📚 Year 10 CIE Additional Mathematics: Case Study Practice | 剑桥IGCSE附加数学案例分析实战演练
Mastering Additional Mathematics requires more than memorising formulas – it demands the ability to apply concepts to real-world problems. This article presents ten case studies that mirror the style of CIE exam questions, covering optimisation, kinematics, exponential growth, trigonometric equations, polynomials, logarithms, binomial expansions, vectors, integration, and coordinate geometry. Each case study is broken down step by step, with paired English and Chinese explanations to strengthen both your mathematical reasoning and bilingual understanding.
掌握附加数学不仅需要熟记公式，更需要把概念运用到实际问题中。本文精选十个与剑桥考试风格贴近的案例，涵盖最优化、运动学、指数增长、三角方程、多项式、对数、二项式展开、向量、积分和坐标几何。每个案例都拆解为详细步骤，并配有中英文对照讲解，帮助你提升数学推理能力和双语理解力。
1. Optimisation of a Rectangular Enclosure | 案例1：矩形围栏的最优化
A farmer has 100 m of fencing and wishes to enclose a rectangular area against a straight wall, with no fence needed along the wall. Let the side parallel to the wall be y metres and the two perpendicular sides each be x metres. The total fencing used is 2x + y = 100. The area A = x y. Express A in terms of x and find the maximum area.
一位农民有100米围栏，想靠着笔直的墙壁围出一个矩形区域，靠墙一侧无需围栏。设平行于墙的边长为 y 米，两条垂直边各为 x 米。所用围栏总量为 2x + y = 100。面积 A = x y。请将 A 用 x 表示，并求最大面积。
From the constraint, y = 100 – 2x. Then A = x(100 – 2x) = 100x – 2x². Differentiate: dA/dx = 100 – 4x. Set dA/dx = 0 gives x = 25. The second derivative d²A/dx² = -4 < 0, confirming a maximum. Substitute x = 25 into y = 100 - 2(25) = 50. Maximum area = 25 × 50 = 1250 m².
由约束条件可得 y = 100 – 2x。于是 A = x(100 – 2x) = 100x – 2x²。求导：dA/dx = 100 – 4x。令导数为零得 x = 25。二阶导数 d²A/dx² = -4 < 0，确认是极大值。将 x = 25 代入 y = 100 - 2(25) = 50。最大面积为 25 × 50 = 1250 m²。
2. Kinematics – Velocity and Acceleration | 案例2：运动学——速度与加速度
A particle moves along a straight line so that its displacement s (m) from a fixed point O at time t (s) is given by s = t³ – 6t² + 9t + 5. Find the velocity and acceleration at time t, the times when the particle is at rest, and the distance travelled in the first 4 seconds.
一质点沿直线运动，其相对固定点 O 的位移 s(米) 与时间 t(秒) 的关系为 s = t³ – 6t² + 9t + 5。求 t 时刻的速度和加速度、质点静止的时刻，以及前4秒内行驶的路程。
Velocity v = ds/dt = 3t² – 12t + 9. Acceleration a = dv/dt = 6t – 12. The particle is at rest when v = 0, so 3t² – 12t + 9 = 0 → t² – 4t + 3 = 0 → (t – 1)(t – 3) = 0, thus t = 1 or t = 3. To find distance travelled, examine displacement: s(0) = 5, s(1) = 1 – 6 + 9 + 5 = 9, s(3) = 27 – 54 + 27 + 5 = 5, s(4) = 64 – 96 + 36 + 5 = 9. Distance = |9-5| + |5-9| + |9-5| = 4 + 4 + 4 = 12 m.
速度 v = ds/dt = 3t² – 12t + 9。加速度 a = dv/dt = 6t – 12。质点静止时 v = 0，即 3t² – 12t + 9 = 0 → t² – 4t + 3 = 0 → (t – 1)(t – 3) = 0，得 t = 1 或 t = 3。求路程需分析位移：s(0)=5，s(1)=1-6+9+5=9，s(3)=27-54+27+5=5，s(4)=64-96+36+5=9。路程 = |9-5| + |5-9| + |9-5| = 4+4+4 = 12 米。
3. Exponential Growth of a Bacterial Culture | 案例3：细菌培养的指数增长
A bacterial culture initially contains 500 cells and grows according to N = 500 e^(0.4t), where t is time in hours. Find the number of cells after 5 hours, the time taken to reach 10 000 cells, and the rate of growth at t = 5.
某细菌培养初始含有500个细胞，按 N = 500 e^(0.4t) 增长，其中 t 为小时数。求5小时后的细胞数、达到10 000个细胞所需时间，以及 t = 5 时的增长率。
After 5 hours: N = 500 e^(0.4×5) = 500 e² ≈ 500 × 7.389 = 3694.5 ≈ 3695 cells. To find t for N = 10000: 10000 = 500 e^(0.4t) → 20 = e^(0.4t) → ln 20 = 0.4t → t = (ln 20)/0.4 ≈ 2.9957/0.4 ≈ 7.49 hours. Rate of growth dN/dt = 0.4 × 500 e^(0.4t) = 200 e^(0.4t). At t = 5, dN/dt = 200 e² ≈ 200 × 7.389 = 1477.8 cells per hour.
5小时后：N = 500 e^(0.4×5) = 500 e² ≈ 500 × 7.389 = 3694.5 ≈ 3695 个细胞。求达到10000的时间：10000 = 500 e^(0.4t) → 20 = e^(0.4t) → ln 20 = 0.4t → t = (ln 20)/0.4 ≈ 2.9957/0.4 ≈ 7.49 小时。增长率 dN/dt = 0.4 × 500 e^(0.4t) = 200 e^(0.4t)。t = 5 时，dN/dt = 200 e² ≈ 200 × 7.389 = 1477.8 个/小时。
4. Trigonometric Equation in Engineering | 案例4：工程中的三角方程
An oscillating mechanism has displacement d = 3 sin(2θ) + 4 cos(2θ). Express d in the form R sin(2θ + α) and solve d = 2 for 0° ≤ θ ≤ 180°.
某振荡机构的位移为 d = 3 sin(2θ) + 4 cos(2θ)。将 d 表示为 R sin(2θ + α) 的形式，并求解在 0° ≤ θ ≤ 180° 范围内 d = 2 的解。
R sin(2θ + α) = R sin(2θ)cos α + R cos(2θ)sin α. Comparing coefficients: R cos α = 3, R sin α = 4. So R = √(3² + 4²) = 5, tan α = 4/3 → α ≈ 53.13° or 0.927 rad. Thus d = 5 sin(2θ + 53.13°). Equation 5 sin(2θ + 53.13°) = 2 → sin(2θ + 53.13°) = 0.4. Let β = 2θ + 53.13°. β = sin⁻¹ 0.4 ≈ 23.58° or 180° – 23.58° = 156.42°. Then 2θ = β – 53.13°. For β = 23.58°, 2θ = -29.55° (reject, negative). For β = 156.42°, 2θ = 103.29° → θ ≈ 51.6°. Also add 360° to β: 156.42° + 360° = 516.42°, 2θ = 463.29° → θ ≈ 231.6° (outside range). Check next cycle: within 0°–180°, solution is θ ≈ 51.6°. Also consider 2θ + 53.13° = 180° – 23.58° = 156.42°, we got that. Another solution from sin positive: β = 23.58° + 360° = 383.58°, 2θ = 330.45° → θ ≈ 165.2°, which is valid. So θ ≈ 51.6°, 165.2°.
R sin(2θ + α) = R sin(2θ)cos α + R cos(2θ)sin α。比较系数：R cos α = 3，R sin α = 4。因此 R = √(3² + 4²) = 5，tan α = 4/3 → α ≈ 53.13°。于是 d = 5 sin(2θ + 53.13°)。方程 5 sin(2θ + 53.13°) = 2 → sin(2θ + 53.13°) = 0.4。令 β = 2θ + 53.13°。β = sin⁻¹ 0.4 ≈ 23.58° 或 180° – 23.58° = 156.42°。则 2θ = β – 53.13°。对于 β = 23.58°，2θ = -29.55°（舍去）；对于 β = 156.42°，2θ = 103.29° → θ ≈ 51.6°。再考虑 β 加 360°：β = 156.42° + 360° = 516.42°，2θ = 463.29° → θ ≈ 231.6°（超出范围）。正弦值正的另一个解来自 β = 23.58° + 360° = 383.58°，2θ = 330.45° → θ ≈ 165.2°，符合范围。因此 θ ≈ 51.6°, 165.2°。
5. Polynomial Remainders and Factorisation | 案例5：多项式余式与因式分解
Given P(x) = 2x³ – 5x² + x + 2. Use the remainder theorem to find the remainder when divided by (x – 2). Factorise P(x) completely and solve P(x) = 0.
已知 P(x) = 2x³ – 5x² + x + 2。使用余式定理求除以 (x – 2) 的余数。将 P(x) 彻底因式分解，并解方程 P(x) = 0。
Remainder = P(2) = 2(8) – 5(4) + 2 + 2 = 16 – 20 + 4 = 0. Since remainder is 0, (x – 2) is a factor. Division gives quadratic 2x² – x – 1. Factorise: 2x² – x – 1 = (2x + 1)(x – 1). Thus P(x) = (x – 2)(2x + 1)(x – 1). Solving P(x) = 0 gives x = 2, x = -1/2, x = 1.
余数 = P(2) = 2(8) – 5(4) + 2 + 2 = 16 – 20 + 4 = 0。由于余数为0，(x – 2) 是一个因式。除法得二次式 2x² – x – 1。因式分解：2x² – x – 1 = (2x + 1)(x – 1)。因此 P(x) = (x – 2)(2x + 1)(x – 1)。解 P(x) = 0 得 x = 2，x = -1/2，x = 1。
6. Solving Logarithmic Equations in Finance | 案例6：金融中的对数方程求解
An investment grows according to A = P(1.05)^n. How many years are required for an initial deposit of $2000 to double? Use logarithms and solve: 4000 = 2000(1.05)^n.
一项投资按照 A = P(1.05)^n 增长。初始存款2000美元翻倍需要多少年？请用对数求解：4000 = 2000(1.05)^n。
Divide both sides by 2000: 2 = (1.05)^n. Take logs: log 2 = n log 1.05. Therefore n = log 2 / log 1.05 ≈ 0.3010 / 0.02119 ≈ 14.2 years. So it takes about 15 years to double in practice (compounding annually). For exact answer, n = ln 2 / ln 1.05 ≈ 14.2067.
两边同除2000：2 = (1.05)^n。取对数：log 2 = n log 1.05。因此 n = log 2 / log 1.05 ≈ 0.3010 / 0.02119 ≈ 14.2 年。实际每年复利，大约15年翻倍。精确值 n = ln 2 / ln 1.05 ≈ 14.2067。
7. Binomial Expansion in Probability | 案例7：二项式展开在概率中的应用
Expand (1 + 2x)^5 up to the term in x³. A biased coin has P(heads) = 2/3, tails = 1/3. Find the probability of getting exactly 3 heads in 5 tosses. Compare the coefficient pattern with the binomial expansion.
将 (1 + 2x)^5 展开至 x³ 项。一枚偏斜硬币出现正面的概率为2/3，反面为1/3。求5次投掷中恰好出现3次正面的概率，并比较二项式展开的系数规律。
Using the binomial theorem: (1 + 2x)^5 = 1 + 5C1(2x) + 5C2(2x)² + 5C3(2x)³ + … = 1 + 5·2x + 10·4x² + 10·8x³ + … = 1 + 10x + 40x² + 80x³ + … Probability of 3 heads in 5 tosses: P(X=3) = 5C3 × (2/3)³ × (1/3)² = 10 × (8/27) × (1/9) = 80/243 ≈ 0.329. Notice the coefficient of x³ in the expansion of (1 + 2x)^5 is 10 × 2³ = 80, matching the combination count times probability weights if we set x = 1/3? Actually, general term shows pattern. The coefficient 80x³ is analogous to the number of favourable outcomes when success probability is doubled relative to a variable.
使用二项式定理：(1 + 2x)^5 = 1 + 5C1(2x) + 5C2(2x)² + 5C3(2x)³ + … = 1 + 5·2x + 10·4x² + 10·8x³ + … = 1 + 10x + 40x² + 80x³ + …。5次中恰好3次正面的概率：P(X=3) = 5C3 × (2/3)³ × (1/3)² = 10 × (8/27) × (1/9) = 80/243 ≈ 0.329。注意在(1+2x)⁵ 展开中 x³ 的系数是 10 × 2³ = 80，与组合数乘以概率权重相似（若设 x=1/3）。其系数规律展示了成功概率翻倍时对应的方式。
8. Vector Displacement in Navigation | 案例8：导航中的向量位移
A boat sails from a harbour on a bearing of 060° for 8 km, then changes course to 150° for 12 km. Find the resultant displacement from the harbour in vector form, and its magnitude and bearing.
一艘船从港口出发，沿方位角060°航行8公里，然后转至150°航向航行12公里。求相对于港口的合位移向量，以及其大小和方位角。
Take unit vectors i east, j north. First leg: bearing 060° means 60° from north towards east, so angle from i axis = 30° east of north? Actually bearing measured clockwise from north. Vector: 8(cos 60° i + sin 60° j) = 8(0.5 i + 0.8660 j) = 4i + 6.928j. Second leg: bearing 150° is 150° from north, i.e. 60° south of east? Angle from positive i: 150° – 90° = 60°? Let’s derive: 150° clockwise from north means direction is 30° south of east? No: East is 090°, so 150° is 60° east of south. Vector: 12(cos 150°? Better: bearing 150° means components: east = 12 sin 150°? Actually standard: bearing angle θ measured clockwise from north. The unit vector: east = sin θ, north = cos θ. So first leg: east = 8 sin 60° = 6.928, north = 8 cos 60° = 4. So 6.928i + 4j. Wait careful: Bearing 060°: north component = 8 cos 60° = 4 km, east component = 8 sin 60° ≈ 6.928 km. So vector 6.928i + 4j. Second leg bearing 150°: north = 12 cos 150° = 12(-√3/2) = -10.392, east = 12 sin 150° = 12(0.5) = 6. So vector 6i – 10.392j. Sum: (6.928+6)i + (4 – 10.392)j = 12.928i – 6.392j. Magnitude = √(12.928² + 6.392²) ≈ √(167.1 + 40.86) = √207.96 ≈ 14.42 km. Bearing: angle from north = arctan(east/north) but note negative north. The vector is in southeast quadrant. Bearing = 180° – arctan(12.928/6.392) = 180° – arctan(2.022) ≈ 180° – 63.7° = 116.3°. So resultant displacement about 14.4 km on bearing 116°.
设单位向量 i 向东，j 向北。第一段：方位角060°表示从北顺时针偏60°，东分量 = 8 sin 60° ≈ 6.928 km，北分量 = 8 cos 60° = 4 km。向量为 6.928i + 4j。第二段方位角150°：北分量 = 12 cos 150° = 12(-√3/2) = -10.392，东分量 = 12 sin 150° = 12(0.5) = 6。向量为 6i – 10.392j。合向量 = (6.928+6)i + (4-10.392)j = 12.928i – 6.392j。模 = √(12.928² + 6.392²) ≈ √207.96 ≈ 14.42 km。方位角：因为北分量为负，东分量为正，位于东南象限。方位角 = 180° – arctan(12.928/6.392) ≈ 180° – 63.7° = 116.3°。因此合位移约14.4公里，方位角116°。
9. Area Under a Curve – Definite Integration | 案例9：曲线下方面积——定积分
Find the area enclosed between the curve y = x² – 4x + 3 and the x-axis from x = 0 to x = 3. The curve crosses the x-axis at x = 1 and x = 3.
求曲线 y = x² – 4x + 3 与 x 轴之间在 x = 0 到 x = 3 之间所围成的面积。该曲线与 x 轴交于 x = 1 和 x = 3。
First, determine sign: For x in [0,1], y positive (e.g., x=0 gives y=3). From 1 to 3, y ≤ 0. So area = ∫₀¹ (x² – 4x + 3) dx + |∫₁³ (x² – 4x + 3) dx|. Compute ∫ (x² – 4x + 3) dx = (1/3)x³ – 2x² + 3x. Evaluate from 0 to 1: [(1/3) – 2 + 3] – 0 = 4/3. From 1 to 3: [(9) – 18 + 9] – [(1/3) – 2 + 3] = 0 – (4/3) = -4/3. Absolute value is 4/3. Total area = 4/3 + 4/3 = 8/3 square units.
首先判断正负：在 [0,1] 区间，y 为正（如 x=0 得 y=3）。从1到3，y ≤ 0。因此面积 = ∫₀¹ (x² – 4x + 3) dx + |∫₁³ (x² – 4x + 3) dx|。计算不定积分 ∫ (x² – 4x + 3) dx = (1/3)x³ – 2x² + 3x。从0到1定积分：[(1/3) – 2 + 3] – 0 = 4/3。从1到3：[(9) – 18 + 9] – [(1/3) – 2 + 3] = 0 – 4/3 = -4/3，绝对值4/3。总面积 = 4/3 + 4/3 = 8/3 平方单位。
10. Coordinate Geometry – Circle and Tangent | 案例10：坐标几何——圆与切线
A circle has equation x² + y² – 6x + 4y – 12 = 0. Find the centre and radius. Determine the equation of the tangent to the circle at the point (6, 2).
一圆的方程为 x² + y² – 6x + 4y – 12 = 0。求圆心和半径，并求该圆在点 (6, 2) 处的切线方程。
Complete the square: (x² – 6x) + (y² + 4y) = 12 → (x – 3)² – 9 + (y + 2)² – 4 = 12 → (x – 3)² + (y + 2)² = 25. So centre C(3, -2), radius r = 5. The gradient of radius to (6,2): (2 – (-2))/(6 – 3) = 4/3. Tangent is perpendicular, so gradient m_tangent = -3/4. Using point-slope form: y – 2 = -3/4 (x – 6) → multiply by 4: 4y – 8 = -3x + 18 → 3x + 4y – 26 = 0.
配方： (x² – 6x) + (y² + 4y) = 12 → (x – 3)² – 9 + (y + 2)² – 4 = 12 → (x – 3)² + (y + 2)² = 25。因此圆心 C(3, -2)，半径 r = 5。半径到点 (6,2) 的斜率为 (2 – (-2))/(6 – 3) = 4/3。切线与之垂直，故斜率 m_tangent = -3/4。用点斜式：y – 2 = -3/4 (x – 6) → 两边乘4：4y – 8 = -3x + 18 → 3x + 4y – 26 = 0。
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