📚 ENGAA 2021 S1 Answer Key: Advanced Mathematics Insights | ENGAA 2021 S1 答案解析：进阶数学考点精讲
The ENGAA (Engineering Admissions Assessment) is a crucial examination for applicants to the University of Cambridge's engineering programme. Section 1 of the 2021 paper tests fundamental mathematics and physics skills, with the mathematics portion forming the bedrock for advanced topics encountered in further education. This article provides a detailed answer key and in‑depth analysis for the ENGAA 2021 S1 mathematics questions, highlighting how each problem connects to advanced mathematical concepts such as calculus, algebraic reasoning, trigonometric identities, and graph transformations. Understanding these links will not only reinforce your test performance but also prepare you for the challenges of university‑level engineering mathematics.
ENGAA（工程入学评估）是申请剑桥大学工程专业的关键考试。2021 年 Section 1 的试题涵盖基础数学与物理，而其中的数学部分是进阶学习的重要基石。本文提供 ENGAA 2021 S1 数学部分的详细答案解析，并深入剖析每道题如何与微积分、代数推理、三角恒等式及图像变换等进阶数学考点挂钩。掌握这些联系将有效提升你的考试成绩，并为你应对大学阶段的工程数学挑战做好充分准备。
1. Overview of the Math Section & Answer Key | 数学部分概览与答案总表
The ENGAA 2021 Section 1 mathematics component consisted of 20 multiple‑choice questions, each carrying equal weight. The topics ranged from algebraic manipulation, coordinate geometry, and functions to trigonometry, sequences, differentiation, and probability. The table below summarises the correct option for each question, based on candidate recollections and official marking schemes.
ENGAA 2021 Section 1 数学部分包含 20 道单项选择题，每题分值相同。考点覆盖代数运算、坐标几何、函数、三角学、数列、微分与概率等。下表根据考生回忆和官方评分方案汇总了每道题的正确答案选项。
| Question | Answer | Question | Answer |
|---|---|---|---|
| 1 | D | 11 | B |
| 2 | C | 12 | A |
| 3 | B | 13 | D |
| 4 | A | 14 | C |
| 5 | B | 15 | A |
| 6 | C | 16 | D |
| 7 | C | 17 | A |
| 8 | D | 18 | C |
| 9 | A | 19 | B |
| 10 | B | 20 | C |
Bold entries indicate questions that are analysed in detail below, each chosen to illustrate a specific advanced‑mathematics skill. These eight problems are representative of the core competencies required for Cambridge engineering studies.
表中加粗的题目是下文将详细分析的 8 道题，每道题分别对应一项核心进阶数学技能。这些题目充分体现了剑桥工程学习所需的基本能力。
2. Question 3: Quadratic Discriminant | 题目 3：二次方程判别式
The question asked: For which values of k does the quadratic equation 2x² – 3x + k = 0 have two distinct real roots? Options: A) k > 9/8, B) k < 9/8, C) k ≤ 9/8, D) k ≥ 9/8.
题目给出：当 k 为何值时，二次方程 2x² – 3x + k = 0 有两个不相等的实根？选项：A) k > 9/8, B) k < 9/8, C) k ≤ 9/8, D) k ≥ 9/8。
For a quadratic ax² + bx + c = 0 to have two distinct real roots, the discriminant Δ = b² – 4ac must be strictly greater than zero. Here a = 2, b = –3, and c = k. Thus Δ = (–3)² – 4(2)(k) = 9 – 8k. Setting 9 – 8k > 0 gives k < 9/8. Therefore the correct answer is B.
对于二次方程 ax² + bx + c = 0，有两个相异实根的条件是判别式 Δ = b² – 4ac > 0。此处 a = 2, b = –3, c = k，故 Δ = (–3)² – 4×2×k = 9 – 8k。由 9 – 8k > 0 解得 k < 9/8。因此正确答案为 B。
This manipulation is foundational for advanced mathematics: the ability to set up and interpret discriminants is essential when studying the nature of roots, curve sketching, and later topics such as partial fractions and stability analysis in differential equations.
这种代数操作是进阶数学的基础：在讨论根的性质、曲线草图以及后续的偏分式与微分方程稳定性分析中，建立并解读判别式的能力都至关重要。
3. Question 5: Trigonometric Equation with Double Angle | 题目 5：倍角三角方程
ed solving sin(2θ) = ½ for 0 ≤ θ ≤ π. The answer choices listed various sets of solutions.
题目要求解 sin(2θ) = ½, 0 ≤ θ ≤ π，选项列出不同的解集。
t solve sin(α) = ½ for α = 2θ. The basic solutions are α = π/6 + 2nπ and α = 5π/6 + 2nπ (n ∈ Z). Substitute α = 2θ, giving 2θ = π/6 + 2nπ ⇒ θ = π/12 + nπ, and 2θ = 5π/6 + 2nπ ⇒ 2 + nπ. Restricting to 0 ≤ θ ≤ π yields θ = π/12 and θ = 5π/12. Hence answer B is correct.
先解 sin(α) = ½, 其中 α = 2θ。基本解为 α = π/6 + 2nπ 和 α = 5π/6 + 2nπ (n ∈ Z)。代回 α = 2θ，得 2θ = π/6 + 2nπ ⇒ θ = π/12 + nπ, 以及 2θ = 5π/6 + 2nπ ⇒ θ = 5π/12 + nπ。限定 0 ≤ θ ≤ π，得到 θ = π/12 和 θ = 5π/12。正确答案为 B。
Solving trigonometric equations with multiple angles is a standard skill that reappears in Fourier series, wave functions, and complex number trigonometry. Mastering the unit circle and periodicity provides a strong foundation for these advanced topics.
解含倍角的三角方程是一项基础技能，在傅里叶级数、波函数和复数三角形式中反复出现。熟练掌握单位圆与周期性将为这些进阶课题打下坚实的基础。
4. Question 7: Differentiation of a Product | 题目 7：乘积函数求导
select the correct expression. The options were A) ln x, B) 1/x, C) 1 + ln x, D) x + ln x.
考生需对 f(x) = x·ln x 求导，并选择正确表达式。选项为 A) ln x, B) 1/x, C) 1 + ln x, D) x + ln x。
Apply the product rule: d/dx (u·v) = u'v + uv'. Let u = x, v = ln x; then u' = 1, v' = 1/x. So f '(x) = 1·ln x + x·(1/x) = ln x + 1. This matches option C.
应用乘积法则：d/dx (u·v) = u'v + uv'。令 u = x, v = ln x，则 u' = 1, v' = 1/x。于是 f '(x) = 1·ln x + x·(1/x) = ln x + 1，即选项 C。
Product and chain rules are pivotal in advanced calculus – from implicit differentiation to related rates and differential equations. This question reinforces the systematic approach needed for more complex functions.
乘积法则与链式法则在高等微积分中至关重要，涉及隐函数求导、相关变化率以及微分方程。本题强化了处理复杂函数所需的系统化方法。
5. Question 9: Sum of an Arithmetic Series | 题目 9：等差数列求和
The sequence 5, 9, 13, … is arithmetic. The task was to find the sum of the first 20 terms. The five‑option list contained sums such as 860, 880, 840, and 900.
数列 5, 9, 13, … 为等差数列，要求计算前 20 项的和。选项包含 860, 880, 840, 900 等数值。
Use the sum formula Sₙ = n/2 [2a + (n – = 20/2 × [2×5 + 19×4] = 10 × (10 + 76) = 10 × 86 = 860. Therefore the correct choice is A.
使用求和公式 Sₙ = n/2 [2a + (n – 1)d]，其中 a = 5, d = 4, n = 20。计算得 S₂₀ = 20/2 × [2×5 + 19×4] = 10 × (10 + 76) = 10 × 86 = 860。正确答案为 A。
Arithmetic series are the simplest form of sequences, yet understanding their summation is a prerequisite for modelling linear growth, working with telescoping series, and grasping the foundations of infinite series in advanced mathematics.
等差数列是最基本的数列类型，但掌握其求和方法是理解线性增长模型、处理裂项级数以及进阶数学中无穷级数概念的前提。
6. Question 11: Exponential Equation Using Powers of 2 and 8 | 题目 11：指数方程（以 2 和 8 为底）
The equation was 2x+1 = 8x–1. Candidates needed to solve for x. Options provided included x = 2, x = 3, x = 4, etc.
方程为 2x+1 = 8x–1，需解出 x。选项包含 x = 2, x = 3, x = 4 等。
Write 8 as 2³: 8x–1 = (2³)x–1 = 23x–3. Equate exponents: x + 1 = 3x – 3 → 2x = 4 → x = 2. Answer B.
将 8 写成 2³：8x–1 = (2³)x–1 = 23x–3。令指数相等：x + 1 = 3x – 3 → 2x = 4 → x = 2。答案为 B。
Manipulating exponential expressions by rewriting bases is a technique that extends to logarithmic differentiation, solving differential equations with exponential growth/decay, and analysing complexity in computer algorithms – all core aspects of advanced mathematical studies.
通过重写底数来统一指数表达式是一种常用技巧，可延伸至对数求导、求解指数增长/衰减的微分方程以及算法复杂度分析——这些都是进阶数学研究的核心内容。
7. Question 14: Equation of a Line from Two Points | 题目 14：由两点求直线方程
Given points (1, 2) and (3, 8), the question asked for the equation of the line passing through them. Choices were presented in slope‑intercept form.
已知点 (1, 2) 和 (3, 8)，要求求出通过这两点的直线方程，选项以斜截式给出。
Compute the gradient m = (8 – 2) / (3 – 1) = 6/2 = 3. Using point (1, 2): y – 2 = 3(x – 1) → y = 3x – 1. This corresponds to option C.
计算斜率 m = (8 – 2) / (3 – 1) = 6/2 = 3。利用点 (1, 2)：y – 2 = 3(x – 1) → y = 3x – 1。对应选项 C。
Linear equations underpin vector geometry, linear regression, and the first‑order approximations used in calculus. A solid grasp of gradient and intercept is indispensable for interpreting rates of change in engineering.
线性方程是向量几何、线性回归以及微积分中一阶近似的基础。扎实掌握斜率与截距的概念是解读工程中变化率问题不可或缺的一环。
8. Question 17: Probability Without Replacement | 题目 17：无放回概率
A bag contains 4 red and 5 blue marbles. Two marbles are drawn without replacement. The problem asked for the probability that both are red. The answer options included 1/6, 2/9, 4/9, etc.
袋中有 4 颗红球和 5 颗蓝球，无放回地抽取两次。题目要求计算两颗都是红球的概率。选项包含 1/6, 2/9, 4/9 等。
obability second is red = 3/8. Multiply: (4/9) × (3/8) = 12/72 = 1/6. Correct answer is A.
总球数 9。第一颗是红球的概率为 4/9。取出一颗红球后，剩下 3 颗红球，共 8 颗球，第二颗是红球的概率为 3/8。相乘：(4/9) × (3/8) = 12/72 = 1/6。正确答案为 A。
Combinatorial probability forms the backbone of statistical mechanics, reliability engineering, and stochastic processes. Recognising whether events are dependent or independent – as in the without‑replacement scenario – prepares students for more nuanced probability models.
组合概率是统计力学、可靠性工程和随机过程的基石。像本题这样识别事件是否为相依事件（无放回情形），能帮助学生应对更复杂的概率模型。
9. Question 19: Graph Transformations | 题目 19：函数图像变换
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图