📚 Year 12 CIE Further Mathematics: Unit Test Mock Paper Walkthrough | 12年级 CIE 进阶数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a typical Year 12 CIE Further Mathematics unit test mock paper. Each question is broken down with step-by-step reasoning, common pitfalls, and key techniques required to master topics such as complex numbers, matrix algebra, hyperbolic functions, and series expansions. Use this as a revision tool to check your understanding and sharpen your problem-solving skills.
本文详细解析了一份典型的12年级CIE进阶数学单元测试模拟卷。每一道题都配以分步推理、常见错误提示以及掌握复数、矩阵代数、双曲函数和级数展开等主题所需的关键技巧。将此作为复习工具，检验你的理解程度并提升解题能力。
1. Complex Numbers – Cartesian and Polar Forms | 第1题：复数 – 直角坐标与极坐标形式
Question: Given z = 3 – 3i, express z in polar form reiθ with –π < θ ≤ π. Hence find z⁵ in Cartesian form a + bi.
题目：已知 z = 3 – 3i，将 z 表示为极坐标形式 reiθ，其中 –π < θ ≤ π。由此求 z⁵ 的直角坐标形式 a + bi。
First compute the modulus: r = |z| = √(3² + (–3)²) = √18 = 3√2. The argument θ satisfies tan θ = –3/3 = –1, so with the point in the fourth quadrant we have θ = –π/4. Thus z = 3√2 e–iπ/4.
首先计算模长：r = |z| = √(3² + (–3)²) = √18 = 3√2。辐角 θ 满足 tan θ = –3/3 = –1，由于点位于第四象限，取 θ = –π/4。因此 z = 3√2 e–iπ/4。
Using de Moivre's theorem: z⁵ = (3√2)⁵ ei(5 × (–π/4)) = (3⁵)(2⁵⁄²) e–i5π/4. Simplify: 3⁵ = 243; 2⁵⁄² = (√2)⁵ = 4√2? Wait: 25/2 = √2⁵ = √32 = 4√2? Actually √32 = 4√2, correct. So modulus = 243 × 4√2 = 972√2. Now –5π/4 is equivalent to 3π/4 (adding 2π). So z⁵ = 972√2 ei3π/4.
利用棣莫弗定理：z⁵ = (3√2)⁵ ei(5 × (–π/4)) = (3⁵)(2⁵⁄²) e–i5π/4。化简：3⁵ = 243；2⁵⁄² = √2⁵ = √32 = 4√2。故模长为 243 × 4√2 = 972√2。–5π/4 加 2π 后等价于 3π/4。所以 z⁵ = 972√2 ei3π/4。
Convert to Cartesian: ei3π/4 = cos(3π/4) + i sin(3π/4) = –√2/2 + i (√2/2). Multiply by 972√2: real part = 972√2 × (–√2/2) = 972 × (–1) = –972; imaginary part = 972√2 × (√2/2) = 972. Hence z⁵ = –972 + 972i.
转换为直角坐标：ei3π/4 = cos(3π/4) + i sin(3π/4) = –√2/2 + i (√2/2)。乘以 972√2 得：实部 = 972√2 × (–√2/2) = –972；虚部 = 972√2 × (√2/2) = 972。因此 z⁵ = –972 + 972i。
Common mistake: forgetting to adjust the argument to the principal range before applying de Moivre. Always add or subtract 2π to get θ inside (–π, π].
常见错误：在应用棣莫弗定理前忘记将辐角调至主值区间。务必加减 2π 使 θ 落在 (–π, π] 内。
2. Roots of a Complex Equation | 第2题：复数方程的根
Question: Solve z³ = 8i, giving all roots in the form reiθ with –π < θ ≤ π. Illustrate the roots on an Argand diagram.
题目：解方程 z³ = 8i，将所有根用 reiθ 的形式表示（–π < θ ≤ π），并在阿干特图上标出这些根。
Write 8i in polar form: modulus 8, argument π/2. Let z = reiθ. Then r³ei3θ = 8ei(π/2 + 2kπ). So r³ = 8 ⇒ r = 2, and 3θ = π/2 + 2kπ ⇒ θ = π/6 + 2kπ/3.
将 8i 写成极坐标形式：模长为 8，辐角为 π/2。设 z = reiθ，则有 r³ei3θ = 8ei(π/2 + 2kπ)。因此 r³ = 8 ⇒ r = 2，且 3θ = π/2 + 2kπ ⇒ θ = π/6 + 2kπ/3。
For k = 0: θ = π/6 → z₁ = 2eiπ/6.
For k = 1: θ = π/6 + 2π/3 = 5π/6 → z₂ = 2ei5π/6.
For k = 2: θ = π/6 + 4π/3 = 3π/2, but 3π/2 is not inside (–π, π]; subtract 2π to get –π/2. So z₃ = 2e–iπ/2.
当 k = 0：θ = π/6 → z₁ = 2eiπ/6。
当 k = 1：θ = π/6 + 2π/3 = 5π/6 → z₂ = 2ei5π/6。
当 k = 2：θ = π/6 + 4π/3 = 3π/2，但 3π/2 不在 (–π, π] 内；减去 2π 得到 –π/2。所以 z₃ = 2e–iπ/2。
On the Argand diagram, these three points lie on a circle of radius 2, separated by angles of 2π/3. Note that symmetry helps: adding 2π/3 repeatedly gives the roots.
在阿干特图上，这三个点位于半径为 2 的圆上，彼此夹角为 2π/3。注意对称性：连续加 2π/3 即可得到所有根。
3. Matrix Algebra – Inverse and Determinant | 第3题：矩阵代数 – 逆矩阵与行列式
Question: Let A =
| 2 | 1 | -1 |
| 0 | 3 | 2 |
| 1 | 0 | 4 |
题目：设 A =
| 2 | 1 | -1 |
| 0 | 3 | 2 |
| 1 | 0 | 4 |
Expand det(A) along the first row: det = 2 × (3×4 – 2×0) – 1 × (0×4 – 2×1) + (–1) × (0×0 – 3×1) = 2×(12) – 1×(–2) – 1×(–3) = 24 + 2 + 3 = 29.
沿第一行展开行列式：det = 2 × (3×4 – 2×0) – 1 × (0×4 – 2×1) + (–1) × (0×0 – 3×1) = 2×(12) – 1×(–2) – 1×(–3) = 24 + 2 + 3 = 29。
Since det ≠ 0, A is invertible. Compute the matrix of cofactors:
C₁₁ = +(3×4 – 2×0) = 12, C₁₂ = –(0×4 – 2×1) = 2, C₁₃ = +(0×0 – 3×1) = –3.
C₂₁ = –(1×4 – (–1)×0) = –4, C₂₂ = +(2×4 – (–1)×1) = 9, C₂₃ = –(2×0 – 1×1) = 1.
C₃₁ = +(1×2 – (–1)×3) = 5, C₃₂ = –(2×2 – (–1)×0) = –4, C₃₃ = +(2×3 – 1×0) = 6.
由于 det ≠ 0，A 可逆。求余子式矩阵：
C₁₁ = +(3×4 – 2×0) = 12，C₁₂ = –(0×4 – 2×1) = 2，C₁₃ = +(0×0 – 3×1) = –3。
C₂₁ = –(1×4 – (–1)×0) = –4，C₂₂ = +(2×4 – (–1)×1) = 9，C₂₃ = –(2×0 – 1×1) = 1。
C₃₁ = +(1×2 – (–1)×3) = 5，C₃₂ = –(2×2 – (–1)×0) = –4，C₃₃ = +(2×3 – 1×0) = 6。
The cofactor matrix is then transposed to get the adjugate: adj(A) =
| 12 | -4 | 5 |
| 2 | 9 | -4 |
| -3 | 1 | 6 |
将余子式矩阵转置得伴随矩阵：adj(A) =
| 12 | -4 | 5 |
| 2 | 9 | -4 |
| -3 | 1 | 6 |
l equal 1, and off-diagonals 0. For example, (row 1) × (col 1): (2)(12/29) + (1)(2/29) + (–1)(–3/29) = (24 + 2 + 3)/29 = 29/29 = 1. The check is left as an exercise.
验证时，计算 A 与 A⁻¹ 的乘积。乘积的主对角线元素应均为 1，其余为 0。例如 (第一行)×(第一列)：(2)(12/29) + (1)(2/29) + (–1)(–3/29) = (24 + 2 + 3)/29 = 29/29 = 1。其余验证留作练习。
4. Hyperbolic Functions – Identities and Differentiation | 第4题：双曲函数 – 恒等式与微分
Question: Prove the identity cosh²x – sinh²x = 1 using the exponential definitions. Hence differentiate y = arsinh(x/2) and express the derivative in algebraic form.
题目：利用指数定义证明恒等式 cosh²x – sinh²x = 1。进而对 y = arsinh(x/2) 求导，并将导数表示为代数形式。
By definition, cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2. Then cosh²x – sinh²x = [ (e²ˣ + 2 + e⁻²ˣ) – (e²ˣ – 2 + e⁻²ˣ) ] / 4 = (4)/4 = 1. The identity holds.
根据定义，cosh x = (eˣ + e⁻ˣ)/2，sinh x = (eˣ – e⁻ˣ)/2。那么 cosh²x – sinh²x = [ (e²ˣ + 2 + e⁻²ˣ) – (e²ˣ – 2 + e⁻²ˣ) ] / 4 = (4)/4 = 1。恒等式成立。
For y = arsinh(x/2), let u = x/2, so y = arsinh u. Recall d/du (arsinh u) = 1 / √u² + 1. Then dy/dx = (1 / √u² + 1) × (1/2) = 1/(2√((x/2)² + 1)). Simplify: √((x²/4) + 1) = √((x² + 4)/4) = √x² + 4/2. Hence dy/dx = 1/(2 × (√x² + 4/2)) = 1/√x² + 4.
对于 y = arsinh(x/2)，令 u = x/2，则 y = arsinh u。已知 d/du (arsinh u) = 1 / √u² + 1。因此 dy/dx = (1 / √u² + 1) × (1/2) = 1/(2√((x/2)² + 1))。化简：√((x²/4) + 1) = √((x² + 4)/4) = √x² + 4/2。故 dy/dx = 1/(2 × (√x² + 4/2)) = 1/√x² + 4。
Note that the derivative of arsinh(x) is often memorised, but deriving it via chain rule with the standard result is safer in an exam.
注意 arsinh(x) 的导数常被直接记忆，但在考试中通过链式法则结合标准结果进行推导更为稳妥。
5. Series – Method of Differences | 第5题：级数 – 差分法
Question: Use the method of differences to find the sum to n terms of
∑r=1n 3/( (3r–1)(3r+2) ).
题目：用差分法求前 n 项和：∑r=1n 3/( (3r–1)(3r+2) )。
First, express the term using partial fractions: 3/( (3r–1)(3r+2) ) = A/(3r–1) + B/(3r+2). Solve: 3 = A(3r+2) + B(3r–1). Equating coefficients: 3A + 3B = 0 (r terms), 2A – B = 3 (constants). From 3A+3B=0 ⇒ B = –A. Then 2A – (–A) = 3A = 3 ⇒ A = 1, B = –1. So the term is 1/(3r–1) – 1/(3r+2).
首先将一般项用部分分式表示：3/( (3r–1)(3r+2) ) = A/(3r–1) + B/(3r+2)。求解：3 = A(3r+2) + B(3r–1)。比较系数：r 项系数 3A + 3B = 0，常数项 2A – B = 3。由 3A+3B=0 得 B = –A。代入 2A – (–A) = 3A = 3 ⇒ A = 1，B = –1。所以项为 1/(3r–1) – 1/(3r+2)。
Now write the sum Sn = ∑r=1n [1/(3r–1) – 1/(3r+2)]. Expanding for r = 1,2,...,n:
r=1: 1/2 – 1/5
r=2: 1/5 – 1/8
r=3: 1/8 – 1/11
...
r=n–1: 1/(3n–4) – 1/(3n–1)
r=n: 1/(3n–1) – 1/(3n+2).
现在写出部分和 Sn = ∑r=1n [1/(3r–1) – 1/(3r+2)]。对 r = 1,2,...,n 展开：
r=1: 1/2 – 1/5
r=2: 1/5 – 1/8
r=3: 1/8 – 1/11
...
r=n–1: 1/(3n–4) – 1/(3n–1)
r=n: 1/(3n–1) – 1/(3n+2)。
Observe the telescoping cancellation: all intermediate terms cancel, leaving Sn = 1/2 – 1/(3n+2).
观察伸缩相消现象：所有中间项相消，剩下 Sn = 1/2 – 1/(3n+2)。
For large n, the sum approaches 1/2. This method is powerful when terms can be split into a difference of successive functions.
当 n 很大时，和趋近于 1/2。当项可拆分为连续函数的差时，此方法非常有效。
6. Matrices – Linear Transformations in the Plane | 第6题：矩阵 – 平面中的线性变换
Question: The matrix M =
| 0 | -1 |
| 1 | 0 |
题目：矩阵 M =
| 0 | -1 |
| 1 | 0 |
M acts on a vector (x, y)ᵀ to give (–y, x)ᵀ. This corresponds to a rotation of 90° anticlockwise about the origin. Check: applying to (1,0) yields (0,1); (0,1) yields (–1,0). The angle is indeed 90°.
M 作用于向量 (x, y)ᵀ 得到 (–y, x)ᵀ。这对应于绕原点逆时针旋转 90°。验证：作用于 (1,0) 得 (0,1)；作用于 (0,1) 得 (–1,0)。确实为 90° 旋转。
The image of the unit square: (0,0) → (0,0); (1,0) → (0,1); (1,1) → (–1,1); (0,1) → (– quare of side length 1, rotated 90°. The area is unchanged: the determinant of M is (0)(0) – (–1)(1) = 1, so area scale factor = 1.
单位正方形的像：(0,0) → (0,0)；(1,0) → (0,1)；(1,1) → (–1,1)；(0,1) → (–1,0)。得到的四边形是一个边长为 1 的正方形，旋转了 90°。面积不变：M 的行列式为 (0)(0) – (–1)(1) = 1，故面积缩放因子为 1。
In general, any pure rotation matrix has determinant +1, preserving both shape and area. Reflection matrices would have determinant –1.
一般来说，任何纯旋转矩阵的行列式均为 +1，既保持形状也保持面积。反射矩阵的行列式为 –1。
7. Differential Equations – Integrating Factor Method | 第7题：微分方程 – 积分因子法
stion: Solve the first-order linear differential equation dy/dx + 3y = 6x, given y(0) = 4.
题目：求解一阶线性微分方程 dy/dx + 3y = 6x，已知 y(0) = 4。
of the form dy/dx + P(x)y = Q(x) with P(x)=3, Q(x)=6x. The integrating factor is μ(x) = e∫P dx = e3x.
该方程形如 dy/dx + P(x)y = Q(x)，其中 P(x)=3，Q(x)=6x。积分因子为 μ(x) = e∫P dx = e3x。
Multiply the ODE by μ: e3x dy/dx + 3e3x y = 6x e3x. The left side is d/dx (y e3x). Integrate both sides: y e3x = ∫ 6x e3x dx.
将常微分方程乘以 μ：e3x dy/dx + 3e3x y = 6x e3x。左边即为 d/dx (y e3x)。两边积分：y e3x = ∫ 6x e3x dx。
Compute the integral by parts: let u = 6x, dv = e3x dx, so du = 6 dx, v = (1/3)e3x. Then ∫ 6x e3x dx = 6x·(1/3)e3x – ∫ (1/3)e3x·6 dx = 2x e3x – 2 ∫ e3x dx = 2x e3x – (2/3)e3x + C.
用分部积分法计算该积分：令 u = 6x，dv = e3x dx，则 du = 6 dx，v = (1/3)e3x。于是 ∫ 6x e3x dx = 6x·(1/3)e3x – ∫ (1/3)e3x·6 dx = 2x e3x – 2 ∫ e3x dx = 2x e3x – (2/3)e3x + C。
Thus y e3x = 2x e3x – (2/3)e3x + C. Divide by e3x: y = 2x – 2/3 + C e–3x. Apply y(0)=4: 4 = 0 – 2/3 + C ⇒ C = 14/3. Final solution: y = 2x – 2/3 + (14/3)e–3x.
因此 y e3x = 2x e3x – (2/3)e3x + C。除以 e3x 得：y = 2x – 2/3 + C e–3x。代入 y(0)=4：4 = 0 – 2/3 + C ⇒ C = 14/3。最终解为 y = 2x – 2/3 + (14/3)e–3x。
Always verify your solution by substituting back into the original ODE and checking the initial condition.
始终应将解代回原微分方程并检查初值条件，以验证正确性。
8. Further Calculus – Maclaurin Series | 第8题：进阶微积分 – 麦克劳林级数
Question: Find the first three non-zero terms of the Maclaurin series for f(x) = arcsin(x). State the range of validity.
题目：求 f(x) = arcsin(x) 的麦克劳林级数前三个非零项，并注明成立范围。
Start with f(0) = arcsin(0) = 0. The first non-zero term will come from derivatives at 0.
f'(x) = 1/√1–x² = (1–x²)⁻¹⁄². So f'(0) = 1.
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