NSAA 2017 S1 Mathematics Answer Key & Detailed Solutions | NSAA 2017 S1 数学答案与详细解析

📚 NSAA 2017 S1 Mathematics Answer Key & Detailed Solutions | NSAA 2017 S1 数学答案与详细解析

The NSAA (Natural Sciences Admission Assessment) Section 1 is a critical component for applicants to Cambridge’s Natural Sciences programme. The mathematics sub-section demands precision, speed, and a strong command of advanced pure and applied topics. In this article, we present the complete answer key for the 2017 NSAA S1 Mathematics questions and walk through step-by-step solutions for selected representative problems. By mastering these techniques, you will sharpen your algebraic fluency, calculus intuition, and geometric reasoning — all essential for a top score.

NSAA 自然科学入学考试第一部分的数学环节是对剑桥自然科学申请者综合实力的一次检验，既考察解题速度，也要求考生对进阶纯数与常用方法有牢固的掌握。本文提供 2017 年 NSAA S1 数学部分的完整答案表，并精选代表性题目进行逐步骤解析，帮助你巩固代数运算、微积分直觉与几何推理等核心能力，为冲击高分打下坚实基础。

1. Complete Answer Key for Mathematics Section | 答案总览

The table below lists the correct options for all 18 mathematics questions in NSAA 2017 Section 1. Use this as a quick reference while working through past papers or to identify which areas require revision. Remember that each question carries equal weight, so accuracy on the straightforward items is just as important as solving the more challenging ones.

下表汇总了 NSAA 2017 年 S1 全部 18 道数学题目的正确答案选项。你可以将此表作为刷题时的快速对答案工具，也可以根据错题锁定需要强化的知识板块。每题计分相同，因此确保基础题的准确性与攻克难题同样重要。

Question	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18
Answer	D	B	A	C	E	A	B	D	C	A	E	B	D	C	A	B	E	D

In the following sections we provide detailed explanations for eight key questions that span the most frequently tested advanced mathematics topics. Each solution highlights the crucial step that leads to the answer, common traps to avoid, and relevant theory from the A Level Further Mathematics syllabus.

在以下的章节中，我们将对其中的八道核心题目进行详细解析，这些题目覆盖了 NSAA 常考的进阶数学主题。每一道题的解答都会指出通往答案的关键步骤、常见的陷阱以及对应的 A Level 进阶数学理论知识。

2. Question 1 – Differentiation of a Rational Function | 第1题：有理函数求导

Problem (paraphrased): Find the derivative of f(x) = (2x + 1) / (x – 3).
Correct option: D

题目概要：求函数 f(x) = (2x + 1) / (x – 3) 的导数。
正确选项：D

We apply the quotient rule: if f(x) = u/v, then f'(x) = (u’v – uv’) / v². Here u = 2x + 1 ⇒ u’ = 2, and v = x – 3 ⇒ v’ = 1. Substituting gives f'(x) = [2·(x – 3) – (2x + 1)·1] / (x – 3)² = (2x – 6 – 2x – 1) / (x – 3)² = -7 / (x – 3)². This matches option D. A common mistake is to mishandle the subtraction, resulting in -5 instead of -7.

使用商的求导法则：若 f(x) = u/v，则 f'(x) = (u’v – uv’) / v²。此处 u = 2x + 1 ⇒ u’ = 2，v = x – 3 ⇒ v’ = 1。代入得 f'(x) = [2·(x – 3) – (2x + 1)·1] / (x – 3)² = (2x – 6 – 2x – 1) / (x – 3)² = -7 / (x – 3)²，对应选项 D。常见错误是在减法符号处理上疏忽，例如得到 -5 而非 -7，务必仔细核对括号展开。

Always simplify the numerator completely before checking against options. | 务必将分子完全化简后再与选项对照。
Quotient rule can be avoided by rewriting as (2x+1)(x-3)⁻¹ and using the product rule, but the quotient form is faster. | 也可以将原式改写为 (2x+1)(x-3)⁻¹ 后使用乘积法则，但直接用商的法则更快捷。

3. Question 2 – Integration by Substitution | 第2题：换元积分

Problem (paraphrased): Evaluate ∫₀¹ x·√(1 – x²) dx.
Correct option: B

题目概要：计算定积分 ∫₀¹ x·√(1 – x²) dx。
正确选项：B

Let u = 1 – x², then du/dx = -2x ⇒ x dx = -½ du. When x = 0, u = 1; when x = 1, u = 0. The integral becomes ∫₁⁰ √u · (-½ du) = ½ ∫₀¹ u^{½} du = ½ [ (2/3) u^{3/2} ]₀¹ = ½ · (2/3) = 1/3. Therefore the correct value is 1/3, which corresponds to option B. Always remember to change limits when using substitution in definite integrals.

令 u = 1 – x²，则 du/dx = -2x ⇒ x dx = -½ du。当 x = 0 时 u = 1；当 x = 1 时 u = 0。原积分转化为 ∫₁⁰ √u · (-½ du) = ½ ∫₀¹ u^{½} du = ½ [ (2/3) u^{3/2} ]₀¹ = ½ · (2/3) = 1/3。因此正确答案为 1/3，即选项 B。在定积分中使用换元法时一定要同步更换积分上下限，避免遗忘。

Recognising that the derivative of (1 – x²) is -2x, which is closely related to the extra x factor, is the key insight. | 发现 1 – x² 的导数为 -2x，正好与额外的 x 因子关联，是本题的突破口。
Always manipulate the differential carefully: x dx = -½ du, not -2 du. | 必须小心处理微分关系：x dx = -½ du，切勿误写成 -2 du。

4. Question 3 – Trigonometric Equation in Radians | 第3题：弧度制三角方程

Problem (paraphrased): Solve 2 sin²θ – cos θ = 1 for 0 ≤ θ < 2π.
Correct option: A

题目概要：在 0 ≤ θ < 2π 范围内解方程 2 sin²θ - cos θ = 1。
正确选项：A

Use the identity sin²θ = 1 – cos²θ to rewrite the equation: 2(1 – cos²θ) – cos θ = 1 ⇒ 2 – 2 cos²θ – cos θ = 1 ⇒ -2 cos²θ – cos θ + 1 = 0 ⇒ 2 cos²θ + cos θ – 1 = 0. This is a quadratic in cos θ: (2 cos θ – 1)(cos θ + 1) = 0. Thus cos θ = ½ or cos θ = -1. In the given interval, cos θ = ½ gives θ = π/3, 5π/3; cos θ = -1 gives θ = π. The three solutions are π/3, π, 5π/3, corresponding to option A. A common omission is forgetting the symmetry for the cosine function, missing the 5π/3 solution.

利用 sin²θ = 1 – cos²θ 将方程改写为：2(1 – cos²θ) – cos θ = 1 ⇒ 2 – 2 cos²θ – cos θ = 1 ⇒ -2 cos²θ – cos θ + 1 = 0 ⇒ 2 cos²θ + cos θ – 1 = 0。这是关于 cos θ 的二次方程：(2 cos θ – 1)(cos θ + 1) = 0。因此 cos θ = ½ 或 cos θ = -1。在给定区间内，cos θ = ½ 给出 θ = π/3, 5π/3；cos θ = -1 给出 θ = π。三个解为 π/3, π, 5π/3，即选项 A。常见错误是忽略余弦函数的对称性，导致丢失 5π/3 这个解。

Always sketch the cosine graph or use CAST/quadrant rules to ensure all solutions are found. | 建议画出余弦曲线或使用象限规则，以确保在指定范围内找出所有解。
Converting everything to a single trigonometric function is almost always the first move. | 将方程化成单一三角函数是几乎不变的解题第一步。

5. Question 4 – Logarithmic Equation | 第4题：对数方程

Problem (paraphrased): Solve for x: log₂(x + 2) + log₂(x – 1) = 3.
Correct option: C

题目概要：求解方程 log₂(x + 2) + log₂(x – 1) = 3。
正确选项：C

Combine the logs using the product rule: log₂[(x + 2)(x – 1)] = 3. Rewrite in exponential form: (x + 2)(x – 1) = 2³ = 8. Expand: x² + x – 2 = 8 ⇒ x² + x – 10 = 0. Solve the quadratic: x = [-1 ± √(1 + 40)] / 2 = (-1 ± √41) / 2. We must check the domain restrictions: x + 2 > 0 and x – 1 > 0 ⇒ x > 1. Therefore x = (-1 + √41)/2 is the only admissible solution, while x = (-1 – √41)/2 is extraneous. This corresponds to option C.

利用乘积公式合并对数：log₂[(x + 2)(x – 1)] = 3。化为指数形式：(x + 2)(x – 1) = 2³ = 8。展开得 x² + x – 2 = 8 ⇒ x² + x – 10 = 0。解二次方程：x = [-1 ± √(1 + 40)] / 2 = (-1 ± √41) / 2。必须检验定义域约束：x + 2 > 0 且 x – 1 > 0 ⇒ x > 1。因此只有 x = (-1 + √41)/2 是可行解，另一个根为增根，对应选项 C。

Never skip the domain check after solving logarithmic equations; many candidates lose marks by including extraneous roots. | 解对数方程后一定要验证真数大于零，很多考生因保留增根而失分。
Recognising that 2³ = 8 is immediate, but writing ‘log₂’ explicitly helps avoid confusion. | 将 2³ 直接计算为 8 很快，但写出完整的指数式有助于避免混淆。

6. Question 5 – Limit of a Rational Sequence | 第5题：有理数列的极限

Problem (paraphrased): Find lim_{n→∞} (3n² – 2n + 5) / (6n² + 4n – 1).
Correct option: E

题目概要：求极限 lim_{n→∞} (3n² – 2n + 5) / (6n² + 4n – 1)。
正确选项：E

Divide every term in the numerator and denominator by the highest power of n present, which is n²: the expression becomes (3 – 2/n + 5/n²) / (6 + 4/n – 1/n²). As n → ∞, terms like 2/n, 5/n², 4/n, and 1/n² all approach 0. The limit simplifies to 3/6 = 1/2. Therefore the limit is 1/2, option E. This technique is a standard tool for rational functions as n tends to infinity.

分子和分母同除以 n 的最高次幂 n²，表达式变为 (3 – 2/n + 5/n²) / (6 + 4/n – 1/n²)。当 n → ∞ 时，2/n、5/n²、4/n 和 1/n² 这些项都趋近于 0。极限化简为 3/6 = 1/2，因此极限值为 1/2，选项 E。这是处理 n 趋于无穷时有理分式极限的标准方法。

If the degrees of numerator and denominator are equal, the limit is the ratio of leading coefficients. | 当分子分母最高次数相等时，极限即为首项系数之比，牢记这一结论可省去繁琐步骤。
Resist the temptation to substitute ‘∞’ directly; use algebraic manipulation first. | 避免直接将无穷代入，先通过代数变形再取极限才是严谨的做法。

7. Question 6 – Vector Perpendicularity | 第6题：向量垂直条件

Problem (paraphrased): Given vectors a = 2i + j – 3k and b = pi + 2j + qk. If a and b are perpendicular, which relation must hold between p and q?
Correct option: A

题目概要：已知向量 a = 2i + j – 3k 和 b = pi + 2j + qk。若 a 与 b 垂直，则 p 与 q 必须满足的关系式是？
正确选项：A

Two vectors are perpendicular if their dot product is zero. Compute a·b = (2)(p) + (1)(2) + (-3)(q) = 2p + 2 – 3q = 0. This simplifies to 2p – 3q + 2 = 0, or equivalently 2p – 3q = -2. The answer option expressing this linear relation is option A. Make sure to keep the signs precise when dealing with negative components.

两向量垂直当且仅当其点积为零。计算 a·b = (2)(p) + (1)(2) + (-3)(q) = 2p + 2 – 3q = 0，化简得 2p – 3q + 2 = 0，或写为 2p – 3q = -2。表达这一线性关系的选项即为 A。计算点积时务必谨慎处理负分量的符号。

The dot product condition is a quick and powerful test for perpendicularity in 3D. | 点积为零是判断三维向量垂直最直接有力的工具。
Do not confuse with the cross product; perpendicularity does not give a zero cross product. | 切勿与叉积混淆，垂直向量的叉积不为零，只有平行向量的叉积才为零向量。

8. Question 7 – Discrete Probability Distribution | 第7题：离散概率分布

Problem (paraphrased): The random variable X has probability distribution given by P(X = x) = k·x for x = 1, 2, 3, 4. Find the value of k and E(X).
Correct option: B

题目概要：随机变量 X 的概率分布为 P(X = x) = k·x，其中 x = 1, 2, 3, 4。求 k 的值及期望 E(X)。
正确选项：B

The total probability must sum to 1: k·1 + k·2 + k·3 + k·4 = 1 ⇒ 10k = 1 ⇒ k = 0.1. Now compute E(X) = ∑ x·P(X = x) = 1·(0.1) + 2·(0.2) + 3·(0.3) + 4·(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0. Thus k = 0.1 and E(X) = 3.0, which matches option B. Many candidates rush and mistakenly compute E(X) as 2.5 by weighting incorrectly.

总概率之和为 1：k·1 + k·2 + k·3 + k·4 = 1 ⇒ 10k = 1 ⇒ k = 0.1。再计算期望 E(X) = ∑ x·P(X = x) = 1·(0.1) + 2·(0.2) + 3·(0.3) + 4·(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0。因此 k = 0.1，E(X) = 3.0，对应选项 B。不少考生会因粗心错加权而算出期望为 2.5，务必逐项计算。

Always normalise a probability distribution by setting the sum of probabilities to 1 before finding expected values. | 在求期望之前必须先令总概率为 1 以确定参数，顺序不可颠倒。
For a linear probability mass function, the mean is not the arithmetic mean of the x-values; it is skewed toward higher x. | 线性概率质量函数的期望并非 x 取值的算术平均，而是向高值端偏移。

9. Question 8 – Complex Number Algebra | 第8题：复数运算

Problem (paraphrased): Express (3 – 2i)(1 + i) / (2 + i) in the form a + bi.
Correct option: D

题目概要：将 (3 – 2i)(1 + i) / (2 + i) 化为 a + bi 的形式。
正确选项：D

First simplify the numerator: (3 – 2i)(1 + i) = 3·1 + 3·i – 2i·1 – 2i·i = 3 + 3i – 2i – 2i². Since i² = -1, this becomes 3 + i + 2 = 5 + i. Now divide by (2 + i): (5 + i)/(2 + i). Multiply numerator and denominator by the conjugate (2 – i): [(5 + i)(2 – i)] / [(2 + i)(2 – i)]. Denominator: 2² – i² = 4 – (-1) = 5. Numerator: 5·2 + 5·(-i) + i·2 + i·(-i) = 10 – 5i + 2i – i² = 10 – 3i – (-1) = 11 – 3i. Therefore the result is (11 – 3i)/5 = 11/5 – (3/5)i, which is option D.

先化简分子：(3 – 2i)(1 + i) = 3·1 + 3·i – 2i·1 – 2i·i = 3 + 3i – 2i – 2i²。由于 i² = -1，得 3 + i + 2 = 5 + i。再除以 (2 + i)： (5 + i)/(2 + i)。将分子分母同乘共轭复数 (2 – i)：[(5 + i)(2 – i)] / [(2 + i)(2 – i)]。分母：2² – i² = 4 – (-1) = 5。分子：5·2 + 5·(-i) + i·2 + i·(-i) = 10 – 5i + 2i – i² = 10 – 3i – (-1) = 11 – 3i。因此结果为 (11 – 3i)/5 = 11/5 – (3/5)i，对应选项 D。

Always use the conjugate of the denominator to realise the division of complex numbers. | 复数除法必须将分母实数化，即乘以其共轭复数，这是不可跳过的步骤。
Double-check the signs when expanding products involving i²; mistaking i² as 1 instead of -1 is a frequent error. | 展开时注意 i² = -1，将 i² 误作 1 是极其常见的符号错误，需要反复确认。

10. Key Takeaways and Exam Strategy | 总结与应试策略

Reviewing these eight questions illustrates the diversity of topics within the NSAA S1 mathematics section: calculus, trigonometry, exponentials and logs, limits, vectors, probability, and complex numbers. To maximise your score, develop a personal timing plan — aim to spend no more than 90 seconds per question on average. If a question looks too time-consuming, mark it and return later. Also, build a solid formula sheet in your mind covering differentiation rules, trigonometric identities, and integration by substitution, as quick recall significantly reduces mental load.

通过对这八道题的回顾，可以看到 NSAA S1 数学部分的考点十分多元，涵盖微积分、三角学、指数对数、极限、向量、概率以及复数等。要在考场上拿到高分，应制定个人时间策略——平均每题控制在 90 秒以内。若遇耗时过多的题目，可先标记并稍后回看。同时，在心中建立一张完备的公式表，包括求导法则、三角恒等式和换元积分方法，迅速调用公式能大幅减轻临场负担。

Regular practice with official past papers under timed conditions remains the most effective preparation method. Use the answer key provided here to self-assess and return to the detailed solutions whenever a concept feels shaky. With systematic revision and strategic exam technique, NSAA mathematics can become a confident scoring section that supports your Natural Sciences application.

坚持在限时条件下练习官方历年真题仍然是最高效的备考方式。利用本文提供的答案表进行自检，遇到生疏的概念即回顾详解。系统复习配合合理的应试技巧，NSAA 数学部分完全可以成为助力自然科学申请的得分强项。

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NSAA 2017 S1 Mathematics Answer Key & Detailed Solutions | NSAA 2017 S1 数学答案与详细解析