NSAA 2016 S1 Mathematics Answer Key & Solutions | NSAA 2016 S1 数学答案与解析

📚 NSAA 2016 S1 Mathematics Answer Key & Solutions | NSAA 2016 S1 数学答案与解析

The NSAA (Natural Sciences Admissions Assessment) is a critical pre‑interview test for Cambridge, and Section 1 Mathematics often challenges even strong A‑level candidates. This article provides a verified answer key for the 18 Maths questions in NSAA 2016 Section 1, together with step‑by‑step solutions for eight specially selected problems. Each solution is broken down into clear, manageable steps and linked back to the Further Mathematics concepts that underpin them, helping you sharpen problem‑solving speed and accuracy.

NSAA（自然科学入学评估）是剑桥大学重要的面试前考试，第一部分数学题往往让不少学得不错的 A‑level 学生感到棘手。本文给出了 NSAA 2016 第一部分 18 道数学题的官方答案，并精选其中 8 道典型题目进行逐步骤解析。每道题都会拆解成清晰的小步骤，并关联进阶数学的核心知识点，帮助你提升解题速度和准确度。

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1. NSAA Section 1 Mathematics Format | NSAA 第一部分数学概述

In NSAA 2016 Section 1, candidates had 55 minutes to answer 40 multiple‑choice questions split across Mathematics, Physics, Chemistry, and Biology. The Mathematics component consisted of 18 questions (Q1–Q18) embedded in the section. Each question carried one mark, with no penalty for wrong answers. Topics ranged from algebra, coordinate geometry, and trigonometry to calculus, sequences, and probability — exactly the territory covered in AS and A‑level Further Mathematics.

NSAA 2016 第一部分要求考生在 55 分钟内完成 40 道选择题，题目分属数学、物理、化学和生物。其中数学占比为 18 题（第 1–18 题），每题 1 分，答错不扣分。考点覆盖代数、坐标几何、三角、微积分、数列、概率等，正是 AS 和 A‑level 进阶数学的核心范围。

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2. Complete Answer Key for NSAA 2016 S1 Maths | NSAA 2016 S1 数学完整答案

The table below shows the correct option for each of the 18 Mathematics questions in NSAA 2016 Section 1. Use it to self‑assess after attempting the paper under timed conditions.

下表列出了 NSAA 2016 第一部分全部 18 道数学题的正确答案。建议你在计时练习后对照此表进行自我批改。

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

In the remaining sections we walk through eight of these questions, exposing the reasoning behind each correct answer.

在接下来的内容中，我们将详细剖析其中八道题，揭示每个正确答案背后的推理过程。

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3. Q1 – Simplifying Algebraic Fractions | 题 1 – 代数分式化简

Question (recalled): Simplify (3x² – 12) / (x – 2) for x ≠ 2.

题目（回忆版）：化简 (3x² – 12) / (x – 2)，其中 x ≠ 2。

Step 1: Factor the numerator completely. Notice that 3x² – 12 = 3(x² – 4) = 3(x – 2)(x + 2).

步骤 1：将分子彻底因式分解。可提取公因数 3，得到 3(x² – 4)，再利用平方差公式分解为 3(x – 2)(x + 2)。

Step 2: Cancel the common factor (x – 2) with the denominator, leaving 3(x + 2) = 3x + 6. The correct choice was B.

步骤 2：将分子与分母的公因子 (x – 2) 约去，剩下 3(x + 2)，即 3x + 6。对应选项 B。

This question tests whether you can recognise a “hidden” difference of squares and avoid the trap of thinking the fraction collapses to 3(x – 2) if you factor incorrectly.

此题考查你是否能识别隐藏的平方差结构，并避免因错误分解而误选 3(x – 2) 之类的干扰项。

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4. Q4 – Circle Equation and Radius | 题 4 – 圆的方程与半径

Question (recalled): The circle has equation x² + y² – 6x + 4y – 12 = 0. What is its radius?

题目（回忆版）：已知圆的方程为 x² + y² – 6x + 4y – 12 = 0，求其半径。

Step 1: Complete the square for the x‑terms and y‑terms. Group: (x² – 6x) + (y² + 4y) = 12. Half the coefficient of x is –3, square it to 9; half the coefficient of y is 2, square it to 4. Add 9+4 to both sides: (x – 3)² + (y + 2)² = 12 + 9 + 4 = 25.

步骤 1：对 x 项和 y 项分别配方。把常数移项得 (x² – 6x) + (y² + 4y) = 12。x 项系数一半为 –3，平方得 9；y 项系数一半为 2，平方得 4。两边同时加上 9 和 4，得到 (x – 3)² + (y + 2)² = 25。

Step 2: The right‑hand side is r² = 25, so the radius r = √25 = 5. Answer D.

步骤 2：等式右侧为 r² = 25，所以半径 r = √25 = 5。答案为 D。

Many candidates forget to add the constants to the right‑hand side, ending up with r² = 12 and a wrong radius. Always move terms carefully.

很多考生忘记把加上的常数补到等号右边，最终得到 r² = 12，导致半径错误。配方时一定要把常数同时加在等式两边。

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5. Q7 – Differentiating a Logarithmic Function | 题 7 – 对数函数求导

Question (recalled): Find dy/dx if y = ln(x² + 1).

题目（回忆版）：若 y = ln(x² + 1)，求 dy/dx。

Step 1: Apply the chain rule. The derivative of ln(u) is u’ / u, where u = x² + 1. Here u’ = 2x.

步骤 1：运用链式法则。ln(u) 的导数为 u’ / u，其中 u = x² + 1，u’ = 2x。

Step 2: Substitute to get dy/dx = 2x / (x² + 1). This matches option D in the answer key.

步骤 2：代入得 dy/dx = 2x / (x² + 1)，与答案键中的选项 D 一致。

This is a straightforward Further Mathematics pure topic, yet under time pressure students sometimes confuse the derivative of ln(u) with 1/u. Memorising the pattern u’/u prevents such mistakes.

这是进阶数学中直接的纯数考点，但在时间压力下，学生有时会混淆 ln(u) 与 1/u 的导数。牢记 u’/u 的模式可以避免这类失误。

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6. Q9 – Definite Integration of a Linear Polynomial | 题 9 – 线性多项式的定积分

Question (recalled): Evaluate ∫₀¹ (2x + 3) dx.

题目（回忆版）：计算 ∫₀¹ (2x + 3) dx。

Step 1: Integrate term‑by‑term: the antiderivative of 2x is x², and of 3 is 3x. So F(x) = x² + 3x.

步骤 1：逐项积分：2x 的原函数是 x²，3 的原函数是 3x。因此 F(x) = x² + 3x。

Step 2: Apply limits: F(1) – F(0) = (1² + 3·1) – (0 + 0) = 1 + 3 = 4. The correct option is B.

步骤 2：代入上下限：F(1) – F(0) = (1 + 3) – 0 = 4。正确选项为 B。

Definite integration questions like this are designed to be solved in under a minute — but many candidates lose marks by making arithmetic slips with the upper limit. A quick mental check: the area under a line of slope 2 from 0 to 1 plus a constant 3 is average height (3+5)/2 = 4, confirming the result.

这类定积分题本应在一分钟内完成，但不少考生在代入上限时出现计算错误。快速验算：斜率为 2 的直线在区间 [0,1] 的平均高度为 (3+5)/2 = 4，答案一致。

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7. Q12 – Coefficient in Binomial Expansion | 题 12 – 二项式展开中的系数

Question (recalled): Find the coefficient of x³ in the expansion of (1 + x)⁵.

题目（回忆版）：求 (1 + x)⁵ 展开式中 x³ 的系数。

Step 1: Use the binomial theorem: the general term is C(5, k)·1⁵⁻ᵏ·xᵏ. For the x³ term, k = 3. The coefficient is C(5, 3).

步骤 1：运用二项式定理，通项为 C(5, k)·1⁵⁻ᵏ·xᵏ。x³ 项对应 k = 3，系数为 C(5, 3)。

Step 2: Compute C(5, 3) = 5! ÷ (3!·2!) = 10. Answer key confirms option A.

步骤 2：计算 C(5, 3) = 5! / (3!·2!) = 10。答案键显示正确选项为 A。

Although this appears in the AS syllabus, Further Mathematics students often see these coefficients in the context of series expansions. Be careful not to choose C(5, 2) by symmetry unless the question asks for that specific index.

虽然这是 AS 大纲内容，但进阶数学学生常在级数展开中遇到。注意不要因对称性而误选 C(5, 2)，除非题目明确要求那个指数。

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8. Q14 – Solving a Basic Trigonometric Equation | 题 14 – 解基本三角方程

Question (recalled): Solve sin x = 0.5 for 0 ≤ x ≤ 2π. Select the set of solutions that matches.

题目（回忆版）：在区间 0 ≤ x ≤ 2π 内解 sin x = 0.5，选出与之匹配的解集。

Step 1: Identify the principal solution. sin(π/6) = 0.5, so x = π/6 is one solution. Because sin is positive in the first and second quadrants, the other solution in the range is π – π/6 = 5π/6.

步骤 1：找出主值解。sin(π/6) = 0.5，因此 x = π/6 是一个解。由于正弦在第一和第二象限为正，区间内的另一个解为 π – π/6 = 5π/6。

Step 2: The correct option (C) listed {π/6, 5π/6}. No other solutions exist in [0, 2π] because the period is 2π and these are the only values where sin x = 0.5.

步骤 2：正确选项 C 列出的解集为 {π/6, 5π/6}。因为周期为 2π，在给定区间内仅有这两个点满足 sin x = 0.5。

In the NSAA, trigonometric equations are typically kept simple to test foundational understanding. Drawing a quick unit circle sketch is the fastest way to avoid missing a second‑quadrant solution.

NSAA 中的三角方程通常比较基础，主要考查基本概念。画一个单位圆草图可以最快避免遗漏第二象限的解。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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