Year 13 CAIE Further Mathematics: Mock Unit Test Walkthrough | 进阶数学单元测试模拟卷解析

📚 Year 13 CAIE Further Mathematics: Mock Unit Test Walkthrough | 进阶数学单元测试模拟卷解析

This article presents a mock unit test covering core topics in the CAIE Year 13 Further Mathematics syllabus. Each section mimics the style of exam questions and provides detailed bilingual solutions. Use this walkthrough to consolidate your understanding of complex numbers, matrices, polar coordinates, hyperbolic functions, differential equations, mechanics collisions, and statistical inference. Careful attention to method marks and final answers is highlighted.

本文基于CAIE Year 13进阶数学大纲设计了一套单元模拟卷，并逐题双语解析。每节模拟典型考题，涵盖复数、矩阵、极坐标、双曲函数、微分方程、力学碰撞和统计推断等核心内容。解析将重点展示得分步骤与关键细节，帮助你在模拟演练中查漏补缺。

1. Complex Numbers: De Moivre’s Theorem | 复数：棣莫弗定理

Mock question: Use de Moivre’s theorem to find all fourth roots of −16. Express each root in the exact polar form r(cosθ + i sinθ).

模拟题：用棣莫弗定理求−16的所有四次方根，以精确极坐标形式 r(cosθ + i sinθ) 表示每个根。

te −16 in polar form. The modulus is 16 and the argument is π. So −16 = 16(cosπ + i sinπ).

首先将−16写成极坐标形式。模为16，辐角为π。因此−16 = 16(cosπ + i sinπ)。

By de Moivre’s theorem, the four fourth roots have modulus 16^1/4 = 2 and arguments given by (π + 2kπ)/4 for k = 0, 1, 2, 3.

根据棣莫弗定理，四个四次方根的模为16^1/4 = 2，辐角为 (π + 2kπ)/4，其中 k = 0, 1, 2, 3。

Substituting the integer values yields the arguments π/4, 3π/4, 5π/4, and 7π/4. The four distinct roots are 2(cos(π/4) + i sin(π/4)), 2(cos(3π/4) + i sin(3π/4)), 2(cos(5π/4) + i sin(5π/4)), and 2(cos(7π/4) + i sin(7π/4)).

代入整数得到辐角 π/4、3π/4、5π/4 和 7π/4。四个互异的根分别为 2(cos(π/4) + i sin(π/4))、2(cos(3π/4) + i sin(3π/4))、2(cos(5π/4) + i sin(5π/4)) 和 2(cos(7π/4) + i sin(7π/4))。

If required in rectangular form, you can evaluate the trigonometric functions to obtain ±√2 ± i√2 variants, but the question asks for the polar format.

若题目要求直角坐标形式，可计算三角函数值得到 ±√2 ± i√2 组合，但本题要求给出极坐标形式。

2. Matrices and Linear Transformations | 矩阵与线性变换

: The matrix M = [ [1, 2, 3], [0, 1, 4], [0, 0, 1] ] represents a linear transformation in ℝ³. Find M⁻¹ and hence solve Mx = b, where b = (5, −2, 1)^T.

模拟题：矩阵 M = [ [1, 2, 3], [0, 1, 4], [0, 0, 1] ] 表示 ℝ³ 中的线性变换。求 M⁻¹，并借此求解 Mx = b，其中 b = (5, −2, 1)^T。

M is an upper-triangular matrix with ones on the main diagonal, so its inverse is also upper-triangular with ones on the diagonal. Perform row operations or use the formula for the inverse of a unit upper-triangular matrix. The inverse is:

M 是主对角线全为1的上三角矩阵，因此其逆矩阵也是对角线为1的上三角阵。通过初等行变换或套用单位上三角矩阵求逆公式，得到逆矩阵：

Now multiply M⁻¹ by the column vector b. x = M⁻¹b = (1×5 + (−2)×(−2) + 5×1, 0×5 + 1×(−2) + (−4)×1, 0×5 + 0×(−2) + 1×1)^T = (5+4+5, −2−4, 1)^T = (14, −6, 1)^T.

用逆矩阵左乘列向量 b。x = M⁻¹b = (1×5 + (−2)×(−2) + 5×1, 0×5 + 1×(−2) + (−4)×1, 0×5 + 0×(−2) + 1×1)^T = (5+4+5, −2−4, 1)^T = (14, −6, 1)^T。

Always check by substituting back into Mx. Here M × (14, −6, 1)^T returns (5, −2, 1)^T, confirming the solution.

务必代回原方程验算。M × (14, −6, 1)^T 得 (5, −2, 1)^T，确认解正确。

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3. Further Calculus: Reduction Formula | 进阶微积分：递推公式

Mock question: Let I_n = ∫₀^π/2 sinⁿx dx for n ≥ 0. Show that I_n = ((n−1)/n) I_n−2 for n ≥ 2, and hence evaluate ∫₀^π/2 sin⁵x dx.

模拟题：对于 n ≥ 0，令 I_n = ∫₀^π/2 sinⁿx dx。证明当 n ≥ 2 时 I_n = ((n−1)/n) I_n−2，并由此计算 ∫₀^π/2 sin⁵x dx。

To derive the reduction formula, use integration by parts with u = sinⁿ⁻¹x and dv = sin x dx. Then du = (n−1) sinⁿ⁻²x cos x dx and v = −cos x. The boundary term [−sinⁿ⁻¹x cos x]₀^π/2 vanishes, leaving I_n = (n−1) ∫₀^π/2 sinⁿ⁻²x cos²x dx.

为推导递推公式，采用分部积分，设 u = sinⁿ⁻¹x，dv = sin x dx，则 du = (n−1) sinⁿ⁻²x cos x dx，v = −cos x。边界项 [−sinⁿ⁻¹x cos x]₀^π/2 为0，化简得 I_n = (n−1) ∫₀^π/2 sinⁿ⁻²x cos²x dx。

Replace cos²x by 1 − sin²x. This yields I_n = (n−1)(I_n−2 − I_n). Solving for I_n gives I_n = ((n−1)/n) I_n−2 as required.

将 cos²x 用 1 − sin²x 替换，得 I_n = (n−1)(I_n−2 − I_n)，解出 I_n 即得公式 I_n = ((n−1)/n) I_n−2。

For the definite integral, apply the reduction: I₅ = (4/5) I₃ = (4/5) × (2/3) I₁. Since I₁ = ∫₀^π/2 sin x dx = 1, we obtain I₅ = (8/15).

对所求定积分使用递推：I₅ = (4/5) I₃ = (4/5) × (2/3) I₁。已知 I₁ = ∫₀^π/2 sin x dx = 1，故 I₅ = 8/15。

4. Polar Coordinates: Area Enclosed by a Curve | 极坐标：曲线围成的面积

lar equation r = 2 + cosθ for 0 ≤ θ ≤ 2π. Find the exact total area enclosed by the curve.

模拟题：曲线的极坐标方程为 r = 2 + cosθ，0 ≤ θ ≤ 2π。求该曲线所围成的精确总面积。

The area in polar coordinates is given by A = (1/2) ∫₀^2π r² dθ. Substitute r:

极坐标下的面积公式为 A = (1/2) ∫₀^2π r² dθ。代入 r：

A = ½ ∫₀^2π (2 + cosθ)² dθ

Expand the square: (2 + cosθ)² = 4 + 4cosθ + cos²θ. Use the double-angle identity cos²θ = (1 + cos2θ)/2. Thus the integrand becomes 4 + 4cosθ + ½ + ½cos2

Published by TutorHao | Year 13 进阶数学 Revision Series | aleveler.com

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