Further Mathematics: ENGAA 2019 Section 1 Answer Key | 进阶数学：ENGAA 2019 S1 答案解析

📚 Further Mathematics: ENGAA 2019 Section 1 Answer Key | 进阶数学：ENGAA 2019 S1 答案解析

Welcome to our detailed answer key for the advanced mathematics questions in ENGAA 2019 Section 1. This resource breaks down selected problems from Part B, covering core further maths topics such as complex numbers, matrices, sequences, calculus, vectors, differential equations and probability. Each solution is explained step‑by‑step to help you master the reasoning required for the Cambridge Engineering Admissions Assessment.

欢迎使用我们为 ENGAA 2019 Section 1 进阶数学题目编写的详细答案解析。本文精选了试卷 Part B 中的关键问题，覆盖复数、矩阵、数列、微积分、向量、微分方程和概率等核心进阶数学主题。每道题均提供逐步解题过程，帮助你掌握剑桥工程入学评估所需的推理技巧。

1. Complex Numbers – Argument of a Quotient | 复数 – 商的辐角

Question 21 asked for the argument of z = (1 + i)/(1 − i). Multiply numerator and denominator by the conjugate of the denominator: z = (1 + i)²/(1² + 1²) = (1 + 2i − 1)/2 = i. Hence z lies on the positive imaginary axis, so arg(z) = π/2. The correct option was B.

第21题要求计算复数 z = (1 + i)/(1 − i) 的辐角。分子分母同乘分母的共轭：z = (1 + i)²/(1 + 1) = i，位于正虚轴上，因此辐角为 π/2。正确答案是 B。

2. Matrices – Inverse of a 2×2 Matrix | 矩阵 – 二阶矩阵的逆

Question 25 gave matrix A = [[2, 1], [4, 3]] and required its inverse. The determinant is det(A) = 2×3 − 1×4 = 2. Using the formula A⁻¹ = (1/det) [[d, −b], [−c, a]] gives A⁻¹ = ½ [[3, −1], [−4, 2]] = [[3/2, −1/2], [−2, 1]]. The matching option was C, where the entries were exactly 3/2, −1/2, −2 and 1.

第25题给出矩阵 A = [[2, 1], [4, 3]]，求其逆矩阵。行列式 det(A) = 2×3 − 1×4 = 2。利用公式 A⁻¹ = (1/det) [[d, −b], [−c, a]]，得 A⁻¹ = ½ [[3, −1], [−4, 2]] = [[3/2, −1/2], [−2, 1]]。对应选项 C，矩阵元素恰好为 3/2, -1/2, -2 和 1。

3. Sequences and Series – Infinite Geometric Sum | 数列与级数 – 无穷等比级数求和

Question 28 presented the infinite geometric series Σ (1/3)ⁿ from n=1 to ∞. The first term a = 1/3 and the common ratio r = 1/3. Since |r| < 1, the sum to infinity is a/(1 − r) = (1/3)/(1 − 1/3) = (1/3)/(2/3) = 1/2. The answer was A.

第28题给出无穷等比级数 Σ_{n=1}^{∞} (1/3)ⁿ。首项 a = 1/3，公比 r = 1/3，|r| < 1，无穷和为 a/(1 − r) = (1/3)/(2/3) = 1/2。答案为 A。

4. Parametric Differentiation – Chain Rule | 参数微分 – 链式法则

Question 31 defined a curve parametrically: x = t² + 1, y = t³ − 3t, and asked for dy/dx at t = 2. Differentiate: dx/dt = 2t, dy/dt = 3t² − 3. Then dy/dx = (dy/dt)/(dx/dt) = (3t² − 3)/(2t). Substituting t = 2 gives (3×4 − 3)/(4) = 9/4. The correct choice was D.

第31题以参数方程定义曲线：x = t² + 1，y = t³ − 3t，求 t = 2 处的 dy/dx。求导：dx/dt = 2t, dy/dt = 3t² − 3。于是 dy/dx = (3t² − 3)/(2t)。代入 t = 2 得 (12 − 3)/4 = 9/4。正确选项为 D。

5. Integration by Substitution – Definite Integral | 换元积分 – 定积分

Question 34 required evaluation of ∫₀¹ 2x e^(x²) dx. Use the substitution u = x², so du = 2x dx. The limits become u = 0 and u = 1. The integral transforms to ∫₀¹ e^u du = [e^u]₀¹ = e − 1. The answer matched option B.

第34题计算定积分 ∫₀¹ 2x e^(x²) dx。采用换元 u = x²，则 du = 2x dx，积分限变为 u = 0 和 u = 1。原式化为 ∫₀¹ e^u du = e − 1。对应选项 B。

6. Vectors – Magnitude and Direction | 向量 – 模与方向

Question 37 gave points A(1, 2, 3) and B(4, 0, −1). The vector AB = (4−1, 0−2, −1−3) = (3, −2, −4). Its magnitude is √(3² + (−2)² + (−4)²) = √(9 + 4 + 16) = √29. The unit vector in the direction of AB is (3/√29, −2/√29, −4/√29). The required answer was C.

第37题给出点 A(1, 2, 3) 和 B(4, 0, −1)。向量 AB = (3, −2, −4)，模为 √(9 + 4 + 16) = √29。AB 方向的单位向量为 (3/√29, −2/√29, −4/√29)。正确选项是 C。

7. Differential Equations – Separation of Variables | 微分方程 – 分离变量法

Question 39 set up the initial value problem dy/dx = 2xy with y(0) = 1. Separating variables gives ∫ dy/y = ∫ 2x dx, so ln|y| = x² + C. Exponentiating yields y = Ae^(x²). Using y(0)=1 forces A = 1, hence y = e^(x²). The answer was A.

第39题给出初值问题 dy/dx = 2xy，y(0)=1。分离变量得 ∫ dy/y = ∫ 2x dx，即 ln|y| = x² + C，从而 y = Ae^(x²)。代入 y(0)=1 得 A=1，故特解为 y = e^(x²)。答案为 A。

8. Binomial Distribution – Probability Calculation | 二项分布 – 概率计算

Question 40 modelled a random variable X ~ B(5, 0.4) and asked for P(X = 3). Using the binomial formula: P(X=3) = C(5,3) × (0.4)³ × (0.6)². Compute C(5,3) = 10, (0.4)³ = 0.064, (0.6)² = 0.36. Multiplying gives 10 × 0.064 × 0.36 = 0.2304. The correct option was D.

第40题假设随机变量 X ~ B(5, 0.4)，求 P(X=3)。运用二项公式：P(X=3) = C(5,3) × (0.4)³ × (0.6)²。计算得 C(5,3)=10，(0.4)³=0.064，(0.6)²=0.36，乘积为 0.2304。对应选项 D。

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