NSAA 2021 S1 Mathematics Answer Key & Further Maths Insights | NSAA 2021 第一部分数学答案与进阶数学解析

📚 NSAA 2021 S1 Mathematics Answer Key & Further Maths Insights | NSAA 2021 第一部分数学答案与进阶数学解析

The NSAA (Natural Sciences Admissions Assessment) is a pre-interview test used by the University of Cambridge for applicants to Natural Sciences. Section 1 Mathematics assesses your ability to apply mathematical reasoning quickly and accurately, often blending GCSE and AS-level content with the kind of logical thinking required in further maths. This article presents the answer key for the 2021 Section 1 Mathematics paper and provides in-depth solutions to selected questions that showcase advanced mathematics techniques, helping you connect core concepts to Further Mathematics skills.

NSAA（自然科学入学评估）是剑桥大学自然科学专业申请者的面试前笔试。第一部分数学考查快速准确应用数学推理的能力，往往混合了GCSE和AS级别内容以及进阶数学中要求的逻辑思维。本文提供2021年第一部分数学答案，并对精选题目进行深入解析，展示进阶数学技巧，帮助你从核心概念延伸到进阶数学能力的培养。

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1. Answer Key Overview | 答案概览

Below is the complete answer key for NSAA 2021 Section 1 Mathematics. Each question is mapped to its primary topic. Use this table to check your responses and identify areas for further study.

以下是NSAA 2021第一部分数学的完整答案。每道题都标注了主要考查主题。你可以对照这份表格检查自己的作答，并找出需要加强学习的内容。

Question	Answer	Topic
1	D	Arithmetic with fractions and percentages
2	B	Standard form and estimation
3	A	Ratio and proportion
4	C	Linear equations in context
5	B	Logarithms and exponent rules
6	D	Area and perimeter of composite shapes
7	A	Pythagoras’ theorem and circles
8	C	Parametric equations and differentiation
9	A	Probability and tree diagrams
10	B	Mean, median, and range
11	C	Inequalities and number lines
12	A	Recurrence sequences and series
13	D	Graphs of quadratic functions
14	B	Volume and surface area of 3D shapes
15	C	Trigonometric equations and identities
16	D	Gradients and equations of lines
17	A	Percentages and compound changes
18	B	Inverse functions and transformations
19	C	Data interpretation from tables
20	D	Polynomials, factor theorem and algebraic division

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2. Solving a Logarithmic Equation (Q5) | 对数方程求解 (第5题)

Question 5 required you to solve a logarithmic equation such as log₂(x+1) − log₂(x−1) = 1. This type of problem appears in Further Mathematics under the topic of exponential and logarithmic functions, and it reinforces the importance of checking domain restrictions.

第5题要求解一个对数方程，例如 log₂(x+1) − log₂(x−1) = 1。这类问题在进阶数学中属于指数与对数函数专题，它强化了检查定义域限制的重要性。

Combine the two logs using the quotient rule: log₂((x+1)/(x−1)) = 1.

利用对数减法法则合并：log₂((x+1)/(x−1)) = 1。

Rewrite in exponential form: (x+1)/(x−1) = 2¹ = 2.

转化为指数形式：(x+1)/(x−1) = 2¹ = 2。

Multiply both sides by (x−1): x + 1 = 2x − 2 ⇒ x = 3. Always verify that x = 3 keeps both arguments positive: 3+1 > 0 and 3−1 > 0, so the solution is valid. Option B was the correct choice.

两边同乘 (x−1)：x + 1 = 2x − 2 ⇒ x = 3。务必验证 x = 3 时两个真数都为正：3+1 > 0 且 3−1 > 0，解有效。正确选项为 B。

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3. Parametric Differentiation (Q8) | 参数方程求导 (第8题)

Question 8 presented a curve defined parametrically, for example x = t², y = 2t. The task was to find the gradient dy/dx at a specific point. This introduces the chain rule in a parametric context — a core Further Maths technique.

第8题给出了一条用参数定义的曲线，例如 x = t², y = 2t，要求在特定点求斜率 dy/dx。这引入了参数条件下的链式法则——这是进阶数学的核心技巧之一。

Compute dy/dt = 2 and dx/dt = 2t. The gradient is dy/dx = (dy/dt) / (dx/dt) = 2 / (2t) = 1/t. At t = 1, dy/dx = 1. The answer matched option C.

计算 dy/dt = 2，dx/dt = 2t。斜率为 dy/dx = (dy/dt) / (dx/dt) = 2 / (2t) = 1/t。当 t = 1 时，dy/dx = 1。答案对应选项 C。

This method can be extended to second derivatives and curve sketching, which frequently appear in Further Pure Mathematics. Understanding the parametric form also lays the foundation for vector calculus and mechanics.

该方法可推广到二阶导数和曲线草图绘制，这在高数纯数部分频繁出现。理解参数形式也为向量微积分和力学打下基础。

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4. Recurrence Sequences (Q12) | 递推数列 (第12题)

Question 12 involved a recurrence relation, possibly uₙ₊₁ = 2uₙ − 1 with u₁ = 3. You had to find a specific term or sum. Recurrence sequences are studied in detail in Further Mathematics under sequences and series, leading to methods like proof by induction and solving linear recurrences.

第12题涉及一个递推关系，可能的形式为 uₙ₊₁ = 2uₙ − 1，u₁ = 3。要求找出某一项的值或前几项和。递推数列在进阶数学的数列与级数专题中有详细研究，进而引出数学归纳法和线性递推的求解方法。

Evaluate term by term: u₁ = 3, u₂ = 2×3 − 1 = 5, u₃ = 2×5 − 1 = 9, u₄ = 2×9 − 1 = 17. For a question asking for u₄, the answer was 17, corresponding to option A. Always check if the question required a sum or a specific term to avoid a careless mistake.

逐项计算：u₁ = 3，u₂ = 2×3 − 1 = 5，u₃ = 2×5 − 1 = 9，u₄ = 2×9 − 1 = 17。若题目要求 u₄，则答案为 17，对应选项 A。务必看清问题是要求和还是求某一项，避免疏忽性错误。

In Further Maths, you would learn to solve the closed form: uₙ = 2ⁿ + 1. This helps verify results quickly and understand the growth of the sequence.

在进阶数学中，你将学会求解通项公式：uₙ = 2ⁿ + 1。这有助于快速验证结果并理解数列的增长趋势。

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5. Trigonometric Identities in Equations (Q15) | 三角恒等式解方程 (第15题)

Question 15 asked you to solve an equation like sin 2x = cos x for 0° ≤ x ≤ 360°. This requires the use of double-angle identities, a topic that sits at the heart of Further Trigonometry.

第15题要求解如 sin 2x = cos x 的方程，区间为 0° ≤ x ≤ 360°。这需要用到倍角公式，是进阶三角学的核心内容。

Rewrite sin 2x as 2 sin x cos x: 2 sin x cos x = cos x ⇒ cos x (2 sin x − 1) = 0. Set each factor to zero: cos x = 0 gives x = 90°, 270°; 2 sin x − 1 = 0 ⇒ sin x = ½ gives x = 30°, 150°.

将 sin 2x 化为 2 sin x cos x：2 sin x cos x = cos x ⇒ cos x (2 sin x − 1) = 0。令每个因式为零：cos x = 0 得 x = 90°, 270°；2 sin x − 1 = 0 ⇒ sin x = ½ 得 x = 30°, 150°。

Collect all solutions within the interval: 30°, 90°, 150°, 270°. This set of solutions matched one of the answer choices (C). This method of factoring after applying identities is essential for solving more complex trigonometric equations in A-level Further Maths.

收集区间内所有解：30°, 90°, 150°, 270°。这一组解与选项 C 对应。这种先应用恒等式再因式分解的方法，对于解决A-Level进阶数学中更复杂的三角方程至关重要。

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6. Inverse Functions and Transformations (Q18) | 反函数与函数变换 (第18题)

Question 18 tested the ability to find an inverse function or apply transformations, for instance given f(x) = √(x + 2) − 1. This topic appears in the Pure Mathematics component of Further Maths and demands a clear understanding of domain and range.

第18题考查求反函数或应用变换的能力，例如给定 f(x) = √(x + 2) − 1。该主题出现在进阶数学纯数部分，要求清晰理解定义域和值域。

To find the inverse, write y = √(x + 2) − 1, swap x and y: x = √(y + 2) − 1. Solve for y: x + 1 = √(y + 2) ⇒ y + 2 = (x + 1)² ⇒ y = (x + 1)² − 2. The domain of f⁻¹ is x ≥ −1 because the original output (range) starts at −1.

求反函数时，设 y = √(x + 2) − 1，交换 x 与 y：x = √(y + 2) − 1。解出 y：x + 1 = √(y + 2) ⇒ y + 2 = (x + 1)² ⇒ y = (x + 1)² − 2。f⁻¹ 的定义域为 x ≥ −1，因为原始输出（值域）从 −1 开始。

The correct expression was f⁻¹(x) = (x + 1)² − 2 for x ≥ −1, which aligned with option B. Remember that the graph of an inverse function is a reflection in the line y = x.

正确表达式为 f⁻¹(x) = (x + 1)² − 2，x ≥ −1，与选项 B 一致。记住反函数的图像是关于直线 y = x 的对称图形。

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7. Polynomials and the Factor Theorem (Q20) | 多项式与因式定理 (第20题)

Question 20 was a typical polynomial problem: given that (x − 2) is a factor of f(x) = 2x³ + kx² − 13x + 6, find k and then factorise completely. The factor theorem is a fundamental tool in Further Mathematics for algebraic manipulation and curve sketching.

第20题是一个典型的多项式问题：已知 (x − 2) 是 f(x) = 2x³ + kx² − 13x + 6 的因子，求 k 并完全分解因式。因式定理是进阶数学中代数运算和曲线绘制的基本工具。

Substitute x = 2 into f(2) = 0: 2(8) + k(4) − 13(2) + 6 = 0 ⇒ 16 + 4k − 26 + 6 = 0 ⇒ 4k − 4 = 0 ⇒ k = 1.

将 x = 2 代入 f(2) = 0：2(8) + k(4) − 13(2) + 6 =

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