AS Further Mathematics Mock Test Paper Analysis | AS 进阶数学单元测试模拟卷解析

📚 AS Further Mathematics Mock Test Paper Analysis | AS 进阶数学单元测试模拟卷解析

This article walks you through a complete mock unit test for CAIE AS Level Further Mathematics (9231), focusing on the core content of Further Pure Mathematics 1. Every question is solved step by step, with full reasoning and examiner-friendly commentary. Use this resource to sharpen your technique, identify common pitfalls, and boost your confidence before the actual exam.

本文为你详细解析一份完整的 CAIE AS 进阶数学（9231）单元测试模拟卷，内容紧扣 Further Pure Mathematics 1 的核心考点。每道题均配有分步解答、完整推理及适合考试的详细说明。利用这份资料打磨解题技巧，发现常见易错点，在真正的大考到来前增强信心。

1. Complex Numbers: Modulus and Argument | 复数：模与辐角计算

The equation (z + 2i)(1 – = 3 + 4i defines a complex number z. Find z in Cartesian form x + iy, and hence determine |z| and arg(z), giving the argument in radians correct to two decimal places.

已知复数方程 (z + 2i)(1 – i) = 3 + 4i，求 z 的代数形式 x + iy，并据此计算 |z| 和 arg(z)，辐角以弧度制给出并精确至两位小数。

First expand the left‑hand side: (z + 2i)(1 – i) = z(1 – i) + 2i(1 – i). Now 2i(1 – i) = 2i – 2i² = 2i + 2 = 2 + 2i. So z(1 – i) + (2 + 2i) = 3 + 4i, giving z(1 – i) = 1 + 2i. Divide by (1 – i): z = (1 + 2i) / (1 – ator and denominator by the conjugate (1 + i): (1 + 2i)(1 + i) = 1 + i + 2i + 2i² = 1 + 3i – 2 = –1 + 3i; denominator (1 – i)(1 + i) = 1 + 1 = 2. Hence z = (–1 + 3i)/2 = –½ + (3/2)i.

首先展开左边：(z + 2i)(1 – i) = z(1 – i) + 2i(1 – i)。其中 2i(1 – i) = 2i – 2i² = 2i + 2 = 2 + 2i。于是 z(1 – i) + (2 + 2i) = 3 + 4i，得 z(1 – i) = 1 + 2i。两边除以 (1 – i) 得 z = (1 + 2i)/(1 – i)。分子分母同乘共轭复数 (1 + i)：分子 (1 + 2i)(1 + i) = 1 + i + 2i + 2i² = 1 + 3i – 2 = –1 + 3i；分母 (1 – i)(1 + i) = 2。因此 z = (–1 + 3i)/2 = –½ + (3/2)i。

Now |z| = √((–½)² + (3/2)²) = √¼ + 9/4 = √10/4 = √2.5 ≈ 1.581 (exact value √10 Re(z) < 0 and Im(z) > 0, so z lies in the second quadrant. The reference angle is arctan(|Im/Re|) = arctan(1.5 / 0.5) = arctan 3 ≈ 1.249 rad. Hence arg(z) = π – 1.249 = 1.892 rad, which is 1.89 rad to two decimal places.

接下来求模：|z| = √((–½)² + (3/2)²) = √¼ + 9/4 = √10/4 = √2.5 ≈ 1.581（精确值为 √10/2）。再求辐角：由于实部为负、虚部为正，z 位于第二象限。基本角 = arctan(1.5/0.5) = arctan 3 ≈ 1.249 rad。故 arg(z) = π – 1.249 = 1.892 rad，保留两位小数得 1.89 rad。

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2. Matrices: Operations, Inverse and Simultaneous Equations | 矩阵：运算、逆矩阵与解线性方程组

Let A =

2	–1
1	3

and B =

1	4
–2	0

. Find (a) 2A + B, (b) AB, (c) A⁻¹, and (d) the solution of the system 2x – y = 5, x + 3y = –1 using A⁻¹.

给定矩阵 A 和 B（见上），求 (a) 2A + B，(b) AB，(c) A⁻¹，(d) 利用 A⁻¹ 求解方程组 2x – y = 5, x + 3y = –1。

(a) 2A =

4	–2
2	6

, so 2A + B =

4+1	–2+4
2–2	6+0

=

5	2
0	6

.

(a) 2A 计算后与 B 相加，结果见上。

(b) AB =

2×1 + (–1)×(–2)	2×4 + (–1)×0
1×1 + 3×(–2)	1×4 + 3×0

=

4	8
–5	4

.

(b) 矩阵乘法 AB 的结果：第一行第一列 2×1+(–1)×(–2)=4，第一行第二列 2×4+0=8；第二行第一列 1×1+3×(–2)= –5，第二行第二列 1×4+0=4。

(c) det A = (2)(3) – (–1)(1) = 6 + 1 = 7. A⁻¹ = 1/7

3	1
–1	2

=

3/7	1/7
–1/7	2/7

.

(c) A 的行列式为 7，伴随矩阵为 [[3,1],[–1,2]]，故逆矩阵见上。

(d) The system can be written as A

x

y

=

5

–1

. Multiply by A⁻¹:

x

y

=

3/7	1/7
–1/7	2/7

5

–1

=

(15/7 – 1/7)

(–5/7 – 2/7)

=

14/7

–7/7

=

2

–1

. Therefore x = 2, y = –1.

(d) 将方程组写成矩阵形式，左乘 A⁻¹ 直接得到 x = 2，y = –1。

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3. Summation of Series: Using Standard Results | 级数求和：标准公式应用

nd an expression in terms of n for the sum of the series 1×2 + 2×3 + 3×4 + ... + n(n + 1).

求级数 1×2 + 2×3 + 3×4 + ... + n(n + 1) 前 n 项和的表达式。

Let S = ∑_r=1ⁿ (n+1)/2] × [ (2n+1)/3 + 1 ] = [n(n+1)/2] × [(2n+1+3)/3] = [n(n+1)/2] × [(2n+4)/3] = n(n+1)(n+2)/3.

令 S = ∑_r=1ⁿ r(r+1) = ∑ r² + ∑ r。代入标准结果 ∑ r = n(n+1)/2 和 ∑ r² = n(n+1)(2n+1)/6，得 S = n(n+1)(2n+1)/6 + n(n+1)/2。提出公因子 n(n+1)/2，化简后得到 S = n(n+1)(n+2)/3。

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4. Roots of Polynomials: Sums and Symmetric Functions | 多项式方程根：和与对称式

The cubic equation 2x³ – Without solving the equation, find the value of α² + β² + γ² and of (α + 1)(β + 1)(γ + 1).

已知三次方程 2x³ – 3x² + 4x + 5 = 0 的三个根为 α, β, γ。不解方程，求 α² + β² + γ² 和 (α + 1)(β + 1)(γ + 1) 的值。

From Vieta's formulas for 2x³ – 3x² + 4x + 5 = 0: α + β + γ = –(–3)/2 = 3/2, αβ + βγ + γα = 4/2 = 2, αβγ = –5/2. Then α² + β² + γ² = (α+β+γ)² – 2(αβ+βγ+γα) = (3/2)² – 2(2) = 9/4 – 4 = –7/4.

由韦达定理：α+β+γ = 3/2，αβ+βγ+γα = 2，αβγ = –5/2。于是 α²+β²+γ² = (3/2)² – 2×2 = 9

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