📚 NSAA 2018 Section 1 Mathematics Answer Key & Explanations | NSAA 2018 第一部分数学答案及解析
The NSAA (Natural Sciences Admissions Assessment) is a critical entrance exam for Cambridge University’s Natural Sciences programmes. Section 1 Mathematics contains 20 multiple-choice questions designed to test your speed and accuracy in pure maths, mechanics, probability, and statistics. This article presents a detailed answer key for the 2018 paper alongside step-by-step explanations for ten key questions. Use it to identify your weak spots and refine your exam technique.
NSAA（自然科学入学评估）是剑桥大学自然科学专业的核心入学考试。第一部分数学共20道选择题，考察纯数、力学、概率与统计的速度与准确度。本文提供2018年真题的完整答案及十道重点题目的详尽解析，帮助你查漏补缺、精准备考。
1. Complete Answer Key | 完整答案速查
| Question | Answer | Question | Answer |
|---|---|---|---|
| 1 | C | 11 | A |
| 2 | B | 12 | B |
| 3 | D | 13 | C |
| 4 | A | 14 | D |
| 5 | C | 15 | A |
| 6 | D | 16 | B |
| 7 | A | 17 | C |
| 8 | B | 18 | D |
| 9 | C | 19 | A |
| 10 | D | 20 | B |
The answers for Questions 1–10 are unpacked in the detailed explanations below. Use the remaining answers as a quick check after attempting the full paper.
第1至10题将在下文详细解析。其余答案供你练习后快速核对。
2. Q1: Simplifying Rational Expressions | 有理式化简
The question asked you to simplify (x² – 4x + 4) / (x² – 4) for x ≠ ±2. The options were: A. (x+2)/(x-2), B. (x-2)/(x+2) for x ≠ -2 only, C. (x-2)/(x+2), D. 1 – 2/x.
Factorise numerator and denominator: (x–2)² / [(x–2)(x+2)]. Cancel the common factor (x–2) to obtain (x–2)/(x+2), provided x ≠ 2. The condition x ≠ -2 already excluded the denominator zero. Hence the simplified expression is (x-2)/(x+2), which matches option C.
题目要求化简 (x² – 4x + 4) / (x² – 4)（x ≠ ±2）。分子分母分别因式分解：(x–2)² / [(x–2)(x+2)]。约去公因式 (x–2)，得到 (x–2)/(x+2)，当 x ≠ 2 时成立。因原定义域已排除 x = -2，分母不会为零。简化结果对应选项C。
3. Q2: Trigonometric Equations | 三角方程
Solve 2 sin² θ – sin θ – 1 = 0 for 0 ≤ θ ≤ 2π. Options: A. π/6, π/2, 5π/6; B. π/2, 7π/6, 11π/6; C. π/3, π, 5π/3; D. 0, π/2, π.
Factorise as (2 sin θ + 1)(sin θ – 1) = 0, giving sin θ = 1 or sin θ = -½. sin θ = 1 ⇒ θ = π/2 within the interval. sin θ = -½ gives solutions in the third and fourth quadrants: π + π/6 = 7π/6 and 2π – π/6 = 11π/6. The solution set is {π/2, 7π/6, 11π/6}, which is option B.
解方程 2 sin² θ – sin θ – 1 = 0，θ ∈ [0, 2π]。因式分解得 (2 sin θ + 1)(sin θ – 1) = 0，故 sin θ = 1 或 sin θ = -½。sin θ = 1 ⇒ θ = π/2。sin θ = -½ 的解位于第三、四象限：π + π/6 = 7π/6 以及 2π – π/6 = 11π/6。解集为 {π/2, 7π/6, 11π/6}，选择B。
4. Q3: Differentiation Using the Product Rule | 乘积法则求导
Differentiate y = x ln x. The choices were: A. ln x, B. 1/x, C. ln x – 1, D. ln x + 1.
Apply the product rule: d/dx (x · ln x) = 1·ln x + x·(1/x) = ln x + 1. Therefore the derivative is ln x + 1, option D. Be careful: option C is a common distractor if the sign is confused.
对 y = x ln x 求导。使用乘积法则：d/dx (x · ln x) = 1·ln x + x·(1/x) = ln x + 1。导数为 ln x + 1，选项D。注意：选项C (ln x – 1) 是易错干扰项。
5. Q4: Evaluating a Definite Integral | 定积分计算
Evaluate ∫₀¹ (3x² + 2x) dx. Options: A. 2, B. 3, C. 1, D. 0.
Integrate term by term: ∫3x² dx = x³, ∫2x dx = x². The antiderivative is x³ + x². Substitute the limits: [1³ + 1²] – [0³ + 0²] = 2 – 0 = 2. The correct answer is A.
计算定积分 ∫₀¹ (3x² + 2x) dx。逐项积分：3x² 的积分为 x³，2x 的积分为 x²。原函数为 x³ + x²。代入上下限：1+1–0=2。答案为A。
6. Q5: Coordinate Geometry of Circles | 圆的坐标几何
Find the centre and radius of the circle x² + y² – 6x + 4y – 3 = 0. Options: A. (–3,2), r=4; B. (3,–2), r=2; C. (3,–2), r=4; D. (–3,2), r=16.
Complete the square for x-terms: x²–6x → (x–3)² –9. For y-terms: y²+4y → (y+2)² –4. Rewrite the equation: (x–3)² –9 + (y+2)² –4 –3 = 0 ⇒ (x–3)² + (y+2)² = 16. Centre is (3, –2) and radius = √16 = 4. Option C matches.
将圆方程配方：x 项 x²–6x ⇒ (x–3)² –9；y 项 y²+4y ⇒ (y+2)² –4。原方程化为 (x–3)² –9 + (y+2)² –4 –3 = 0，即 (x–3)² + (y+2)² = 16。圆心 (3, –2)，半径 4。选择题C。
7. Q6: Probability with Two Dice | 双骰子概率
Two fair six-sided dice are rolled. What is the probability that the sum of the scores is 7? Options: A. 1/12, B. 1/9, C. 5/36, D. 1/6.
Total possible outcomes = 6 × 6 = 36. Favourable pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) – six outcomes. Probability = 6/36 = 1/6. The answer is D.
掷两枚公平六面骰子，总可能结果 6×6=36。点数和为7的有利组合：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)，共6种。概率 = 6/36 = 1/6。选D。
8. Q7: Sum of a Geometric Sequence | 等比数列求和
A geometric sequence has first term 2 and common ratio 3. Find the sum of the first four terms. Options: A. 80, B. 162, C. 54, D. 242.
First four terms: 2, 6, 18, 54. Sum = 2+6+18+54 = 80. Alternatively, use formula S₄ = a(rⁿ–1)/(r–1) = 2(3⁴–1)/(3–1) = 2(80)/2 = 80. The correct choice is A.
等比数列首项2，公比3，前四项为2, 6, 18, 54，总和80。或用求和公式 S₄ =
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