Cross-disciplinary Integrated Problem Training for CAIE GCSE Further Mathematics | GCSE CAIE 进阶数学：跨学科综合题型训练

📚 Cross-disciplinary Integrated Problem Training for CAIE GCSE Further Mathematics | GCSE CAIE 进阶数学：跨学科综合题型训练

In the CAIE IGCSE Additional Mathematics (0606) course, many real-world problems require combining mathematical techniques with concepts from physics, biology, economics, and other sciences. This article presents a series of cross-disciplinary integrated problems that help you practise applying quadratic functions, calculus, trigonometry, logarithms, and other topics in authentic contexts. Each problem is followed by a step-by-step solution, bridging pure maths and practical applications.

在 CAIE IGCSE 附加数学 (0606) 课程中，许多现实世界的问题需要将数学技巧与物理、生物、经济和其他科学的概念相结合。本文提供了一系列跨学科的综合题型训练，帮助你在真实情境中练习应用二次函数、微积分、三角学、对数等主题。每个问题都配有逐步解答，在纯粹数学和实际应用之间搭建桥梁。

1. Kinematics and Quadratic Functions | 运动学与二次函数

A ball is thrown vertically upwards with an initial speed of 20 m s⁻¹ from the ground. Its height s (in metres) after t seconds is given by s = 20t − 5t². (a) Find the maximum height reached. (b) Determine the times when the ball is at a height of 15 m. (c) How long is the ball more than 15 m above the ground?

一个球以 20 m s⁻¹ 的初速度从地面竖直向上抛出。t 秒后，其高度 s（米）由 s = 20t − 5t² 给出。(a) 求到达的最大高度。(b) 确定球在 15 m 高度时的时刻。(c) 球在 15 m 高度以上的时间有多长？

Solution: The equation is a quadratic function. (a) Maximum height occurs at vertex t = −b/(2a) = −20/(2 × (−5)) = 2 s. Then s = 20×2 − 5×2² = 20 m. (b) Solve s = 15: 20t − 5t² = 15 ⇒ 5t² − 20t + 15 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t−1)(t−3)=0, so t = 1 s and t = 3 s. (c) The ball is above 15 m between t = 1 and t = 3, so duration = 2 s.

解答：这是一个二次函数。(a) 最大高度在顶点 t = −b/(2a) = −20/(2 × (−5)) = 2 s 处取得。然后 s = 20×2 − 5×2² = 20 m。(b) 解 s = 15：20t − 5t² = 15 ⇒ 5t² − 20t + 15 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t−1)(t−3)=0，故 t = 1 s 和 t = 3 s。(c) 球在 t = 1 到 t = 3 之间高于 15 m，所以持续时间为 2 s。

2. Optimisation and Differential Calculus | 最优化与微分

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