NSAA 2019 Section 1 Maths Answer Key & Solutions | NSAA 2019 S1 数学答案与解析

📚 NSAA 2019 Section 1 Maths Answer Key & Solutions | NSAA 2019 S1 数学答案与解析

The NSAA (Natural Sciences Admissions Assessment) is a critical entrance exam for applicants to Natural Sciences at the University of Cambridge. Section 1 features multiple-choice questions in mathematics, alongside physics, chemistry, and biology. This article provides a comprehensive answer key with detailed worked solutions for the mathematics portion of NSAA 2019 Section 1. Each question is explained step by step, mirroring the reasoning needed to secure top marks and helping you master the problem-solving techniques typical of the assessment.

NSAA（自然科学入学评估）是申请剑桥大学自然科学专业的关键入学考试。第 1 部分包含数学、物理、化学和生物的选择题。本文提供 2019 年 NSAA 第 1 部分数学部分的完整答案与详细解析。每道题都配有逐步推理，还原获取高分所需的思维过程，助你掌握该考试中的典型解题技巧。

1. Question 1 – Simplifying Surd Expressions | 根式化简

Simplify √48 + √27 – √75.
Options: A: 0, B: 2√3, C: 4√3, D: 6√3

化简 √48 + √27 – √75。
选项：A: 0, B: 2√3, C: 4√3, D: 6√3

Break each surd into a product of a perfect square and a radical: √48 = √(16×3) = 4√3; √27 = √(9×3) = 3√3; √75 = √(25×3) = 5√3. The expression becomes 4√3 + 3√3 – 5√3 = (4+3–5)√3 = 2√3. Hence the correct answer is B.

将每个根式拆分为完全平方数与根式的乘积：√48 = √(16×3) = 4√3；√27 = √(9×3) = 3√3；√75 = √(25×3) = 5√3。原式变为 4√3 + 3√3 – 5√3 = (4+3–5)√3 = 2√3。因此正确答案为 B。

2. Question 2 – Solving Exponential Equations | 解指数方程

Solve 2²ˣ⁺¹ = 8ˣ⁻¹.
Options: A: x = 2, B: x = 3, C: x = 4, D: x = 5

解方程 2²ˣ⁺¹ = 8ˣ⁻¹。
选项：A: x = 2, B: x = 3, C: x = 4, D: x = 5

Express both sides with the same base 2. Since 8 = 2³, the right-hand side becomes (2³)ˣ⁻¹ = 2³ˣ⁻³. Equating exponents gives 2x+1 = 3x–3. Rearranging: 1+3 = 3x–2x, so x = 4. The answer is C.

两边同写成以 2 为底的幂。因为 8 = 2³，右边化为 (2³)ˣ⁻¹ = 2³ˣ⁻³。比较指数得 2x+1 = 3x–3。移项：1+3 = 3x–2x，解得 x = 4。答案为 C。

3. Question 3 – Logarithmic Equations | 对数方程

Solve log₂(x+1) + log₂(x–1) = 3.
Options: A: x = 2, B: x = 3, C: x = ±3, D: x = 4

解 log₂(x+1) + log₂(x–1) = 3。
选项：A: x = 2, B: x = 3, C: x = ±3, D: x = 4

Combine the logs: log₂[(x+1)(x–1)] = log₂(x²–1) = 3. Rewrite in exponential form: x²–1 = 2³ = 8 → x² = 9 → x = ±3. The domain requires x+1 > 0 and x–1 > 0, so x > 1. Thus x = 3 is the only valid solution. Answer: B.

合并对数：log₂[(x+1)(x–1)] = log₂(x²–1) = 3。改写为指数式：x²–1 = 2³ = 8 → x² = 9 → x = ±3。定义域要求 x+1 > 0 且 x–1 > 0，即 x > 1。因此唯一有效解为 x = 3。答案：B。

4. Question 4 – Trigonometric Equations | 三角方程

Solve 2 sin²θ – sinθ – 1 = 0 for 0 ≤ θ ≤ 2π.
Options: A: θ = π/2, 7π/6, 11π/6 B: θ = π/6, 5π/6 C: θ = π/3, 4π/3 D: θ = π/2, 3π/2

解方程 2 sin²θ – sinθ – 1 = 0，其中 0 ≤ θ ≤ 2π。
选项：A: θ = π/2, 7π/6, 11π/6 B: θ = π/6, 5π/6 C: θ = π/3, 4π/3 D: θ = π/2, 3π/2

Factor the quadratic in sinθ: (2 sinθ + 1)(sinθ – 1) = 0. Hence sinθ = –½ or sinθ = 1. For sinθ = 1 in [0, 2π], θ = π/2. For sinθ = –½, the solutions are θ = 7π/6 and θ = 11π/6. The complete solution set is {π/2, 7π/6, 11π/6}. Correct option is A.

将 sinθ 的二次式分解因式：(2 sinθ + 1)(sinθ – 1) = 0。因此 sinθ = –½ 或 sinθ = 1。在 [0, 2π] 内，sinθ = 1 得 θ = π/2；sinθ = –½ 得 θ = 7π/6 与 θ = 11π/6。完整解集为 {π/2, 7π/6, 11π/6}，正确选项为 A。

5. Question 5 – Differentiation Using Product Rule | 乘积法则求导

Differentiate y = x² e³ˣ.
Options: A: dy/dx = 2x e³ˣ + 3x² e³ˣ B: dy/dx = 2x e³ˣ C: dy/dx = x e³ˣ (2 + 3x) D: Both A and C

求 y = x² e³ˣ 的导数。
选项：A: dy/dx = 2x e³ˣ + 3x² e³ˣ B: dy/dx = 2x e³ˣ C: dy/dx = x e³ˣ (2 + 3x) D: A 和 C 均正确

Apply the product rule with u = x² (u’ = 2x) and v = e³ˣ (v’ = 3e³ˣ). Then dy/dx = u’v + uv’ = 2x e³ˣ + x²·3e³ˣ = 2x e³ˣ + 3x² e³ˣ. Factorising gives x e³ˣ (2 + 3x). Options A and C represent the same expression, so the answer is D.

使用乘积法则：设 u = x² (u’ = 2x)，v = e³ˣ (v’ = 3e³ˣ)。则 dy/dx = u’v + uv’ = 2x e³ˣ + x²·3e³ˣ = 2x e³ˣ + 3x² e³ˣ。提取公因式得 x e³ˣ (2 + 3x)。选项 A 和 C 等价，故答案为 D。

6. Question 6 – Integration by Recognition | 观察法积分

Evaluate ∫ (2x+1) eˣ²⁺ˣ dx.
Options: A: eˣ²⁺ˣ + C B: (x² + x) eˣ²⁺ˣ + C C: 2eˣ²⁺ˣ + C D: e²ˣ⁺¹ + C

计算 ∫ (2x+1) eˣ²⁺ˣ dx。
选项：A: eˣ²⁺ˣ + C B: (x² + x) eˣ²⁺ˣ + C C: 2eˣ²⁺ˣ + C D: e²ˣ⁺¹ + C

Notice that the derivative of the exponent (x² + x) is

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