📚 Further Maths: NSAA 2020 S1 Mathematics Answer Key & Solutions | 进阶数学：NSAA 2020 S1 数学答案与解析
The NSAA (Natural Sciences Admissions Assessment) Section 1 Mathematics paper is a critical exam for Cambridge applicants. The 2020 paper tested core A-level and Further Maths reasoning through multiple-choice questions. This article shares the complete answer key and delivers detailed, bilingual step-by-step solutions, helping you master the techniques and avoid typical traps.
NSAA（自然科学入学评估）第一部分数学试卷是剑桥申请者的关键考试。2020 年的试卷以选择题形式考查了核心 A-level 和进阶数学推理。本文分享完整答案并给出详尽的双语分步解析，帮助你掌握解题技巧并避开常见陷阱。
1. Algebraic Simplification (Q1, Answer D) | 代数化简（第1题，答案 D）
Question 1 asks to simplify (x² – 4)/(x – 2) for x ≠ 2. The correct answer is D: x + 2.
第1题要求化简 (x² – 4)/(x – 2)，x ≠ 2。正确答案是 D：x + 2。
Factor the numerator as a difference of squares: x² – 4 = (x – 2)(x + 2). Since the denominator is x – 2 and x ≠ 2, you can cancel the common factor, leaving x + 2.
将分子因式分解为平方差：x² – 4 = (x – 2)(x + 2)。分母为 x – 2 且 x ≠ 2，因此可以约去公因子，得到 x + 2。
d at x = 2, but the simplified form hides this. The cancellation is only valid when x ≠ 2.
始终牢记定义域限制：原表达式在 x = 2 处无定义，但化简后的形式掩盖了这一点。只有在 x ≠ 2 时约分才有效。
2. Solving Quadratic Equations (Q2, Answer B) | 解二次方程（第2题，答案 B）
Q2 requires solving x² – 5x + 6 = 0. The correct choice is B: x = 2, x = 3.
第2题要求解 x² – 5x + 6 = 0。正确选项是 B：x = 2，x = 3。
Factor the quadratic directly: x² – 5x + 6 = (x – 2)(x – 3). Setting each factor to zero gives the roots 2 and 3.
直接对二次式因式分解：x² – 5x + 6 = (x – 2)(x – 3)。令每个因子为零即得根 2 和 3。
You could also use the quadratic formula: x = [5 ± √25 – 24 nfirming the same two solutions. Be careful with signs when substituting into the formula.
也可使用二次公式：x = [5 ± √25 – 24]/2 = (5 ± 1)/2，同样得到这两个解。代入公式时须注意符号。
3. Inequality with Modulus (Q3, Answer C) | 含绝对值不等式（第3题，答案 C）
Q3 deals with |2x – 3| < 5. The answer is C: –1 < x < 4.
第3题考查 |2x – 3| < 5。答案是 C：–1 < x < 4。
Rewrite the absolute inequality as a compound inequality: –5 < 2x – 3 < 5. Add 3 to all parts to get –2 < 2x < 8, then divide by 2 to obtain –1 < x < 4.
将绝对值不等式改写为复合不等式：–5 < 2x – 3 < 5。所有部分加 3 得 –2 < 2x < 8，再除以 2 得 –1 < x < 4。
A common mistake is to solve |2x – 3| < 5 as 2x – 3 < 5 or 2x – 3 > –5 separately, but you must ensure the solution satisfies both conditions simultaneously.
常见错误是分开解得 2x – 3 < 5 或 2x – 3 > –5，但必须确保解同时满足两个条件。
4. Coordinate Geometry – Distance (Q4, Answer A) | 坐标几何 – 距离（第4题，答案 A）
Q4 asks for the distance between points (1, 2) and (4, 6). The correct answer is A: 5.
第4题求点 (1, 2) 与 (4, 6) 之间的距离。正确答案为 A：5。
Use the distance formula: d = √[(x₂ – x₁)² + (y₂ – y₁)²] = √[(4 – 1)² + (6 – 2)²] = √3² + 4² = √9 + 16 = √25 = 5.
使用距离公式：d = √[(x₂ – x₁)² + (y₂ – y₁)²] = √[(4 – 1)² + (6 – 2)²] = √3² + 4² = √9 + 16 = √25 = 5。
This is a classic 3-4-5 right triangle. Recognizing Pythagorean triples can save time in the NSAA.
这是一个经典的 3-4-5 直角三角形。识别勾股数可以节省 NSAA 答题时间。
5. Functions and Graphs (Q5, Answer C) | 函数与图像（第5题，答案 C）
swer is C (a curve passing through (1,0), increasing, with a vertical asymptote at x = 0).
第5题涉及识别 y = ln x 的图像。答案是 C（曲线过 (1,0) 点，递增，在 x = 0 处有垂直渐近线）。
The natural logarithm function is defined for x > 0, has an x-intercept at (1,0), and increases slowly. Its domain and asymptotic behaviour must be distinguished from exponential graphs.
自然对数函数的定义域为 x > 0，在 (1,0) 处有 x 轴截距，并缓慢增加。必须将其定义域和渐近行为与指数函数图像区分开。
tch out for graphs of y = ln(x + a) where horizontal shifts change the asymptote position.
注意 y = ln(x + a) 这类图像，其水平移动会改变渐近线位置。
6. Trigonometric Equations (Q6, Answer A) | 三角方程（第6题，答案 A）
Q6: Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°. The correct answer is A: θ = 30°, 150°.
第6题：在 0° ≤ θ ≤ 360° 内解 sin θ = 1/2。正确答案为 A：θ = 30°，150°。
econd quadrants. The principal value is 30° (π/6 rad), and the supplementary angle is 180° – 30° = 150°.
正弦在第一和第二象限为正。主值为 30°（π/6 弧度），补角为 180° – 30° = 150°。
Using the CAST diagram or the unit circle helps avoid missing the second solution. Neglecting the quadrant rule is a frequent error.
使用 CAST 图或单位圆有助于避免遗漏第二个解。忽略象限法则是常见错误。
7. Exponential Equations and Logarithms (Q7, Answer B) | 指数方程与对数（第7题，答案 B）
Q7: Solve e²ˣ = 5. The answer is B: x = ½ ln 5.
第7题：解 e²ˣ = 5。答案为 B：x = ½ ln 5。
Take the natural logarithm of both sides: ln(e²ˣ) = ln 5. Since ln(e²ˣ) = 2x, we get 2x = ln 5, so x = (ln 5)/2.
对两边取自然对数：ln(e²ˣ) = ln 5。由于 ln(e²ˣ) = 2x，得 2x = ln 5，因此 x = (ln 5)/2。
Some students mistakenly write x = ln(5/2). The logarithm law ln(e²ˣ) = 2x is fundamental; always apply the inverse function carefully.
有些学生错误地写成 x = ln(5/2)。基本法则 ln(e²ˣ) = 2x 不可混淆；应用反函数需谨慎。
8. Differentiation – Product Rule (Q8, Answer D) | 导数 – 乘积法则（第8题，答案 D）
Q8: Differentiate x³ sin x. The correct answer is D: 3x² sin x + x³ cos x.
第8题：求 x³ sin x 的导数。正确答案是 D：3x² sin x + x³ cos x。
Apply the product rule: if u = x³ and v = sin x, then u' = 3x² and v' = cos x. The derivative is u'v + uv' = 3x² sin x + x³ cos x.
应用乘积法则：设 u = x³，v = sin x，则 u' = 3x²，v' = cos x。导数为 u'v + uv' = 3x² sin x + x³ cos x。
Do not forget to differentiate both factors; a common slip is writing only 3x² sin x or x³ cos x.
不要忘记对两个因子分别求导；常见失误是只写出 3x² sin x 或 x³ cos x。
9. Definite Integration (Q9, Answer C) | 定积分（第9题，答案 C）
Q9: Evaluate ∫₀² (2x + 1) dx. The result is C: 6.
第9题：计算 ∫₀² (2x + 1) dx。结果为 C：6。
Find the antiderivative: ∫(2x + 1) dx = x² + x + C. Then apply the limits: [x² +
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