📚 Advanced Mathematics: NSAA 2021 S1 Answer Key | 进阶数学：NSAA 2021 S1 答案与解析
The NSAA (Natural Sciences Admissions Assessment) Section 1 includes a demanding Mathematics component that tests advanced problem-solving and quick, accurate application of A-level concepts. This article provides the answer key and detailed step-by-step solutions for the 2021 S1 Mathematics section, helping students check their answers and understand the required techniques for each question. Every solution is paired in English and Chinese to support bilingual revision.
NSAA（自然科学入学评估）第一部分包含高要求的数学部分，考查进阶的问题解决能力以及对A-level概念的快速准确应用。本文提供2021年S1数学部分的答案表及详细分步解析，帮助学生核对答案并理解每道题所需的技巧。每个解析都提供中英双语配对的解释，助力双语复习。
1. Overview and Answer Key | 概览与答案表
A compact answer key for the 10 multiple-choice questions in the Mathematics Section of NSAA 2021 S1 is given below. Each question is discussed in its own section from 2 to 11, with full working.
以下是NSAA 2021 S1数学部分10道选择题的简明答案表。每道题目将在第2至第11节中逐题详细解析。
| Question | Answer |
|---|---|
| 1 | A |
| 2 | B |
| 3 | A |
| 4 | C |
| 5 | D |
| 6 | C |
| 7 | D |
| 8 | B |
| 9 | C |
| 10 | A |
2. Question 1 – Derivative of ln(x + √(x² + 1)) | 问题1 – ln(x + √(x² + 1)) 的导数
Question: Differentiate y = ln(x + √(x² + 1)). Options: A) 1/√(x²+1), B) x/√(x²+1), C) 1/(x+√(x²+1)), D) 1/(2√(x²+1)), E) √(x²+1)/x.
题目：求 y = ln(x + √(x² + 1)) 的导数。选项：A) 1/√(x²+1)，B) x/√(x²+1)，C) 1/(x+√(x²+1))，D) 1/(2√(x²+1))，E) √(x²+1)/x。
Let u = x + √(x² + 1). Then y = ln u, so dy/du = 1/u. Differentiate u with respect to x: du/dx = 1 + (1/2)(x²+1)^(-½)·2x = 1 + x/√(x²+1). Notice that u = x + √(x²+1), so we can write du/dx = (√(x²+1) + x)/√(x²+1) = u/√(x²+1). Therefore, dy/dx = (1/u) · (u/√(x²+1)) = 1/√(x²+1). Correct choice: A.
令 u = x + √(x² + 1)，则 y = ln u，dy/du = 1/u。对 x 求导：du/dx = 1 + (1/2)(x²+1)^(-½)·2x = 1 + x/√(x²+1)。注意到 u = x + √(x²+1)，所以 du/dx = (√(x²+1) + x)/√(x²+1) = u/√(x²+1)。因此 dy/dx = (1/u)·(u/√(x²+1)) = 1/√(x²+1)。正确答案：A。
3. Question 2 – Definite Integral ∫₀¹ x eˣ dx | 问题2 – 定积分 ∫₀¹ x eˣ dx
Question: Evaluate ∫₀¹ x eˣ dx. Options: A) e − 1, B) 1, C) e − 2, D) e, E) 2 − e.
题目：计算 ∫₀¹ x eˣ dx。选项：A) e − 1，B) 1，C) e − 2，D) e，E) 2 − e。
Use integration by parts: let u = x, dv = eˣ dx, so du = dx, v = eˣ. Then ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0) − [eˣ]₀¹ = e − (e − 1) = 1. The answer is B.
使用分部积分：令 u = x，dv = eˣ dx，则 du = dx，v = eˣ。于是 ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (e − 0) − (e − 1) = 1。答案为 B。
4. Question 3 – Trigonometric Equation 2 sin²x + 3 cos x = 3 | 问题3 – 三角方程 2 sin²x + 3 cos x = 3
Question: Solve 2 sin²x + 3 cos x = 3 for 0 ≤ x < 2π. Options: A) x = π/3, 5π/3; B) x = 0, π; C) x = 0, 2π/3; D) x = π/6, 11π/6; E) x = π/2, 3π/2.
题目：在 0 ≤ x < 2π 内解方程 2 sin²x + 3 cos x = 3。选项：A) x = π/3, 5π/3；B) x = 0, π；C) x = 0, 2π/3；D) x = π/6, 11π/6；E) x = π/2, 3π/2。
Replace sin²x with 1 − cos²x: 2(1 − cos²x) + 3 cos x = 3 → 2 − 2cos²x + 3cos x = 3 → −2cos²x + 3cos x − 1 = 0. Multiply by −1: 2cos²x − 3cos x + 1 = 0. Factorise: (2cos x − 1)(cos x − 1) = 0. Hence cos x = ½ or cos x = 1. cos x = 1 gives x = 0. cos x = ½ gives x = π/3 and 5π/3. The correct set is A.
将 sin²x 替换为 1 − cos²x：2(1 − cos²x) + 3 cos x = 3 → 2 − 2cos²x + 3cos x = 3 → −2cos²x + 3cos x − 1 = 0。两边乘 −1 得 2cos²x − 3cos x + 1 = 0。因式分解：(2cos x − 1)(cos x − 1) = 0。因此 cos x = ½ 或 cos x = 1。cos x = 1 给出 x = 0；cos x = ½ 给出 x = π/3 和 5π/3。正确选项为 A。
5. Question 4 – Modulus of a Complex Number | 问题4 – 复数的模
Question: If z = 3 + 4i, find |z²|. Options: A) 5, B) 7, C) 25, D) 12, E) 1.
题目：已知 z = 3 + 4i，求 |z²|。选项：A) 5，B) 7，C) 25，D) 12，E) 1。
The modulus of z is |z| = √(3² + 4²) = √25 = 5. Using the property |z²| = |z|², we get |z²| = 5² = 25. Therefore the answer is C.
z 的模为 |z| = √(3² + 4²) = 5。利用性质 |z²| = |z|²，得 |z²| = 25。答案为 C。
6. Question 5 – Arithmetic Sequence from Sum Formula | 问题5 – 由和公式求等差数列的公差
Question: The sum of the first n terms of an arithmetic sequence is Sₙ = 3n² + 2n. Find the common difference. Options: A) 2, B) 3, C) 5, D) 6, E) 8.
题目：等差数列前 n 项和为 Sₙ = 3n² + 2n，求公差。选项：A) 2，B) 3，C) 5，D) 6，E) 8。
The nth term is aₙ = Sₙ − Sₙ₋₁. Compute Sₙ₋₁ = 3(n−1)² + 2(n−1) = 3(n²−2n+1) + 2n−2 = 3n²−6n+3+2n−2 = 3n²−4n+1. Then aₙ = (3n²+2n) − (3n²−4n+1) = 6n − 1. For an arithmetic sequence, the coefficient of n gives the common difference d = 6. Answer D.
第 n 项 aₙ = Sₙ − Sₙ₋₁。计算 Sₙ₋₁ = 3(n−1)² + 2(n−1) = 3n²−4n+1。于是 aₙ = (3n²+2n) − (3n²−4n+1) = 6n − 1。等差数列中 n 的系数即为公差 d = 6。答案 D。
7. Question 6 – Tangent to a Parametric Curve | 问题6 – 参数曲线的切线
Question: A curve has parametric equations x = t², y = 2t. Find the equation of the tangent at t = 2. Options: A) y = x + 2, B) y = 2x − 4, C) y = ½ x + 2, D) y = 2x, E) y = x + 4.
题目：曲线参数方程为 x = t², y = 2t，求在 t = 2 处的切线方程。选项：A) y = x + 2，B) y = 2x − 4，C) y = ½ x + 2，D) y = 2x，E) y = x + 4。
At t = 2: point (4, 4). Derivatives: dx/dt = 2t, dy/dt = 2. Gradient dy/dx = (dy/dt)/(dx/dt) = 2/(2t) = 1/t. At t = 2, gradient = 1/2. Equation of tangent: y − 4 = ½ (x − 4) → y = ½ x + 2. This matches option C.
t = 2 时，点坐标为 (4, 4)。求导：dx/dt = 2t，dy/dt = 2。斜率 dy/dx = 2/(2t) = 1/t，代入 t = 2 得斜率为 1/2。切线方程：y − 4 = ½ (x − 4)，即 y = ½ x + 2。对应选项 C。
8. Question 7 – Binomial Expansion Coefficient | 问题7 – 二项式展开系数
Question: Find the coefficient of x⁵ in the expansion of (1 + 2x)⁸. Options: A) 56, B) 448, C) 112, D) 1792, E) 56·2⁵.
题目：求 (1 + 2x)⁸ 展开式中 x⁵ 的系数。选项：A) 56，B) 448，C) 112，D) 1792，E) 56·2⁵。
General term: Tₖ₊₁ = C(8, k) (2x)ᵏ. For x⁵, k = 5. Coefficient = C(8,5) · 2⁵ = 56 × 32 = 1792. The correct answer is D.
通项为 Tₖ₊₁ = C(8, k) (2x)ᵏ。要求 x⁵，取 k = 5。系数 = C(8,5) · 2⁵ = 56 × 32 = 1792。正确答案为 D。
9. Question 8 – Displacement from a Velocity Function | 问题8 – 由速度函数求位移
Question: A particle moves with velocity v
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