Year 12 CIE Further Mathematics: Mock Unit Test Paper Solutions Explained | Year 12 CIE 进阶数学：单元测试模拟卷解析

📚 Year 12 CIE Further Mathematics: Mock Unit Test Paper Solutions Explained | Year 12 CIE 进阶数学：单元测试模拟卷解析

This article provides detailed, step-by-step solutions to a mock unit test paper designed for Year 12 CIE Further Mathematics (9231). The paper covers core topics including complex numbers, matrices, vectors, hyperbolic functions, and differential equations. Each solution is presented with a clear English explanation immediately followed by its Chinese counterpart, helping you reinforce both conceptual understanding and examination technique. Use these worked examples to identify common pitfalls, consolidate methods, and build confidence for your actual tests.

本文为Year 12 CIE进阶数学(9231)的一套单元测试模拟卷提供详细的逐步解析。试卷涵盖复数、矩阵、向量、双曲函数和微分方程等核心主题。每个解答都先给出清晰的英文解释，紧接着提供对应的中文说明，帮助你同时强化概念理解和考试技巧。通过这些范例解析，你可以识别常见陷阱、巩固方法并增强应试信心。

1. Complex Numbers: Polar Form and De Moivre's Theorem | 复数：极坐标形式与棣莫弗定理

Question from mock paper: Express z = 1 - i√3 in polar form. Hence find z⁵ and determine the smallest positive integer n such that zⁿ is real.

模拟卷题目：将 z = 1 - i√3 表示为极坐标形式。由此求出 z⁵ 并确定使 zⁿ 为实数的最小正整数 n。

Step 1: Calculate the modulus r = √(1² + (√3)²) = √1+3 = 2.

步骤1：计算模 r = √(1² + (√3)²) = √1+3 = 2。

Determine the argument θ. Since the complex number is in the fourth quadrant, θ = arctan(-√3 / 1) = -π/3. The polar form is therefore

z = 2[cos(-π/3) + i sin(-π/3)].

步骤2：确定辐角 θ。因为该复数位于第四象限，θ = arctan(-√3 / 1) = -π/3。极坐标形式为

z = 2[cos(-π/3) + i sin(-π/3)].

Step 3: Apply De Moivre's theorem to find z⁵. Raise the modulus to the power 5 and multiply the argument by 5:

z⁵ = 2⁵[cos(5 × -π/3) + i sin(5 × -π/3)] = 32[cos(-5π/3) + i sin(-5π/3)].

步骤3：应用棣莫弗定理求 z⁵。将模升五次方，并将辐角乘以5：

z⁵ = 2⁵[cos(5 × -π/3) + i sin(5 × -π/3)] = 32[cos(-5π/3) + i sin(-5π/3)].

ine and sine, -5π/3 is coterminal with π/3, so z⁵ = 32[cos(π/3) + i sin(π/3)] = 32(1/2 + i √3/2) = 16 + 16√3 i.

利用余弦和正弦的周期性，-5π/3 与 π/3 同界，因此 z⁵ = 32[cos(π/3) + i sin(π/3)] = 32(1/2 + i √3/2) = 16 + 16√3 i。

must be zero, i.e. sin(nθ) = 0. Using θ = -π/3, we require n(-π/3) = kπ for some integer k ⇒ -1, giving n = 3. (Equivalently, using a positive argument 5π/3 leads to the same result.)

步骤4：为使 zⁿ 为实数，虚部必须为零，即 sin(nθ) = 0。利用 θ = -π/3，要求 n(-π/3) = kπ，其中 k 为整数 ⇒ n = -3k。当 k = -1 时得到最小正整数 n = 3。（等价地，采用正辐角 5π/3 也会得到相同结果。）

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2. Roots of Polynomial Equations with Complex Coefficients | 复数系数多项式方程的根

Question: z = i is a root of the equation z³ - (2+i)z² + (1+2i)z - i = 0, find the other two roots.

题目：已知 z = i 是方程 z³ - (2+i)z² + (1+2i)z - i = 0 的一个根，求另外两个根。

we obtain the quadratic factor z² - 2z + 1. Setting this equal to zero gives (z - 1)² = 0 ⇒ z = 1 (repeated root). Thus the roots are z = i and z = 1 (multiplicity 2).

因为 i 是一个根，所以 (z - i) 是一个因式。用长除法将三次多项式除以 (z - i)，得到二次因式 z² - 2z + 1。令其等于零得 (z - 1)² = 0 ⇒ z = 1（重根）。因此所有根为 z = i 和 z = 1（二重根）。

i: i³ - (2+i)i² + (1+2i)i - i = -i - (2+i)(-1) + i(1+2i) - i = -i + 2 + i + i - 2 - i = 0.

验证：代入 z = i：i³ - (2+i)i² + (1+2i)i - i = -i - (2+i)(-1) + i(1+2i) - i = -i + 2 + i + i - 2 - i = 0。

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3. Matrix Algebra: Determinant and Inverse of a 3x3 Matrix | 矩阵代数：3x3矩阵的行列式与逆矩阵

Question: For matrix A shown below, find det(A) and A⁻¹.

题目：对于如下所示的矩阵 A，求 det(A) 和 A⁻¹。

2	1	1
0	1	3
1	2	1

Expand det(A) along the first row: det(A) = 2|1 3; 2 1| - 1|0 3; 1 1| + 1|0 1; 1 2| 3×1) + 1(0×2 - 1×1) = 2(1-6) - 1(0-3) + 1(0-1) = 2(-5) - 1(-3) + 1(-1) = -10 + 3 - 1 = -8.

沿第一行展开 det(A)：det(A) = 2|1 3; 2 1| - 1|0 3; 1 1| + 1|0 1; 1 2| = 2(1 - 6) - 1(0 - 3) + 1(0 - 1) = 2(-5) - 1(-3) + 1(-1) = -10 + 3 - 1 = -8。

The matrix of cofactors is:

(1-6)= -5	-(0-3)= 3	(0-1)= -1
-(1-2)= 1	(2-1)= 1	-(4-1)= -3
(3-1)= 2	-(6-0)= -6	(2-0)= 2

Transpose to get the adjugate matrix and divide by det(A) = -8:

A⁻¹ = (1/-8) × adj(A) =

5/8	-1/8	-1/4
-3/8	-1/8	3/4
1/8	3/8	-1/4

转置得到伴随矩阵并除以行列式 -8 即可得逆矩阵。

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4. Solving Linear Systems using the Inverse Matrix | 利用逆矩阵求解线性方程组

Question: Using the inverse matrix from Question 3, solve the system:

2x + y + z = 5
y + 3z = 7
x + 2y + z = 4

题目：利用第3题求出的逆矩阵解方程组：

2x + y + z = 5
y + 3z = 7
x + 2y + z = 4

Write the system as AX = B, where X = [x y z]ᵀ and B = [5 7 4]ᵀ. The solution is X = A⁻¹ B.

将方程组写作 AX = B，其中 X = [x y z]ᵀ，B = [5 7 4]ᵀ。解为 X = A⁻¹ B。

Multiply the inverse matrix by the constants column:

x = (5/8)×5 + (-1/8)×7 + (-1/4)×4 = 25/8 - 7/8 - 1 = (25-7-8)/8 = 10/8 = 5/4,

y = (-3/8)×5 + (-1/8)×7 + (3/4)×4 = -15/8 - 7/8 + 3 = (-22/8) + 3 = -11/4 + 12/4 = 1/4,

z = (1/8)×5 + (3/8)×7 + (-1/4)×4 = 5/8 + 21/8 - 1 = 26/8 - 1 = 13/4 - 4/4 = 9/4.

用逆矩阵乘以常数列：

x = (5/8)×5 + (-1/8)×7 + (-1/4)×4 = 25/8 - 7/8 - 1 = (25-7-8)/8 = 10/8 = 5/4,

y = (-3/8)×5 + (-1/8)×7 + (3/4)×4 = -15/8 - 7/8 + 3 = (-22/8) + 3 = -11/4 + 12/4 = 1/4,

z = (1/8)×5 + (3/8)×7 + (-1/4)×4 = 5/8 + 21/8 - 1 = 26/8 - 1 = 13/4 - 4/4 = 9/4.

Hence the solution is x = 1.25, y = 0.25, z = 2.25.

因此解为 x = 1.25，y = 0.25，z = 2.25。

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5. Vector Geometry: Intersection of a Line and a Plane | 向量几何：直线与平面的交点

Question: int where the line r = (i + 2j - k) + λ(2i - j + 3k) meets the plane r · (i + j + 2k) = 3.

题目：求直线 r = (i + 2j - k) + λ(2i - j + 3k) 与平面 r · (i + j + 2k) = 3 的交点坐标。

r in component form: r = (1+2λ)i + (2-λ)j + (-1+3λ)k. Substitute into the plane equation:

(1+2λ) + (2-λ) + 2(-1+3λ) = 3.

将 r 用分量形式表示：r = (1+2λ)i + (2-λ)j + (-1+3λ)k。代入平面方程：

(1+2λ) + (2-λ) + 2(-1+3λ) = 3.

Simplify: 1+2λ+2-λ-2+6λ = 3 → (1+2-2) + (2λ - λ + 6λ) = 3 → 1 + 7λ = 3 → 7λ = 2 → λ = 2/7.

化简：1+2λ+2-λ-2+6λ = 3 → (1+2-2) + (2λ - λ + 6λ) = 3 → 1 + 7λ = 3 → 7λ = 2 → λ = 2/7。

2 - 2/7 = 12/7, z = -1 + 3(2/7) = -1 + 6/7 = -1/7. Intersection point: (11/7, 12/7, -1/7).

代回 λ：x = 1+2(2/7) = 11/7，y = 2 - 2/7 = 12/7，z = -1 + 6/7 = -1/7。交点为 (11/7, 12/7, -1/7)。

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6. Hyperbolic Functions: Proving Identities and Solving Equations | 双曲函数：证明恒等式与解方程

Question: (a) Using the definitions cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ - e⁻ˣ = 1. (b) Solve the equation 2 cosh x + sinh x = 5, giving your answer in logarithmic form.

题目：(a) 利用定义 cosh x = (eˣ + e⁻ˣ)/2 和 sinh x = (eˣ - e⁻ˣ)/2，证明 cosh²x - sinh²x = 1。(b) 解方程 2 cosh x + sinh x = 5，并将答案写成对数形式。

(a) cosh²x - sinh²x = [(eˣ + e⁻ˣ)/2]² - [(eˣ - e⁻ˣ)/2]² = (e²ˣ + 2 + e⁻²ˣ - e²ˣ + 2 - e⁻²ˣ)/4 = 4/4 = 1.

(a) cos

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