Case Study Practice in Pre-U Further Mathematics | Pre-U 进阶数学案例分析实战演练

📚 Case Study Practice in Pre-U Further Mathematics | Pre-U 进阶数学案例分析实战演练

In Pre-U Further Mathematics, the ability to dissect complex problems and apply rigorous reasoning is what distinguishes top performers. This article walks you through eight carefully chosen case studies drawn from pure and applied topics, with every step explained in both English and Chinese. Use these as templates for your own exam practice.

在Pre-U进阶数学中，拆分复杂问题并运用严谨推理的能力是高分的关键。本文精选八个来自纯数学与应用数学的案例，每一步均提供中英文双语讲解，可作为你备考训练的模板。

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1. Proof by Induction | 数学归纳法证明

Case: Prove that for all positive integers n, the sum of the squares of the first n natural numbers is ∑(r=1 to n) r² = n(n+1)(2n+1)/6.

案例： 证明对所有正整数n，前n个自然数的平方和公式为 ∑(r=1 至 n) r² = n(n+1)(2n+1)/6。

Base case (n=1): LHS = 1² = 1. RHS = 1 × 2 × 3 / 6 = 1, so the statement holds.

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

Inductive hypothesis: Assume the formula is true for some integer k ≥ 1, i.e., ∑(r=1 to k) r² = k(k+1)(2k+1)/6.

归纳假设： 假设对某个整数k≥1公式成立，即 ∑(r=1 至 k) r² = k(k+1)(2k+1)/6。

Inductive step (k → k+1): Consider the sum to k+1 terms. Then ∑(r=1 to k+1) r² = ∑(r=1 to k) r² + (k+1)². Substitute the inductive hypothesis and combine over a common denominator 6.

归纳步骤 (k → k+1)： 考虑前k+1项的和，有 ∑(r=1 至 k+1) r² = ∑(r=1 至 k) r² + (k+1)²，代入归纳假设并通分分母6。

Factorising gives (k+1)/6 [k(2k+1) + 6(k+1)] = (k+1)/6 (2k² + 7k + 6) = (k+1)(k+2)(2k+3)/6, which is exactly the formula with n = k+1. Hence, by the principle of mathematical induction, the result is true for all positive integers n.

提取公因式(k+1)/6 [k(2k+1)+6(k+1)] = (k+1)/6 (2k²+7k+6) = (k+1)(k+2)(2k+3)/6，这正是n=k+1时的公式。因此由数学归纳法原理，结论对所有正整数n成立。

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2. Complex Numbers and de Moivre’s Theorem | 复数与棣莫弗定理

Case: Using de Moivre’s theorem, express cos 3θ in terms of cos θ. Hence evaluate (1 + i√3)⁶ in the form a + ib.

案例： 利用棣莫弗定理将cos 3θ用cos θ表示，并由此计算(1 + i√3)⁶为a+ib形式。

Let z = cos θ + i sin θ. Then by de Moivre, z³ = (cos θ + i sin θ)³ = cos 3θ + i sin 3θ. Expand the left side using the binomial theorem: cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ. Equating real parts: cos 3θ = cos³θ – 3 cos θ sin²θ. Replace sin²θ by 1 – cos²θ to obtain cos 3θ = 4 cos³θ – 3 cos θ.

令z = cos θ + i sin θ。由棣莫弗定理，z³ = (cos θ + i sin θ)³ = cos 3θ + i sin 3θ。用二项式定理展开左边：cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ。比较实部得cos 3θ = cos³θ – 3 cos θ sin²θ。用sin²θ = 1 – cos²θ代入即得cos 3θ = 4 cos³θ – 3 cos θ。

For the evaluation, write 1 + i√3 in polar form. The modulus is √(1² + (√3)²) = 2, and the argument is arctan(√3/1) = π/3. Thus 1 + i√3 = 2 (cos π/3 + i sin π/3). Raising to the sixth power: (1 + i√3)⁶ = 2⁶ [cos(6 × π/3) + i sin(6 × π/3)] = 64 (cos 2π + i sin 2π) = 64(1 + 0i) = 64.

现在计算(1 + i√3)⁶。将其写为极坐标形式：模长√(1²+(√3)²)=2，辐角为arctan(√3/1)=π/3。所以1 + i√3 = 2(cos π/3 + i sin π/3)。六次方后得2⁶[cos(6×π/3) + i sin(6×π/3)] = 64(cos 2π + i sin 2π) = 64。

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3. Matrices and Linear Transformations | 矩阵与线性变换

Case: Given the matrix A = [[2, 1], [1, 2]], find its eigenvalues and corresponding eigenvectors. Describe the transformation represented by A.

案例： 已知矩阵A = [[2, 1], [1, 2]]，求其特征值及对应的特征向量，并描述A所代表的线性变换。

The characteristic equation is det(A – λI) = 0 ⇒ (2-λ)² – 1 = 0 ⇒ λ² – 4λ + 3 = 0. Solving gives eigenvalues λ₁ = 1, λ₂ = 3.

特征方程为det(A – λI)=0 ⇒ (2-λ)² – 1 = 0 ⇒ λ² – 4λ + 3 = 0。解得特征值λ₁ = 1，λ₂ = 3。

For λ = 1, solve (A – I)v = 0: [[1, 1], [1, 1]] [x; y] = [0; 0], giving x + y = 0, so an eigenvector is (1, -1) or any scalar multiple.

对于λ=1，解(A-I)v=0：[[1,1],[1,1]][x; y]=[0;0]，得x+y=0，特征向量可取(1, -1)或其倍数。

For λ = 3, (A – 3I)v = 0 gives [[-1, 1], [1, -1]] [x; y] = [0; 0], so x – y = 0, eigenvector (1, 1) etc.

对于λ=3，(A-3I)v=0得[[-1,1],[1,-1]][x; y]=[0;0]，故x-y=0，特征向量为(1, 1)等。

Geometrically, A represents a stretch by factor 3 in the direction (1,1) and a stretch by factor 1 (i.e., no change) in the orthogonal direction (1,-1). This is a shear-like scaling along two perpendicular axes.

几何上，A表示沿方向(1,1)以因子3拉长，沿与之垂直的方向(1,-1)以因子1保持不变（即无变化），相当于沿两条垂直轴的缩放。

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4. First-Order Differential Equation Modelling | 一阶微分方程建模

Case: Newton’s Law of Cooling states that the rate of change of temperature T of an object is proportional to the difference between T and the ambient temperature Tₐ = 20°C. A hot object at 100°C is placed in a room at 20°C, and after 5 minutes its temperature is 60°C. Find its temperature after 10 minutes.

案例： 牛顿冷却定律：物体温度T的变化率与T和环境温度Tₐ=20°C之差成正比。一个100°C的热物体放在20°C的房间里，5分钟后温度降为60°C。求10分钟后的温度。

The differential equation is dT/dt = -k(T – 20), with k > 0. Separate variables: ∫ dT/(T-20) = ∫ -k dt ⇒ ln|T-20| = -kt + C. Thus T = 20 + A e^(-kt), where A is constant.

微分方程为dT/dt = -k(T-20)，k>0。分离变量得∫ dT/(T-20)=∫ -k dt ⇒ ln|T-20| = -kt + C。因此T = 20 + A e^(-kt)，A为常数。

Using T(0) = 100: 100 = 20 + A ⇒ A = 80. So T = 20 + 80 e^(-kt). At t=5, T=60: 60 = 20 + 80 e^(-5k) ⇒ e^(-5k) = 40/80 = 0.5. Therefore -5k = ln(1/2) ⇒ k = (ln 2)/5.

由初始条件T(0)=100得A=80，故T=20+80 e^(-kt)。代入t=5时T=60：60=20+80 e^(-5k) ⇒ e^(-5k)=0.5，因此k = (ln 2)/5。

At t=10: T = 20 + 80 e^(-10k) = 20 + 80 (e^(-5k))² = 20 + 80 (0.5)² = 20 + 20 = 40°C. The temperature after 10 minutes is 40°C.

t=10时：T = 20 + 80 e^(-10k) = 20 + 80 (e^(-5k))² = 20 + 80 (0.5)² = 20 + 20 = 40°C。10分钟后温度为40°C。

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5. Polar Coordinates: Area Enclosed by a Curve | 极坐标：曲线围成的面积

Case: Find the area enclosed by the cardioid r = 2(1 + cos θ).

案例： 求心形线 r = 2(1 + cos θ) 所围区域的面积。

The area in polar coordinates is A = ½ ∫₀^(2π) r² dθ. Here r² = 4(1 + cos θ)² = 4(1 + 2 cos θ + cos²θ). Use the double-angle identity cos²θ = (1 + cos 2θ)/2. Then r² = 4[1 + 2 cos θ + ½ + ½ cos 2θ] = 4[3/2 + 2 cos θ + ½ cos 2θ] = 6 + 8 cos θ + 2 cos 2θ.

极坐标下的面积公式为 A = ½ ∫₀^(2π) r² dθ。此处r² = 4(1+cos θ)² = 4(1+2cos θ+cos²θ)。利用倍角公式cos²θ=(1+cos 2θ)/2，则r² = 4[1+2cos θ+½+½ cos 2θ] = 6+8cos θ+2cos 2θ。

Thus A = ½ ∫₀^(2π) (6 + 8 cos θ + 2 cos 2θ) dθ = ½ [6θ + 8 sin θ + sin 2θ] from 0 to 2π. Since sin 2π = sin 0 = 0, all sine terms vanish, leaving A = ½ (6 × 2π) = 6π. The total area is 6π square units.

因此 A = ½ ∫₀^(2π) (6+8cos θ+2cos 2θ) dθ = ½ [6θ+8 sin θ+ sin 2θ]₀^(2π)。因sin 2π = sin 0 = 0，正弦项均为零，得A = ½ (12π) = 6π 平方单位。

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6. Series Expansion: Maclaurin Series | 级数展开：麦克劳林级数

Case: Obtain the Maclaurin series for f(x) = eˣ sin x up to and including the term in x⁵.

案例： 求函数 f(x) = eˣ sin x 的麦克劳林展开式，写到含x⁵的项为止。

It is efficient to multiply the series for eˣ and sin x. We have eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + … and sin x = x – x³/3! + x⁵/5! – … . Multiply term by term, retaining only powers up to x⁵.

直接相乘eˣ与sin x的级数最高效。已知 eˣ = 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120 + … ，sin x = x – x³/6 + x⁵/120 – … 。逐项相乘并仅保留至x⁵次幂。

Term by term: 1 × (x – x³/6 + x⁵/120) = x – x³/6 + x⁵/120. x × (x – x³/6) = x² – x⁴/6. x²/2 × (x – x³/6) = x³/2 – x⁵/12. x³/6 × (x) = x⁴/6. x⁴/24 × (x) = x⁵/24. Summing coefficients for each power:

逐项相乘：1×(x – x³/6 + x⁵/120) = x – x³/6 + x⁵/120；x×(x – x³/6) = x² – x⁴/6；x²/2×(x – x³/6) = x³/2 – x⁵/12；x³/6×(x) = x⁴/6；x⁴/24×(x) = x⁵/24。合并同次幂系数：

x: 1; x²: 1; x³: -1/6 + 1/2 = 1/3; x⁴: -1/6 + 1/6 = 0; x⁵: 1/120 – 1/12 + 1/24 = 1/120 – 10/120 + 5/120 = -4/120 = -1/30. Hence, f(x) ≈ x + x² + (1/3)x³ + 0·x⁴ – (1/30)x⁵.

x: 1；x²: 1；x³: -1/6+1/2=1/3；x⁴: -1/6+1/6=0；x⁵: 1/120 – 1/12 + 1/24 = -1/30。因此 f(x) ≈ x + x² + (1/3)x³ – (1/30)x⁵。

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7. Vectors: Distances and Intersections | 向量：距离与交线

Case: Find the perpendicular distance from the point P(1, 2, 3) to the plane 2x – y + 2z = 5. Also find the line of intersection of this plane with the plane x + y – z = 2.

案例： 求点P(1,2,3)到平面2x – y + 2z = 5的垂直距离，并求出该平面与平面x + y – z = 2的交线。

The distance d from point (x₁,y₁,z₁) to plane ax+by+cz=d₀ is |ax₁+by₁+cz₁ – d₀| / √(a²+b²+c²). Here a=2, b=-1, c=2, d₀=5. Substituting P: numerator = |2(1) – 2 + 2(3) – 5| = |2 – 2 + 6 – 5| = |1| = 1. Denominator = √(4+1+4) = √9 = 3. So d = 1/3 units.

点到平面的距离公式 d = |ax₁+by₁+cz₁ – d₀| / √(a²+b²+c²)。此处a=2,b=-1,c=2,d₀=5。代入P得分子=|2(1)-2+2(3)-5|=|1|=1，分母=√(4+1+4)=3。故d=1/3单位。

For the line of intersection, solve the system: 2x – y + 2z = 5 and x + y – z = 2. Add the equations to eliminate y: 3x + z = 7 ⇒ z = 7 – 3x. Substitute into the second equation: x + y – (7 – 3x) = 2 ⇒ 4x + y – 7 = 2 ⇒ y = 9 – 4x. So parametric equations: let x = λ, then y = 9 – 4λ, z = 7 – 3λ. The line can be written as r = (0, 9, 7) + λ(1, -4, -3).

求交线：解方程组2x – y + 2z=5 与 x + y – z=2。相加消去y得3x + z = 7 ⇒ z = 7 – 3x。代入第二个方程得x + y – (7 – 3x) = 2 ⇒ y = 9 – 4x。参数形式：令x = λ，则y = 9 – 4λ，z = 7 – 3λ。交线可写为 r = (0,9,7) + λ(1,-4,-3)。

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8. Probability Generating Functions | 概率生成函数

Case: The probability generating function (PGF) of a discrete random variable X is G_X(t) = (0.2 + 0.8t)³. Find the probability distribution of X, and hence its mean and variance.

案例： 离散随机变量X的概率生成函数为 G_X(t) = (0.2 + 0.8t)³。求X的概率分布，并由此求出其均值与方差。

Recognise that the PGF is of the binomial form (q + pt)^n where n=3, p=0.8, q=0.2. Thus X ~ B(3, 0.8). The probability mass function is P(X = k) = C(3,k) (0.8)^k (0.2)^(3-k) for k = 0,1,2,3. Explicitly: P(X=0)=0.008, P(X=1)=0.096, P(X=2)=0.384, P(X=3)=0.512.

该PGF具有二项分布的形式 (q + pt)^n，其中n=3，p=0.8，q=0.2。因此X ~ B(3, 0.8)。概率质量函数为 P(X=k) = C(3,k) (0.8)^k (0.2)^(3-k)，k=0,1,2,3。具体数值：P(X=0)=0.008，P(X=1)=0.096，P(X=2)=0.384，P(X=3)=0.512。

Using known results for a binomial distribution: mean E(X) = np = 3 × 0.8 = 2.4. Variance Var(X) = npq = 3 × 0.8 × 0.2 = 0.48. Alternatively, these can be derived from the PGF: G'(1) = mean, G”(1) + G'(1) – [G'(1)]² = variance. Both methods yield the same results.

利用二项分布的已知结果：均值E(X) = np = 2.4，方差Var(X) = npq = 0.48。也可通过PGF求得：G'(1) = 均值，G”(1) + G'(1) – [G'(1)]² = 方差，两种方法结果一致。

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