorisation is key.
- 清晰写出三要素：基础、假设与归纳结论，分解因式是关键。
3. Matrices: Inverting a 3×3 Matrix and Determinants | 矩阵：求逆3×3与行列式
Question: Find the inverse of matrix A = [[1,2,0], [0,-1,3], [2,1,-1]].
题目：求矩阵 A = [[1,2,0], [0,-1,3], [2,1,-1]] 的逆矩阵。
([[0,3],[2,-1]]) + 0. = 1×((-1)(-1)-3×1) - 2×(0×(-1)-3×2) = (1-3) - 2×(0-6) = -2 +12 = 10.
步骤1：计算行列式。沿第一行展开：det(A) = 1×( (-1)(-1)-3×1 ) - 2×( 0×(-1)-3×2 ) = -2 + 12 = 10。
M₂₁ = -2, M₂₂ = -1, M₂₃ = -3, M₃₁ = 6, M₃₂ = 3, M₃₃ = -1. Cofactor matrix C has entries Cᵢⱼ = (-1)ⁱ⁺ʲ Mᵢⱼ.
步骤2：求余子式矩阵并添加正负号得代数余子式矩阵。余子式：M₁₁=1, M₁₂=-6, M₁₃=2; M₂₁=-2, M₂₂=-1, M₂₃=-3; M₃₁=6, M₃₂=3, M₃₃=-1。代数余子式 C₁₁=1, C₁₂=6, C₁₃=2, C₂₁=2, C₂₂=-1, C₂₃=3, C₃₁=6, C₃₂=-3, C₃₃=-1。
,-3], [2,3,-1]]. Finally, A⁻¹ = (1/det(A)) adj(A) = 1/10 × [[1,2,6], [6,-1,-3], [2,3,-1]].
步骤3：转置代数余子式矩阵得伴随矩阵 adj(A)，然后 A⁻¹ = (1/10) adj(A) = [[0.1,0.2,0.6], [0.6,-0.1,-0.3], [0.2,0.3,-0.1]]。
- rs → cofactors → transpose → multiply by 1/det. A common error is forgetting to transpose.
- 牢记顺序：余子式 → 代数余子式 → 转置 → 乘 1/det。常犯错误是忘记转置。
4. Summing Series: Using Standard Results for ∑r² | 级数求和：利用∑r²标准结果
Question: Find an expression for ∑r=1n r(r+2). Simplify your answer.
题目：求 ∑r=1n r(r+2) 的表达式并化简。 Published by TutorHao | Year 12 进阶数学 Revision Series | aleveler.com
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