📚 Year 13 CAIE Further Mathematics Case Study Practice | Year 13 CAIE 进阶数学案例分析实战演练
Welcome to this case study practice article designed for Year 13 CAIE Further Mathematics students. Through a series of contextual problems, you will reinforce key topics such as complex numbers, matrices, differential equations, hyperbolic functions, polar coordinates, vectors, induction, Maclaurin series, mechanics, and statistics. Each case is broken into clear steps with bilingual explanations to sharpen your problem-solving skills and exam technique.
欢迎来到为 Year 13 CAIE 进阶数学学生设计的案例分析实战演练。通过一系列实际情境问题，你将巩固复数、矩阵、微分方程、双曲函数、极坐标、向量、归纳法、麦克劳林级数、力学和统计等核心专题。每个案例都拆分为逐步解析，并配有中英双语解释，助你提升解题能力与应试技巧。
1. Complex Numbers and Transformations | 复数与变换
Consider the transformation w = (1 + i)z + 3. We first write the complex multiplier in polar form: 1 + i = √2 eiπ/4. Multiplication by this factor therefore rotates a point by π/4 radians and stretches its distance from the origin by a factor of √2. Adding 3 then translates every image by the vector (3, 0) in the complex plane.
考虑变换 w = (1 + i)z + 3。先将乘数写成极坐标形式：1 + i = √2 eiπ/4。因此乘以该因子意味着将点绕原点旋转 π/4 弧度，并将其到原点的距离拉伸 √2 倍。随后加上 3 相当于在复平面内将所有像点平移向量 (3, 0)。
To find the image of the circle |z| = 2, rearrange the transformation: z = (w - 3)/(1 + i). Substituting into the condition gives |w - 3| / |1 + i| = 2, so |w - 3| = 2√2. Hence the image is a circle centred at 3 with radius 2√2. This neatly illustrates how multiplication scales and rotates while addition shifts the centre.
为求圆 |z| = 2 的像，变换式整理得 z = (w - 3)/(1 + i)。代入条件得 |w - 3| / |1 + i| = 2，故 |w - 3| = 2√2。因此像是一个以 3 为圆心、半径为 2√2 的圆。这清晰地展示了乘法如何缩放和旋转，而加法如何平移圆心。
2. Matrices and Linear Transformations | 矩阵与线性变换
A linear transformation T on the plane is represented by the matrix M = [[2, 1], [0, 3]]. Find the image of the line y = 4x + 2 under T. We start by writing a general point on the line as (t, 4t+2). Applying T gives image coordinates (x', y') = M (t, 4t+2)T.
平面上的线性变换 T 由矩阵 M = [[2, 1], [0, 3]] 表示。求直线 y = 4x + 2 在 T 下的像。首先将直线上的一般点写为 (t, 4t+2)。应用 T 得到像点坐标 (x', y') = M (t, 4t+2)T。
Alternatively, it is often simpler to use the inverse transformation. The inverse matrix is M-1 = [[1/2, -1/6], [0, 1/3]]. Expressing the pre-image in terms of the image gives x = (1/2)x' - (1/6)y' and y = (1/3)y'. Substituting into the original line equation yields (1/3)y' = 4[(1/2)x' - (1/6)y'] + 2. Simplifying, we obtain y' = 2x' + 6 after clearing fractions. So the image is the line y = 2x + 6 (dropping primes).
此外，使用逆变换往往更简单。逆矩阵为 M-1 = [[1/2, -1/6], [0, 1/3]]。用像坐标表示原像坐标得 x = (1/2)x' - (1/6)y' 且 y = (1/3)y'。代入原直线方程得 (1/3)y' = 4[(1/2)x' - (1/6)y'] + 2。化简并消去分母后得到 y' = 2x' + 6，所以像直线为 y = 2x + 6（去掉撇号）。
3. Differential Equations and Modelling | 微分方程与建模
A hot coffee at 90 °C is left in a room at a steady 20 °C. Newton's law of cooling states dT/dt = -k(T - 20) with k > 0, where T is the temperature in °C and t is time in minutes. Separating variables and integrating: ∫ dT/(T - 20) = ∫ -k dt gives ln|T - 20| = -kt + C. Hence T - 20 = Ae-kt. Using T(0) = 90 determines A = 70, so the solution is T = 20 + 70e-kt.
一杯 90 °C 的热咖啡放置在恒温 20 °C 的房间内。牛顿冷却定律表示为 dT/dt = -k(T - 20)，k > 0，其中 T 为温度 (°C)，t 为时间 (分钟)。分离变量并积分：∫ dT/(T - 20) = ∫ -k dt 得到 ln|T - 20| = -kt + C。于是 T - 20 = Ae-kt。代入初始条件 T(0) = 90 确定 A = 70，故解为 T = 20 + 70e-kt。
If after 5 minutes the temperature has dropped to 60 °C, we find k. Plugging in: 60 = 20 + 70e-5k → e-5k = 40/70 = 4/7, so k = - (1/5) ln(4/7). Using this, predict the temperature after 10 minutes: T(10) = 20 + 70e
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