Year 12 AQA Further Mathematics: Case Study Practice | AQA 进阶数学 Year 12：案例分析实战演练

📚 Year 12 AQA Further Mathematics: Case Study Practice | AQA 进阶数学 Year 12：案例分析实战演练

This article presents a series of carefully selected case studies that mirror the style and depth of AQA Year 12 Further Mathematics exam questions. Each section focuses on a core topic, showing step‑by‑step reasoning, just as you would need to demonstrate in an assessment. The worked examples build the analytical skills and mathematical fluency required for high marks.

本文呈现一系列精心挑选的案例分析，模拟 AQA Year 12 进阶数学考试的题型与深度。每个部分聚焦一个核心主题，展示逐步推理的过程，正如你在考试中需要展现的那样。这些例题有助于培养获得高分所需的分析能力与数学流畅度。

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1. Complex Roots of Quadratic Equations | 二次方程的复数根

uation z² − 6z + 34 = 0 over the complex numbers and express the roots in the form a ± bi.

在复数范围内求解方程 z² − 6z + 34 = 0，并将根表示为 a ± bi 的形式。

Using the quadratic formula, z = [6 ± √36 − 136] / 2 = [6 ± √−100] / 2 = [6 ± 10i] / 2, giving z = 3 ± 5i. Notice the conjugate pair.

使用求根公式，z = [6 ± √36 − 136] / 2 = [6 ± √−100] / 2 = [6 ± 10i] / 2，得 z = 3 ± 5i。注意共轭对的出现。

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2. Matrix Transformations in the Plane | 平面上的矩阵变换

d the image of the point (4, 1) and verify that the distance from the origin is preserved.

设矩阵 M 表示绕原点逆时针旋转 90°。已知 M = [[0, -1], [1, 0]]，求点 (4, 1) 的像并验证其到原点的距离不变。

e column vector [4; 1] gives [0×4 + (-1)×1; 1×4 + 0×1] = [-1; 4]. The original distance is √4²+1²=√17; the image distance is √((-1)²+4²)=√17, confirming an isometry.

将 M 乘以列向量 [4; 1] 得 [0×4 + (-1)×1; 1×4 + 0×1] = [-1; 4]。原点到原点的距离为 √4²+1²=√17；像的距离为 √((-1)²+4²)=√17，验证了等距变换。

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3. Roots of Polynomials and Vieta's Formulas | 多项式根与韦达定理

as roots α, β, γ. Without solving the equation, find the value of α² + β² + γ² and αβ + βγ + γα.

三次方程 x³ − 7x² + 14x − 8 = 0 的三个根为 α, β, γ。不解方程，求 α² + β² + γ² 和 αβ + βγ + γα 的值。

+β²+γ² = (Σα)² − 2Σαβ = 49 − 28 = 21. The sum of products in pairs is already given as 14.

由韦达定理：Σα = 7, Σαβ = 14, αβγ = 8。于是 α²+β²+γ² = (Σα)² − 2Σαβ = 49 − 28 = 21。两两乘积之和已给出为 14。

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4. Summation of Series by Induction | 用归纳法求级数和

rove by induction that for all positive integers n, Σ (from 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。

= (k+1)(k+2)(2k+3)/6, which is the RHS with n=k+1. Hence, by induction the formula holds.

基础情况 n=1：左边=1²=1，右边=1×2×3/6=1。假设 n=k 成立。则 n=k+1 时，左边 = Σ r² (到 k+1) = k(k+1)(2k+1)/6 + (k+1)² = (k+1)[k(2k+1)/6 + (k+1)] = (k+1)(k+2)(2k+3)/6，这正是 n=k+1 时的右边。因此，由归纳法公式成立。

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5. Hyperbolic Identities Proofs | 双曲函数恒等式证明

Starting from the definitions cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ − e⁻ˣ)/2, prove that cosh² x − sinh² x ≡ 1.

由定义 cosh x = (eˣ + e⁻ˣ)/2 和 sinh x = (eˣ − e⁻ˣ)/2 出发，证明 cosh² x − sinh² x ≡ 1。

Compute cosh² x = (e²ˣ + 2 + e⁻²ˣ)/4 and sinh² x = (e²ˣ − 2 + e⁻²ˣ)/4. Subtracting gives [(e²ˣ + 2 + e⁻²ˣ) − (e²ˣ − 2 + e⁻²ˣ)] / 4 = (4)/4 = 1. This identity is analogous to the trigonometric Pythagorean identity.

计算 cosh² x = (e²ˣ + 2 + e⁻²ˣ)/4，sinh² x = (e²ˣ − 2 + e⁻²ˣ)/4。相减得 [(e²ˣ + 2 + e⁻²ˣ) − (e²ˣ − 2 + e⁻²ˣ)] / 4 = 4/4 = 1。这一恒等式类似于三角恒等式中的勾股恒等式。

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6. Polar Coordinates and Area Calculation | 极坐标与面积计算

Find the area enclosed by the polar curve r = 2 + cos θ for 0 ≤ θ ≤ 2π.

求极坐标曲线 r = 2 + cos θ 在 0 ≤ θ ≤ 2π 内所围成的面积。

s θ + 0.5cos 2θ) dθ. Integrating term by term over a full period, the cosine terms vanish, leaving A = ½ × 4.5 × 2π = 4.5π ≈ 14.14.

面积为 A = ½ ∫₀²π (2 + cos θ)² dθ = ½ ∫₀²π (4 + 4cos θ + cos²θ) dθ。利用 cos²θ = (1+cos 2θ)/2。则 A = ½ ∫₀²π [4 + 4cos θ + ½ + ½cos 2θ] dθ = ½ ∫₀²π (4.5 + 4cos θ + 0.5cos 2θ) dθ。逐项积分，在一个完整周期内余弦项积分为零，得 A = ½ × 4.5 × 2π = 4.5π ≈ 14.14。

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7. First-Order Differential Equations | 一阶微分方程

Solve the differential equation dy/dx + 2y = 4e⁻ˣ, with the initial condition y(0) = 3. Interpret the long‑term behaviour.

求解微分方程 dy/dx + 2y = 4e⁻ˣ，满足初始条件 y(0) = 3，并解释长期行为。

The integrating factor is e^{∫2 dx} = e²ˣ. Multiply the equation: e²ˣ dy/dx + 2e²ˣ y = 4eˣ. The left side becomes d/dx (e²ˣ y). Integrate: e²ˣ y = ∫4eˣ dx = 4eˣ + C. Thus y = 4e⁻ˣ + Ce⁻²ˣ. Using y(0)=3 gives 3 = 4 + C ⇒ C = −1. So y = 4e⁻ˣ − e⁻²ˣ. As x → ∞, both terms decay to 0, so y → 0.

积分因子为 e^{∫2 dx} = e²ˣ。方程两边同乘：e²ˣ dy/dx + 2e²ˣ y = 4eˣ。左边成为 d/dx (e²ˣ y)。积分：e²ˣ y = ∫4eˣ dx = 4eˣ + C。因此 y = 4e⁻ˣ + Ce⁻²ˣ。利用 y(0)=3 得 3 = 4 + C ⇒ C = −1。故 y = 4e⁻ˣ − e⁻²ˣ。当 x → ∞ 时，两项均衰减至 0，因此 y → 0。

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8. Geometric Applications of Complex Numbers | 复数在几何中的应用

The points A, B, C in the complex plane represent the numbers 1+2i, 4+5i, and 7+2i respectively. Show that triangle ABC is right‑angled and find its area.

复平面上的点 A、B、C 分别表示复数 1+2i，4+5i，7+2i。证明三角形 ABC 为直角三角形并求其面积。

(3,3)·(6,0) = 18+0=18. However, check BC = 3−3i. We find AB·BC = (3)(3)+(3)(−3)=9−9=0, so AB ⟂ BC, meaning ∠B = 90°. The lengths: |AB| = √3²+3²=√18=3√2, |BC| = √(3²+(−3)²)=3√2. Area = ½ × 3√2 × 3√2 = 9.

计算向量：AB = (4+5i) − (1+2i) = 3+3i；AC = (7+2i) − (1+2i) = 6。实数点积对应：检查 AB·BC = (3)(3)+(3)(−3)=0，故 AB ⟂ BC，∠B = 90°。长度：|AB|=√3²+3²=√18=3√2，|BC|=√(3²+(−3)²)=3√2。面积 = ½ × 3√2 × 3√2 = 9。

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9. Solving Matrix Equations | 求解矩阵方程

trices A = [[2, 1], [1, 2]] and B = [[1, -1], [0, 3]], find the matrix X such that AX = B.

已知矩阵 A = [[2, 1], [1, 2]]，B = [[1, -1], [0, 3]]，求满足 AX = B 的矩阵 X。

×3]] = (1/3)[[2, -2−3], [-1, 1+6]] = (1/3)[[2, -5], [-1, 7]] = [[2/3, -5/3], [-1/3, 7/3]].

先求 A⁻¹。det(A)=4−1=3。因此 A⁻¹ = (1/3)[[2, -1], [-1, 2]]。于是 2×3]] = (1/3)[[2, -2−3], [-1, 1+6]] = (1/3)[[2, -5], [-1, 7]] = [[2/3, -5/3], [-1/3, 7/3]]。

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10. Further Induction: Divisibility | 进阶归纳法：整除性

Prove that 7ⁿ − 1 is divisible by 6 for all positive integers n using mathematical induction.

用数学归纳法证明对所有正整数 n，7ⁿ − 1 可被 6 整除。

− 1 = 42m + 6 = 6(7m+1), which is a multiple of 6. Hence the statement holds by induction.

基础情况 n=1：7¹ − 1 = 6，能被 6 整除。假设 n=k 时成立，即 7ᵏ − 1 = 6m。对于 n=k+1，7ᵏ⁺¹ − 1 = 7·7ᵏ − 1 = 7(6m+1) − 1 = 42m + 7 − 1 = 42m + 6 = 6(7m+1)，为 6 的倍数。因此由归纳法命题成立。

Published by TutorHao | Further Mathematics Revision Series | aleveler.com

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