AS OCR Further Mathematics: Interdisciplinary Problem-Solving Practice | AS OCR 进阶数学：跨学科综合题型训练

📚 AS OCR Further Mathematics: Interdisciplinary Problem-Solving Practice | AS OCR 进阶数学：跨学科综合题型训练

In AS Level Further Mathematics, the ability to connect ideas from different strands of the syllabus is essential for tackling the most challenging examination questions. This article provides targeted practice on interdisciplinary problems that blend pure mathematics with mechanics, statistics, and other applied topics. Each section presents a typical question, a step-by-step solution, and commentary to help you build fluency in moving between concepts such as calculus and kinematics, series and probability, matrices and geometry, and vectors and forces.

在 AS 进阶数学中，将不同模块的思想联系起来是解决高难度考题的关键能力。本文提供跨学科综合题型的专项训练，融合纯数学与力学、统计等应用领域。每一节展示一个典型题目、逐步求解过程与评注，帮助你在微积分与运动学、级数与概率、矩阵与几何、向量与力等概念之间流畅转换。

---

1. Kinematics and Calculus | 运动学与微积分

t from a fixed point is s = 2 m. Determine the velocity and the displacement when t = 4 s.

质点沿直线运动，加速度为 a(t) = 6t − 4 m s⁻²。在 t = 0 时，速度 v = 3 m s⁻¹，相对于固定点的位移 s = 2 m。求 t = 4 s 时的速度和位移。

ing v(0) = 3 gives C = 3, so v(t) = 3t² − 4t + 3. At t = 4, v = 3(16) − 16 + 3 = 35 m s⁻¹.

对加速度积分求速度：v = ∫(6t − 4) dt = 3t² − 4t + C。利用 v(0) = 3 得 C = 3，故 v(t) = 3t² − 4t + 3。当 t = 4 时，v = 3(16) − 16 + 3 = 35 m s⁻¹。

s(0) = 2 gives D = 2, so s(t) = t³ − 2t² + 3t + 2. At t = 4, s = 64 − 32 + 12 + 2 = 46 m.

再对速度积分求位移：s = ∫(3t² − 4t + 3) dt = t³ − 2t² + 3t + D。代入 s(0) = 2 得 D = 2，故 s(t) = t³ − 2t² + 3t + 2。当 t = 4 时，s = 64 − 32 + 12 + 2 = 46 m。

This straightforward combination of polynomials, integration and initial conditions is common in mechanics problems that rely on core pure skills.

这种多项式、积分与初始条件的简单组合常见于依赖核心纯数学技能的力学问题。

---

2. Differential Equations in Mechanics | 力学中的微分方程

nstant. The equation of motion is dv/dt = g − kv, with g = 9.8 m s⁻². Solve for v(t) given that the skydiver starts from rest, and find the speed after 5 seconds when k = 0.2 s⁻¹.

一名跳伞者质量为 m，竖直下落，受重力与空气阻力作用，阻力大小为 mkv，其中 v 为速度，k 为正常数。运动方程为 dv/dt = g − kv，g = 9.8 m s⁻²。跳伞者从静止开始，求解 v(t)，并求当 k = 0.2 s⁻¹ 时，5 秒后的速度。

variables: 1/(g − kv) dv = dt. Integrate both sides: ∫(1/(g − kv)) dv = ∫1 dt → (−1/k) ln |g − kv| = t + C. Using v(0) = 0 gives C = (−1/k) ln g.

分离变量：1/(g − kv) dv = dt。两边积分：∫(1/(g − kv)) dv = ∫1 dt → (−1/k) ln |g − kv| = t + C。利用 v(0) = 0 得 C = (−1/k) ln g。

(g/k)(1 − e⁻ᵏᵗ). With k = 0.2, at t = 5: v = (9.8/0.2)(1 − e⁻¹) ≈ 49 × 0.632 = 31.0 m s⁻¹.

整理得：ln(g − kv) = ln g − kt → g − kv = g e⁻ᵏᵗ → v = (g/k)(1 − e⁻ᵏᵗ)。代入 k = 0.2，t = 5 时 v = (9.8/0.2)(1 − e⁻¹) ≈ 49 × 0.632 = 31.0 m s⁻¹。

This problem links first-order linear ODEs with an exponential model, reinforcing the integration of 1/(linear) and the use of initial conditions.

此题将一阶线性常微分方程与指数模型联系起来，强化了 “1/(线性函数)” 的积分及初始条件的应用。

---

3. Probability Generating Functions and Series | 概率生成函数与级数

A discrete random variable X follows the distribution P(X = x) = (1/3)(2/3)ˣ t the probability generating function is G(t) = t/(3 − 2t) and hence find E(X) and Var(X).

离散随机变量 X 服从分布 P(X = x) = (1/3)(2/3)ˣ⁻¹，x = 1, 2, 3, ... 证明其概率生成函数为 G(t) = t/(3 − 2t)，并由此求 E(X) 与 Var(X)。

The PGF is defined by G(t) = Σ tˣ P(X = x) for x from 1 to ∞. Substitute P(X=x): G(t) = Σ tˣ (1/3)(2/3)ˣ⁻¹ = (1/3) t Σ (2t/3)ˣ⁻¹. The sum is an infinite geometric series with first term 1 and common ratio (2t/3) for |2t/3| < 1. So Σ (2t/3)ˣ⁻¹ = 1/(1 − 2t/3) = 3/(3 − 2t). Multiplying by (1/3)t gives G(t) = t/(3 − 2t).

概率生成函数定义为 G(t) = Σ tˣ P(X = x)，x 从 1 到无穷。代入分布：G(t) = Σ tˣ (1/3)(2/3)ˣ⁻¹ = (1/3) t Σ (2t/3)ˣ⁻¹。该和为无限几何级数，首项 1，公比 (2t/3)（|2t/3| < 1）。故 Σ (2t/3)ˣ⁻¹ = 1/(1 − 2t/3) = 3/(3 − 2t)。乘以 (1/3)t 得 G(t) = t/(3 − 2t)。

Differentiate: G'(t) = (3 − 2t)⁻² × 3 − t(−2)/(3 − 2t)²? Wait, use quotient or product rule: G(t) = t (3 − 2t)⁻¹. G'(t) = (3 − 2t)⁻¹ + t × 2 (3 − 2t)⁻² = (3 − 2t + 2t)/(3 − 2t)² = 3/(3 − 2t)². Then E(X) = G'(1) = 3/(1)² = 3. Second derivative: G''(t) = 12/(3 − 2t)³, so G''(1) = 12. Var(X) = G''(1) + G'(1) − [G'(1)]² = 12 + 3 − 9 = 6.

求导：G(t) = t (3 − 2t)⁻¹。G'(t) = (3 − 2t)⁻¹ + t × 2(3 − 2t)⁻² = (3 − 2t + 2t)/(3 − 2t)² = 3/(3 − 2t)²。因此 E(X) = G'(1) = 3。二阶导数为 G''(t) = 12/(3 − 2t)³，G''(1) = 12。Var(X) = G''(1) + G'(1) − [G'(1)]² = 12 + 3 − 9 = 6。

This exercise blends geometric series and differentiation, showing how pure techniques unlock statistical measures.

本题综合了几何级数与微分，展现纯数学技巧如何解锁统计度量。

---

广告 / Sponsored

4. Matrices and Composite Transformations | 矩阵与复合变换

matrix C = BA and describe the single transformation represented by C. Determine the image of the point (2, 3) under C.

矩阵 A 表示绕原点逆时针旋转 90°，矩阵 B 表示关于 x 轴的反射。写出 A 和 B。求矩阵 C = BA，并描述 C 所表示的单一变换。求点 (2, 3) 在 C 下的像。

sponds to a reflection in the line y = −x. Applying C to (2,3) gives [[0,−1],[−1,0]] (2,3)ᵀ = (−3, −2).

A = [[0, −1], [1, 0]]；B = [[1, 0], [0, −1]]。则 C = B A = [[1,0],[0,−1]] [[0,−1],[1,0]] = [[0,−1],[−1,0]]。该矩阵对应关于直线 y = −x 的反射。将 C 作用到 (2,3) 上得 (−3, −2)。

(0×0)−(−1×−1) = −1, we confirm the transformation is a reflection (orientation-reversing).

计算行列式 det(C) = (0×0)−(−1×−1) = −1，可确认此变换是反射（改变方向）。

---

5. Complex Numbers and Vector Geometry | 复数与向量几何

hat represents the vector from A to B, and determine the angle between the vector AB and the positive real axis.

在 Argand 图中，点 A 和 B 分别表示复数 z₁ = 3 + 4i 与 z₂ = −1 + 2i。求表示从 A 到 B 的向量的复数，并求向量 AB 与正实轴之间的夹角。

rd quadrant, so the principal argument is −(π − arctan(0.5)) or equivalently π + arctan(0.5) if measured from positive real axis. The acute angle to the negative real axis is arctan(0.5). The angle from the positive real axis is therefore 180° + arctan(0.5) or about 206.6°. For the direction of the vector, we often take the angle to the horizontal as arctan(1/2) but in the third quadrant, so 180° + 26.6° = 206.6°. A more refined analysis uses the argument: arg(−4 − 2i) = −(π − arctan(0.5)) = −153.4° (or 206.6°).

向量 AB 由 z₂ − z₁ = (−1 + 2i) − (3 + 4i) = −4 − 2i 给出。−4 − 2i 的辐角即所求夹角。tan θ = (−2)/(−4) = 1/2，但点位于第三象限，故主辐角为 −(π − arctan(0.5)) 或相对于正实轴的 180° + arctan(0.5) ≈ 206.6°。锐角为 arctan(0.5) ≈ 26.6°，故夹角为 180° + 26.6° = 206.6°。

Thus complex subtraction directly yields both magnitude and direction of a vector, linking pure algebra to geometry.

因此复数减法直接给出向量的模与方向，将纯代数与几何联系起来。

---

6. Optimisation Using Differentiation | 利用微分进行优化

An open box is made from a square sheet of cardboard of side 30 cm by cutting out squares of side x cm from each corner and folding up the sides. Express the volume V in terms of x, and find the value of x that maximises V.

从一个边长为 30 cm 的正方形纸板四角各切去边长为 x cm 的小正方形，折起四边制成一个无盖盒子。用 x 表示体积 V，并求使 V 最大的 x 值。

= 12(x² − 20x + 75). Set derivative to zero: x² − 20x + 75 = 0 → (x − 5)(x − 15) = 0. So x = 5 or x = 15. Second derivative: d²V/dx² = 24x − 240. At x = 5, d²V/dx² = 120 − 240 = −120 < 0 → maximum. At x = 15, positive → minimum. The feasible domain is 0 < x < 15, so x = 5 cm gives maximum volume.

底面是边长为 (

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

更多咨询请联系16621398022（同微信）