📚 Unit Test Mock Paper Analysis | 单元测试模拟卷解析
This mock paper analysis is designed for CIE Pre‑U Further Mathematics students who want to consolidate core techniques before a formal assessment. Each question targets a key topic from the pure stream, with full worked solutions and commentary on common pitfalls.
本模拟卷解析面向准备正式评估的 CIE Pre‑U 进阶数学学生。每一题都紧扣纯数部分的核心课题，提供完整解答步骤并指出常见错误，帮助巩固技巧。
1. Complex Numbers and Polynomial Roots | 复数与多项式根
Question: Given that 1 + i is a root of the equation z³ – (2+i)z² + (2i)z – 1 = 0, find the other two roots.
题目：已知 1 + i 是方程 z³ – (2+i)z² + (2i)z – 1 = 0 的一个根，求另外两个根。
Since the coefficients are not all real, the conjugate is not necessarily a root. Instead, use polynomial division by (z – (1+i)).
因为系数并非全为实数，共轭复数不一定是根。我们改用 (z – (1+i)) 做多项式除法。
Perform synthetic division with 1 + i. The quotient is z² – z + (1 – i).
用 1 + i 进行综合除法，得到商式 z² – z + (1 – i)。
Now solve z² – z + (1 – i) = 0 using the quadratic formula: z = [1 ± √(1 – 4(1 – i))]/2 = [1 ± √(–3 + 4i)]/2.
现在用二次公式解 z² – z + (1 – i) = 0：z = [1 ± √(1 – 4(1 – i))]/2 = [1 ± √(–3 + 4i)]/2。
To find √(–3 + 4i), let (x + iy)² = –3 + 4i. Equating real and imaginary parts gives x² – y² = –3 and 2xy = 4.
为计算 √(–3 + 4i)，令 (x + iy)² = –3 + 4i。比较实部和虚部得 x² – y² = –3 和 2xy = 4。
Solving yields x = 1, y = 2 or x = –1, y = –2, so √(–3 + 4i) = ±(1 + 2i).
解得 x = 1, y = 2 或 x = –1, y = –2，因此 √(–3 + 4i) = ±(1 + 2i)。
Thus the remaining roots are z = [1 + (1 + 2i)]/2 = 1 + i (the given root) and z = [1 – (1 + 2i)]/2 = –i. The three roots are 1 + i, 1 + i, –i (note the repeated root).
于是另外两个根为 z = [1 + (1 + 2i)]/2 = 1 + i（已知根）以及 z = [1 – (1 + 2i)]/2 = –i。三个根为 1 + i（二重根）和 –i。
Common mistake: Forgetting to check for repeated roots or assuming complex conjugates appear automatically.
常见错误：忘记检查重根，或错误地认为共轭根自动出现。
2. Matrix Transformations | 矩阵变换
Question: The matrix A =
| 2 | -1 |
| 1 | 2 |
represents a linear transformation. Describe this transformation geometrically and find the image of the unit square.
题目：矩阵 A =
| 2 | -1 |
| 1 | 2 |
表示一个线性变换。试描述该变换的几何意义并求单位正方形的像。
Check the determinant: det(A) = (2)(2) – (–1)(1) = 5. The transformation is area-preserving with scale factor 5.
计算行列式：det(A) = (2)(2) – (–1)(1) = 5。变换带面积缩放因子 5。
Find eigenvalues: λ satisfies (2-λ)(2-λ) – (–1)(1) = (λ-2)² + 1 = 0 ⇒ λ = 2 ± i. No real eigenvectors, so the transformation is a rotation combined with a uniform scaling.
求特征值：λ 满足 (2-λ)(2-λ) – (–1)(1) = (λ-2)² + 1 = 0 ⇒ λ = 2 ± i。没有实数特征向量，所以变换是一个旋转加上一致缩放。
Express A as √5 × (1/√5)
| 2 | -1 |
| 1 | 2 |
. The matrix R = (1/√5)A has orthogonal columns and det(R)=1, so it is a rotation by angle θ where cosθ = 2/√5, sinθ = 1/√5.
将 A 写作 √5 × (1/√5)
| 2 | -1 |
| 1 | 2 |
。矩阵 R = (1/√5)A 的列正交且 det(R)=1，所以它是旋转角 θ 的旋转矩阵，其中 cosθ = 2/√5，sinθ = 1/√5。
Geometrically, A represents a rotation by θ ≈ 26.6° followed by an enlargement with factor √5 (or the other order, since scaling and rotation commute).
几何上，A 表示先旋转约 26.6°，再进行因子为 √5 的放大（或顺序互换，因为缩放与旋转可交换）。
The unit square with vertices (0,0), (1,0), (0,1), (1,1) maps to (0,0), (2,1), (–1,2), (1,3). The area of the image is 5.
单位正方形顶点 (0,0), (1,0), (0,1), (1,1) 映射为 (0,0), (2,1), (–1,2), (1,3)。像的面积为 5。
3. Hyperbolic Functions | 双曲函数
Question: Solve the equation 5 cosh x + 3 sinh x = 7, giving your answer in logarithmic form.
题目：解方程 5 cosh x + 3 sinh x = 7，答案用对数形式表示。
Use definitions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2. Substitute and multiply by 2: 5(eˣ + e⁻ˣ) + 3(eˣ – e⁻ˣ) = 14.
利用定义：cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2。代入并乘以 2：5(eˣ + e⁻ˣ) + 3(eˣ – e⁻ˣ) = 14。
Simplify: 8eˣ + 2e⁻ˣ = 14 ⇒ divide by 2: 4eˣ + e⁻ˣ = 7.
化简：8eˣ + 2e⁻ˣ = 14 ⇒ 除以 2：4eˣ + e⁻ˣ = 7。
Let u = eˣ > 0, then 4u + 1/u = 7 ⇒ 4u² – 7u + 1 = 0.
令 u = eˣ > 0，则 4u + 1/u = 7 ⇒ 4u² – 7u + 1 = 0。
Solve: u = [7 ± √(49 – 16)]/8 = [7 ± √33]/8. Both are positive, but we need u > 0; both are fine. Choose the one that makes physical sense or check.
求解：u = [7 ± √(49 – 16)]/8 = [7 ± √33]/8。两者均正，都可取。
Then x = ln u. The solutions are x = ln((7 + √33)/8) or x = ln((7 – √33)/8). However, check original equation: (7 – √33)/8 ≈ 0.155, valid.
于是 x = ln u。解为 x = ln((7 + √33)/8) 或 x = ln((7 – √33)/8)。检验原方程均成立。
Key point: Always verify derived exponential values, especially when squaring may introduce extraneous solutions.
关键点：务必检验导出的指数值，尤其平方可能引入增根。
4. Polar Coordinates and Area | 极坐标与面积
Question: The curve has polar equation r = 2 + cos θ for 0 ≤ θ ≤ 2π. Find the total area enclosed by the curve.
题目：曲线极坐标方程为 r = 2 + cos θ，0 ≤ θ ≤ 2π。求曲线所围总面积。
Area formula: A = (1/2)∫₀²π r² dθ = (1/2)∫₀²π (2 + cos θ)² dθ.
面积公式：A = (1/2)∫₀²π r² dθ = (1/2)∫₀²π (2 + cos θ)² dθ。
Expand: (2 + cos θ)² = 4 + 4 cos θ + cos²θ. Use cos²θ = (1 + cos 2θ)/2.
展开：(2 + cos θ)² = 4 + 4 cos θ + cos²θ。利用 cos²θ = (1 + cos 2θ)/2。
Thus integrand = 4 + 4 cos θ + ½ + (1/2) cos 2θ = 9/2 + 4 cos θ + (1/2) cos 2θ.
于是被积函数 = 4 + 4 cos θ + ½ + (1/2) cos 2θ = 9/2 + 4 cos θ + (1/2) cos 2θ。
Integrate termwise from 0 to 2π: ∫₀²π (9/2) dθ = 9π; ∫₀²π 4 cos θ dθ = 0; ∫₀²π (1/2) cos 2θ dθ = 0.
逐项积分从 0 到 2π：∫₀²π (9/2) dθ = 9π；∫₀²π 4 cos θ dθ = 0；∫₀²π (1/2) cos 2θ dθ = 0。
Therefore A = (1/2) × 9π = (9π)/2.
因此 A = (1/2) × 9π = (9π)/2。
Common error: Forgetting the factor 1/2 in the area formula or mis-evaluating the integral of cos²θ.
常见错误：忘记面积公式中的 1/2 因子，或错误计算 cos²θ 的积分。
5. First‑Order Differential Equations | 一阶微分方程
Question: Solve the differential equation dy/dx + (2/x) y = 4x, given y(1) = 3.
题目：求解微分方程 dy/dx + (2/x) y = 4x，已知 y(1) = 3。
Identify the integrating factor: μ(x) = e^(∫ (2/x) dx) = e^(2 ln x) = x².
确定积分因子：μ(x) = e^(∫ (2/x) dx) = e^(2 ln x) = x²。
Multiply through: x² dy/dx + 2x y = 4x³. The left side is d/dx (x² y).
乘以积分因子：x² dy/dx + 2x y = 4x³。左边可写成 d/dx (x² y)。
Integrate both sides: x² y = ∫ 4x³ dx = x⁴ + C.
两边积分：x² y = ∫ 4x³ dx = x⁴ + C。
Hence y = x² + C x⁻².
于是 y = x² + C x⁻²。
Apply initial condition y(1) = 3: 3 = 1² + C × 1⁻² ⇒ C = 2.
代入初值 y(1) = 3：3 = 1² + C × 1⁻² ⇒ C = 2。
Final solution: y = x² + 2/x².
最终解：y = x² + 2/x²。
Tip: Always check the domain; here x > 0 because of the coefficient (2/x).
提示：注意定义域；由于系数含 2/x，这里 x > 0。
6. Second‑Order Linear Differential Equations | 二阶线性微分方程
Question: Find the general solution of d²y/dx² – 5 dy/dx + 6y = e^(2x).
题目：求微分方程 d²y/dx² – 5 dy/dx + 6y = e^(2x) 的通解。
Auxiliary equation: m² – 5m + 6 = 0 ⇒ (m – 2)(m – 3) = 0 ⇒ m = 2, 3.
辅助方程：m² – 5m + 6 = 0 ⇒ (m – 2)(m – 3) = 0 ⇒ m = 2, 3。
Complementary function: y_c = A e^(2x) + B e^(3x).
余函数：y_c = A e^(2x) + B e^(3x)。
For the particular integral, the rhs is e^(2x). Since 2 is a root of the auxiliary equation, try y_p = C x e^(2x).
特解部分，右边为 e^(2x)。由于 2 是辅助方程的根，尝试 y_p = C x e^(2x)。
Differentiate: y_p’ = C e^(2x)(1 + 2x); y_p” = C e^(2x)(4 + 4x).
求导：y_p’ = C e^(2x)(1 + 2x)；y_p” = C e^(2x)(4 + 4x)。
Substitute into LHS: C e^(2x)[(4+4x) –5(1+2x) +6x] = C e^(2x)[4+4x–5–10x+6x] = C e^(2x)[–1] = –C e^(2x).
代入左边：C e^(2x)[(4+4x) –5(1+2x) +6x] = C e^(2x)[4+4x–5–10x+6x] = C e^(2x)[–1] = –C e^(2x)。
Set equal to e^(2x) ⇒ –C = 1 ⇒ C = –1. So y_p = –x e^(2x).
令其等于 e^(2x) ⇒ –C = 1 ⇒ C = –1。故 y_p = –x e^(2x)。
General solution: y = A e^(2x) + B e^(3x) – x e^(2x).
通解：y = A e^(2x) + B e^(3x) – x e^(2x)。
7. Vector Equations of Lines and Planes | 直线与平面的向量方程
Question: The line L has equation r = (i + 2j – k) + t(–i + 3j + k). The plane Π has equation 2x – y + 3z = 7. Find the intersection point of L and Π.
题目：直线 L 方程为 r = (i + 2j – k) + t(–i + 3j + k)。平面 Π 方程为 2x – y + 3z = 7。求 L 与 Π 的交点。
Parametric coordinates from L: x = 1 – t, y = 2 + 3t, z = –1 + t.
由 L 的参数式：x = 1 – t, y = 2 + 3t, z = –1 + t。
Substitute into Π: 2(1 – t) – (2 + 3t) + 3(–1 + t) = 7.
代入 Π：2(1 – t) – (2 + 3t) + 3(–1 + t) = 7。
Simplify: 2 – 2t – 2 – 3t – 3 + 3t = 7 ⇒ (2–2–3) + (–2t–3t+3t) = –3 – 2t = 7.
化简：2 – 2t – 2 – 3t – 3 + 3t = 7 ⇒ –3 – 2t = 7。
Solve: –2t = 10 ⇒ t = –5.
求解：–2t = 10 ⇒ t = –5。
Intersection point: x = 1 – (–5) = 6, y = 2 + 3(–5) = –13, z = –1 + (–5) = –6. So point is (6, –13, –6).
交点：x = 1 – (–5) = 6, y = 2 + 3(–5) = –13, z = –1 + (–5) = –6，坐标为 (6, –13, –6)。
Verify by plugging into plane equation: 2(6) – (–13) + 3(–6) = 12 + 13 – 18 = 7, correct.
代入平面方程验证：2(6) – (–13) + 3(–6) = 12 + 13 – 18 = 7，正确。
8. Proof by Induction | 数学归纳法证明
Question: Prove by induction that for all positive integers n, ∑ᵣ₌₁ⁿ r(r+1) = (n/3)(n+1)(n+2).
题目：用归纳法证明对所有正整数 n，∑ᵣ₌₁ⁿ r(r+1) = (n/3)(n+1)(n+2)。
Base case n=1: LHS = 1·2 = 2; RHS = (1/3)(2)(3) = 2. True.
基础步 n=1：左边 = 1·2 = 2；右边 = (1/3)(2)(3) = 2。成立。
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