A-Level CIE Further Mathematics Unit Test Paper | A-Level CIE 进阶数学：单元测试卷

📚 A-Level CIE Further Mathematics Unit Test Paper | A-Level CIE 进阶数学：单元测试卷

This unit test paper is designed for students preparing for the CIE A-Level Further Mathematics examination. It covers essential topics from the Further Pure Mathematics 1 (FP1) syllabus, including complex numbers, roots of polynomials, summation of series, matrices, proof by induction, and vectors. Each question is followed by a detailed, step-by-step solution to reinforce key concepts and common techniques. Use this resource to test your understanding, identify weak areas, and build confidence before the real exam.

本单元测试卷专为准备 CIE A-Level 进阶数学考试的学生设计。试卷覆盖进阶纯数学1 (FP1) 的核心主题，包括复数、多项式根、级数求和、矩阵、归纳法证明以及向量。每道题均附有详细的逐步解答，以巩固关键概念与常用技巧。请使用本资源检验自己的理解程度、发现薄弱环节，并在正式考试前树立信心。

1. Complex Numbers – Solving an Equation | 问题1：复数 – 解方程

Solve the equation (2 + i)z + 1 − 3i = (1 − 3i)z, giving your answer in the form a + bi where a, b ∈ ℝ.

解方程 (2 + i)z + 1 − 3i = (1 − 3i)z，并将答案表示为 a + bi 的形式，其中 a, b 为实数。

Solution:

解答：

Collect the z‑terms on one side: (2 + i)z − (1 − 3i)z = 3i − 1.

将含 z 的项移到一边： (2 + i)z − (1 − 3i)z = 3i − 1。

Simplify the coefficient: (2 + i − 1 + 3i)z = −1 + 3i ⇒ (1 + 4i)z = −1 + 3i.

化简系数： (2 + i − 1 + 3i)z = −1 + 3i ⇒ (1 + 4i)z = −1 + 3i。

Thus z = (−1 + 3i) / (1 + 4i). Multiply numerator and denominator by the complex conjugate of the denominator (1 − 4i).

于是 z = (−1 + 3i) / (1 + 4i)。将分子与分母同乘以分母的共轭复数 (1 − 4i)。

z = [(−1 + 3i)(1 − 4i)] / [(1 + 4i)(1 − 4i)] = [−1×(1) + (−1)×(−4i) + 3i×1 + 3i×(−4i)] / [1² + 4²]

z = [(−1 + 3i)(1 − 4i)] / [(1 + 4i)(1 − 4i)] = [−1×1 + (−1)×(−4i) + 3i×1 + 3i×(−4i)] / [1² + 4²]

Compute numerator: −1 + 4i + 3i −12i². Since i² = −1, −12i² = 12. So numerator = (−1 + 12) + (4i + 3i) = 11 + 7i.

计算分子： −1 + 4i + 3i −12i²。因为 i² = −1， −12i² = 12。所以分子 = (−1 + 12) + (4i + 3i) = 11 + 7i。

Denominator: 1 + 16 = 17. Hence z = (11 + 7i) / 17 = 11/17 + (7/17)i.

分母： 1 + 16 = 17。因此 z = (11 + 7i) / 17 = 11/17 + (7/17)i。

Answer: z = 11/17 + (7/17)i.

答案： z = 11/17 + (7/17)i。

2. Roots of Polynomial Equations – Finding Unknown Coefficients | 问题2：多项式方程的根 – 求未知系数

The equation x³ + px² + qx + 5 = 0, where p and q are real, has a root 1 + 2i. Determine p and q and find all the roots.

方程 x³ + px² + qx + 5 = 0 中 p 与 q 为实数，且有一个根为 1 + 2i。求 p 和 q 的值以及所有根。

Solution:

解答：

Since the coefficients are real, the complex conjugate 1 − 2i is also a root.

由于系数均为实数，共轭复数 1 − 2i 也是该方程的根。

The cubic has three roots. Let the third root be r (real).

该三次方程有三个根，设第三个根为实数 r。

Sum of roots = −p: (1+2i) + (1−2i) + r = 2 + r = −p.

根之和 = −p： (1+2i) + (1−2i) + r = 2 + r = −p。

Product of roots taken two at a time = q: (1+2i)(1−2i) + (1+2i)r + (1−2i)r = (1²+2²) + r(2) = 5 + 2r = q.

两两根之积的和 = q： (1+2i)(1−2i) + (1+2i)r + (1−2i)r = (1+4) + 2r = 5 + 2r = q。

Product of all three roots = −5: (1+2i)(1−2i) × r = 5r = −5 ⇒ r = −1.

三根之积 = −5： (1+2i)(1−2i) × r = 5r = −5 ⇒ r = −1。

Then 2 + r = 2 − 1 = 1, so −p = 1 ⇒ p = −1.

于是 2 + r = 2 − 1 = 1，即 −p = 1 ⇒ p = −1。

And q = 5 + 2(−1) = 3.

并且 q = 5 + 2(−1) = 3。

The roots are 1 + 2i, 1 − 2i, and −1.

三个根分别为 1 + 2i、1 − 2i 和 −1。

3. Summation of Series – Using Standard Results | 问题3：级数求和 – 使用标准结果

Find an expression for Σ(r=1 to n) r(r + 2) in terms of n, simplifying your answer.

求 Σ(r=1 到 n) r(r + 2) 的表达式，用 n 表示并化简。

Solution:

解答：

Expand the term: r(r + 2) = r² + 2r.

展开通项： r(r + 2) = r² + 2r。

Hence Σ r(r+2) = Σ r² + 2 Σ r (summations from r=1 to n).

因此 Σ r(r+2) = Σ r² + 2 Σ r (求和范围均为 r=1 到 n)。

Use standard results: Σ r² = n(n+1)(2n+1)/6, Σ r = n(n+1)/2.

代入标准结果： Σ r² = n(n+1)(2n+1)/6， Σ r = n(n+1)/2。

So Σ r(r+2) = n(n+1)(2n+1)/6 + 2 × n(n+1)/2 = n(n+1)(2n+1)/6 + n(n+1).

所以 Σ r(r+2) = n(n+1)(2n+1)/6 + 2 × n(n+1)/2 = n(n+1)(2n+1)/6 + n(n+1)。

Combine into a single fraction: write n(n+1) as 6n(n+1)/6.

通分：将 n(n+1) 写成 6n(n+1)/6。

Sum = [n(n+1)(2n+1) + 6n(n+1)] / 6 = n(n+1)[(2n+1) + 6] / 6 = n(n+1)(2n+7)/6.

和式 = [n(n+1)(2n+1) + 6n(n+1)] / 6 = n(n+1)[(2n+1) + 6] / 6 = n(n+1)(2n+7)/6。

Therefore Σ(r=1 to n) r(r+2) = (1/6)n(n+1)(2n+7).

答案为 Σ(r=1 到 n) r(r+2) = (1/6)n(n+1)(2n+7)。

4. Matrices – Inverse and Linear Equations | 问题4：矩阵 – 逆矩阵与线性方程组

Given the matrix M = [[2, -1], [0, 3]]. Find M⁻¹ and hence solve the system of equations: 2x − y = 4, 3y = 6.

已知矩阵 M = [[2, -1], [0, 3]]。求 M⁻¹，并以此解方程组： 2x − y = 4, 3y = 6。

Solution:

解答：

For a 2×2 matrix [[a, b], [c, d]], the inverse is (1/det)[[d, -b], [-c, a]], provided det ≠ 0.

对于 2×2 矩阵 [[a, b], [c, d]]，其逆矩阵为 (1/det)[[d, -b], [-c, a]]，前提是行列式不为零。

Here det(M) = 2×3 − (−1)×0 = 6. Since it is non‑zero, M is invertible.

此处 det(M) = 2×3 − (−1)×0 = 6。行列式不为零，故 M 可逆。

Thus M⁻¹ = (1/6) [[3, 1], [0, 2]] = [[1/2, 1/6], [0, 1/3]].

因此 M⁻¹ = (1/6)[[3, 1], [0, 2]] = [[1/2, 1/6], [0, 1/3]]。

The system can be written as M[[x], [y]] = [[4], [6]]. Premultiply both sides by M⁻¹.

方程组可写为 M[[x], [y]] = [[4], [6]]。两边左乘 M⁻¹。

[[x], [y]] = M⁻¹ [[4], [6]] = [[1/2, 1/6], [0, 1/3]] [[4], [6]] = [[ (1/2)×4 + (1/6)×6 ], [ 0×4 + (1/3)×6 ]] = [[2 + 1], [2]] = [[3], [2]].

[[x], [y]] = M⁻¹ [[4], [6]] = [[1/2, 1/6], [0, 1/3]] [[4], [6]] = [[ (1/2)×4 + (1/6)×6 ], [ 0×4 + (1/3)×6 ]] = [[2 + 1], [2]] = [[3], [2]]。

Solution: x = 3, y = 2.

解得 x = 3, y = 2。

5. Proof by Induction – Sum of Squares | 问题5：数学归纳法证明 – 平方和公式

Prove, by mathematical induction, that for all positive integers n, Σ(r=1 to n) r² = (1/6)n(n+1)(2n+1).

用数学归纳法证明，对所有正整数 n，均有 Σ(r=1 到 n) r² = (1/6)n(n+1)(2n+1)。

Solution:

解答：

Let P(n) be the statement Σ(r=1 to n) r² = n(n+1)(2n+1)/6.

令 P(n) 表示命题 Σ(r=1 到 n) 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.

基础情况 n = 1：左边 = 1² = 1。右边 = 1×2×3/6 = 1，故 P(1) 成立。

Inductive step: Assume P(k) true for some k ≥ 1, i.e. Σ(r=1 to k) r² = k(k+1)(2k+1)/6.

归纳步骤：假设对某个 k ≥ 1，P(k) 成立，即 Σ(r=1 到 k) r² = k(k+1)(2k+1)/6。

Consider Σ(r=1 to k+1) r² = Σ(r=1 to k) r² + (k+1)². Substitute the assumption.

考虑 Σ(r=1 到 k+1) r² = Σ(r=1 到 k) r² + (k+1)²，代入假设。

This equals k(k+1)(2k+1)/6 + (k+1)².

它等于 k(k+1)(2k+1)/6 + (k+1)²。

Factor (k+1)/6: = (k+1)/6 [k(2k+1) + 6(k+1)].

提取公因子 (k+1)/6： = (k+1)/6 [k(2k+1) + 6(k+1)]。

Simplify the bracket: 2k² + k + 6k + 6 = 2k² + 7k + 6 = (k+2)(2k+3).

化简括号内： 2k² + k + 6k + 6 = 2k² + 7k + 6 = (k+2)(2k+3)。

Hence sum = (k+1)(k+2)(2k+3)/6, which is exactly the RHS of P(k+1).

因此和式 = (k+1)(k+2)(2k+3)/6，这正是 P(k+1) 的右边。

Thus P(k) true ⇒ P(k+1) true.

所以 P(k) 成立 ⇒ P(k+1) 成立。

By mathematical induction, P(n) is true for all n ∈ ℕ.

根据数学归纳法，对所有自然数 n，P(n) 成立。

6. Vector Geometry – Line Equation and Angle | 问题6：向量几何 – 直线方程与夹角

The position vectors of points A and B are a = i + 2j − 3k and b = 4i − j + 2k. Find the vector AB and the Cartesian equation of the line through A and B. Also compute the acute angle between AB and the vector c = i + j + k.

点 A 和点 B 的位置向量分别为 a = i + 2j − 3k 和 b = 4i − j + 2k。求向量 AB、过 A 与 B 的直线的笛卡尔方程，并计算 AB 与向量 c = i + j + k 之间的锐角。

Solution:

解答：

Vector AB = b − a = (4−1)i + (−1−2)j + (2−(−3))k = 3i − 3j + 5k.

向量 AB = b − a = (4−1)i + (−1−2)j + (2−(−3))k = 3i − 3j + 5k。

The line through A with direction AB can be written in parametric form: r = a + λ AB, but for Cartesian form we use (x − x₀)/3 = (y − y₀)/(−3) = (z − z₀)/5. Taking A as (1, 2, −3):

过 A 且方向为 AB 的直线的笛卡尔方程可由 (x − x₀)/3 = (y − y₀)/(−3) = (z − z₀)/5 给出。取 A 的坐标 (1, 2, −3)：

(x − 1)/3 = (y − 2)/(−3) = (z + 3)/5.

即为直线方程。

Angle θ between AB = 3i − 3j + 5k and c = i + j + k: cos θ = |AB·c| / (|AB||c|).

AB 与 c 的夹角： cos θ = |AB·c| / (|AB||c|)。

AB·c = 3×1 + (−3)×1 + 5×1 = 3 − 3 + 5 = 5.

AB·c = 3×1 + (−3)×1 + 5×1 = 5。

|AB| = √(3² + (−3)² + 5²) = √(9 + 9 + 25) = √43.

|AB| = √(9+9+25) = √43。

|c| = √(1²+1²+1²) = √3.

|c| = √3。

Thus cos θ = 5 / (√43 × √3) = 5/√129. The acute angle is θ = arccos(5/√129).

于是 cos θ = 5 / (√43 × √3) = 5/√129。锐角 θ = arccos(5/√129)。

7. Complex Loci – Perpendicular Bisector | 问题7：复数轨迹 – 垂直平分线

Sketch the locus of z satisfying |z − 2i| = |z − 4|. Describe the locus in Cartesian terms and find its equation.

画出满足 |z − 2i| = |z − 4| 的 z 的轨迹。用笛卡尔坐标描述该轨迹，并求其方程。

Solution:

解答：

Let z = x + yi. Then |(x + yi) − 2i| = |(x + yi) − 4|.

令 z = x + yi，则 |(x + yi) − 2i| = |(x + yi) − 4|。

Left side: |x + i(y − 2)| = √(x² + (y−2)²).

左边： |x + i(y − 2)| = √(x² + (y−2)²)。

Right side: |(x − 4) + yi| = √((x−4)² + y²).

右边： |(x − 4) + yi| = √((x−4)² + y²)。

Squaring both sides: x² + (y−2)² = (x−4)² + y².

两边平方： x² + (y−2)² = (x−4)² + y²。

Expand: x² + y² − 4y + 4 = x² − 8x + 16 + y².

展开： x² + y² − 4y + 4 = x² − 8x + 16 + y²。

Cancel x², y²: −4y + 4 = −8x + 16 ⇒ 8x − 4y = 12 ⇒ 2x − y = 3.

消去 x²、y²： −4y + 4 = −8x + 16 ⇒ 8x − 4y = 12 ⇒ 2x − y = 3。

The locus is the straight line y = 2x − 3. Geometrically, it is the perpendicular bisector of the line segment joining the points (0,2) and (4,0) in the complex plane.

该轨迹为直线 y = 2x − 3。从几何意义上看，它是复平面上连接点 (0,2) 与 (4,0) 的线段的垂直平分线。

8. Roots and Coefficients – Symmetric Functions | 问题8：根与系数 – 对称函数

The quadratic equation 2x² − 3x + 4 = 0 has roots α and β. Without solving the equation, find the values of α² + β² and α³ + β³.

二次方程 2x² − 3x + 4 = 0 的根为 α 和 β。不解方程，求 α² + β² 和 α³ + β³ 的值。

Solution:

解答：

From the equation: α + β = 3/2, αβ = 4/2 = 2.

由方程得： α + β = 3/2， αβ = 4/2 = 2。

Alpha² + β² = (α + β)² − 2αβ = (3/2)² − 2×2 = 9/4 − 4 = 9/4 − 16/4 = −7/4.

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

For the sum of cubes, use α³ + β³ = (α + β)³ − 3αβ(α + β).

立方和公式： α³ + β³ = (α + β)³ − 3αβ(α + β)。

Substitute: (3/2)³ − 3×2×(3/2) = 27/8 − 9 = 27/8 − 72/8 = −45/8.

代入： (3/2)³ − 3×2×(3/2) = 27/8 − 9 = −45/8。

Thus α² + β² = −7/4 and α³ + β³ = −45/8.

因此 α² + β² = −7/4， α³ + β³ = −45/8。

9. Matrix Algebra – Transformations and Determinants | 问题9：矩阵代数 – 变换与行列式

A transformation T : ℝ² → ℝ² is represented by the matrix A = [[0, -2], [2, 0]]. Find the image of the point (3, 1) under T and explain the geometric transformation. Also compute det(A) and state whether T is area‑preserving.

变换 T : ℝ² → ℝ² 由矩阵 A = [[0, -2], [2, 0]] 表示。求点 (3, 1) 在

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A-Level CIE Further Mathematics Unit Test Paper | A-Level CIE 进阶数学：单元测试卷