📚 A-Level Edexcel Further Mathematics: Unit Test Mock Paper Walkthrough | A-Level Edexcel 进阶数学：单元测试模拟卷解析
This article provides a full breakdown of a mock unit test for Edexcel A-Level Further Mathematics, covering key topics from the Further Pure 1 (FP1) specification. Each question is solved step by step with clear reasoning, helping you consolidate your understanding and exam technique.
本文为 Edexcel A-Level 进阶数学提供一套完整的单元测试模拟卷解析，涵盖进一步纯数 1（FP1）的核心考点。每道题目均配有详细的步骤与思路讲解，帮助你巩固知识点并提升应试技巧。
1. Complex Numbers – Simplification, Modulus and Argument | 复数化简、模与辐角
Question: Express (3 + 4i)/(1 − 2i) in the form a + bi, where a, b ∈ ℝ. Hence find its modulus and argument, and sketch the point on an Argand diagram.
题目：将 (3 + 4i)/(1 − 2i) 表示为 a + bi 的形式，其中 a, b ∈ ℝ。由此求其模和辐角，并在 Argand 图上标出该点。
Begin by multiplying the numerator and the denominator by the complex conjugate of the denominator, 1 + 2i.
首先，将分子和分母同时乘以分母的共轭复数 1 + 2i。
(3 + 4i)(1 + 2i) / ((1 − 2i)(1 + 2i)) = (3 + 6i + 4i + 8i²) / (1 + 4)
Since i² = −1, the numerator becomes 3 + 10i − 8 = −5 + 10i. The denominator is 5.
因为 i² = −1，分子变为 3 + 10i − 8 = −5 + 10i，分母为 5。
z = (−5 + 10i) / 5 = −1 + 2i
The modulus is |z| = √((−1)² + 2²) = √5. The argument θ satisfies tan θ = |2/(−1)| he complex number lies in the second quadrant, so the principal argument is π − arctan(2).
模为 |z| = √((−1)² + 2²) = √5。辐角 θ 满足 tan θ = |2/(−1)| = 2，但该复数位于第二象限，因此主辐角为 π − arctan(2)。
On an Argand diagram, the point (−1, 2) is plotted with the distance √5 from the origin and the angle measured anticlockwise from the positive real axis.
在 Argand 图上，标出点 (−1, 2)，该点到原点的距离为 √5，辐角从正实轴逆时针测量。
2. Roots of Quadratic Equations – Forming New Equations | 二次方程的根 – 构造新方程
Question: on x² − 5x + 7 = 0 has roots α and β. Find a quadratic equation whose roots are α² and β².
题目：已知二次方程 x² − 5x + 7 = 0 的根为 α 和 β，求以 α² 和 β² 为根的二次方程。
From the original equation, sum of roots α + β = 5, product αβ = 7.
由原方程得，根的和 α + β = 5，根的积 αβ = 7。
For the new roots α² and β², the new sum is α² + β² = (α + β)² − 2αβ = 25 − 14 = 11.
对于新根 α² 和 β²，新和 α² + β² = (α + β)² − 2αβ = 25 − 14 = 11。
The new product is α²β² = (αβ)² = 49.
新积为 α²β² = (αβ)² = 49。
Required equation: x² − (sum)x + product = x² − 11x + 49 = 0
This method avoids finding α and β explicitly and relies entirely on the symmetry of roots, a core skill in FP1.
该方法避免显式求出 α 和 β，完全依赖根的对称性，这是 FP1 的核心技能之一。
3. Matrix Algebra – Multiplication and Inverse | 矩阵代数 – 乘法与逆矩阵
Question: Let A = [2, 1; 3, 4] and B = [5, −2; −1, 3]. Calculate AB and find the inverse matrix A⁻¹.
题目：设 A = [2, 1; 3, 4]，B = [5, −2; −1, 3]。计算 AB，并求逆矩阵 A⁻¹。
Matrix multiplication is performed by taking rows of the first matrix and columns of the second.
矩阵乘法通过将第一个矩阵的行与第二个矩阵的列对应相乘求和完成。
| 2×5 + 1×(−1) = 9 | 2×(−2) + 1×3 = −1 |
| 3×5 + 4×(−1) = 11 | 3×(−2) + 4×3 = 6 |
AB = [9, −1; 11, 6]
, the formula is A⁻¹ = (1/(ad − bc)) [d, −b; −c, a], provided the determinant is non-zero.
对于 A = [a, b; c, d]，其逆矩阵公式为 A⁻¹ = (1/(ad − bc)) [d, −b; −c, a]，前提是行列式不为零。
Here det(A) = 2×4 − 1×3 = 5.
此处的行列式 det(A) = 2×4 − 1×3 = 5。
A⁻¹ = 1/5 [4, −1; −3, 2] = [4/5, −1/5; −3/5, 2/5]
4. Series Summation – Using Standard Results | 级数求和 – 使用标准结果
Question: Find an expression in terms of n for the sum ∑r=1n r(r + 2).
题目：用 n 表示求和 ∑r=1n r(r + 2) 的表达式。
Expand the general term: r(r + 2) = r² + 2r.
展开通项：r(r + 2) = r² + 2r。
The sum splits into ∑r=1n r² + 2∑r=1n r. Use the standard formulae:
求和式拆分为 ∑r=1n r² + 2∑r=1n r。使用标准公式：
∑ r² = n(n+1)(2n+1)/6, ∑ r = ½ n(n+1)
Therefore the sum S = n(n+1)(2n+1)/6 + 2 × ½ n(n+1) = n(n+1)(2n+1)/6 + n(n+1).
因此，总和 S = n(n+1)(2n+1)/6 + 2 × ½ n(n+1) = n(n+1)(2n+1)/6 + n(n+1)。
Combine the terms by factoring n(n+1):
合并同类项，提取 n(n+1)：
S = n(n+1) [ (2n+1)/6 + 1 ] = n(n+1) (2n+1 + 6) / 6 = n(n+1)(2n+7)/6
This compact expression can be used to evaluate the sum for any positive integer n quickly.
该简洁表达式可用于快速计算任意正整数 n 的求和结果。
5. Proof by Induction – Sum of Squares | 数学归纳法证明 – 平方和公式
Question: Prove by mathematical induction that for all positive integers n, ∑r=1n r² = n(n+1)(2n+1)/6.
题目：用数学归纳法证明，对所有正整数 n，有 ∑r=1n r² = n(n+1)(2n+1)/6。
Basis step: For n = 1, LHS = 1² = 1; RHS = 1×2×3/6 = 1. True.
基础步骤：当 n = 1 时，左式 = 1² = 1；右式 = 1×2×3/6 = 1。成立。
Induction hypothesis: Assume the formula holds for n = k, i.e. ∑r=1k r² = k(k+1)(2k+1)/6.
归纳假设：假设当 n = k 时公式成立，即 ∑r=1k r² = k(k+1)(2k+1)/6。
Inductive step: Consider n = k+1: ∑r=1k+1 r² = ∑r=1k r² + (k+1)².
归纳递推：考虑 n = k+1：∑r=1k+1 r² = ∑r=1k r² + (k+1)²。
Substitute the hypothesis: = k(k+1)(2k+1)/6 + (k+1)².
代入归纳假设：= k(k+1)(2k+1)/6 + (k+1)²。
= (k+1) [ k(2k+1)/6 + (k+1) ] = (k+1) [ (2k² + k)/6 + (6k+6)/6 ] = (k+1)(2k² + 7k + 6)/6.
提取公因式 (k+1)：= (k+1) [ k(2k+1)/6 + (k+1) ] = (k+1) [ (2k² + k)/6 + (6k+6)/6 ] = (k+1)(2k² + 7k + 6)/6。
The quadratic 2k² + 7k + 6 factorises to (k+2)(2k+3).
二次式 2k² + 7k + 6 可分解为 (k+2)(2k+3)。
Result: (k+1)(k+2)(2(k+1)+1)/6
This matches the formula with n = k+1, completing the proof by induction.
这与 n = k+1 时的公式一致，至此归纳法证明完成。
6. Numerical Methods – Linear Interpolation | 数值方法 – 线性插值
Question: s a root in the interval [1, 2]. Use linear interpolation to find an approximation to this root, performing three iterations.
题目：f(x) = x³ − x − 1。方程 f(x) = 0 在区间 [1, 2] 内有一个根。使用线性插值求该根的近似值，进行三次迭代。
te f(1) = 1 − 1 − 1 = −1, f(2) = 8 − 2 − 1 = 5. There is a sign change, confirming a root.
首先计算 f(1) = 1 − 1 − 1 = −1，f(2) = 8 − 2 − 1 = 5。符号改变，确认存在一个根。
The linear interpolation formula is xnew = a − (b − a)f(a)/(f(b) − f(a)).
线性插值公式为 x_新 = a − (b − a)f(a)/(f(b) − f(a))。
Iteration 1: 6 ≈ 1.1667. f(x₁) ≈ (1.1667)³ − 1.1667 − 1 ≈ −0.5787 (negative). Root lies in [1.1667, 2].
第一次迭代：a=1, b=2。x₁ = 1 − (1)(−1)/(5 − (−1)) = 1 + 1/6 ≈ 1.1667。f(x₁) ≈ (1.1667)³ − 1.1667 − 1 ≈ −0.5787（负）。根位于 [1.1667, 2]。
Iteration 2: 0.5787)) ≈ 1.1667 + 0.0866 = 1.2533. f(x₂) ≈ −0.2853 (negative). Root lies in [1.2533, 2].
第二次迭代：a=1.1667, b=2。x₂ = 1.1667 − (0.8333)(−0.5787)/(5 − (−0.5787)) ≈ 1.1667 + 0.0866 = 1.2533。f(x₂) ≈ −0.2853（负）。根位于 [1.2533, 2]。
Iteration 3: .0403 = 1.2936. f(x₃) is still negative (≈ −0.126), so the root is slightly beyond 1.2936.
第三次迭代：a=1.2533, b=2。x₃ = 1.2533 − (0.7467)(−0.2853)/(5 − (−0.2853)) ≈ 1.2533 + 0.0403 = 1.2936。f(x₃) 仍为负 (≈ −0.126)，因此根略大于 1.2936。
After three iterations the approximate root is 1.294 (rounded). Further iterations would home in on the true value near 1.3247.
三次迭代后，根的近似值为 1.294（四舍五入）。进一步迭代将逼近真实值约 1.3247。
7. Complex Locus – Cartesian Equation from Modulus Condition | 复数轨迹 – 由模条件求直角坐标方程
Question: The complex number z satisfies |z − 3i| = 2|z + i|. Find the Cartesian equation of the locus of z and describe the curve geometrically.
题目：复数 z 满足 |z − 3i| = 2|z + i|。求 z 的轨迹的直角坐标方程，并描述该曲线的几何特征。
Let z = x + yi. Then |x + (y−3)i| = 2|x + (y+1)i|.
设 z = x + yi。则 |x + (y−3)i| = 2|x + (y+1)i|。
Square both sides to eliminate the modulus signs: x² + (y−3)² = 4[ x² + (y+1)² ].
两边平方消去模：x² + (y−3)² = 4[ x² + (y+1)² ]。
Expand: x² + y² − 6y + 9 = 4x² + 4(y² + 2y + 1) = 4x² + 4y² + 8y + 4.
展开：x² + y² − 6y + 9 = 4x² + 4y² + 8y + 4。
Bring all terms to one side: 0 = 3x² + 3y² + 14y − 5.
将所有项移到一边：0 = 3x² + 3y² + 14y − 5。
and complete the square: x² + y² + (14/3)y − 5/3 = 0 → x² + (y + 7/3)² − 49/9 − 15/9 = 0.
除以 3 并配方：x² + y² + (14/3)y − 5/3 = 0 → x² + (y + 7/3)² − 49/9 − 15/9 = 0。
x² + (y + 7/3)² = 64/9
This is a circle with centre (0, −7/3) and radius 8/3. The locus is a circle in the Argand plane.
这是一个圆，圆心为 (0, −7/3)，半径为 8/3。该轨迹是 Argand 平面上的一个圆。
8. Matrix Transformations – Image and Geometric Description | 矩阵变换 – 像
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