NSAA 2018 S1 Mathematics Answer Key | NSAA 2018 S1 进阶数学答案与解析

📚 NSAA 2018 S1 Mathematics Answer Key | NSAA 2018 S1 进阶数学答案与解析

The NSAA (Natural Sciences Admissions Assessment) Section 1 Mathematics paper tests a range of advanced mathematical skills essential for science courses. This article provides the complete answer key for the 2018 sitting, together with detailed bilingual solutions for every question. The objective is to help students verify their performance, understand common pitfalls, and reinforce key techniques required for top scores.

NSAA（自然科学入学评估）第一部分的数学试卷旨在考查一系列对科学课程至关重要的进阶数学技能。本文提供了2018年考试的完整答案表，并为每道题目配上了详细的中英双语解析。目的是帮助学生核对成绩、理解常见易错点，并巩固获取高分所需的关键解题方法。

1. Overview and Full Answer Table | 试卷概览与完整答案表

The 2018 NSAA Section 1 Mathematics part comprises 20 multiple-choice questions covering algebra, geometry, trigonometry, calculus, probability, and logarithms. Below is the quick-reference answer table for all questions.

2018年NSAA第一部分数学卷包含20道选择题，覆盖代数、几何、三角、微积分、概率和对数等知识。以下是全部题目的快捷答案参考表。

Question	Answer	Question	Answer
1	16	11	10
2	m = 1	12	2a + b
3	820	13	5
4	x = -2 or 3	14	5√3
5	2x + 3	15	√3/2
6	14/45	16	32
7	2√5	17	6
8	-3/4	18	0.8
9	max (0,2), min (2,-2)	19	4 m
10	40	20	-1/x²

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2. Exponents and Surds | 指数与根式

Q1. Simplify 2⁴⁰ ÷ 4¹⁸. Write 4 as 2², so 4¹⁸ = (2²)¹⁸ = 2³⁶. Then 2⁴⁰ / 2³⁶ = 2⁴ = 16.

Q1. 化简 2⁴⁰ ÷ 4¹⁸。将 4 写成 2²，因此 4¹⁸ = (2²)¹⁸ = 2³⁶。于是 2⁴⁰ / 2³⁶ = 2⁴ = 16。

Q14. Simplify √48 + √27 − √12. Factor out perfect squares: √48 = √16×3=4√3, √27 = 3√3, √12 = 2√3. Sum: 4√3+3√3−2√3 = 5√3.

Q14. 化简 √48 + √27 − √12。提取完全平方数：√48 = √16×3=4√3，√27 = 3√3，√12 = 2√3。相加得 4√3+3√3−2√3 = 5√3。

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3. Algebra and Equations | 代数与方程

Q2. A line y = mx + c passes through (1,3) and (4,6). Find m. Gradient m = (6−3)/(4−1) = 3/3 = 1.

Q2. 已知直线 y = mx + c 经过 (1,3) 和 (4,6)，求 m。斜率 m = (6−3)/(4−1) = 3/3 = 1。

Q4. Solve |2x − 1| = 5. Two cases: 2x−1 = 5 → x = 3; 2x−1 = −5 → x = −2. Thus x = −2 or 3.

Q4. 解方程 |2x − 1| = 5。分两种情况：2x−1 = 5 → x = 3；2x−1 = −5 → x = −2。因此 x = −2 或 3。

Q16. Solve log₂(x) = 5. Rewrite in exponential form: x = 2⁵ = 32.

Q16. 解 log₂(x) = 5。转化为指数形式：x = 2⁵ = 32。

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4. Functions and Graphs | 函数与图像

Q5. f(x) = x² + 2x. Find f(x+1) − f(x). +1) = (x+1)² + 2(x+1) = x²+2x+1+2x+2 = x²+4x+3. Subtract f(x): (x²+4x+3) − (x²+2x) = 2x+3.

Q5. f(x) = x² + 2x，求 f(x+1) − f(x)。计算 f(x+1) = (x+1)² + 2(x+1) = x²+2x+1+2x+2 = x²+4x+3。减去 f(x) 得 (x²+4x+3) − (x²+2x) = 2x+3。

Q20. Differentiate f(x) = 1/x. Rewrite as x⁻¹, then f'(x) = −1·x⁻² = −1/x².

Q20. 求 f(x) = 1/x 的导数。将其写作 x⁻¹，则 f'(x) = −1·x⁻² = −1/x²。

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5. Coordinate Geometry | 坐标几何

Q7. A circle centred at (0,0) has tangent x + 2y = 10. Find the radius. The radius equals the perpendicular distance from centre to tangent: r = |0+0−10| / √1²+2² = 10/√5 = 2√5.

Q7. 圆心在 (0,0) 的圆，切线为 x + 2y = 10，求半径。半径等于圆心到切线的垂直距离：r = |0+0−10| / √1²+2² = 10/√5 = 2√5。

Q13. Find the length of segment AB with A(1,2) and B(5,5). Distance = √[(5−1)² + (5−2)²] = √4²+3² = √25 = 5.

Q13. 求 A(1,2) 与 B(5,5) 之间的线段长度。距离 = √[(5−1)² + (5−2)²] = √4²+3² = √25 = 5。

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6. Sequences and Series | 数列与级数

Q3. Find the sum of the first 20 terms of the arithmetic sequence 3, 7, 11, ... rst term a = 3, common difference d = 4. Sum S₂₀ = 20/2 [2×3 + (20−1)×4] = 10(6+76) = 820.

Q3. 求等差数列 3, 7, 11, ... 的前 20 项和。首项 a = 3，公差 d = 4。求和 S₂₀ = 20/2 [2×3 + (20−1)×4] = 10(6+76) = 820。

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7. Trigonometry | 三角学

Q8. If sinθ = 0.6 and cosθ is negative, find tanθ. sinθ = 3/5, so cosθ = −√1 − 9/25 = −4/5. Hence tanθ = (3/5) / (−4/5) = −3/4.

Q8. 已知 sinθ = 0.6 且 cosθ 为负，求 tanθ。sinθ = 3/5，故 cosθ = −√1 − 9/25 = −4/5。因此 tanθ = (3/5) / (−4/5) = −3/4。

Q15. Evaluate sin(π/3). sin(60°) = √3/2.

Q15. 求 sin(π/3) 的值。sin(60°) = √3/2。

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8. Differentiation and Applications | 微分及其应用

Q9. Find the stationary points of y = x³ − 3x² + 2 and determine their nature. 6. At x=0, d²y/dx² = −6 < 0 → local max (0,2). At x=2, d²y/dx² = 6 > 0 → local min (2,−2).

Q9. 求曲线 y = x³ − 3x² + 2 的驻点并判断其性质。求导 dy/dx = 3x² − 6x = 3x(x−2)=0 → x=0, 2。二阶导 d²y/dx² = 6x−6。当 x=0 时 < 0 → 极大值 (0,2)；x=2 时 > 0 → 极小值 (2,−2)。

Q17. Find the gradient of the tangent to y = x² at x = 3. dy/dx = 2x, at x=3 gradient = 6.

Q17. 求曲线 y = x² 在 x = 3 处的切线斜率。dy/dx = 2x，代入 x=

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