Year 12 CAIE Further Maths: Mock Unit Test Walkthrough | CAIE 进阶数学单元测试模拟卷解析

📚 Year 12 CAIE Further Maths: Mock Unit Test Walkthrough | CAIE 进阶数学单元测试模拟卷解析

Welcome to this in‑depth walkthrough of a typical Year 12 CAIE Further Mathematics unit test. The mock paper is designed around the core content of the first term, especially Further Pure Mathematics 1 (FP1). We will tackle problems on roots of polynomials, complex numbers, matrix algebra, proof by induction, polar coordinates and rational functions. Every solution is broken down into small, exam‑friendly steps, with English and Chinese explanations side by side so that you can reinforce your understanding in both languages.

欢迎阅读这份面向 CAIE 进阶数学 Year 12 单元测试的模拟卷逐题精讲。试卷围绕第一学期的核心内容，特别是进阶纯数 1（FP1）。我们将处理多项式根、复数、矩阵代数、数学归纳法、极坐标以及有理函数等典型题目。每道题的解答都被拆分成适合考试的小步骤，中英对照，帮助你用两种语言巩固理解。

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1. Question 1: Sum and Product of Roots | 问题1：根的和与积

The quadratic equation x² − 6x + 4 = 0 has roots α and β. Without solving the equation, find α+β, αβ, α²+β² and (α−β)².

已知二次方程 x² − 6x + 4 = 0 的两个根为 α 和 β。不解方程，求 α+β、αβ、α²+β² 以及 (α−β)²。

Using Vieta’s formulas, the sum of roots α+β = 6 and the product αβ = 4.

由韦达定理，根之和 α+β = 6，根之积 αβ = 4。

Then α²+β² = (α+β)² − 2αβ = 36 − 8 = 28.

于是 α²+β² = (α+β)² − 2αβ = 36 − 8 = 28。

Also (α−β)² = (α+β)² − 4αβ = 36 − 16 = 20.

同理 (α−β)² = (α+β)² − 4αβ = 36 − 16 = 20。

These symmetric expressions are extremely common in FP1; always convert them into sums and products.

这些对称式在 FP1 中非常常见，务必将它们转化为根的和与积。

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2. Question 2: Forming a New Quadratic Equation | 问题2：构造新二次方程

For the same roots α and β from x² − 6x + 4 = 0, find a quadratic equation whose roots are 2α+1 and 2β+1.

沿用上题方程 x² − 6x + 4 = 0 的根 α 和 β，求以 2α+1 和 2β+1 为根的二次方程。

Sum of the new roots S = (2α+1)+(2β+1) = 2(α+β) + 2 = 2×6 + 2 = 14.

新根之和 S = (2α+1)+(2β+1) = 2(α+β) + 2 = 2×6 + 2 = 14。

Product P = (2α+1)(2β+1) = 4αβ + 2(α+β) + 1 = 4×4 + 2×6 + 1 = 29.

新根之积 P = (2α+1)(2β+1) = 4αβ + 2(α+β) + 1 = 16 + 12 + 1 = 29。

The required equation is x² − Sx + P = 0 → x² − 14x + 29 = 0.

因此所求方程为 x² − Sx + P = 0 → x² − 14x + 29 = 0。

Always remember the structure: x² − (sum of roots)x + (product of roots) = 0.

牢记这一结构：x² − (根之和)x + (根之积) = 0。

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3. Question 3: Solving a Quadratic with Complex Roots | 问题3：求解根的二次方程

Solve z² − 4z + 13 = 0, giving your answers in the form a + b i.

解方程 z² − 4z + 13 = 0，答案写成 a + b i 的形式。

Compute the discriminant Δ = b² − 4ac = (−4)² − 4×1×13 = 16 − 52 = −36.

计算判别式 Δ = b² − 4ac = (−4)² − 4×1×13 = 16 − 52 = −36。

Since Δ is negative, the roots are complex: √Δ = √(−36) = 6 i.

因为 Δ 为负数，根为复数：√Δ = √(−36) = 6 i。

Using the quadratic formula, z = [4 ± 6 i] / 2 = 2 ± 3 i.

代入求根公式，z = (4 ± 6 i) / 2 = 2 ± 3 i。

Thus the two roots are z₁ = 2 + 3 i and z₂ = 2 − 3 i. They form a conjugate pair.

因此两根为 z₁ = 2 + 3 i 以及 z₂ = 2 − 3 i，它们是一对共轭复数。

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4. Question 4: Argand Diagram, Modulus and Argument | 问题4：阿尔冈图、模与辐角

For the complex number z = 2 + 3 i, plot it on an Argand diagram and find its modulus |z| and argument arg(z).

对于复数 z = 2 + 3 i，在阿尔冈图上描点，并求其模 |z| 与辐角 arg(z)。

The point representing z is (2, 3). The modulus is the distance from the origin:

该复数对应坐标为 (2, 3)。模为到原点的距离：

|z| = √(2² + 3²) = √(4 + 9) = √13.

|z| = √(2² + 3²) = √(4 + 9) = √13。

The argument is the angle with the positive real axis: arg(z) = arctan(3/2). Since the point is in the first quadrant, no adjustment is needed.

辐角是与正实轴之间的夹角：arg(z) = arctan(3/2)。因点在第一象限，无需调整。

In radians, arg(z) ≈ 0.983 rad. Always check the quadrant when determining the argument.

用弧度表示，arg(z) ≈ 0.983 rad。求辐角时务必检查象限。

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5. Question 5: Inverse of a 2×2 Matrix | 问题5：二阶矩阵的逆

Find the inverse of the matrix A = [2 -1; 3 5].

求矩阵 A = [2 -1; 3 5] 的逆矩阵。

First calculate the determinant: det(A) = (2)(5) − (−1)(3) = 10 + 3 = 13. Since det(A) ≠ 0, the inverse exists.

先计算行列式：det(A) = (2)(5) − (−1)(3) = 10 + 3 = 13。因为 det(A) ≠ 0，逆矩阵存在。

Swap the elements on the main diagonal, change the signs of the off‑diagonal elements to obtain the adjugate:

交换主对角线元素，变更次对角线元素的符号，得到伴随矩阵：

5	1
-3	2

Now multiply by 1/det: A⁻¹ = (1/13) × adj(A).

再乘以 1/det：A⁻¹ = (1/13) × adj(A)。

Thus A⁻¹ =

5/13	1/13
-3/13	2/13

因此 A⁻¹ =

5/13	1/13
-3/13	2/13

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6. Question 6: Solving Simultaneous Equations with Matrices | 问题6：矩阵解联立方程

Use the inverse matrix from Question 5 to solve the system:
2x − y = 7
3x + 5y = 2.

利用第5题求得的逆矩阵解方程组：
2x − y = 7
3x + 5y = 2。

Write the system in matrix form AX = B, where A = [2 -1; 3 5], X = [x; y], B = [7; 2].

将方程组写成矩阵格式 AX = B，其中 A = [2 -1; 3 5]，X = [x; y]，B = [7; 2]。

Since A⁻¹ exists, X = A⁻¹ B.

由于 A⁻¹ 存在，X = A⁻¹ B。

Compute: x = (5/13)×7 + (1/13)×2 = 35/13 + 2/13 = 37/13.

计算：x = (5/13)×7 + (1/13)×2 = 35/13 + 2/13 = 37/13。

y = (−3/13)×7 + (2/13)×2 = −21/13 + 4/13 = −17/13.

y = (−3/13)×7 + (2/13)×2 = −21/13 + 4/13 = −17/13。

Solution: x = 37/13, y = −17/13. Always check by substituting back.

解得：x = 37/13，y = −17/13。务必代回原方程检验。

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7. Question 7: Proof by Induction – Summation | 问题7：数学归纳法证明——求和

Prove by induction that for all positive integers n, ∑_r=1ⁿ r = n(n+1)/2.

用数学归纳法证明，对所有正整数 n，有 ∑_r=1ⁿ r = n(n+1)/2。

Base case n = 1: LHS = 1, RHS = 1×2/2 = 1. True.Published by TutorHao | Year 12 进阶数学 Revision Series | aleveler.com

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