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

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

This mock paper analysis covers eight essential topics from the Cambridge AS Further Mathematics syllabus, including complex numbers, matrices, roots of polynomials, induction, series, and transformations. Each question is broken down with detailed solutions, highlighting common pitfalls and efficient methods. Use this revision resource to consolidate your understanding and boost your exam confidence.

本模拟卷解析涵盖剑桥AS进阶数学大纲中八个核心主题，包括复数、矩阵、多项式根、归纳法、级数和变换。每道题都配有详细解答，强调常见错误与高效方法。通过这份复习资料，巩固理解并提升考试信心。

1. Complex Numbers Arithmetic | 复数运算

2i, compute (a) z₁ + z₂, (b) z₁z₂, and (c) z₁/z₂. Express all results in the form a + bi.

题目：已知 z₁ = 3 + 4i 和 z₂ = 1 − 2i，计算 (a) z₁ + z₂, (b) z₁z₂, 以及 (c) z₁/z₂。所有结果均以 a + bi 形式表示。

ply combining real and imaginary parts: (3 + 4i) + (1 − 2i) = (3 + 1) + (4 − 2)i = 4 + 2i.

(a) 解答：加法直接合并实部与虚部：(3 + 4i) + (1 − 2i) = (3 + 1) + (4 − 2)i = 4 + 2i。

(4i)(1) + (4i)(−2i) = 3 − 6i + 4i − 8i². Since i² = −1, this becomes 3 − 2i + 8 = 11 − 2i.

(b) 解答：使用分配律相乘：(3)(1) + (3)(−2i) + (4i)(1) + (4i)(−2i) = 3 − 6i + 4i − 8i²。因为 i² = −1，结果为 3 − 2i + 8 = 11 − 2i。

+ 2i)/(1 + 2i) = (3 + 6i + 4i + 8i²)/(1² + 2²) = (3 + 10i − 8)/5 = (−5 + 10i)/5 = −1 + 2i.

(c) 解答：进行除法时，分子分母同乘分母的共轭复数 1 + 2i： (1 + 2i)/(1 + 2i) = (3 + 6i + 4i + 8i²)/(1 + 4) = (3 + 10i − 8)/5 = (−5 + 10i)/5 = −1 + 2i。

---

2. Matrix Multiplication and Inverses | 矩阵乘法与逆矩阵

Given matrices A = [[2, 1], [0, −3]] and B = [[4, −2], [1, 1]], find (a) AB and (b) A⁻¹.

给定矩阵 A = [[2, 1], [0, −3]] 和 B = [[4, −2], [1, 1]]，求 (a) AB 和 (b) A⁻¹。

0×4 + (−3)×1, 0×(−2) + (−3)×1]] = [[8 + 1, −4 + 1], [0 − 3, 0 − 3]] = [[9, −3], [−3, −3]].

(a) 解答：用A的行乘B的列： [0×4 + (−3)×1, 0×(−2) + (−3)×1]] = [[8 + 1, −4 + 1], [0 − 3, 0 − 3]] = [[9, −3], [−3, −3]]。

by 1/det(A) × [[d, −b], [−c, a]] = (1/−6) × [[−3, −1], [0, 2]] = [[1/2, 1/6], [0, −1/3]].

(b) 解答：对于 A = [[2, 1], [0, −3]]，行列式 det(A) = (2)(−3) − (1)(0) = −6。逆矩阵公式为 1/det(A) × [[d, −b], [−c, a]] = (1/−6) × [[−3, −1], [0, 2]] = [[1/2, 1/6], [0, −1/3]]。

---

3. Sums and Products of Roots | 根的和与积

The quadratic equation 3x² + 2x − 7 = 0 has roots α and β. Find α + β, αβ, and α² + β².

二次方程 3x² + 2x − 7 = 0 有根 α 和 β。求 α + β, αβ 以及 α² + β²。

m of roots = −b/a, product = c/a. Here a = 3, b = 2, c = −7. Thus α + β = −2/3, αβ = −7/3.

对于 ax² + bx + c = 0，根的和 = −b/a，积 = c/a。此处 a = 3, b = 2, c = −7。因此 α + β = −2/3，αβ = −7/3。

Then α² + β² = (α + β)² − 2αβ = (−2/3)² − 2(−7/3) = 4/9 + 14/3 = 4/9 + 42/9 = 46/9.

进而 α² + β² = (α + β)² − 2αβ = (−2/3)² − 2(−7/3) = 4/9 + 14/3 = 4/9 + 42/9 = 46/9。

---

广告 / Sponsored

4. Proof by Mathematical Induction | 数学归纳法证明

Prove by induction that for all positive integers n: ∑_r=1ⁿ r² = n(n + 1)(2n + 1)/6.

用归纳法证明对所有正整数 n 有：∑_r=1ⁿ r² = n(n + 1)(2n + 1)/6。

Base case n = 1: LHS = 1² = 1, RHS = 1×2×3/6 = 1, so P(1) is true.

基准情况 n = 1：左边 = 1² = 1，右边 = 1×2×3/6 = 1，因此 P(1) 成立。

Inductive hypothesis: Assume P(k) true, i.e., ∑_r=1^k r² = k(k + 1)(2k + 1)/6.

归纳假设：假设 P(k) 成立，即 ∑_r=1^k r² = k(k + 1)(2k + 1)/6。

Inductive step: For n = k + 1, ∑_r=1^k+1 r² = ∑_r=1^k )/6 (2k² + 7k + 6) = (k + 1)(k + 2)(2k + 3)/6, which is exactly the form for n = k + 1. Hence P(k) ⇒ P(k + 1). By induction, the statement holds for all n.

归纳步骤：当 n = k + 1 时，∑_r=1^k+1 r² = ∑_r=1^k r² + (k + 1)²。利用假设，等于 k(k + 1)(2k + 1)/6 + (k + 1)²。提取公因式 (k + 1)/6： + 1)] = (k + 1)/6 [2k² + k + 6k + 6] = (k + 1)/6 (2k² + 7k + 6) = (k + 1)(k + 2)(2k + 3)/6，恰好是 n = k + 1 时的形式。因此 P(k) ⇒ P(k + 1)。由归纳法，该命题对所有 n 成立。

---

5. Summation of Series | 级数求和

Evaluate ∑_r=1ⁿ (r² + 2r) in terms of n, simplifying your answer.

求 ∑_r=1ⁿ (r² + 2r) 关于 n 的表达式，并化简。

Split the sum: ∑r² + 2∑r. Using standard results, ∑_r=1ⁿ r² = n(n + 1)(2n + 1)/6 and ∑_r=1ⁿ xpression becomes [n(n + 1)(2n + 1)/6] + 2 × [n(n + 1)/2] = n(n + 1)(2n + 1)/6 + n(n + 1).

拆分求和：∑r² + 2∑r。利用标准结果，∑_r=1ⁿ r² = n(n + 1)(2n + 1)/6，∑_r=1ⁿ r = n(n + 1)/2。因此表达式变为 [n(n + 1)(2n + 1)/6] + 2 × [n(n + 1)/2] = n(n + 1)(2n + 1)/6 + n(n + 1)。

common factor: n(n + 1) [(2n + 1)/6 + 1] = n(n + 1) [(2n + 1 + 6)/6] = n(n + 1)(2n + 7)/6.

合并项，提取 n(n + 1) 作为公因子：n(n + 1) [(2n + 1)/6 + 1] = n(n + 1) [(2n + 1 + 6)/6] = n(n + 1)(2n + 7)/6。

---

6. Geometric Representation of Complex Numbers | 复数的几何表示

Describe the locus of points z satisfying |z − (2 + i)| = 3, and determine whether the point z = 5 + i lies on this locus.

描述满足 |z − (2 + i)| = 3 的点 z 的轨迹，并判断点 z = 5 + i 是否在该轨迹上。

The equation |z − a| ius r. Here a = 2 + i and r = 3, so the locus is a circle centred at (2, 1) with radius 3.

方程 |z − a| = r 表示以复数点 a 为圆心、半径为 r 的圆。此处 a = 2 + i，r = 3，故轨迹是以 (2, 1) 为圆心、半径为 3 的圆。

To check z = 5 + i, compute |(5 + i) − (2 + i)| = |3 + 0i| = 3. Since the distance equals the radius, the point lies exactly on the circle.

检验 z = 5 + i，计算 |(5 + i) − (2 + i)| = |3 + 0i| = 3。由于该距离等于半径，该点恰好位于圆上。

---

7. Matrix Transformations | 矩阵变换

ents a geometric transformation in the plane. Identify this transformation and find the image of the point (3, 2) under M.

矩阵 M = [[0, −1], [1, 0]] 表示平面上的一个几何变换。识别该变换，并求点 (3, 2) 在 M 下的像。

Applying M to a general vector [x; y] gives [0·x − 1·y; 1·x + 0·y] =

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

更多咨询请联系16621398022（同微信）