📚 Year 12 Edexcel Further Maths: Unit Test Mock Paper Walkthrough | Year 12 Edexcel 进阶数学：单元测试模拟卷解析
This article provides a detailed, step-by-step walkthrough of a typical Year 12 Edexcel Further Mathematics unit test. Each question has been designed to mirror the style of Core Pure 1 and Core Pure 2 topics, including complex numbers, matrices, series, hyperbolic functions, polar coordinates, first‑order differential equations, vectors and roots of polynomials. Every solution step is presented in English immediately followed by its Chinese translation to support bilingual learning.
本文为一份典型的 Year 12 Edexcel 进阶数学单元测试提供了详细的分步解析。每道题目均模仿 Core Pure 1 和 Core Pure 2 的命题风格，涵盖复数、矩阵、级数、双曲函数、极坐标、一阶微分方程、向量以及多项式根等内容。所有解题步骤均以英文与中文配对呈现，便于双语学习。
1. Complex Numbers: Quadratic with Complex Coefficients | 复数：带复数系数的二次方程
Solve the equation z² + (1 – i)z – i = 0. Give your answers in the form a + bi.
解方程 z² + (1 – i)z – i = 0，答案写成 a + bi 的形式。
We use the quadratic formula z = [–b ± √(b² – 4ac)] / (2a) with a = 1, b = 1 – i and c = –i.
我们使用二次公式 z = [–b ± √(b² – 4ac)] / (2a)，其中 a = 1, b = 1 – i, c = –i。
First compute the discriminant Δ = b² – 4ac = (1 – i)² – 4(1)(–i).
首先计算判别式 Δ = b² – 4ac = (1 – i)² – 4(1)(–i)。
Expand (1 – i)²: 1 – 2i + i² = 1 – 2i – 1 = –2i. Then –4(1)(–i) = +4i. Hence Δ = –2i + 4i = 2i.
展开 (1 – i)²：1 – 2i + i² = 1 – 2i – 1 = –2i。而 –4(1)(–i) = +4i。因此 Δ = –2i + 4i = 2i。
Δ = 2i
We now need √(2i). Notice that (1 + i)² = 1 + 2i + i² = 2i, so √(2i) = ±(1 + i).
现在我们需要求 √(2i)。注意到 (1 + i)² = 1 + 2i + i² = 2i，因此 √(2i) = ±(1 + i)。
Apply the formula: z = [–(1 – i) ± (1 + i)] / 2.
代入公式：z = [–(1 – i) ± (1 + i)] / 2。
Take the ‘+’ sign: z = [–1 + i + 1 + i] / 2 = (2i) / 2 = i.
取 ‘+’ 号：z = [–1 + i + 1 + i] / 2 = (2i) / 2 = i。
Take the ‘–’ sign: z = [–1 + i – 1 – i] / 2 = (–2) / 2 = –1.
取 ‘–’ 号：z = [–1 + i – 1 – i] / 2 = (–2) / 2 = –1。
Therefore the solutions are z = –1 and z = i.
因此解为 z = –1 和 z = i。
2. Matrices: Inverse and Linear Transformations | 矩阵：逆矩阵与线性变换
Given matrix M = [2, -1; 3, 4], find det(M) and M⁻¹. Hence find the image of the unit square under M.
给定矩阵 M = [[2, -1], [3, 4]]，求 det(M) 与 M⁻¹。进而求单位正方形在 M 下的像。
det(M) = (2)(4) – (–1)(3) = 8 + 3 = 11.
det(M) = (2)(4) – (–1)(3) = 8 + 3 = 11。
det(M) = 11
For a 2×2 matrix [a, b; c, d], the inverse is (1/det) [d, -b; -c, a].
对于 2×2 矩阵 [a, b; c, d]，其逆矩阵为 (1/行列式) [d, -b; -c, a]。
Thus M⁻¹ = (1/11) [4, 1; -3, 2] = [4/11, 1/11; -3/11, 2/11].
因此 M⁻¹ = (1/11) [4, 1; -3, 2] = [4/11, 1/11; -3/11, 2/11]。
The unit square has vertices O(0,0), A(1,0), B(1,1), C(0,1). Apply M to each position vector.
单位正方形的顶点为 O(0,0), A(1,0), B(1,1), C(0,1)。将 M 作用于各位置向量。
O’ = M(0,0) = (0,0). A’ = M(1,0) = (2,3). B’ = M(1,1) = (1,7). C’ = M(0,1) = (-1,4).
O’ = M(0,0) = (0,0)。A’ = M(1,0) = (2,3)。B’ = M(1,1) = (1,7)。C’ = M(0,1) = (-1,4)。
The image is a parallelogram with vertices (0,0), (2,3), (1,7) and (-1,4). Its area is |det(M)| × 1 = 11.
像为一个平行四边形，顶点为 (0,0), (2,3), (1,7) 和 (-1,4)。其面积为 |det(M)| × 1 = 11。
3. Series: Method of Differences | 级数：差分法
Find an expression for Σr=1n 1/(r(r+1)).
求 Σr=1n 1/(r(r+1)) 的表达式。
Use partial fractions: 1/(r(r+1)) = 1/r – 1/(r+1).
使用部分分式：1/(r(r+1)) = 1/r – 1/(r+1)。
Write out the terms: (1/1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + … + (1/n – 1/(n+1)).
写出各项：(1/1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + … + (1/n – 1/(n+1))。
Almost all terms cancel, leaving the first and the last: 1 – 1/(n+1).
几乎所有项相消，只剩首项与末项：1 – 1/(n+1)。
Σr=1n 1/(r(r+1)) = 1 – 1/(n+1) = n/(n+1)
Hence, for n=100, the sum is 100/101 ≈ 0.9901.
因此，当 n=100 时，和为 100/101 ≈ 0.9901。
4. Hyperbolic Functions: Proving Identities | 双曲函数：证明恒等式
Prove that cosh²x – sinh²x ≡ 1.
证明 cosh²x – sinh²x ≡ 1。
Recall the definitions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2.
回顾定义：cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2。
cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4, sinh²x = (e²ˣ – 2 + e⁻²ˣ)/4.
cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4，sinh²x = (e²ˣ – 2 + e⁻²ˣ)/4。
Subtract: cosh²x – sinh²x = [(e²ˣ + 2 + e⁻²ˣ) – (e²ˣ – 2 + e⁻²ˣ)] / 4 = (4)/4 = 1.
相减：cosh²x – sinh²x = [(e²ˣ + 2 + e⁻²ˣ) – (e²ˣ – 2 + e⁻²ˣ)] / 4 = (4)/4 = 1。
This is the fundamental hyperbolic identity, analogous to cos²θ + sin²θ = 1.
这是基本双曲恒等式，对应于三角恒等式 cos²θ + sin²θ = 1。
5. Polar Coordinates: Area and Tangent | 极坐标：面积与切线
The curve C has polar equation r = 1 + cos θ, 0 ≤ θ ≤ 2π. Find the area enclosed by C and the slope of the tangent at θ = π/2.
曲线 C 的极坐标方程为 r = 1 + cos θ, 0 ≤ θ ≤ 2π。求 C 围成的面积以及在 θ = π/2 处切线的斜率。
The area is A = ½ ∫ 02π r² dθ = ½ ∫ 02π (1 + cos θ)² dθ.
面积为 A = ½ ∫ 02π r² dθ = ½ ∫ 02π (1 + cos θ)² dθ。
Expand: (1 + cos θ)² = 1 + 2cos θ + cos²θ = 1 + 2cos θ + (1 + cos 2θ)/2.
展开：(1 + cos θ)² = 1 + 2cos θ + cos²θ = 1 + 2cos θ + (1 + cos 2θ)/2。
Simplify to 3/2 + 2cos θ + ½ cos 2θ. Integrate: A = ½ [ (3/2)θ + 2sin θ + (1/4) sin 2θ ]02π = ½ [ (3/2)(2π) + 0 ] = (3π)/2.
化简为 3/2 + 2cos θ + ½ cos 2θ。积分：A = ½ [ (3/2)θ + 2sin θ + (1/4) sin 2θ ]02π = ½ [ (3/2)(2π) + 0 ] = (3π)/2。
For the slope, convert to Cartesian: x = r cos θ, y = r sin θ. Use dy/dx = (dy/dθ)/(dx/dθ).
对于斜率，转化为直角坐标：x = r cos θ, y = r sin θ。用 dy/dx = (dy/dθ)/(dx/dθ)。
At θ = π/2, r = 1 + cos(π/2) = 1. Then x = 0, y = 1. Compute derivatives: dx/dθ = r’ cos θ – r sin θ, dy/dθ = r’ sin θ + r cos θ, with r’ = –sin θ.
在 θ = π/2 处，r = 1 + cos(π/2) = 1。则 x = 0, y = 1。计算导数：dx/dθ = r’ cos θ – r sin θ, dy/dθ = r’ sin θ + r cos θ，其中 r’ = –sin θ。
At θ = π/2: r’ = –1, cos θ = 0, sin θ = 1. So dx/dθ = (–1)(0) – (1)(1) = –1, dy/dθ = (–1)(1) + (1)(0) = –1.
在 θ = π/2 处：r’ = –1, cos θ = 0, sin θ = 1。故 dx/dθ = (–1)(0) – (1)(1) = –1，dy/dθ = (–1)(1) + (1)(0) = –1。
Therefore dy/dx = (–1)/(–1) = 1. The tangent has slope 1.
因此 dy/dx = (–1)/(–1) = 1。切线斜率为 1。
6. First-Order Differential Equations | 一阶微分方程
Solve dy/dx + 2y = e⁻ˣ, subject to y(0) = 3.
解微分方程 dy/dx + 2y = e⁻ˣ，满足 y(0) = 3。
This is a linear first-order ODE. The integrating factor IF = e∫ 2 dx = e2x.
这是一阶线性常微分方程。积分因子 IF = e∫ 2 dx = e2x。
Multiply both sides: e2x dy/dx + 2e2x y = ex.
两边同乘：e2x dy/dx + 2e2x y = ex。
The left side is the derivative of (e2x y): d/dx (e2x y) = ex.
左边为 (e2x y) 的导数：d/dx (e2x y) = ex。
Integrate: e2x y = ex + C.
积分：e2x y = ex + C。
Hence y = e–x + C e–2x.
因此 y = e–x + C e–2x。
Use y(0)=3: 3 = 1 + C ⇒ C = 2.
代入 y(0)=3：3 = 1 + C ⇒ C = 2。
The particular solution is y = e–x + 2 e–2x.
特解为 y = e–x + 2 e–2x。
7. Vectors: Cross Product and Area | 向量：叉积与面积
Given vectors a = i + 2j + 2k and b = 2i + j – 2k, find a × b and the area of the triangle formed by a and b.
给定向量 a = i + 2j + 2k 和 b = 2i + j – 2k，求 a × b 以及由 a 和 b 构成的三角形面积。
Compute the cross product using the determinant: a × b = |i j k; 1 2 2; 2 1 -2|.
用行列式计算叉积：a × b = |i j k; 1 2 2; 2 1 -2|。
Expand: i(2×(-2) – 2×1) – j(1×(-2) – 2×2) + k(1×1 – 2×2).
展开：i(2×(-2) – 2×1) – j(1×(-2) – 2×2) + k(1×1 – 2×2)。
Simplify: i(–4 – 2) – j(–2 – 4) + k(1 – 4) = –6i + 6j – 3k.
化简：i(–4 – 2) – j(–2 – 4) + k(1 – 4) = –6i + 6j – 3k。
Thus a × b = –6i + 6j – 3k. Its magnitude is √(36+36+9) = √81 = 9.
因此 a × b = –6i + 6j – 3k。其模为 √(36+36+9) = √81 = 9。
The area of the triangle is (1/2)|a × b| = 9/2 = 4.5 square units.
三角形面积为 (1/2)|a × b| = 9/2 = 4.5 平方单位。
8. Roots of Polynomials: Complex Conjugates | 多项式根：共轭复根
The equation x⁴ – 4x³ + 9x² – 16x + 20 = 0 has a root 1 + 2i. Find the other three roots.
方程 x⁴ – 4x³ + 9x² – 16x + 20 = 0 有一根为 1 + 2i。求其他三个根。
Since coefficients are real, the conjugate 1 – 2i is also a root.
由于系数为实数，共轭根 1 – 2i 也是根。
The quadratic factor from these roots is (x – (1+2i))(x – (1–2i)) = x² – 2x + 5.
由这两根构成的二次因式为 (x – (1+2i))(x – (1–2i)) = x² – 2x + 5。
Divide the quartic by x² – 2x + 5. Perform polynomial long division:
用 x² – 2x + 5 去除四次多项式。进行多项式长除：
Quotient: x² – 2x + 4. Check: (x² – 2x + 5)(x² – 2x + 4) = x⁴ – 4x³ + 9x² – 16x + 20, which matches.
商式为 x² – 2x + 4。检验：(x² – 2x + 5)(x² – 2x + 4) = x⁴ – 4x³ + 9x² – 16x + 20，吻合。
Now solve x² – 2x + 4 = 0. Discriminant Δ = 4 – 16 = –12. Roots: x = [2 ± √(–12)]/2 = 1 ± i√3.
现在解 x² – 2x + 4 = 0。判别式 Δ = 4 – 16 = –12。根为 x = [2 ± √(–12)]/2 = 1 ± i√3。
The four roots are 1 + 2i, 1 – 2i, 1 + i√3, 1 – i√3.
四个根为 1 + 2i, 1 – 2i, 1 + i√3, 1 – i√3。
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