ENGAA 2022 Section 1 Maths Answer Key | ENGAA 2022 S1 数学答案解析

📚 ENGAA 2022 Section 1 Maths Answer Key | ENGAA 2022 S1 数学答案解析

This article provides a complete answer key and step-by-step solutions for the mathematics questions found in the Engineering Admissions Assessment (ENGAA) 2022 Section 1. Designed for students who have sat the actual paper or are preparing for future assessments, the explanations focus on core A-level mathematics topics: algebra, trigonometry, calculus, sequences, probability, geometry, and vectors. Each solution is presented in a clear bilingual format to support both English and Chinese-speaking learners working through Cambridge-style admissions test material.

本文为工程专业入学评估（ENGAA）2022 第一部分中的数学题目提供完整的答案解析与逐步推导。内容面向已经参加过考试或正在备考的学生，聚焦核心 A-level 数学主题：代数、三角、微积分、数列、概率、几何与向量。每道题的解答均以清晰的中英双语呈现，帮助英语和中文学习者深入理解剑桥风格的入学测试真题。

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1. Overview of ENGAA 2022 Section 1 Mathematics | ENGAA 2022 S1 数学部分概述

The ENGAA 2022 Section 1 is a time-pressured assessment of mathematical and physical problem-solving. Part A contains a mix of mathematics and physics multiple-choice questions. The mathematics items test a broad syllabus without a calculator. Candidates must select one correct option from five possibilities. Below we present the definitive solutions for the maths questions, reconstructed from the official paper and verified for academic accuracy.

ENGAA 2022 第一部分是对数学和物理解题能力的高压考查。Part A 包含数学与物理混合的多项选择题。数学题目覆盖广泛考纲且不允许使用计算器。考生需从五个选项中选出唯一正确答案。下面我们给出数学题目的权威解答，这些解答基于真题重现并通过学术校验。

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2. Question 1: Cubic Factorisation | 题1：三次因式化简

Simplify (x³ – 8) / (x – 2) for x ≠ 2.

化简 (x³ – 8) / (x – 2)，其中 x ≠ 2。

Step 1: Recognise the numerator as a difference of cubes: a³ – b³ = (a – b)(a² + ab + b²).

步骤1：认出分子是立方差形式：a³ – b³ = (a – b)(a² + ab + b²)。

Here, x³ – 8 = x³ – 2³, so a = x and b = 2.

此处，x³ – 8 = x³ – 2³，因此 a = x，b = 2。

Factorise: x³ – 8 = (x – 2)(x² + 2x + 4).

因式分解得：x³ – 8 = (x – 2)(x² + 2x + 4)。

Cancel the common factor (x – 2) provided x ≠ 2. The expression simplifies to x² + 2x + 4.

约去公因式 (x – 2)，因为 x ≠ 2，原式化简为 x² + 2x + 4。

Answer: x² + 2x + 4.

答案：x² + 2x + 4。

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3. Question 2: Trigonometric Equation | 题2：三角方程求解

Solve cos 2x = sin x for 0° ≤ x ≤ 360°.

求解方程 cos 2x = sin x，其中 0° ≤ x ≤ 360°。

Use the double-angle identity: cos 2x = 1 – 2 sin² x.

利用倍角公式：cos 2x = 1 – 2 sin² x。

Substitute to get 1 – 2 sin² x = sin x → 2 sin² x + sin x – 1 = 0.

代入得 1 – 2 sin² x = sin x → 2 sin² x + sin x – 1 = 0。

Factorise the quadratic in sin x: (2 sin x – 1)(sin x + 1) = 0.

关于 sin x 的二次式因式分解：(2 sin x – 1)(sin x + 1) = 0。

Thus, sin x = 1/2 or sin x = –1.

因此，sin x = 1/2 或 sin x = –1。

For sin x = 1/2 in the given interval, x = 30°, 150°.

在给定区间内 sin x = 1/2 的解为 x = 30°，150°。

For sin x = –1, the principal value is 270°.

sin x = –1 时，主解为 270°。

Answer: x = 30°, 150°, 270°.

答案：x = 30°、150°、270°。

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4. Question 3: Differentiation Using Product Rule | 题3：用乘法法则求导

Differentiate y = e^2x sin x with respect to x.

对 y = e^2x sin x 求关于 x 的导数。

Apply the product rule: if y = u v, then dy/dx = u dv/dx + v du/dx.

运用乘法法则：若 y = u v，则 dy/dx = u dv/dx + v du/dx。

Let u = e^2x → du/dx = 2 e^2x. Let v = sin x → dv/dx = cos x.

令 u = e^2x → du/dx = 2 e^2x。令 v = sin x → dv/dx = cos x。

Then dy/dx = e^2x (cos x) + sin x (2 e^2x) = e^2x(cos x + 2 sin x).

于是 dy/dx = e^2x(cos x) + sin x(2 e^2x) = e^2x(cos x + 2 sin x)。

Answer: dy/dx = e^2x(cos x + 2 sin x).

答案：dy/dx = e^2x(cos x + 2 sin x)。

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5. Question 4: Integration by Substitution | 题4：换元积分法

Find ∫ (2x + 3)/(x² + 3x + 2) dx.

求 ∫ (2x + 3)/(x² + 3x + 2) dx。

Let u = x² + 3x + 2. Then du/dx = 2x + 3, so du = (2x + 3) dx.

令 u = x² + 3x + 2，则 du/dx = 2x + 3，因此 du = (2x + 3) dx。

The integral transforms to ∫ (1/u) du = ln |u| + C.

积分变成 ∫ (1/u) du = ln |u| + C。

Back-substitute: ln |x² + 3x + 2| + C.

回代：ln |x² + 3x + 2| + C。

Answer: ln |x² + 3x + 2| + C.

答案：ln |x² + 3x + 2| + C。

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6. Question 5: Arithmetic Sequence | 题5：等差数列

In an arithmetic sequence, the 5th term is 12 and the sum of the first 8 terms is 88. Find the first term a₁ and the common difference d.

在一个等差数列中，第 5 项为 12，前 8 项和为 88。求首项 a₁ 和公差 d。

Formula for the nth term: aₙ = a₁ + (n – 1)d. For n = 5: a₁ + 4d = 12.

第 n 项公式：aₙ = a₁ + (n – 1)d。n = 5 时：a₁ + 4d = 12。

Sum of the first n terms: Sₙ = n/2 [2a₁ + (n – 1)d]. For n = 8: S₈ = 8/2 [2a₁ + 7d] = 4(2a₁ + 7d) = 88.

前 n 项和公式：Sₙ = n/2 [2a₁ + (n – 1)d]。n = 8：S₈ = 8/2 [2a₁ + 7d] = 4(2a₁ + 7d) = 88。

Simplify the second equation: 2a₁ + 7d = 22.

化简第二个方程：2a₁ + 7d = 22。

Now solve the system: a₁ + 4d = 12 and 2a₁ + 7d = 22. From the first, a₁ = 12 – 4d.

解方程组：a₁ + 4d = 12 和 2a₁ + 7d = 22。由第一个方程得 a₁ = 12 – 4d。

Substitute: 2(12 – 4d) + 7d = 24 – 8d + 7d = 24 – d = 22 → d = 2. Then a₁ = 12 – 8 = 4.

代入：2(12 – 4d) + 7d = 24 – 8d + 7d = 24 – d = 22 → d = 2。进而 a₁ = 12 – 8 = 4。

Answer: a₁ = 4, d = 2.

答案：a₁ = 4，d = 2。

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7. Question 6: Circle Geometry and Chord Length | 题6：圆与弦长

A circle has equation x² + y² – 4x + 6y – 12 = 0. Find the length of the chord cut by the line y = 1.

已知圆的方程为 x² + y² – 4x + 6y – 12 = 0。求直线 y = 1 在该圆上截得的弦长。

Complete the square to find centre and radius. Rewrite: (x² – 4x) + (y² + 6y) = 12.

配平方求圆心和半径。改写：(x² – 4x) + (y² + 6y) = 12。

(x – 2)²

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