Year 13 CCEA Further Mathematics: Unit Test Mock Paper Solutions | 英国 CCEA 考试局 Year 13 进阶数学单元测试模拟卷解析

📚 Year 13 CCEA Further Mathematics: Unit Test Mock Paper Solutions | 英国 CCEA 考试局 Year 13 进阶数学单元测试模拟卷解析

This article presents detailed step-by-step solutions to a mock unit test designed for CCEA Year 13 Further Mathematics. The content mirrors a typical Further Pure Mathematics 1 (FPM1) assessment, covering complex numbers, matrices, roots of polynomials, series summation, proof by induction, hyperbolic functions, matrix transformations and inequalities. Each solution is accompanied by relevant tips and common pitfalls to help you consolidate key skills for the AS-level unit test.

本文为 CCEA Year 13 进阶数学单元模拟测试提供详细的逐步解法。内容模拟典型的进阶纯数一 (FPM1) 试题，涵盖复数、矩阵、多项式的根、级数求和、归纳法证明、双曲函数、矩阵变换及不等式。每个解法都附有相关提示和常见错误，帮助你巩固 AS 单元考试所需的关键技能。

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1. Complex Numbers: Solving a Cubic Equation | 复数：解三次方程

oots exactly in polar form r(cos θ + i sin θ). This is a standard application of de Moivre's theorem for roots of a complex number.

题目要求：解方程 z³ = 8i，并用精确的极坐标形式 r(cos θ + i sin θ) 给出所有根。这是棣莫弗定理对复数根的标准应用。

First, express 8i in modulus-argument form. The modulus is 8 and the argument is π/2, so we write 8(cos π/2 + i sin π/2). For the cube roots, the modulus of z will be 8^1/3 = 2. The argument becomes (π/2 + 2kπ)/3 for k = 0, 1, 2.

首先，将 8i 表示为模-幅角形式。模为 8，幅角为 π/2，所以写成 8(cos π/2 + i sin π/2)。对于三次根，z 的模将是 8^1/3 = 2。幅角分别为 (π/2 + 2kπ)/3，其中 k = 0, 1, 2。

z = 2[cos((π/2 + 2kπ)/3) + i sin((π/2 + 2kπ)/3)] for k = 0, 1, 2

Now evaluate each root:

现在计算每个根：

• k = 0: z₀ = 2(cos π/6 + i sin π/6) = 2(√3/2 + ½ i) = √3 + i

  当 k=0 时：z₀ = 2(cos π/6 + i sin π/6) = 2(√3/2 + ½ i) = √3 + i

• k = 1: z₁ = 2(cos 5π/6 + i sin 5π/6) = 2(-√3/2 + ½ i) = -√3 + i

  当 k=1 时：z₁ = 2(cos 5π/6 + i sin 5π/6) = 2(-√3/2 + ½ i) = -√3 + i

• k = 2: z₂ = 2(cos 3π/2 + i sin 3π/2) = 2(0 - i) = -2i

  当 k=2 时：z₂ = 2(cos 3π/2 + i sin 3π/2) = 2(0 - i) = -2i

A common mistake is forgetting to add 2kπ before dividing by 3, which would give only one root. Always remember that an nth-degree equation must have exactly n roots in the complex plane.

一个常见错误是在除以 3 之前忘记加上 2kπ，这会导致只得到一个根。务必记住 n 次方程在复平面内恰好有 n 个根。

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2. Matrices: Inverse and Solving a Linear System | 矩阵：逆矩阵与解线性方程组

= 7. This tests the core technique of representing simultaneous equations in matrix form.

题目给出矩阵 A = [[2,1],[1,3]]，要求求 A⁻¹，再运用其解方程组 2x + y = 5, x + 3y = 7。这考察了用矩阵形式表示联立方程的核心方法。

inverse exists. The inverse of a 2×2 matrix [[a,b],[c,d]] is (1/det)[[d, -b],[-c, a]], so:

首先计算行列式：det A = 2×3 - 1×1 = 6 - 1 = 5。由于行列式不为零，逆矩阵存在。2×2 矩阵 [[a,b],[c,d]] 的逆为 (1/det)[[d, -b],[-c, a]]，因此：

A⁻¹ = 1/5 [[3, -1],[-1, 2]]

rite the system as A[x; y] = [5; 7]. Then multiply both sides by A⁻¹: [x; y] = A⁻¹ [5; 7].

将方程组写为 A[x; y] = [5; 7]。然后两边乘以 A⁻¹： [x; y] = A⁻¹ [5; 7]。

(-1)×7) = (1/5)(15 - 7) = 8/5 for x, and (1/5)((-1)×5 + 2×7) = (1/5)(-5 + 14) = 9/5 for y.

执行乘法运算：x = (1/5)(3×5 + (-1)×7) = (1/5)(15 - 7) = 8/5；y = (1/5)((-1)×5 + 2×7) = (1/5)(-5 + 14) = 9/5。

nal equations: 2×(8/5) + 9/5 = 25/5 = 5, and 8/5 + 3×(9/5) = 35/5 = 7. Both are satisfied.

务必代回原方程检验：2×(8/5) + 9/5 = 25/5 = 5，8/5 + 3×(9/5) = 35/5 = 7，均满足。

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3. Roots of Polynomials: Relationships and Symmetric Sums | 多项式的根：关系与对称和

The cubic equation x³ – 6x² + 11x – s to state the values of Σα, Σαβ and αβγ, and then find α² + β² + γ² without solving the equation.

三次方程 x³ – 6x² + 11x – 6 = 0 的根为 α, β, γ。题目要求写出 Σα、Σαβ 和 αβγ 的值，然后不用解方程求出 α² + β² + γ²。

Using Vieta' ading coefficient 1), we have Σα = -p, Σαβ = q, and αβγ = -r. Here p = -6, q = 11, r = -6.

使用三次方程 x³ + px² + qx + r = 0（首项系数为 1）的韦达定理，有 Σα = -p，Σαβ = q，αβγ = -r。这里 p = -6，q = 11，r = -6。

Σα = 6, Σαβ = 11, αβγ = 6

To find α² + β² + γ², recall the identity: α² + β² + γ² = (α+β+γ)² – 2(αβ + βγ + γα).

要求 α² + β² + γ²，回忆恒等式：α² + β² + γ² = (α+β+γ)² – 2(αβ + βγ + γα)。

Substitute the known values: (6)² – 2×11 = 36 – 22 = 14.

代入已知数值：(6)² – 2×11 = 36 – 22 = 14。

This method is much faster than finding the roots (which are 1, 2, 3) individually. Ensure the signs of the coefficients are handled correctly – a very common mistake is to confuse -p with p.

这种方法比单独求出根 (1, 2, 3) 要快得多。务必正确处理系数的符号——一个极常见的错误是混淆 -p 和 p。

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4. Summation of Series: Evaluating Σ r(r+1) | 级数求和：计算 Σ r(r+1)

erms of n. This requires knowledge of standard series formulas and algebraic manipulation.

任务是用 n 表示和式 Σ (r=1 到 n) r(r+1)。这需要掌握标准级数公式和代数运算。

Expand the general term: r(r+1) = r² + r. Therefore the sum separates into Σ r² + Σ r.

展开通项：r(r+1) = r² + r。因此求和拆分为 Σ r² + Σ r。

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

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

Add the two expressions: n(n+1)(2n+1)/6 + n(n+1)/2. Write the second term with denominator 6: 3n(n+1)/6.

将两式相加：n(n+1)(2n+1)/6 + n(n+1)/2。将第二项写成分母为 6 的分数：3n(n+1)/6。

Sum = [n(n+1)(2n+1) + 3n(n+1)] / 6 = n(n+1)[(2n+1)+3]/6 = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3

value, such as n=2: r=1→2, r=2→6, total=8; the formula gives 2×3×4/3=8, which is correct.

用一个小值检验公式会很有用，例如 n=2：r=1→2，r=2→6，总和=8；公式给出 2×3×4/3=8，正确。

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

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

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

Base case: When n=1, LHS = 1² = 1, RHS = 1×2×3/6 = 1. The statement holds for n=1.

基础情形：当 n=1 时，左边 = 1² = 1，右边 = 1×2×3/6 = 1。命题对 n=1 成立。

Inductive step: Assume the statement is true for n=k, i.e. Σ r² = k(k+1)(2k+1)/6.

归纳步骤：假设命题对 n=k 成立，即 Σ r² = k(k+1)(2k+1)/6。

Now consider the sum up to k+1: Σ (r=1 to k+1) r² = Σ (r=1 to k) r² + (k+1)².

现在考虑直到 k+1 的和：Σ (r=1 到 k+1) r² = Σ (r=1 到 k) r² + (k+1)²。

Substitute the assumption: k(k+1)(2k+1)/6 + (k+1)². Factor out (k+1): (k+1)[k(2k+1)/6 + (k+1)].

代入假设：k(k+1)(2k+1)/6 + (k+1)²。提出公因式 (k+1)：(k+1)[k(2k+1)/6 + (k+1)]。

Published by TutorHao | Year 13 进阶数学 Revision Series | aleveler.com

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