📚 Advanced Mathematics: NSAA 2019 Section 1 Question Paper | 进阶数学：NSAA 2019 S1 试卷解析
The NSAA (Natural Sciences Admissions Assessment) is a pre-interview test used by the University of Cambridge for applicants to Natural Sciences. Section 1 consists of multiple-choice questions in Mathematics and one of Physics, Chemistry, or Biology. The 2019 Mathematics paper demands strong analytical skills, covering topics from algebra, functions, trigonometry, exponentials, and probability. In this article, we will break down typical questions, offering clear solution strategies and highlighting common pitfalls. This will help you refine your problem-solving technique and time management for this challenging assessment.
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测试。第一部分包含数学以及物理、化学或生物中的一科的选择题。2019年数学试卷对分析能力要求很高，涵盖代数、函数、三角学、指数与对数以及概率等主题。本文将拆解典型题目，提供清晰的解题策略并指出常见误区。这将帮助你完善解题技巧和时间管理，以应对这一富有挑战性的评估。
1. Overview of NSAA Section 1 Mathematics | NSAA 第一部分数学概述
The Mathematics section of NSAA 2019 S1 contains 20 multiple-choice questions to be completed in 60 minutes. Questions range from straightforward calculations to multi-step problems that require deep conceptual understanding. Each question has five options, with no penalty for incorrect answers, so guessing is encouraged if time is short. You need to be proficient in algebraic manipulation, function transformation, geometric reasoning, trigonometric identities, and basic probability.
NSAA 2019 S1 数学部分包含20道选择题，需在60分钟内完成。题目从直接计算到需要深刻概念理解的多步骤问题。每道题有五个选项，答错不扣分，因此时间不够时鼓励猜测。你需要熟练掌握代数运算、函数变换、几何推理、三角恒等式和基础概率。
2. Algebra and Equations: Radical Equation | 代数与方程：根式方程
Consider a typical radical equation: Solve √(x + 3) = x − 3. We must isolate the radical and square both sides, but always check for extraneous solutions.
考虑一个典型的根式方程：解 √(x + 3) = x − 3。我们需要分离根式并两边平方，但务必检验增根。
Square both sides: (√(x + 3))² = (x − 3)² → x + 3 = x² − 6x + 9.
两边平方：(√(x + 3))² = (x − 3)² → x + 3 = x² − 6x + 9。
Rearrange to form a quadratic: 0 = x² − 7x + 6 → x² − 7x + 6 = 0.
整理得二次方程：0 = x² − 7x + 6 → x² − 7x + 6 = 0。
Factorise: (x − 6)(x − 1) = 0, so potential solutions are x = 6 and x = 1.
因式分解：(x − 6)(x − 1) = 0，所以可能的解为 x = 6 和 x = 1。
Check each in the original equation: For x = 6, LHS √(6+3) = √9 = 3, RHS 6−3 = 3, valid. For x = 1, LHS √(1+3) = 2, RHS 1−3 = −2, invalid. Therefore the only solution is x = 6.
代入原方程检验：x = 6 时，左边 √(6+3) = √9 = 3，右边 6−3 = 3，成立。x = 1 时，左边 √(1+3) = 2，右边 1−3 = −2，不成立。因此唯一解为 x = 6。
In multiple-choice context, quickly spotting the extraneous root saves time. Remember to always verify if squaring introduced false solutions.
在选择题中，快速识别增根能节省时间。记住总是检验平方是否引入了假解。
3. Functions and Composition | 函数与复合
A common NSAA topic is function composition and domain. Let f(x) = 1 / (x − 1). Find the expression for f(f(x)) and determine its domain.
NSAA 常见主题是函数复合与定义域。设 f(x) = 1 / (x − 1)。求 f(f(x)) 的表达式并确定其定义域。
First, compute f(f(x)) = f(1/(x−1)) = 1 / ( [1/(x−1)] − 1 ).
首先，计算 f(f(x)) = f(1/(x−1)) = 1 / ( [1/(x−1)] − 1 )。
Simplify the denominator: [1/(x−1)] − 1 = (1 − (x−1)) / (x−1) = (2 − x) / (x−1).
简化分母：[1/(x−1)] − 1 = (1 − (x−1)) / (x−1) = (2 − x) / (x−1)。
Thus f(f(x)) = 1 / ( (2−x)/(x−1) ) = (x−1) / (2−x).
因此 f(f(x)) = 1 / ( (2−x)/(x−1) ) = (x−1) / (2−x)。
Now find the domain: The inner function f(x) requires x ≠ 1. Also f(f(x)) is undefined when its own denominator 2−x = 0, i.e., x = 2. Additionally, the input to the outer f is 1/(x−1) which must not equal 1, because f(x) is undefined at x=1. So set 1/(x−1) ≠ 1 → x−1 ≠ 1 → x ≠ 2. This matches the previous restriction. Hence domain is all real x except x = 1 and x = 2.
现在求定义域：内层函数 f(x) 要求 x ≠ 1。同时 f(f(x)) 自身分母 2−x = 0 时无定义，即 x = 2。此外，外层的 f 的输入 1/(x−1) 不能等于 1，因为 f(x) 在 x=1 无定义。令 1/(x−1) ≠ 1 → x−1 ≠ 1 → x ≠ 2。这与之前的限制一致。因此定义域为所有实数，除去 x = 1 和 x = 2。
Notice that the simplified form (x−1)/(2−x) might misleadingly suggest only x ≠ 2, but the original composition carries the restriction from the inner function. Always track step-by-step restrictions.
注意简化后的形式 (x−1)/(2−x) 可能误导为只要求 x ≠ 2，但原始复合携带了内层函数的限制。务必逐步跟踪限制条件。
4. Trigonometry: Double-Angle Formula | 三角学：倍角公式
Given sinθ = 3/5 and θ is acute, find the value of cos 2θ. This is a direct application of the double-angle identity.
已知 sinθ = 3/5 且 θ 为锐角，求 cos 2θ 的值。这是倍角公式的直接应用。
Use the identity cos 2θ = 1 − 2 sin²θ. Plug in sinθ = 3/5.
使用恒等式 cos 2θ = 1 − 2 sin²θ。代入 sinθ = 3/5。
Calculate sin²θ = (3/5)² = 9/25. Then cos 2θ = 1 − 2×(9/25) = 1 − 18/25 = 7/25.
计算 sin²θ = (3/5)² = 9/25。那么 cos 2θ = 1 − 2×(9/25) = 1 − 18/25 = 7/25。
Alternatively, using cos 2θ = cos²θ − sin²θ requires finding cosθ first. Since θ is acute, cosθ = √(1 − (3/5)²) = √(1 − 9/25) = √(16/25) = 4/5. Then cos 2θ = (4/5)² − (3/5)² = 16/25 − 9/25 = 7/25. Both give the same result.
或者，使用 cos 2θ = cos²θ − sin²θ 需要先求 cosθ。由于 θ 是锐角，cosθ = √(1 − (3/5)²) = √(1 − 9/25) = √(16/25) = 4/5。然后 cos 2θ = (4/5)² − (3/5)² = 16/25 − 9/25 = 7/25。两种方法结果一致。
This problem highlights the importance of knowing multiple trigonometric identities and choosing the one that uses given data directly to minimise errors.
这题凸显了掌握多种三角恒等式并选择直接利用已知数据的方法以最小化错误的重要性。
5. Exponentials and Logarithms | 指数与对数
Solve for x: 3²ˣ = 5ˣ⁺¹. Since the bases are different, taking logarithms of both sides is the standard approach.
解 x：3²ˣ = 5ˣ⁺¹。由于底数不同，两边取对数是标准方法。
Take natural logarithm on both sides: ln(3²ˣ) = ln(5ˣ⁺¹). Apply the power rule: 2x ln 3 = (x+1) ln 5.
两边取自然对数：ln(3²ˣ) = ln(5ˣ⁺¹)。应用幂法则：2x ln 3 = (x+1) ln 5。
Expand the right side: 2x ln 3 = x ln 5 + ln 5. Collect x terms: 2x ln 3 − x ln 5 = ln 5.
展开右边：2x ln 3 = x ln 5 + ln 5。合并含 x 的项：2x ln 3 − x ln 5 = ln 5。
Factor x: x (2 ln 3 − ln 5) = ln 5. Thus x = ln 5 / (2 ln 3 − ln 5).
提取 x：x (2 ln 3 − ln 5) = ln 5。于是 x = ln 5 / (2 ln 3 − ln 5)。
We can simplify further using logarithm properties: 2 ln 3 = ln(3²) = ln 9. Then denominator becomes ln 9 − ln 5 = ln(9/5). So x = ln 5 / ln(9/5). This can be expressed as log base (9/5) of 5, which is acceptable.
我们可以利用对数性质进一步化简：2 ln 3 = ln(3²) = ln 9。分母变为 ln 9 − ln 5 = ln(9/5)。因此 x = ln 5 / ln(9/5)。这可以表示为以 (9/5) 为底 5 的对数，这是可以接受的。
Always check if the multiple-choice options present equivalent forms. Understanding change of base formula is crucial.
务必检查选择题选项是否提供了等价形式。理解换底公式至关重要。
6. Probability Without Replacement | 无放回概率
A box contains 4 red and 6 blue balls. Two balls are drawn at random without replacement. Find the probability that at least one is red.
一个盒子装有 4 个红球和 6 个蓝球。随机抽取两球且不放回。求至少一球是红色的概率。
It is easier to compute the complement: the probability that no red ball is drawn, i.e., both are blue.
计算其补集更简单：没有抽到红球的概率，即两球都是蓝色。
P(both blue) = (6/10) × (5/9) = 30/90 = 1/3.
P（两球都是蓝色）= (6/10) × (5/9) = 30/90 = 1/3。
Therefore, P(at least one red) = 1 − 1/3 = 2/3.
因此，P（至少一红）= 1 − 1/3 = 2/3。
Alternatively, directly calculate sum of P(exactly one red) and P(two reds): (4/10 × 6/9) + (6/10 × 4/9) + (4/10 × 3/9) = 48/90 + 12/90 = 60/90 = 2/3.
或者，直接计算 P（恰好一红）与 P（两红）之和：(4/10 × 6/9) + (6/10 × 4/9) + (4/10 × 3/9) = 48/90 + 12/90 = 60/90 = 2/3。
The complement method is often faster and reduces arithmetic mistakes in exam settings. Recognise when to use it.
补集法通常更快，并在考试中减少算术错误。学会识别何时使用它。
7. Coordinate Geometry: Circle and Tangent | 坐标几何：圆与切线
A typical NSAA question might read: A circle has equation x² + y² = 25. Find the gradient of the tangent at the point (3, 4).
典型的 NSAA 题目可能是：圆的方程为 x² + y² = 25。求点 (3, 4) 处切线的斜率。
For the circle x² + y² = r², the radius to point (x₁, y₁) has gradient y₁/x₁. The tangent is perpendicular to the radius, so its gradient m_tangent = −x₁/y₁.
对于圆 x² + y² = r²，过点 (x₁, y₁) 的半径斜率为 y₁/x₁。切线与半径垂直，因此其斜率 m_切线 = −x₁/y₁。
Plug in (3,4): m = −3/4.
代入 (3,4)：m = −3/4。
Alternatively, use implicit differentiation: 2x + 2y (dy/dx) = 0 → dy/dx = −x/y. At (3,4), dy/dx = −3/4.
或者，使用隐函数微分：2x + 2y (dy/dx) = 0 → dy/dx = −x/y。在 (3,4) 处，dy/dx = −3/4。
Such problems require swift recall of geometric properties of circles. Make sure you know that radius and tangent are perpendicular.
此类题目需要快速回忆圆的几何性质。确保你知道半径与切线垂直。
8. Sequences and Series: Arithmetic Progression | 数列与级数：等差数列
An arithmetic progression has first term 5 and common difference 3. Find the sum of the first 20 terms.
一个等差数列首项为 5，公差为 3。求前 20 项之和。
Use the formula S_n = n/2 [2a + (n−1)d]. Here n=20, a=5, d=3.
使用公式 S_n = n/2 [2a + (n−1)d]。这里 n=20，a=5，d=3。
Compute 2a + (n−1)d = 2×5 + 19×3 = 10 + 57 = 67.
计算 2a + (n−1)d = 2×5 + 19×3 = 10 + 57 = 67。
Then S_20 = 20/2 × 67 = 10 × 67 = 670.
因此 S_20 = 20/2 × 67 = 10 × 67 = 670。
Quick checks: The 20th term is a + 19d = 5 + 57 = 62. Average of first and last terms is (5+62)/2 = 33.5. Multiply by n gives 33.5 × 20 = 670. Consistent.
快速检验：第 20 项为 a + 19d = 5 + 57 = 62。首末项平均值为 (5+62)/2 = 33.5。乘以 n 得 33.5 × 20 = 670。一致。
Arithmetic series problems often appear; memorising both sum formulas is advised.
等差数列问题经常出现；建议记住两种求和公式。
9. Problem-Solving Techniques for NSAA Mathematics | NSAA 数学解题技巧
Effective strategies include: identify the relevant topic quickly, eliminate obviously wrong options, back-substitute answer choices when appropriate, and use dimensional analysis. When stuck, draw diagrams and annotate key values.
有效策略包括：迅速识别相关主题、排除明显错误选项、在适当时回代答案选项、以及使用量纲分析。卡住时，画图并标注关键数值。
Time allocation: Each question should average 3 minutes. If a question seems too lengthy, flag it and return later. Do not get stuck; the test is designed to stretch even strong candidates.
时间分配：每道题平均 3 分钟。如果某题似乎耗时过长，标记它并稍后回来。不要卡住；测试设计本就要让优秀考生也感到有挑战。
Practice with timed sections and review mistakes. Focus on speed without compromising accuracy. For multiple-choice, sometimes estimating or testing boundary values can lead to the correct answer faster than a full algebraic solution.
进行限时练习并复习错误。在保证准确度的同时追求速度。对于选择题，有时估算或代入边界值会比完整的代数解法更快得到正确答案。
10. Common Mistakes to Avoid | 应避免的常见错误
Students often forget to check domain restrictions in functions, mishandle signs when squaring equations, or misapply trigonometric identities. Another frequent error is reading ‘without replacement’ as ‘with replacement’ in probability.
学生常常忘记检查函数的定义域限制、在方程平方时错误处理符号、或者误用三角恒等式。另一个常见错误是将概率中的“不放回”当成“放回”处理。
In logs, be careful not to assume ln(a+b) = ln a + ln b. Always verify your solution in the original equation, especially for log and radical equations. Also, double-check whether the question asks for gradient or normal, sum or nth term.
在处理对数时，小心不要假设 ln(a+b) = ln a + ln b。务必在原方程中验证你的解，对于对数方程和根式方程尤其重要。同时，仔细核对题目是求斜率还是法线、求和还是第 n 项。
11. Time Management and Exam Day Tips | 时间管理与考试日建议
Begin with your strongest subject to build confidence. For mathematics, quickly scan the paper and start with straightforward algebra or probability questions. Use spare time to re-check marked questions. Bring a calculator that is permitted and ensure you are familiar with its functions.
从你最擅长的科目开始以建立信心。对于数学，快速浏览试卷并从简单的代数或概率题入手。利用剩余时间复查标记的题目。携带允许的计算器，并确保你熟悉其功能。
Read each question carefully; NSAA often includes distractors that match common mistakes. If you have to guess, eliminate unreasonable options to improve odds. Stay calm and manage stress through deep breathing.
仔细阅读每道题；NSAA 经常包含与常见错误匹配的干扰项。如果不得不猜测，先排除不合理选项以提高几率。通过深呼吸保持冷静并管理压力。
12. Summary and Final Review | 总结与最终回顾
The 2019 NSAA S1 Mathematics paper tests a broad spectrum of your mathematical foundation. By thoroughly understanding the underlying concepts and practising with past-paper style questions, you can significantly improve your performance. Remember to balance speed with precision, use smart strategies, and learn from every mistake. Best of luck in your preparation!
2019 年 NSAA 第一部分数学试卷测试了你广泛的数学基础。通过深入理解基本概念并用往届风格的题目进行练习，你可以显著提高表现。记住平衡速度与准确度，运用聪明的策略，并从每一个错误中学习。祝你备考顺利！
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