📚 Advanced Mathematics: ENGAA 2021 S1 Question Paper Analysis | 进阶数学：ENGAA 2021 S1 试卷解析
The ENGAA (Engineering Admissions Assessment) 2021 Section 1 is a fast-paced multiple-choice test that assesses mathematical fluency and problem-solving ability essential for engineering at the University of Cambridge. This article walks through a selection of advanced mathematics questions from the paper, offering detailed solutions and strategic approaches. Each worked example is presented in a bilingual format to reinforce both conceptual understanding and academic terminology.
ENGAA 2021第1部分是一场时间紧凑的选择题测试，考察剑桥工程学习所必需的数学流畅度与问题解决能力。本文选取试卷中的进阶数学题目，提供详细解题过程与策略分析。每个例题以中英双语形式呈现，帮助读者在理解概念的同时掌握学术用语。
1. Simplifying Algebraic Expressions | 代数表达式化简
Simplify (3x³y⁻²)⁻² × (2x⁻¹y⁴)³. Start by expanding each bracket using the rule (am)n = amn.
化简 (3x³y⁻²)⁻² × (2x⁻¹y⁴)³。首先利用指数律 (aᵐ)ⁿ = aᵐⁿ 展开每一个括号。
(3x³y⁻²)⁻² = 3⁻² · x⁻⁶ · y⁴ = (1/9) x⁻⁶ y⁴.
(3x³y⁻²)⁻² = 3⁻² · x⁻⁶ · y⁴ = (1/9) x⁻⁶ y⁴。
(2x⁻¹y⁴)³ = 2³ · x⁻³ · y¹² = 8 x⁻³ y¹².
(2x⁻¹y⁴)³ = 2³ · x⁻³ · y¹² = 8 x⁻³ y¹²。
the two results: (1/9)×8 x⁻⁶⁻³ y⁴⁺¹² = (8/9) x⁻⁹ y¹⁶. This can be expressed as 8y¹⁶/(9x⁹).
将两部分相乘：(1/9)×8 x⁻⁶⁻³ y⁴⁺¹² = (8/9) x⁻⁹ y¹⁶。结果可写为 8y¹⁶/(9x⁹)。
2. Solving Equations with Absolute Value | 含绝对值的方程求解
Solve for x: |2x − 3| + 2. Consider two cases depending on the sign of the expression inside the absolute value.
解方程 |2x − 3| = x + 2。根据绝对值内部表达式的符号，分两种情况讨论。
Case 1: 2x − 3 ≥ 0 (x ≥ 1.5). Then |2x − 3| = 2x − 3. Equation becomes 2x − 3 = x + 2 ⇒ x = 5. Since 5 ≥ 1.5, it is valid.
情况一：2x − 3 ≥ 0（即 x ≥ 1.5），则有 |2x − 3| = 2x − 3。方程化为 2x − 3 = x + 2 ⇒ x = 5。5 ≥ 1.5，满足条件，解有效。
Case 2: 2x − 3 < 0 (x < 1.5). Then |2x − 3| = −(2x − 3) = 3 − 2x. Equation becomes 3 − 2x = x + 2 ⇒ 1 = 3x ⇒ x = 1/3. Since 1/3 < 1.5, it is also valid.
情况二：2x − 3 < 0（即 x < 1.5），则有 |2x − 3| = −(2x − 3) = 3 − 2x。方程化为 3 − 2x = x + 2 ⇒ 1 = 3x ⇒ x = 1/3。1/3 < 1.5，也有效。
The solution set is x = 1/3 or x = 5. Always verify solutions by substitution.
解集为 x = 1/3 或 x = 5。务必代入原方程验证解的正确性。
3. Quadratic Discriminant Analysis | 二次判别式分析
Find the range of values of k for which x² + (k − 2)x + 4 = 0 has two distinct real roots.
求实数 k 的取值范围，使方程 x² + (k − 2)x + 4 = 0 有两个不相等的实根。
to have two distinct real roots, the discriminant Δ = b² − 4ac must be greater than zero.
二次方程 ax² + bx + c = 0 有两个不相等实根的条件是判别式 Δ = b² − 4ac > 0。
a = 1, b = k − 2, c = 4. Compute Δ = (k − 2)² − 4(1)(4) = k² − 4k + 4 − 16 = k² − 4k − 12.
此处 a = 1, b = k − 2, c = 4。计算 Δ = (k − 2)² − 4×1×4 = k² − 4k + 4 − 16 = k² − 4k − 12。
Test intervals: k < −2, −2 < k < 6, k > 6. The quadratic is positive when k < −2 or k > 6.
解 Δ > 0：k² − 4k − 12 > 0，因式分解得 (k − 6)(k + 2) > 0。临界值为 k = −2 和 k = 6。分区讨论：当 k < −2 或 k > 6 时不等式成立。
Thus, for two distinct real roots, k < −2 or k > 6.
因此，方程有两个不相等实根的条件是 k < −2 或 k > 6。
4. Trigonometric Equations | 三角方程
lve sin(2x) = cos x for 0 ≤ x ≤ 2π. Use the double-angle identity sin(2x) = 2 sin x cos x.
解方程 sin(2x) = cos x，其中 0 ≤ x ≤ 2π。利用二倍角公式 sin(2x) = 2 sin x cos x。
The equation becomes 2 sin x cos x = cos x. Rearrange: 2 sin x cos x − cos x = 0 ⇒ cos x (2 sin x − 1) = 0.
方程化为 2 sin x cos x = cos x，移项得 2 sin x cos x − cos x = 0 ⇒ cos x (2 sin x − 1) = 0。
Hence, either cos x = 0 or sin x = 1/2.
于是，cos x = 0 或 sin x = 1/2。
For cos x = 0 in [0, 2π]: x = π/2 and x = 3π/2.
在 [0, 2π] 内 cos x = 0 的解为 x = π/2 和 x = 3π/2。
For sin x = 1/2 in [0, 2π]: x = π/6 and x = 5π/6.
sin x = 1/2 在给定区间的解为 x = π/6 和 x = 5π/6。
/6, π/2, 5π/6, 3π/2. Always check for extraneous roots; all satisfy the original equation.
最终解集为 x = π/6, π/2, 5π/6, 3π/2。检查表明所有解均满足原方程。
5. Exponential and Logarithmic Equations | 指数与对数方程
Solve 2e²ˣ − 5eˣ + 2 = 0. This is a hidden quadratic in eˣ.
解方程 2e²ˣ − 5eˣ + 2 = 0。这是隐含的关于 eˣ 的二次方程。
Let y = eˣ, then e²ˣ = (eˣ)² = y². The equation transforms to 2y² − 5y + 2 = 0.
设 y = eˣ，则 e²ˣ = y²。方程变为 2y² − 5y + 2 = 0。
Factorise: 2y² − 5y + 2 = (2y − 1)(y − 2) = 0 ⇒ y = 1/2 or y = 2.
因式分解得 (2y − 1)(y − 2) = 0 ⇒ y = 1/2 或 y = 2。
Recall y = eˣ > 0, so both values are acceptable. For y = 1/2: eˣ = 1/2 ⇒ x = ln(1/2) = −ln 2. For y = 2: eˣ = 2 ⇒ x = ln 2.
由于 eˣ > 0，两解均可接受。由 eˣ = 1/2 得 x = ln(1/2) = −ln 2；由 eˣ = 2 得 x = ln 2。
The solutions are x = ± ln 2.
解为 x = ± ln 2。
6. Differentiation of Composite Functions | 复合函数求导
Differentiate y = ln(sin x) and hence find the gradient of the tangent at x = π/4.
对 y = ln(sin x) 求导，并求曲线在 x = π/4 处切线的斜率。
rule: if y = ln u where u = sin x, then dy/dx = (1/u) · du/dx = (1/sin x) · cos x = cot x.
应用链式法则：令 y = ln u，u = sin x，则 dy/dx = (1/u)·(du/dx) = (1/sin x)·cos x = cot x。
Thus, dy/dx = cot x. Evaluate at x = π/4: cot(π/4) = cos(π/4)/sin(π/4) = (√2/2)/(√2/2) = 1.
因此导数 dy/dx = cot x。在 x = π/4 处，cot(π/4) = cos(π/4)/sin(π/4) = (√2/2)/(√2/2) = 1。
s 1. A quick check: the derivative of ln(sin x) exists when sin x > 0, which holds at π/4.
该点切线斜率为 1。检查定义域：当 sin x > 0 时导数存在，π/4 满足条件。
7. Integration by Parts | 分部积分法
Evaluate the definite integral I = ∫₀π/2 x cos x dx. Use integration by parts.
计算定积分 I = ∫₀π/2 x cos x dx。采用分部积分法。
= cos x dx. Then du = dx, v = sin x. The formula ∫ u dv = uv − ∫ v du gives I = [x sin x]₀π/2 − ∫₀π/2 sin x dx.
设 u = x，dv = cos x dx，则 du = dx，v = sin x。分部积分公式 ∫ u dv = uv − ∫ v du 给出 I = [x sin x]₀π/2 − ∫₀π/2 sin x dx。
the first part: at π/2, x sin x = (π/2)×1 = π/2; at 0, it is 0. So the difference is π/2.
第一部分计算：在 π/2 处，x sin x = (π/2)×1 = π/2；在 0 处为 0。差值为 π/2。
The remaining integral: ∫₀π/2 sin x dx = [−cos x]₀π/2 = (−cos(π/2)) − (−cos 0) = 0 − (−1) = 1.
剩余的积分：∫₀π/2 sin x dx = [−cos x]₀π/2 = (−0) − (−1) = 1。
Thus, I = π/2 − 1. It is often useful to check via numerical approximation.
因此 I = π/2 − 1。可用数值近似检验，cos x 的积分无误。
8. Summation of a Quadratic Sequence | 二次型数列求和
The nth term of a sequence is given by aₙ = 3n² − 2n. Find the sum S₂₀ = Σn=120 aₙ.
某数列通项为 aₙ = 3n² − 2n，求前 20 项和 S₂₀ = Σn=120 aₙ。
Use the standard summation formulas: Σn=1N n = N(N+1)/2, Σn=1N n² = N(N+1)(2N+1)/6.
使用标准求和公式：Σn=1N n = N(N+1)/2，Σn=1N n² = N(N+1)(2N+1)/6。
from 1 to 20. Substitute N=20: Σ n² = 20×21×41/6 = (17220)/6 = 2870; Σ n = 20×21/2 = 210.
故 S₂₀ = 3 Σ n² − 2 Σ n。代入 N=20：Σ n² = 20×21×41/6 = 17220/6 = 2870；Σ n = 20×21/2 = 210。
Then S₂₀ = 3×2870 − 2×210 = 8610 − 420 = 8190.
于是 S₂₀ = 3×2870 − 2×210 = 8610 − 420 = 8190。
The sum of the first 20 terms is 8190. These formulas are essential for solving ENGAA sequence problems efficiently.
前 20 项和为 8190。掌握这些求和公式可高效解答 ENGAA 数列题。
9. Coordinate Geometry: Tangent Equation | 坐标几何：切线方程
Find the equation of the tangent to the curve y = x³ − 3x² + 2x at the point where x = 1.
求曲线 y = x³ − 3x² + 2x 在 x = 1 处的切线方程。
First, find the y-coordinate: y(1) = 1³ − 3×1² + 2×1 = 1 − 3 + 2 = 0. The point is (1, 0).
先求 y 坐标：y(1) = 1 − 3 + 2 = 0，切点为 (1, 0)。
The gradient is given by dy/dx = 3x² − 6x + 2. At x = 1, dy/dx = 3 − 6 + 2 = −1.
切线斜率即导数 dy/dx = 3x² − 6x + 2。在 x = 1 处，斜率为 3 − 6 + 2 = −1。
Using point-gradient form: y − 0 = −1(x − 1) ⇒ y = −x + 1.
运用点斜式：y − 0 = −1(x − 1) ⇒ y = −x + 1。
x + 1. In ENGAA, quick gradient evaluation is often required; double-check the derivative.
切线方程为 y = −x + 1。ENGAA 常要求快速计算斜率，务必仔细求导。
10. Probability Without Replacement | 无放回概率
A bag contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. Find the probability that the balls are of different colours.
袋中有 4 个红球和 6 个蓝球，随机无放回抽取两球。求两球颜色不同的概率。
ways to choose 2 balls from 10 is C(10,2) = 45. Alternatively, use sequential probability.
从 10 个球中任取 2 球的总组合数为 C(10,2) = 45。也可用乘法概率求解。
Different colours can happen in two mutually exclusive orders: Red then Blue, or Blue then Red.
颜色不同可分为两个互斥顺序：先红后蓝，或先蓝后红。
P(Blue then Red) = (6/10) × (4/9) = 24/90 = 4/15. Total probability = 4/15 + 4/15 = 8/15.
P(先红后蓝) = (4/10)×(6/9) = 24/90 = 4/15；P(先蓝后红) = (6/10)×(4/9) = 4/15。总概率 = 4/15 + 4/15 = 8/15。
one red and one blue = C(4,1)×C(6,1) = 24. Probability = 24/45 = 8/15. Both methods agree.
组合方法：取一红一蓝的组合数为 C(4,1)×C(6,1) = 24，概率 = 24/45 = 8/15。两种方法结果一致。
The answer is 8/15. In multiple-choice settings, cross-checking with a different approach builds confidence.
答案为 8/15。在选择题中，采用不同方法交叉验证可增强信心。
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