Year 11 OCR Further Mathematics: Interdisciplinary Integrated Problem Training

📚 Year 11 OCR Further Mathematics: Interdisciplinary Integrated Problem Training | 跨学科综合题型训练

In Year 11 OCR Further Mathematics, the ability to apply mathematical methods across different subjects is a key skill that distinguishes high-performing students. Interdisciplinary problems require you to recognise mathematical structures in physics, economics, engineering and finance, and to use calculus, vectors, sequences and matrices as versatile tools. This article provides a collection of integrated problem-training exercises, each linking a core Further Maths topic to a real-world context, complete with step-by-step reasoning.

在 Year 11 OCR 进阶数学中，能够将数学方法应用于不同学科是区分高分学生的关键技能。跨学科问题要求你在物理、经济学、工程和金融等领域识别出数学结构，并把微积分、向量、数列和矩阵当作通用工具来使用。本文提供了一组综合题型训练，每一道题都将一个核心进阶数学主题与一个现实情境连接起来，并配有逐步推理过程。

1. Applying Calculus to Kinematics | 微积分在运动学中的应用

An object moves along a straight line with velocity v(t) = 6t − 2t² m/s for t ≥ 0. Find the displacement between t = 0 and t = 4 seconds, and determine the maximum speed during this interval.

一物体沿直线运动，t ≥ 0 时速度 v(t) = 6t − 2t² 米/秒。求 t = 0 至 t = 4 秒间的位移，并确定该时间段内的最大速率。

To obtain displacement, we integrate the velocity function: s = ∫₀⁴ (6t − 2t²) dt = [3t² − (2/3)t³] from 0 to 4, giving 3×16 − ⅔×64 = 48 − 42.67 ≈ 5.33 m. The speed |v(t)| is maximum when its derivative is zero: a(t) = 6 − 4t = 0 ⇒ t = 1.5 s. Checking |v(1.5)| = |6×1.5 − 2×(1.5)²| = |9 − 4.5| = 4.5 m/s, which is greater than the speed at t = 0 (0) and at t = 4 (|24 − 32| = 8 → actually 8 m/s). Wait: at t = 4, v = 6×4−2×16=24−32=−8, so speed is 8 m/s. Thus the maximum speed is 8 m/s, occurring at t = 4.

为了得到位移，我们对速度函数积分：s = ∫₀⁴ (6t − 2t²) dt = [3t² − ⅔t³]₀⁴ = 48 − 42.67 ≈ 5.33 米。速率 |v(t)| 的最大值出现在其导数为零时：a(t) = 6 − 4t = 0 ⇒ t = 1.5 秒。比较 |v(1.5)| = 4.5 米/秒与端点 t=0 时 0 米/秒、t=4 时 |−8| = 8 米/秒，因此最大速率为 8 米/秒。

2. Marginal Analysis in Economics | 经济学中的边际分析

A company’s total revenue function is R(q) = 120q − 0.5q², and its total cost function is C(q) = 200 + 20q + 0.2q². Use differentiation to find the profit-maximising output q and the corresponding maximum profit.

某公司的总收入函数为 R(q) = 120q − 0.5q²，总成本函数为 C(q) = 200 + 20q + 0.2q²。用微分法求出利润最大化的产量 q 以及相应的最大利润。

Profit π(q) = R(q) − C(q) = (120q − 0.5q²) − (200 + 20q + 0.2q²) = −200 + 100q − 0.7q². Setting the derivative equal to zero: dπ/dq = 100 − 1.4q = 0 gives q = 100/1.4 ≈ 71.43 units. The second derivative is −1.4 < 0, confirming a maximum. Substituting back: π_max = −200 + 100×71.43 − 0.7×(71.43)² ≈ −200 + 7143 − 3571 ≈ 3372. The maximum profit is roughly £3372.

利润 π(q) = R(q) − C(q) = (120q − 0.5q²) − (200 + 20q + 0.2q²) = −200 + 100q − 0.7q²。令导数为零：dπ/dq = 100 − 1.4q = 0 得 q ≈ 71.43 件。二阶导数为 −1.4 < 0 确认为极大值。代入后得最大利润约 £3372。

3. Resolving Forces with Vectors | 向量分解与力学平衡

A particle is in equilibrium under the action of three forces: F₁ = (3i + 4j) N, F₂ = (−5i + 2j) N, and an unknown force F₃. Determine F₃ and its magnitude.

一个质点在三个力作用下保持平衡：F₁ = (3i + 4j) N，F₂ = (−5i + 2j) N 以及未知力 F₃。求 F₃ 及其大小。

For equilibrium, resultant force = 0, so F₁ + F₂ + F₃ = 0 ⇒ F₃ = −(F₁ + F₂). Compute F₁ + F₂ = (3−5)i + (4+2)j = −2i + 6j, thus F₃ = 2i − 6j N. Magnitude = √(2² + (−6)²) = √(4 + 36) = √40 = 2√10 ≈ 6.32 N.

平衡时合力为零，因此 F₁ + F₂ + F₃ = 0 ⇒ F₃ = −(F₁ + F₂)。计算得 F₁ + F₂ = −2i + 6j，所以 F₃ = 2i − 6j N。大小 = √(2² + (−6)²) = √40 = 2√10 ≈ 6.32 N。

4. Geometric Sequences in Finance | 金融中的等比数列

An investor deposits £5000 into an account offering 4% compound interest per annum. After how many full years will the balance first exceed £7000? Use logarithms to solve.

一位投资者将 £5000 存入年利率 4% 的复利账户。经过多少整年后，余额将首次超过 £7000？用对数求解。

The balance after n years is 5000 × (1.04)ⁿ. Set up the inequality: 5000 × 1.04ⁿ > 7000 ⇒ 1.04ⁿ > 1.4. Taking logs: n log(1.04) > log(1.4). Hence n > log(1.4) / log(1.04) ≈ 0.146128 / 0.017033 ≈ 8.58. So after 9 full years the balance exceeds £7000.

n 年后的余额为 5000 × 1.04ⁿ。建立不等式：5000 × 1.04ⁿ > 7000 ⇒ 1.04ⁿ > 1.4。取对数：n log(1.04) > log(1.4)，得 n > 8.58。因此经过 9 整年后余额超过 £7000。

5. Exponential Growth and Decay | 指数增长与衰减

The number of bacteria in a culture grows according to N(t) = 200 e^(0.15 t), where t is in hours. Find the time needed for the population to double and the rate of growth when t = 5 hours.

某培养皿中的细菌数量按 N(t) = 200 e^(0.15t) 增长，t 以小时计。求出种群翻倍所需的时间，以及 t = 5 小时时的增长率。

Doubling means N = 400: 200 e^(0.15t) = 400 ⇒ e^(0.15t) = 2 ⇒ 0.15t = ln 2 ⇒ t = (ln 2)/0.15 ≈ 4.62 hours. The rate of growth is dN/dt = 200 × 0.15 e^(0.15t) = 30 e^(0.15t). At t = 5, rate = 30 e^(0.75) ≈ 30 × 2.117 = 63.5 bacteria per hour.

翻倍即 N = 400：200 e^(0.15t) = 400 ⇒ e^(0.15t) = 2 ⇒ t = (ln 2)/0.15 ≈ 4.62 小时。增长率 dN/dt = 30 e^(0.15t)，在 t = 5 时约为 30 × 2.117 = 63.5 个细菌/小时。

6. Matrix Transformations in Computer Graphics | 计算机图形学中的矩阵变换

A triangle has vertices at coordinates A(1,0), B(3,0) and C(2,2). Apply a rotation of 90° anticlockwise about the origin followed by a reflection in the line y = x. Find the final coordinates using matrix multiplication.

一个三角形的顶点坐标为 A(1,0), B(3,0), C(2,2)。先绕原点逆时针旋转 90°，再关于直线 y = x 反射。用矩阵乘法求出最终坐标。

Rotation 90° matrix: R = [[0, −1], [1, 0]]. Reflection in y = x matrix: S = [[0, 1], [1, 0]]. Combined transformation T = SR = [[0,1],[1,0]] × [[0,−1],[1,0]] = [[1,0],[0,−1]]. So each point (x,y) maps to (x, −y). Applying to vertices: A(1,0)→(1,0); B(3,0)→(3,0); C(2,2)→(2,−2). The final triangle has vertices (1,0), (3,0), (2,−2).

旋转 90° 矩阵 R = [[0, −1], [1, 0]]。反射 S = [[0, 1], [1, 0]]。组合 T = SR = [[1,0],[0,−1]]，即 (x,y) → (x, −y)。应用于各顶点：A(1,0)→(1,0); B(3,0)→(3,0); C(2,2)→(2,−2)。最终三角形顶点为 (1,0), (3,0), (2,−2)。

7. Probability and Decision Making | 概率与决策制定

A game offers two strategies: Strategy X gives a 60% chance of winning £50 and 40% chance of losing £20. Strategy Y gives a 30% chance of winning £120 and 70% chance of losing £30. By calculating expected values, decide which strategy is better. Also find the minimal win probability needed for Strategy Y to match Strategy X’s expected gain.

某游戏有两种策略：策略 X 有 60% 机会赢 £50，40% 机会输 £20；策略 Y 有 30% 机会赢 £120，70% 机会输 £30。通过计算期望值来决定哪种策略更好，并求要使策略 Y 的期望收益等于策略 X，其获胜概率至少应为多少。

Expected gain for X: E(X) = 0.6×50 + 0.4×(−20) = 30 − 8 = £22. For Y: E(Y) = 0.3×120 + 0.7×(−30) = 36 − 21 = £15. Thus Strategy X is better. To match E(X)=£22 with Y, let p be win probability: p×120 + (1−p)×(−30) = 22 ⇒ 120p − 30 + 30p = 22 ⇒ 150p = 52 ⇒ p ≈ 0.3467. So Y needs a winning chance above 34.7% to be preferable.

策略 X 期望收益 = 0.6×50 + 0.4×(−20) = £22。策略 Y 期望收益 = 0.3×120 + 0.7×(−30) = £15。因此策略 X 更好。令策略 Y 的获胜概率为 p，则 120p − 30(1−p) = 22 ⇒ 150p = 52 ⇒ p ≈ 0.3467。即获胜概率需高于 34.7% 才与 X 等效。

8. Optimising Volume with Derivatives | 导数优化体积问题

An open box is made from a 20 cm by 20 cm square sheet by cutting out equal squares of side x cm from each corner and folding up the flaps. Find x that maximises the volume and calculate that maximum volume.

从一个 20 厘米 × 20 厘米的正方形薄板四角各剪去边长为 x 厘米的小正方形，然后折起成开口盒。求使体积最大的 x 值，并计算最大体积。

The resulting length and width are (20 − 2x) each, height = x. Volume V = x(20 − 2x)² = 4x(10 − x)². Expand: V = 4x(100 − 20x + x²) = 400x − 80x² + 4x³. Differentiate: dV/dx = 400 − 160x + 12x². Set to zero: 12x² − 160x + 400 = 0 ÷4 ⇒ 3x² − 40x + 100 = 0. Solving: x = [40 ± √(1600 − 1200)]/(6) = [40 ± √400]/6 = [40 ± 20]/6. Possible solutions: x = 60/6 = 10 (invalid as it gives zero base) or x = 20/6 = 10/3 ≈ 3.33 cm. Second derivative test: d²V/dx² = −160 + 24x; at x = 10/3 it is −160 + 80 = −80 < 0, confirming a maximum. Max volume V = 4×(10/3)×(10 − 10/3)² = (40/3)×(20/3)² = (40/3)×(400/9) = 16000/27 ≈ 592.6 cm³.

盒子长宽均为 20−2x，高为 x，体积 V = x(20−2x)² = 4x(10−x)²。展开得 V = 400x − 80x² + 4x³。令 dV/dx = 400 − 160x + 12x² = 0，化简为 3x² − 40x + 100 = 0。解得 x = 10/3 ≈ 3.33 厘米（x=10 无效）。二阶导数为负，确为极大值。最大体积 V = (40/3)×(20/3)² = 16000/27 ≈ 592.6 立方厘米。

9. Work Done by a Variable Force | 变力做功的积分计算

A force acting on an object varies with position according to F(x) = 12 − x² newtons, where x is in meters. Calculate the work done by this force when moving the object from x = 0 m to x = 3 m.

作用在一个物体上的力随位置变化为 F(x) = 12 − x² 牛顿，x 单位为米。求物体从 x = 0 米移动到 x = 3 米过程中该力所做的功。

Work = ∫₀³ F(x) dx = ∫₀³ (12 − x²) dx = [12x − (1/3)x³] from 0 to 3 = (12×3 − 27/3) − 0 = 36 − 9 = 27 joules. This definite integral illustrates how calculus connects a variable physical quantity to energy transfer.

功 = ∫₀³ F(x) dx = ∫₀³ (12 − x²) dx = [12x − ⅓x³]₀³ = 36 − 9 = 27 焦耳。这个定积分演示了微积分如何将变化的物理量与能量传递联系起来。

10. Amortisation with Sequences | 数列与分期偿还

A loan of £8000 is repaid in equal annual instalments over 5 years with an interest rate of 5% per annum. Using the formula for the present value of an annuity, calculate the annual payment. Assume payments are made at the end of each year.

一笔 £8000 的贷款分 5 年以等额年金方式偿还，年利率 5%。利用年金现值公式计算每年的还款额。假设每年年末还款。

Let the annual payment be P. The present value of the series of payments is P × (1 − (1.05)⁻⁵) / 0.05 = 8000. Compute the annuity factor: (1 − 1.05⁻⁵)/0.05. 1.05⁻⁵ = 1/(1.05⁵). 1.05⁵ ≈ 1.2762816, so 1.05⁻⁵ ≈ 0.783526. Then 1 − 0.783526 = 0.216474. Divide by 0.05 gives 4.32948. Therefore P = 8000 / 4.32948 ≈ £1847.58. The repayment schedule can be modelled using geometric sequences where each payment’s present value forms a decreasing geometric progression.

设年还款额为 P，各年还款的现值之和为 P × (1 − 1.05⁻⁵) / 0.05 = 8000。年金因子 = (1 − 1.05⁻⁵)/0.05。计算 1.05⁻⁵ ≈ 0.7835，因子 ≈ 4.3295。因此 P ≈ £1847.58。还款计划可借助等比数列建模，每笔还款的现值构成递减的等比级数。

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

更多咨询请联系16621398022（同微信）

Year 11 OCR Further Mathematics: Interdisciplinary Integrated Problem Training | 跨学科综合题型训练