📚 AS CCEA Further Mathematics: Case Study Analysis & Practice | AS CCEA 进阶数学：案例分析实战演练
This article presents a series of worked case studies covering the core topics of the AS CCEA Further Mathematics specification. Each case study is designed to reinforce key concepts and problem-solving techniques, with detailed solutions provided in a bilingual format. By working through these examples, students can deepen their understanding and gain confidence for the examination.
本文通过一系列案例分析，覆盖了AS CCEA进阶数学大纲的核心主题。每个案例旨在巩固关键概念和解题技巧，并以双语形式提供详细解答。通过练习这些例题，学生可以加深理解，增强考试信心。
1. Complex Numbers in Polar Form and De Moivre’s Theorem | 复数极坐标形式与德莫弗定理
Case Study 1: Express z = 1 + i√3 in modulus-argument form, then use de Moivre’s theorem to compute z⁵ in Cartesian form a + ib.
案例1：将 z = 1 + i√3 写成模-辐角形式，然后利用德莫弗定理计算 z⁵ 的笛卡尔形式 a + ib。
Step 1 – Modulus: r = |z| = √(1² + (√3)²) = √(1 + 3) = 2.
步骤1 – 模：r = |z| = √(1² + (√3)²) = √(1 + 3) = 2。
Step 2 – Argument: θ = tan⁻¹(√3/1) = π/3. Thus z = 2(cos(π/3) + i sin(π/3)).
步骤2 – 辐角：θ = tan⁻¹(√3/1) = π/3。所以 z = 2(cos(π/3) + i sin(π/3))。
Step 3 – Apply de Moivre’s theorem: z⁵ = 2⁵ [cos(5·π/3) + i sin(5·π/3)] = 32 [cos(5π/3) + i sin(5π/3)].
步骤3 – 应用德莫弗定理：z⁵ = 2⁵ [cos(5·π/3) + i sin(5·π/3)] = 32 [cos(5π/3) + i sin(5π/3)]。
Step 4 – Evaluate: cos(5π/3) = cos(300°) = 1/2, sin(5π/3) = -√3/2. Therefore z⁵ = 32(1/2 – i√3/2) = 16 – 16i√3.
步骤4 – 计算：cos(5π/3) = cos(300°) = 1/2，sin(5π/3) = -√3/2。因此 z⁵ = 32(1/2 – i√3/2) = 16 – 16i√3。
Key point: Always check the quadrant of the argument; here the complex number lies in the first quadrant, so θ is positive acute.
关键点：务必检查辐角象限；这里复数位于第一象限，因此 θ 为正锐角。
2. Matrix Inversion and Solving a Linear System | 矩阵求逆与解线性方程组
Case Study 2: A system of equations is given: 2x + y – z = 3, x – y + 2z = 1, 3x + 2y + z = 4. Express the system in matrix form AX = B, find the inverse of A and hence solve for x, y, z.
案例2：给定方程组：2x + y – z = 3，x – y + 2z = 1，3x + 2y + z = 4。将方程组写成矩阵形式 AX = B，求出 A 的逆矩阵，并由此解出 x, y, z。
Matrix A = [[2, 1, -1], [1, -1, 2], [3, 2, 1]], X = [x, y, z]ᵀ, B = [3, 1, 4]ᵀ.
矩阵 A = [[2, 1, -1], [1, -1, 2], [3, 2, 1]]，X = [x, y, z]ᵀ，B = [3, 1, 4]ᵀ。
Determinant of A: det(A) = 2(-1·1 – 2·2) – 1(1·1 – 2·3) + (-1)(1·2 – (-1)·3) = 2(-1 – 4) – 1(1 – 6) – 1(2 + 3) = 2(-5) – 1(-5) – 1(5) = -10 + 5 – 5 = -10.
A 的行列式：det(A) = 2(-1·1 – 2·2) – 1(1·1 – 2·3) + (-1)(1·2 – (-1)·3) = 2(-1 – 4) – 1(1 – 6) – 1(2 + 3) = 2(-5) – 1(-5) – 1(5) = -10 + 5 – 5 = -10。
Matrix of cofactors and adjugate lead to A⁻¹ = (1/det(A)) adj(A). After computation we obtain A⁻¹ = [[-0.5, -0.3, 0.1], [0.5, 0.5, -0.5], [0.5, -0.1, -0.3]]. (Decimal representation for clarity; exact fractions available.)
伴随矩阵与余子式矩阵给出 A⁻¹ = (1/det(A)) adj(A)。计算后得到 A⁻¹ = [[-0.5, -0.3, 0.1], [0.5, 0.5, -0.5], [0.5, -0.1, -0.3]]。（为清晰使用小数；精确分数可自行转换。）
Solution: X = A⁻¹ B = [[-0.5, -0.3, 0.1], [0.5, 0.5, -0.5], [0.5, -0.1, -0.3]] [3, 1, 4]ᵀ = [ -1.5 – 0.3 + 0.4, 1.5 + 0.5 – 2, 1.5 – 0.1 – 1.2 ]ᵀ = [-1.4, 0, 0.2]ᵀ. Hence x = -1.4, y = 0, z = 0.2.
解：X = A⁻¹ B = [[-0.5, -0.3, 0.1], [0.5, 0.5, -0.5], [0.5, -0.1, -0.3]] [3, 1, 4]ᵀ = [ -1.5 – 0.3 + 0.4, 1.5 + 0.5 – 2, 1.5 – 0.1 – 1.2 ]ᵀ = [-1.4, 0, 0.2]ᵀ。因此 x = -1.4，y = 0，z = 0.2。
Verification: 2(-1.4)+0-0.2 = -2.8-0.2=-3 ≠3? Wait, check arithmetic: actually 2(-1.4)+0 -0.2 = -2.8-0.2 = -3, but RHS is 3, so error. Let’s recompute carefully. (This illustrates the importance of checking; the exact fractional inverse is A⁻¹ = [[-0.5, -0.3, 0.1] correctly? Let’s recompute.) The correct exact inverse: A⁻¹ = (1/-10) * [[ -5, -3, 1 ], [5, 5, -5], [5, -1, -3]] = [[0.5, 0.3, -0.1], [-0.5, -0.5, 0.5], [-0.5, 0.1, 0.3]]. Then X = A⁻¹ B = [0.5*3+0.3*1-0.1*4 = 1.5+0.3-0.4=1.4; -0.5*3 -0.5*1+0.5*4 = -1.5-0.5+2=0; -0.5*3+0.1*1+0.3*4 = -1.5+0.1+1.2=-0.2]. So x=1.4, y=0, z=-0.2. Check: 2(1.4)+0-(-0.2)=2.8+0.2=3, correct. So correct solution x=1.4, y=0, z=-0.2.
核对：正确的逆矩阵为 A⁻¹ = (1/-10) * [[-5, -3, 1], [5, 5, -5], [5, -1, -3]]，因此 x=1.4, y=0, z=-0.2。代入验证正确。展示了求逆时务必仔细计算行列式与伴随矩阵。
3. Summation of Series (∑r² and ∑r³) | 级数求和（∑r² 与 ∑r³）
Case Study 3: Evaluate ∑ⁿᵣ₌₁ (r+1)(2r-3) and express it as a polynomial in n. Use standard formulae for ∑r, ∑r².
案例3：计算 ∑ⁿᵣ₌₁ (r+1)(2r-3)，并将其表示为 n 的多项式。使用标准公式 ∑r、∑r²。
First expand the term: (r+1)(2r-3) = 2r² – 3r + 2r – 3 = 2r² – r – 3.
首先展开通项：(r+1)(2r-3) = 2r² – 3r + 2r – 3 = 2r² – r – 3。
Sum from r=1 to n: S = 2∑r² – ∑r – ∑3 = 2∑r² – ∑r – 3n.
对 r=1 到 n 求和：S = 2∑r² – ∑r – ∑3 = 2∑r² – ∑r – 3n。
Recall ∑r = n(n+1)/2, ∑r² = n(n+1)(2n+1)/6.
回忆 ∑r = n(n+1)/2，∑r² = n(n+1)(2n+1)/6。
S = 2 [n(n+1)(2n+1)/6] – [n(n+1)/2] – 3n = [n(n+1)(2n+1)/3] – [n(n+1)/2] – 3n.
S = 2 [n(n+1)(2n+1)/6] – [n(n+1)/2] – 3n = [n(n+1)(2n+1)/3] – [n(n+1)/2] – 3n。
Get common denominator 6: = [2n(n+1)(2n+1) – 3n(n+1) – 18n] / 6. Factor n: = n [2(n+1)(2n+1) – 3(n+1) – 18] /6.
取公分母 6：= [2n(n+1)(2n+1) – 3n(n+1) – 18n] / 6。提取因子 n：= n [2(n+1)(2n+1) – 3(n+1) – 18] /6。
Simplify inside: 2(2n²+3n+1) – 3n – 3 – 18 = 4n²+6n+2 -3n -21 = 4n²+3n -19. So S = n(4n²+3n-19)/6.
化简内部：2(2n²+3n+1) – 3n – 3 – 18 = 4n²+6n+2 -3n -21 = 4n²+3n -19。所以 S = n(4n²+3n-19)/6。
Check for n=1: term = (1+1)(2-3)=2*(-1)=-2. Formula gives 1*(4+3-19)/6 = ( -12)/6 = -2. Correct.
检验 n=1：通项 = (1+1)(2-3)=2*(-1)=-2。公式给出 1*(4+3-19)/6 = -12/6 = -2。正确。
4. Implicit Differentiation and Parametric Equations | 隐函数微分与参数方程
Case Study 4: The curve C has parametric equations x = t² + 2, y = 4t – t³. Find dy/dx in terms of t, and the equation of the tangent at t = 1.
案例4：曲线 C 的参数方程为 x = t² + 2，y = 4t – t³。求用 t 表示的 dy/dx，并求 t = 1 处的切线方程。
Use chain rule: dy/dx = (dy/dt) / (dx/dt). dx/dt = 2t, dy/dt = 4 – 3t². So dy/dx = (4 – 3t²) / (2t).
使用链式法则：dy/dx = (dy/dt) / (dx/dt)。dx/dt = 2t，dy/dt = 4 – 3t²。因此 dy/dx = (4 – 3t²) / (2t)。
At t = 1: dy/dx = (4 – 3)/2 = 1/2. Point coordinates: x = 1²+2=3, y = 4 – 1 = 3.
当 t = 1：dy/dx = (4 – 3)/2 = 1/2。点坐标：x = 1²+2=3，y = 4 – 1 = 3。
Equation of tangent: y – 3 = (1/2)(x – 3) => y = (1/2)x + 3 – 3/2 = (1/2)x + 3/2.
切线方程：y – 3 = (1/2)(x – 3) => y = (1/2)x + 3/2。
For implicit differentiation: consider 3x² + 2xy – y² = 4. Differentiate both sides w.r.t x: 6x + 2y + 2x dy/dx – 2y dy/dx = 0. Rearr: dy/dx = -(6x+2y)/(2x-2y) = -(3x+y)/(x-y).
对于隐函数微分，例如 3x² + 2xy – y² = 4。两边对 x 求导：6x + 2y + 2x dy/dx – 2y dy/dx = 0。整理得 dy/dx = -(6x+2y)/(2x-2y) = -(3x+y)/(x-y)。
5. Vector Dot and Cross Products with Applications | 向量的点乘与叉乘及其应用
Case Study 5: Given vectors a = i + 2j – k, b = -i + j + 2k. Find a·b, a×b, and the angle between a and b. Also find a unit vector perpendicular to both a and b.
案例5：已知向量 a = i + 2j – k，b = -i + j + 2k。求 a·b，a×b，以及 a 与 b 的夹角。并求出与 a 和 b 均垂直的单位向量。
Dot product: a·b = (1)(-1) + (2)(1) + (-1)(2) = -1 + 2 – 2 = -1.
点乘：a·b = (1)(-1) + (2)(1) + (-1)(2) = -1 + 2 – 2 = -1。
Cross product: a×b = det |i j k; 1 2 -1; -1 1 2| = i(2·2 – (-1)·1) – j(1·2 – (-1)·(-1)) + k(1·1 – 2·(-1)) = i(4+1) – j(2 – 1) + k(1 + 2) = 5i – j + 3k.
叉乘：a×b = det |i j k; 1 2 -1; -1 1 2| = i(2·2 – (-1)·1) – j(1·2 – (-1)·(-1)) + k(1·1 – 2·(-1)) = i(4+1) – j(2 – 1) + k(1 + 2) = 5i – j + 3k。
Angle θ: cosθ = (a·b)/(|a||b|). |a| = √(1+4+1)=√6, |b| = √(1+1+4)=√6. cosθ = -1/6, θ = arccos(-1/6) ≈ 99.6°.
夹角 θ：cosθ = (a·b)/(|a||b|)。|a| = √(1+4+1)=√6，|b| = √(1+1+4)=√6。cosθ = -1/6，θ = arccos(-1/6) ≈ 99.6°。
Unit vector perpendicular: n = (a×b)/|a×b|. |a×b| = √(25+1+9)=√35. So unit vector = (5i – j + 3k)/√35.
垂直单位向量：n = (a×b)/|a×b|。|a×b| = √(25+1+9)=√35。所以单位向量 = (5i – j + 3k)/√35。
6. Binomial Distribution and Poisson Approximation | 二项分布与泊松近似
Case Study 6: A factory produces components with a 2% defect rate. A random sample of 150 components is taken. Use a Poisson approximation to find the probability of observing at most 2 defectives.
案例6：某工厂生产的产品有 2% 的缺陷率。随机抽取 150 个样品。使用泊松近似计算最多观察到 2 个次品的概率。
Binomial: n=150, p=0.02. Mean λ = np = 3. Poisson approximation is valid as n large, p small.
二项分布：n=150，p=0.02。均值 λ = np = 3。n 大 p 小，泊松近似适用。
X ~ Po(3). P(X ≤ 2) = P(X=0)+P(X=1)+P(X=2) = e⁻³(1 + 3 + 3²/2!) = e⁻³(1+3+4.5) = e⁻³ × 8.5. e⁻³ ≈ 0.0498, so probability ≈ 0.423.
X ~ Po(3)。P(X ≤ 2) = P(X=0)+P(X=1)+P(X=2) = e⁻³(1 + 3 + 3²/2!) = e⁻³(1+3+4.5) = e⁻³ × 8.5。e⁻³ ≈ 0.0498，概率 ≈ 0.423。
Exact binomial: P(X ≤ 2) = (0.98)¹⁵⁰ + 150·0.02·(0.98)¹⁴⁹ + C(150,2)·(0.02)²·(0.98)¹⁴⁸ ≈ 0.422, very close.
精确二项计算：P(X ≤ 2) ≈ 0.422，与近似值非常接近。
Use of conditions: np < 5 or n > 50, p < 0.1 is typical for Poisson approximation.
使用条件：np < 5 或 n > 50，p < 0.1 是泊松近似的典型条件。
7. Projectile Motion in Mechanics | 力学中的抛射体运动
Case Study 7: A particle is projected from ground level with speed 20 m/s at an angle 30° above the horizontal. Find the time of flight, maximum height, and horizontal range. (Take g = 9.8 m/s²)
案例7：一质点从地面以 20 m/s 的速率、与水平面成 30° 角发射。求飞行时间、最大高度和水平射程。（取 g = 9.8 m/s²）
Resolve velocity: uₓ = 20 cos30° = 10√3 ≈ 17.32 m/s, uᵧ = 20 sin30° = 10 m/s.
分解速度：uₓ = 20 cos30° = 10√3 ≈ 17.32 m/s，uᵧ = 20 sin30° = 10 m/s。
Time of flight: t_flight = 2uᵧ/g = 20/9.8 ≈ 2.04 s.
飞行时间：t_flight = 2uᵧ/g = 20/9.8 ≈ 2.04 s。
Maximum height: H = uᵧ²/(2g) = 100/(19.6) ≈ 5.10 m.
最大高度：H = uᵧ²/(2g) = 100/(19.6) ≈ 5.10 m。
Range: R = uₓ × t_flight = 17.32 × 2.04 ≈ 35.33 m. Or use formula R = u² sin(60°)/g = 400 × (√3/2)/9.8 ≈ 35.33 m.
射程：R = uₓ × 飞行时间 = 17.32 × 2.04 ≈ 35.33 m，或用公式 R = u² sin(60°)/g = 400 × (√3/2)/9.8 ≈ 35.33 m。
Trajectory equation: y = x tanθ – (g x²)/(2u² cos²θ). Confirm at x=R, y=0.
轨迹方程：y = x tanθ – (g x²)/(2u² cos²θ)。验证当 x=R 时 y=0。
8. Hypothesis Testing (One-Tailed Binomial Test) | 假设检验（二项单尾检验）
Case Study 8: A die is believed to be biased towards 6. In 50 rolls, 14 sixes appear. Test at the 5% significance level whether there is evidence of bias. Use binomial distribution with p=1/6 under H₀.
案例8：怀疑一枚骰子偏向于 6 点。在 50 次投掷中，出现 14 次 6 点。以 5% 显著性水平检验是否有偏差的证据。零假设下 X~B(50, 1/6)。
H₀: p = 1/6; H₁: p > 1/6 (one-tailed). Test statistic: X = number of sixes.
H₀: p = 1/6；H₁: p > 1/6（单尾）。检验统计量：X = 出现 6 的次数。
Under H₀, X ~ B(50, 1/6). Find P(X ≥ 14). Use normal approximation or exact binomial. Exact: P(X ≥ 14) = 1 – P(X ≤ 13). Using tables or calculator: P(X ≤ 13) ≈ 0.9682, so P(X ≥ 14) ≈ 0.0318.
在 H₀ 下，X ~ B(50, 1/6)。求 P(X ≥ 14)。精确计算：P(X ≤ 13) ≈ 0.9682，因此 P(X ≥ 14) ≈ 0.0318。
Since 0.0318 < 0.05, we reject H₀. There is sufficient evidence at 5% level to suggest the die is biased towards 6.
由于 0.0318 < 0.05，拒绝 H₀。在 5% 的显著性水平下有充分证据表明骰子偏向 6 点。
Check critical region: For one-tailed 5%, we need smallest x such that P(X ≥ x) ≤ 0.05. Here x=14 gives 0.0318, so critical region is X ≥ 14.
检查临界域：单尾 5% 下，需要最小的 x 使得 P(X ≥ x) ≤ 0.05。此处 x=14 得到 0.0318，因此临界域为 X ≥ 14。
9. Separable Differential Equations | 可分离变量的微分方程
Case Study
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