📚 ENGAA 2019 Section 1 Mathematics: Essential Problems Explained | ENGAA 2019 第一部分数学核心题解
The ENGAA 2019 Section 1 Mathematics section contains 20 multiple-choice questions designed to test speed, accuracy, and a deep understanding of advanced mathematical topics. This article presents a detailed breakdown of eight representative question types, offering clear solutions and highlighting the key concepts you need to master. Each problem is explained with step-by-step reasoning that mirrors the thought process required under timed conditions.
ENGAA 2019 第一部分数学包含20道选择题，旨在考查解题速度、准确性以及对进阶数学知识的深刻理解。本文精选了八道代表性题型，提供清晰解答，并强调你需要掌握的核心概念。每道题都配有逐步推理过程，模拟限时条件下的思维路径，帮助你高效备考。
1. Discriminant and Nature of Roots | 判别式与根的性质
The quadratic equation 2x² + kx + 8 = 0 has two distinct real roots. Find the set of possible values of k.
已知二次方程 2x² + kx + 8 = 0 有两个不相等的实根，求 k 的取值范围。
For a quadratic ax² + bx + c = 0, the discriminant Δ = b² − 4ac determines the nature of the roots. Two distinct real roots require Δ > 0. Here a = 2, b = k, c = 8, so Δ = k² − 4 × 2 × 8 = k² − 64. Setting k² − 64 > 0 gives (k − 8)(k + 8) > 0, hence k < −8 or k > 8.
对于二次方程 ax² + bx + c = 0，判别式 Δ = b² − 4ac 决定了根的性质。两个不相等实根要求 Δ > 0。本题 a=2, b=k, c=8，计算 Δ = k² − 4×2×8 = k² − 64。解不等式 k² − 64 > 0，得 (k − 8)(k + 8) > 0，因此 k < −8 或 k > 8。
The key is to apply the strict inequality for distinct real roots. A common error is to forget that a positive discriminant excludes the case of equal roots (Δ = 0). Always check the direction of the inequality and express the final answer in interval notation or as a pair of inequalities.
关键在于正确使用严格不等式以表示不等实根。常见的错误是忘记判别式大于零排除了等根情形（Δ=0）。务必核对不等号方向，并用区间或不等式对的形式给出最终答案。
2. Logarithmic Equations | 对数方程
Solve the equation log₂(x + 3) − log₂(x − 2) = 1.
解方程 log₂(x + 3) − log₂(x − 2) = 1。
Use the logarithm subtraction rule: log₂ A − log₂ B = log₂(A / B). The equation becomes log₂[(x + 3)/(x − 2)] = 1. Rewrite in exponential form: (x + 3)/(x − 2) = 2¹ = 2. Multiply both sides by (x − 2): x + 3 = 2(x − 2) ⇒ x + 3 = 2x − 4 ⇒ 7 = x. Check the domain: the arguments must be positive, so x + 3 > 0 and x − 2 > 0, which gives x > 2. The solution x = 7 satisfies this domain.
利用对数减法法则：log₂ A − log₂ B = log₂(A / B)。方程化为 log₂[(x + 3)/(x − 2)] = 1。写成指数形式：(x + 3)/(x − 2) = 2¹ = 2。两边同乘 (x − 2)：x + 3 = 2x − 4 ⇒ x = 7。需要对真数进行定义域检验：x + 3 > 0 且 x − 2 > 0，得到 x > 2。解 x = 7 满足定义域。
Always remember to check the domain when solving logarithmic equations. Extraneous solutions can arise from algebraic manipulation, so verifying that the result makes all log arguments positive is essential to avoid losing marks.
解对数方程时务必检验定义域。代数变形可能产生增根，因此验证结果是否使所有对数的真数为正是避免失分的关键。
3. Trigonometric Equations and Identities | 三角方程与恒等式
Solve 2 sin²θ − 3 sin θ + 1 = 0 for 0 ≤ θ ≤ 2π.
在 0 ≤ θ ≤ 2π 范围内解方程 2 sin²θ − 3 sin θ + 1 = 0。
Let u = sin θ. The equation becomes 2u² − 3u + 1 = 0, which factorises as (2u − 1)(u − 1) = 0. So u = ½ or u = 1. Substituting back: sin θ = ½ gives θ = π/6, 5π/6; sin θ = 1 gives θ = π/2. All three solutions lie within the interval.
设 u = sin θ，方程化为 2u² − 3u + 1 = 0，因式分解得 (2u − 1)(u − 1) = 0，故 u = ½ 或 u = 1。代回原函数：sin θ = ½ 时，θ = π/6, 5π/6；sin θ = 1 时，θ = π/2。三个解均在指定区间内。
For trigonometric quadratics, treat the trigonometric function as a single variable. After finding the basic angles, use the unit circle to locate all solutions within the given interval. Double-check that none of the values fall outside the domain, especially when the equation involves a restricted range.
解三角二次方程时，可将三角函数视为单个变量。求出基本角后，利用单位圆确定指定区间内的所有解。务必复查是否有解越界，尤其当方程涉及有限值域时。
4. Differentiation: Velocity and Acceleration | 微分：速度与加速度
A particle moves along a straight line with displacement s(t) = t³ − 5t² + 6t metres after t seconds. Find the acceleration of the particle at the instant when its velocity is zero.
一质点沿直线运动，t 秒后的位移为 s(t) = t³ − 5t² + 6t 米。求速度为瞬时为零时的加速度。
Velocity v(t) is the first derivative of displacement: v(t) = s'(t) = 3t² − 10t + 6. Acceleration a(t) is the second derivative: a(t) = v'(t) = 6t − 10. Set v(t) = 0: 3t² − 10t + 6 = 0. Solving this quadratic gives t = (10 ± √(100 − 72))/6 = (10 ± √28)/6 = (10 ± 2√7)/6 = (5 ± √7)/3. Substitute each t into a(t). For t = (5 + √7)/3, a = 6((5+√7)/3) − 10 = 2(5+√7) − 10 = 10 + 2√7 − 10 = 2√7. For t = (5 − √7)/3, a = 2(5−√7) − 10 = 10 − 2√7 − 10 = −2√7. Thus the acceleration magnitudes are 2√7 m/s², with opposite signs depending on the time.
速度 v(t) 是位移的一阶导数：v(t) = s'(t) = 3t² − 10t + 6。加速度 a(t) 是二阶导数：a(t) = v'(t) = 6t − 10。令 v(t) = 0：3t² − 10t + 6 = 0。解二次方程得 t = (10 ± √(100 − 72))/6 = (10 ± √28)/6 = (5 ± √7)/3。将两个 t 值分别代入 a(t)。t = (5 + √7)/3 时，a = 6((5+√7)/3) − 10 = 2√7；t = (5 − √7)/3 时，a = −2√7。因此加速度大小为 2√7 m/s²，符号因时刻不同而异。
This problem illustrates the need to handle two solutions for time when velocity is zero, each giving a different acceleration. In mechanics contexts, always interpret sign as direction. The question did not ask for magnitude alone, so presenting both values is essential for completeness.
本题说明在速度为零时可能得到两个时刻，对应不同的加速度。在力学中，符号代表方向，因此应完整给出两个值，而非仅给出大小。
5. Integration: Area Between Curves | 积分：曲线间的面积
Find the area of the region bounded by the curve y = x² and the line y = 2x + 3.
求曲线 y = x² 与直线 y = 2x + 3 所围成区域的面积。
First find intersection points by setting x² = 2x + 3 ⇒ x² − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0, so x = −1 and x = 3. For x in [−1, 3], the line y = 2x + 3 lies above y = x². The area A = ∫₋₁³ [(2x + 3) − x²] dx. Compute the integral: ∫ (2x + 3 − x²) dx = x² + 3x − (x³)/3. Evaluate from −1 to 3: at x = 3, value = 9 + 9 − 27/3 = 9 + 9 − 9 = 9. At x = −1, value = 1 − 3 − (−1/3) = 1 − 3 + 1/3 = −5/3. Therefore A = 9 − (−5/3) = 9 + 5/3 = 32/3 square units.
首先求交点，令 x² = 2x + 3 ⇒ x² − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0，得 x = −1 和 x = 3。在区间 [−1, 3] 上，直线 y = 2x + 3 位于曲线 y = x² 上方。面积 A = ∫₋₁³ [(2x + 3) − x²] dx。计算积分：∫ (2x + 3 − x²) dx = x² + 3x − x³/3。从 −1 到 3 代入：x = 3 时，值为 9 + 9 − 27/3 = 9；x = −1 时，值为 1 − 3 − (−1/3) = −5/3。故 A = 9 − (−5/3) = 32/3 平方单位。
Always sketch the curves to identify which function is the upper one over the integration interval. Use the correct subtraction order (top minus bottom). This straightforward approach avoids sign errors and ensures the area is positive.
永远先画出曲线草图以确定在积分区间内哪条曲线是上边界。使用正确的减法顺序（上减下），这样可以避免符号错误，确保面积为正值。
6. Sum of n Terms and the nth Term | 前 n 项和与第 n 项
The sum of the first n terms of a sequence is given by Sₙ = 3n² + 2n. Determine the 5th term of the sequence.
已知某数列的前 n 项和为 Sₙ = 3n² + 2n，求该数列的第 5 项。
For any sequence, the nth term uₙ can be found using uₙ = Sₙ − Sₙ₋₁ for n ≥ 2, provided S₀ = 0. Here S₅ = 3(5)² + 2(5) = 75 + 10 = 85. S₄ = 3(4)² + 2(4) = 48 + 8 = 56. Therefore u₅ = S₅ − S₄ = 85 − 56 = 29.
对于任意数列，当 n ≥ 2 时，第 n 项 uₙ 可通过 uₙ = Sₙ − Sₙ₋₁ 求得（假设 S₀ = 0）。本题中 S₅ = 3×25 + 10 = 85，S₄ = 3×16 + 8 = 56，因此 u₅ = 85 − 56 = 29。
This technique works for both arithmetic and geometric sequences, as well as sequences defined by non-linear Sₙ formulas. It is often faster than finding a general formula for the nth term, especially when only a specific term is required.
这一技巧适用于等差、等比数列，以及由非线性 Sₙ 公式定义的数列。当只需要某个特定项时，这种方法往往比推导通项公式更快捷。
7. Probability Without Replacement | 不放回抽样概率
A bag contains 4 red, 5 blue and 1 green marble. Two marbles are drawn randomly without replacement. Find the probability that both marbles are the same colour.
一个袋子里有 4 颗红球、5 颗蓝球和 1 颗绿球，随机不放回地抽取两颗。求两颗球颜色相同的概率。
Total marbles = 10. Number of ways to choose 2 marbles = C(10,2) = 45. Favourable outcomes: both red (C(4,2)=6) + both blue (C(5,2)=10) + both green (C(1,2)=0, impossible). So favourable = 6 + 10 = 16. Probability = 16/45.
总球数为 10。选 2 颗的总方法数为 C(10,2) = 45。有利结果：两红（C(4,2)=6）+ 两蓝（C(5,2)=10）+ 两绿（C(1,2)=0，不可能），合计 16 种。概率为 16/45。
When events are independent only in the sense of combination counting, it is efficient to use the combinations formula. Always confirm the sample space and ensure you do not count any impossible colour pairs twice. The answer can be simplified only if numerator and denominator share common factors; here 16/45 is already in simplest form.
在不放回抽样且事件通过组合来计数时，使用组合公式最有效。务必确认样本空间，避免重复计算不可能的颜色对。本题分数 16/45 已是最简形式。
8. Coordinate Geometry: Tangent to a Circle | 坐标几何：圆的切线
A circle has equation x² + y² = 10. The line y = mx − 5 is a tangent to the circle. Find
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