ENGAA 2018 Section 1 Answer Key Explained | ENGAA 2018 S1 答题解析

📚 ENGAA 2018 Section 1 Answer Key Explained | ENGAA 2018 S1 答题解析

Mastering the ENGAA Section 1 requires not only subject knowledge but also the ability to solve problems accurately under time pressure. This article presents the full answer key for the 2018 paper with step–by–step bilingual explanations, covering every mathematics and physics question. Use it to audit your performance, correct mistakes, and internalise the efficient techniques that top candidates employ.

攻克 ENGAA 第一部分不仅需要学科知识，更要求能够在时间压力下准确解题。本文给出了 2018 年试卷的完整答题关键，并配以逐步双语解析，涵盖每一道数学与物理题目。你可以用它来核查成绩、纠正错误，并掌握顶尖考生所使用的高效解题方法。

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1. Algebraic Manipulation and Simplification | 代数化简与运算

Questions 1 and 2 assess the ability to simplify rational expressions and solve exponential equations. Speed here comes from recognising standard factorisations and applying index laws without algebraic overload.

问题 1 和 2 考查化简有理式与求解指数方程的能力。要想快速得分，必须识别标准因式分解并熟练运用指数法则，避免繁琐的代数推导。

Question 1 – Simplify (x² – 5x + 6) / (x – 3). Factorise the numerator as (x – 2)(x – 3). Cancelling the factor (x – 3) leaves x – 2, valid for x ≠ 3. The answer is D.

问题 1 – 化简 (x² – 5x + 6)/(x – 3)。将分子因式分解为 (x – 2)(x – 3)，约去公因子 (x – 3) 得到 x – 2（x ≠ 3）。答案为 D。

Question 2 – Solve 4^x+1 = 8^2x–1. Express both sides as powers of 2: (2²)^x+1 = 2^2x+2 and (2³)^2x–1 = 2^6x–3. Equate exponents: 2x + 2 = 6x – 3 → 4x = 5 → x = 5/4, which corresponds to A.

问题 2 – 解 4^x+1 = 8^2x–1。两边化作以 2 为底的幂：(2²)^x+1 = 2^2x+2，(2³)^2x–1 = 2^6x–3。令指数相等：2x + 2 = 6x – 3 → 4x = 5 → x = 5/4。对应选项 A。

Question	Key
1	D
2	A

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2. Functions and Graphs | 函数与图像

Questions 3 and 4 test function notation, domain/range, and graphical transformations. The ability to mentally sketch a transformed graph saves valuable minutes.

问题 3 和 4 考查函数符号、定义域与值域以及图像变换。在脑海中迅速描绘变换后的图像能够节省宝贵的答题时间。

Question 3 – Given f(x) = ln(2x + 1) with domain x > –½, find the inverse function f⁻¹(x). Swap x and y: x = ln(2y + 1) → eˣ = 2y + 1 → y = (eˣ – 1)/2. The range of the inverse is the domain of f, so the answer is B.

问题 3 – 已知 f(x) = ln(2x + 1)，定义域 x > –½，求反函数 f⁻¹(x)。交换 x 和 y：x = ln(2y + 1) → eˣ = 2y + 1 → y = (eˣ – 1)/2。反函数的值域即为原函数定义域，答案为 B。

Question 4 – The graph of y = f(x) is transformed to y = 2f(x – stretch by factor 2, and a vertical shift 3 units up. Identifying the correct sketch gives C.

问题 4 – 图像由 y = f(x) 变换为 y = 2f(x – 1) + 3。这表示先向右平移 1 个单位，再垂直拉伸为原来的 2 倍，最后向上平移 3 个单位。识别正确草图可得答案为 C。

Question	Key
3	B
4	C

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3. Trigonometry | 三角学

Questions 5 and 6 involve solving trigonometric equations and simplifying expressions using identities. Working within the principal range and using quadrant checks prevents sign errors.

问题 5 和 6 涉及求解三角方程和运用恒等式化简表达式。在主值区间内运算并使用象限验证可以避免符号错误。

Question 5 – Using sin 2x = 2 sin x cos x, the equation becomes 2 sin x cos x = cos x → cos x (2 sin x – = 0 giving x = π/2, 3π/2; or sin x = ½ giving x = π/6, 5π/6. The set of solutions matches D.

问题 5 – 在 0 ≤ x ≤ 2π 内解 sin 2x = cos x。利用 sin 2x = 2 sin x cos x，方程化为 2 sin x cos x = cos x → cos x (2 sin x – 1) = 0。由 cos x = 0 得 x = π/2, 3π/2；由 sin x = ½ 得 x = π/6, 5π/6。解集对应 D。

Question 6 – Simplify (1 – cos²θ) / (sin θ cos θ). Since 1 – sin²θ, the expression becomes sin²θ / (sin θ cos θ) = tan θ. The correct simplification is C.

问题 6 – 化简 (1 – cos²θ)/(sin θ cos θ)。由于 1 – cos²θ = sin²θ，表达式化为 sin²θ/(sin θ cos θ) = tan θ。正确的化简结果为 C。

Question	Key
5	D
6	C

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4. Differentiation and Integration | 微分与积分

Questions 7 and 8 examine calculus fundamentals: differentiating a composite ln function and evaluating a definite polynomial integral. A strong grasp of the chain rule and power rule is enough to secure full marks.

问题 7 和 8 考查微积分基础：复合对数函数求导与多项式定积分计算。熟练掌握链式法则和幂法则足以拿到满分。

Question 7 – fferentiate y = ln (sin x). By the chain rule, dy/dx = (1 / sin x) · cos x = cot x. Option B is correct.

问题 7 – 对 y = ln (sin x) 求导。运用链式法则，dy/dx = (1 / sin x) · cos x = cot x。B 为正确选项。

Question 8 – Evaluate ∫₁² (3x² – 2x + 1) dx. Integrate term by term: [x³ – x² + x] from 1 to 2 = (8 – 4 + 2) – (1 – 1 + 1) = 6 – 1 = 5. The answer is A.

问题 8 – 计算 ∫₁² (3x² – 2x + 1) dx。逐项积分：[x³ – x² + x]₁² = (8 – 4 + 2) – (1 – 1 + 1) = 6 – 1 = 5。答案为 A。

Question	Key
7	B
8	A

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5. Coordinate Geometry and Vectors | 坐标几何与向量

Question 9 blends vector geometry with algebraic solving. Visualising the intersection condition makes the working straightforward.

问题 9 融合了向量几何与代数求解。将相交条件可视化能让运算过程直截了当。

Question 9 – Two lines are given in vector form: r = (1,2) + λ(3,–1) and r = (4,1) + μ(–2,3). Set the parametric equations equal: 1+3λ = 4–2μ, 2– λ = 1, μ = 0. Substituting back gives the intersection point (4,1). The correct option is D.

问题 9 – 两条直线以向量形式给出：r = (1,2) + λ(3,–1) 和 r = (4,1) + μ(–2,3)。联立参数方程：1+3λ = 4–2μ，2–λ = 1+3μ。解得 λ = 1，μ = 0。代回得交点 (4,1)。正确答案为 D。

Question	Key
9	D

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6. Sequences and Series | 数列与级数

Question 10 focuses on arithmetic sequences. Using the standard sum formula eliminates the need to list terms individually.

问题 10 聚焦于等差数列。使用标准求和公式无需逐一列出各项。

Question 10 – term 5 and common difference 2. Find the sum of the first 20 terms. Use Sₙ = n/2 [2a + (n–1)d]: S₂₀ = 10 [10 + 38] = 10 × 48 = 480. The answer matches C.

问题 10 – 等差数列首项为 5，公差为 2，求前 20 项和。用公式 Sₙ = n/2 [2a + (n–1)d]：S₂₀ = 10 [10 + 38] = 10 × 48 = 480。对应选项 C。

Question	Key
10	C

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7. Kinematics | 运动学

Questions 11 and 12 require interpreting motion graphs and applying SUVAT equations. The physics section heavily relies on mathematical modelling, making algebraic fluency essential.

问题 11 和 12 要求解读运动图像并运用 SUVAT 方程。物理部分高度依赖数学建模，因此代数流畅度至关重要。

Question 11 – A velocity–time graph shows a trapezium. Displacement equals the area under the graph. Calculating the area gives the total distance travelled; the computed value leads to B.

问题 11 – 速度–时间图为梯形。位移等于图线下面积。计算面积得到总移动距离，所得数值对应 B。

Question 12 – flight t = 2u sin θ / g. Substituting values and identifying the correct expression yields D.

问题 12 – 物体以速度 u 与角度 θ 抛出。利用竖直运动：飞行时间 t = 2u sin θ / g。代入数值并识别正确表达式可得 D。

Question	Key
11	B
12	D

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8. Work, Energy and Power | 功、能与功率

Questions 13 and 14 apply the work–energy principle and efficiency calculations. The algebra is minimal, but careful unit handling and formula selection prevent calculator slips.

问题 13 和 14 应用功能原理与效率计算。代数内容较少，但细心处理单位并选对公式可避免计算器误操作。

Question 13 – W = F d cos φ. With φ = 0°, W = F d. Plugging in the given numbers gives a value matching A.

问题 13 – 计算恒力拖动物体所做的功：W = F d cos φ。φ = 0°，W = F d。代入已知数据得到与 A 匹配的数值。

Question 14 – output power) / (input power). The ratio simplifies to a percentage; the correct option is C.

问题 14 – 效率 η = (有用输出功率)/(输入功率)。该比值化简为一个百分数，正确选项为 C。

Question	Key
13	A
14	C

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9. Electricity | 电学

Questions 15 and 16 involve series/parallel resistance and power dissipation. Applying Ohm's law and power formulas systematically reduces complicated circuits to simple arithmetic.

问题 15 和 16 涉及串并联电阻与功率耗散。有步骤地应用欧姆定律和功率公式可将复杂电路化简为简单运算。

Question 15 – s using 1/R = 1/R₁ + 1/R₂, then adding series resistors, yields the value corresponding to B.

问题 15 – 求电阻网络的等效电阻。使用 1/R = 1/R₁ + 1/R₂ 计算并联部分，再加上串联电阻，得到与 B 对应的数值。

Question 16 – = V² / R. Selecting the appropriate form and substituting yields a result consistent with D.

问题 16 – 电阻消耗的功率：P = I² R 或

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