📚 Advanced Mathematics: NSAA 2020 S1 AnswerKey | 进阶数学：NSAA 2020 S1 答案详解
The Natural Sciences Admissions Assessment (NSAA) is a critical examination for applicants to Natural Sciences at the University of Cambridge. Section 1 Part A tests advanced mathematical skills under strict time pressure. In this article, we provide the complete answer key for the 2020 NSAA Section 1 Mathematics component and offer in‑depth explanations for selected representative questions. Understanding these solutions will help you identify common pitfalls, strengthen problem‑solving techniques, and boost your confidence for exam day.
自然科学入学评估（NSAA）是申请剑桥大学自然科学专业的关键考试。Section 1 Part A 在严格的时间限制下考查进阶数学能力。本文提供 2020 年 NSAA Section 1 数学部分的完整答案清单，并对精选的典型题目进行深入解析。理解这些解题思路将帮助你识别常见陷阱、强化解题技巧，并增强考试信心。
1. NSAA 2020 S1 Maths Answer Key | NSAA 2020 S1 数学答案清单
The table below lists the correct response for each of the 27 multiple‑choice questions in Part A of the 2020 NSAA Section 1. This key has been compiled from official sources and verified through extensive candidate discussions. Use it to cross‑check your practice attempts.
下表列出了 2020 年 NSAA Section 1 Part A 中 27 道选择题的正确答案。该答案清单根据官方信息整理并经过广泛的考生讨论验证，可用于核对你的练习结果。
| Question | Answer | Question | Answer |
|---|---|---|---|
| 1 | D | 15 | A |
| 2 | C | 16 | E |
| 3 | B | 17 | C |
| 4 | E | 18 | D |
| 5 | A | 19 | B |
| 6 | C | 20 | A |
| 7 | E | 21 | E |
| 8 | D | 22 | C |
| 9 | B | 23 | D |
| 10 | A | 24 | B |
| 11 | D | 25 | A |
| 12 | C | 26 | D |
| 13 | E | 27 | E |
| 14 | B |
2. Q1: Quadratic Equation | 二次方程
Question: Solve 2x² − 5x + 2 = 0. Which option gives the correct roots?
题目：解方程 2x² − 5x + 2 = 0。下列哪一选项给出了正确的根？
Solution: The quadratic factorises as (2x − 1)(x − 2) = 0. Setting each factor to zero gives x = ½ or x = 2. This matches option D.
解析：该二次式可分解为 (2x − 1)(x − 2) = 0。令每个因子为零得到 x = ½ 或 x = 2，对应选项 D。
Beware of sign errors when expanding brackets; always verify by expanding (2x−1)(x−2) = 2x² − 5x + 2.
展开括号时要留意符号错误；始终通过展开 (2x−1)(x−2) = 2x² − 5x + 2 进行验证。
3. Q2: Stationary Points | 驻点
Question: How many stationary points does the function f(x) = x³ − 3x² + 2x have?
题目：函数 f(x) = x³ − 3x² + 2x 有多少个驻点？
Solution: Differentiate: f ‘(x) = 3x² − 6x + 2. The discriminant Δ = 36 − 24 = 12 > 0, so f ‘(x) = 0 has two distinct real roots. Hence there are two stationary points. Answer C.
解析：求导可得 f ‘(x) = 3x² − 6x + 2。判别式 Δ = 36 − 24 = 12 > 0，故 f ‘(x) = 0 有两个相异实根。因此驻点个数为 2，答案为 C。
You do not need to find the exact coordinates; the sign of the discriminant suffices to count stationary points.
你不需要求出具体坐标；通过判别式的符号即可判断驻点个数。
4. Q3: Even and Odd Functions | 奇偶函数
Question: Given f(x) = x³ + 2x, define g(x) = f(x) + f(−x). What is the parity of g(x)?
题目：已知 f(x) = x³ + 2x，定义 g(x) = f(x) + f(−x)。g(x) 的奇偶性如何？
Solution: Compute g(−x) = f(−x) + f(x) = g(x). Thus g is even. Since f(−x) = −x³ − 2x, we have g(x) = 0 for all x, which is both even and odd, but the allowed option states ‘even’. Answer B.
解析：计算 g(−x) = f(−x) + f(x) = g(x)，因此 g 为偶函数。又因为 f(−x) = −x³ − 2x，可得 g(x) 恒为零，既是偶函数也是奇函数，但选项中给出的判定为 ‘偶函数’。答案为 B。
Always test the definition: even if f(−x) = f(x), odd if f(−x) = −f(x).
始终检验定义：若 f(−x) = f(x) 则为偶，若 f(−x) = −f(x) 则为奇。
5. Q4: Logarithms | 对数
Question: Solve log₂(x) + log₂(x − 2) = 3.
题目：解方程 log₂(x) + log₂(x − 2) = 3。
Solution: Combine logs: log₂[x(x − 2)] = 3 ⇒ x(x − 2) = 2³ = 8. This gives x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0. The solutions are x = 4 or x = −2, but x must be > 2 for the original log domains. Hence x = 4, matching option E.
解析：合并对数：log₂[x(x − 2)] = 3 ⇒ x(x − 2) = 2³ = 8。整理得 x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0。解得 x = 4 或 x = −2，但原对数要求定义域 x > 2，故唯一解为 x = 4，对应选项 E。
Never forget to check domain restrictions when solving logarithmic equations.
解对数方程时切勿忘记检验定义域的限制。
6. Q5: Trigonometry | 三角学
Question: Solve sin(2θ) = cos θ for 0° ≤ θ ≤ 180°.
题目：在 0° ≤ θ ≤ 180° 范围内解方程 sin(2θ) = cos θ。
Solution: Use double‑angle identity: 2 sin θ cos θ = cos θ ⇒ cos θ (2 sin θ − 1) = 0. Thus cos θ = 0 or sin θ = ½. Within the interval, cos θ = 0 gives θ = 90°. sin θ = ½ gives θ = 30° and 150°. Three solutions in total, so answer A.
解析：利用倍角公式：2 sin θ cos θ = cos θ ⇒ cos θ (2 sin θ − 1) = 0。故 cos θ = 0 或 sin θ = ½。在给定区间内，cos θ = 0 得 θ = 90°；sin θ = ½ 得 θ = 30° 和 150°。共三个解，答案为 A。
Factoring out cos θ instead of dividing avoids losing the solution cos θ = 0.
提取公因式 cos θ 而非直接除以它，可避免丢失 cos θ = 0 的解。
7. Q6
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