ENGAA 2023 S1 Maths AnswerKey & Detailed Solutions | ENGAA 2023 S1 数学答案与精讲

📚 ENGAA 2023 S1 Maths AnswerKey & Detailed Solutions | ENGAA 2023 S1 数学答案与精讲

The ENGAA 2023 Section 1 Mathematics component featured a blend of pure and applied problems, testing speed, accuracy, and sophisticated problem‑solving. This article presents a curated answer key with step‑by‑step explanations for ten representative questions, based on post‑exam recollections and official specification topics. Use these solutions to sharpen your advanced mathematics skills and build confidence for future admissions tests.

2023 年 ENGAA 第一部分的数学部分融合了纯数学与应用数学题目，重点考查解题速度、正确率与高阶思维能力。本文根据考后回忆与官方大纲，精选十道代表性题目，给出答案解析并逐步拆解思路。希望这份精讲帮助读者巩固进阶数学考点，为今后的入学笔试做好充足准备。

1. Quadratic Inequalities | 二次不等式

The problem required solving the inequality x² – 4x – 5 ≤ 0 over the real numbers.

题目要求解实数范围内的不等式 x² – 4x – 5 ≤ 0。

Factorising the quadratic gives (x – 5)(x + 1) ≤ 0. The critical values are x = –1 and x = 5. A sign‑chart or test‑point method shows the product is non‑positive precisely between the roots, including the endpoints.

将二次式因式分解得 (x – 5)(x + 1) ≤ 0，临界值为 x = –1 和 x = 5。通过符号表或取值检验可知，乘积在两根之间（含端点）时非正。

Hence the solution set is x ∈ [–1, 5].

因此解集为 x ∈ [–1, 5]。

2. Trigonometric Equation | 三角方程

Candidates needed to solve 2 sin²θ – cosθ – 1 = 0 for 0 ≤ θ < 2π.

考生需在 0 ≤ θ < 2π 范围内解方程 2 sin²θ – cosθ – 1 = 0。

Use the identity sin²θ = 1 – cos²θ to rewrite the equation as 2(1 – cos²θ) – cosθ – 1 = 0, which simplifies to –2 cos²θ – cosθ + 1 = 0. Multiply by –1 to obtain 2 cos²θ + cosθ – 1 = 0.

利用恒等式 sin²θ = 1 – cos²θ，原方程化为 2(1 – cos²θ) – cosθ – 1 = 0，整理得 –2 cos²θ – cosθ + 1 = 0。两边乘以 –1 得到 2 cos²θ + cosθ – 1 = 0。

Factorisation yields (2 cosθ – 1)(cosθ + 1) = 0, so cosθ = 1/2 or cosθ = –1. Within the given interval, the solutions are θ = π/3, π, and 5π/3.

因式分解得 (2 cosθ – 1)(cosθ + 1) = 0，故 cosθ = 1/2 或 cosθ = –1。在指定区间内，解为 θ = π/3、π 和 5π/3。

3. Differentiation & Tangent Line | 微分与切线

The task was to find the equation of the tangent to the curve y = x eˣ at the point where x = 0.

题目要求求曲线 y = x eˣ 在 x = 0 处的切线方程。

Differentiate using the product rule: dy/dx = eˣ + x eˣ = eˣ(1 + x). At x = 0, y = 0⋅1 = 0 and the gradient is e⁰(1+0) = 1.

用乘法法则求导：dy/dx = eˣ + x eˣ = eˣ(1 + x)。在 x = 0 处，y = 0⋅1 = 0，斜率等于 e⁰(1+0) = 1。

The tangent line passes through (0,0) with slope 1, so its equation is y = x.

切线过点 (0,0) 且斜率为 1，因此方程为 y = x。

4. Integration by Substitution | 换元积分法

Evaluate the definite integral ∫₀¹ 2x/(x²+1)² dx.

计算定积分 ∫₀¹ 2x/(x²+1)² dx。

Let u = x² + 1. Then du = 2x dx, and when x = 0, u = 1; when x = 1, u = 2. The integral transforms to ∫₁² u⁻² du.

令 u = x² + 1，则 du = 2x dx。当 x = 0 时 u = 1，x = 1 时 u = 2。积分变为 ∫₁² u⁻² du。

∫ u⁻² du = –u⁻¹, so the value is [–1/u]₁² = (–1/2) – (–1) = 1/2.

∫ u⁻² du = –u⁻¹，代入上下限得 [–1/u]₁² = (–1/2) – (–1) = 1/2。

5. Coordinate Geometry: Circle Tangents | 坐标几何：圆的切线

A circle is given by x² + y² – 6x + 8y = 0. Determine the centre and radius, then find the tangent at the origin.

已知圆方程为 x² + y² – 6x + 8y = 0，求圆心和半径，并写出过原点的切线方程。

Complete the square: (x² – 6x + 9) + (y² + 8y + 16) = 9 + 16 → (x – 3)² + (y + 4)² = 25. Thus the centre is C(3, –4) and the radius is 5. The origin (0,0) lies on the circle because 0² + 0² – 0 + 0 = 0.

配方得：(x² – 6x + 9) + (y² + 8y + 16) = 9 + 16 → (x – 3)² + (y + 4)² = 25。因此圆心为 C(3, –4)，半径为 5。原点 (0,0) 满足方程，故在圆上。

The radius OC has slope (–4 – 0)/(3 – 0) = –4/3, so the tangent slope is the negative reciprocal, 3/4. The tangent through (0,0) is y = (3/4)x, or 3x – 4y = 0.

半径 OC 的斜率为 (–4 – 0)/(3 – 0) = –4/3，故切线斜率为其负倒数 3/4。过原点的切线为 y = (3/4)x，即 3x – 4y = 0。

6. Arithmetic Sequences | 等差数列

In an arithmetic progression, the 4th term is 14 and the 9th term is 34. Find the sum of the first 20 terms.

等差数列中，第 4 项为 14，第 9 项为 34。求前 20 项之和。

Let the first term be a and common difference d. We have a + 3d = 14 and a + 8d = 34. Subtracting gives 5d = 20 → d = 4. Then a = 14 – 12 = 2.

设首项为 a，公差为 d。有 a + 3d = 14 与 a + 8d = 34。两式相减得 5d = 20 → d = 4，进而 a = 14 – 12 = 2。

The sum of the first n terms is Sₙ = n/2 [2a + (n–1)d]. For n = 20, S₂₀ = 10 [2×2 + 19×4] = 10 × (4 + 76) = 800.

前 n 项和公式为 Sₙ = n/2 [2a + (n–1)d]。代入 n = 20 得 S₂₀ = 10 [2×2 + 19×4] = 10 × (4 + 76) = 800。

7. Vector Angle Calculation | 向量夹角计算

Given vectors p = 3i – j + 2k and q = i + 4j – k, find the acute angle between them.

已知向量 p = 3i – j + 2k 与 q = i + 4j – k，求它们的锐角夹角。

Compute the dot product: p·q = 3×1 + (–1)×4 + 2×(–1) = 3 – 4 – 2 = –3. Magnitudes: |p| = √(9 + 1 + 4) = √14; |q| = √(1 + 16 + 1) = √18 = 3√2.

计算点积：p·q = 3×1 + (–1)×4 + 2×(–1) = 3 – 4 – 2 = –3。模长：|p| = √(9 + 1 + 4) = √14；|q| = √(1 + 16 + 1) = √18 = 3√2。

cosθ = (–3) / (√14 × 3√2) = –1 / √28 = –1 / (2√7). Since the dot product is negative, θ is obtuse. The acute angle between the lines of action is φ = 180° – θ, so cosφ = 1/(2√7). The acute angle is arccos(1/(2√7)).

cosθ = (–3) / (√14 × 3√2) = –1 / √28 = –1 / (2√7)。点积为负说明 θ 为钝角，两条直线方向的锐角 φ = 180° – θ，因此 cosφ = 1/(2√7)。锐角为 arccos(1/(2√7))。

8. Complex Numbers: Modulus & Argument | 复数：模与辐角

Express the modulus and argument of the complex number z = –2 + 2i.

写出复数 z = –2 + 2i 的模与辐角。

Modulus: |z| = √((–2)² + 2²) = √(4 + 4) = √8 = 2√2. The number lies in the second quadrant, so arg(z) = π – arctan(2/2) = π – π/4 = 3π/4. (Principal argument).

模：|z| = √((–2)² + 2²) = √(4 + 4) = √8 = 2√2。该数位于第二象限，故辐角主值为 arg(z) = π – arctan(2/2) = π – π/4 = 3π/4。

9. Probability Without Replacement | 不放回概率

A bag contains 4 red, 5 blue, and 1 yellow ball. Two balls are drawn at random without replacement. Calculate the probability that both balls are blue.

袋中有 4 红、5 蓝、1 黄球。随机不放回地抽取两球，求两球均为蓝色的概率。

Total number of balls = 10. Probability first is blue = 5/10. After drawing one blue, 4 blue remain out of 9 balls. Probability second is blue = 4/9. Multiply: P(both blue) = (5/10) × (4/9) = 20/90 = 2/9.

总球数为 10。第一次抽到蓝色的概率 = 5/10。抽掉一个蓝球后，剩下 4 蓝在 9 球中，第二次抽到蓝色的概率 = 4/9。相乘得 P(两蓝) = (5/10) × (4/9) = 20/90 = 2/9。

10. Exponential Equations | 指数方程

Solve the equation 4ˣ – 3·2ˣ⁺¹ + 8 = 0 for real x.

解指数方程 4ˣ – 3·2ˣ⁺¹ + 8 = 0。

Write 4ˣ as (2²)ˣ = 2²ˣ = (2ˣ)². Also 2ˣ⁺¹ = 2·2ˣ. Substituting y = 2ˣ (with y > 0) transforms the equation into y² – 3·2·y + 8 = 0, i.e. y² – 6y + 8 = 0.

将 4ˣ 写成 (2²)ˣ = 2²ˣ = (2ˣ)²，而 2ˣ⁺¹ = 2·2ˣ。令 y = 2ˣ (y > 0)，方程化为 y² – 6y + 8 = 0。

Factorising gives (y – 2)(y – 4) = 0, so y = 2 or y = 4. Returning to x: 2ˣ = 2 ⇒ x = 1; 2ˣ = 4 ⇒ x = 2. Both solutions are valid.

因式分解得 (y – 2)(y – 4) = 0，故 y = 2 或 y = 4。代回：2ˣ = 2 ⇒ x = 1；2ˣ = 4 ⇒ x = 2。两解均成立。

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ENGAA 2023 S1 Maths AnswerKey & Detailed Solutions | ENGAA 2023 S1 数学答案与精讲