NSAA 2023 S1 Answer Key | NSAA 2023 第一部分答案详解

📚 NSAA 2023 S1 Answer Key | NSAA 2023 第一部分答案详解

The NSAA (Natural Sciences Admissions Assessment) Section 1 Mathematics component challenges students with 20 multiple-choice questions covering advanced algebra, geometry, calculus, and more. This article provides the full answer key and step-by-step explanations for every question in the 2023 S1 paper. Mastering these solutions will sharpen your problem-solving skills and boost your confidence for the real exam.

NSAA（自然科学入学评估）第一部分数学部分包含 20 道选择题，涵盖代数、几何、微积分等进阶数学考点。本文将给出 2023 年 S1 试卷的完整答案，并逐题提供详细解析。掌握这些解题思路能够提升你的应试能力，为正式考试做好准备。

1. Answer Key Overview | 答案概览

The table below lists the correct answers for all 20 mathematics questions in NSAA 2023 Section 1. Use it to check your work before diving into the detailed explanations.

下表列出了 NSAA 2023 第一部分所有 20 道数学题的正确选项。在阅读详细解析之前，可以先核对答案。

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

Each question is discussed in detail in the sections below, grouped by topic. The walkthroughs highlight common pitfalls and efficient strategies.

下文按专题分组，逐题进行详细讨论。解析中将突出常见失分点与高效解题策略。

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2. Algebra and Functions | 代数与函数 (Questions 1-3)

Question 1: Solve 2ˣ = 8² for x. The right-hand side gives 8² = 64 = 2⁶, so equating exponents yields x = 6. The correct choice is E. Many candidates miscount the power of 2, but writing 64 explicitly as 2⁶ avoids errors.

题目 1：求解 2ˣ = 8²。右边 8² = 64 = 2⁶，因此指数相等得 x = 6。正确答案为 E。很多考生会数错 2 的幂次，直接将 64 写成 2⁶ 能有效避免失误。

Question 2: Find the domain of f(x) = (x-3)/(x+2). The denominator cannot be zero, so x ≠ −2. There are no other restrictions (e.g., no square roots or logarithms). Thus the domain is all real numbers except −2. Answer: A.

题目 2：求 f(x) = (x-3)/(x+2) 的定义域。分母不能为零，故 x ≠ −2。该分式无其他限制（如根号或对数），因此定义域为除 −2 外的全体实数。答案为 A。

Question 3: Simplify √48 + √27 − √12. Break each surd into its prime factors: √48 = 4√3, √27 = 3√3, √12 = 2√3. Summing gives 4√3 + 3√3 − 2√3 = 5√3. The answer is B.

题目 3：化简 √48 + √27 − √12。将每个根式分解：√48 = 4√3, √27 = 3√3, √12 = 2√3。合并得 4√3 + 3√3 − 2√3 = 5√3。答案为 B。

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3. Coordinate Geometry and Trigonometry | 坐标几何与三角学 (Questions 4-6)

Question 4: A line passes through (2,5) and (4,9). Its gradient is (9−5)/(4−2) = 4/2 = 2. Answer C.

题目 4：一条直线过点 (2,5) 和 (4,9)，斜率为 (9−5)/(4−2) = 4/2 = 2。答案为 C。

Question 5: In triangle ABC, angle A = 30°, side a = 10, side b = 20. Use the sine rule: sin B / 20 = sin 30° / 10. Since sin 30° = 0.5, we get sin B / 20 = 0.05, so sin B = 1. Hence B = 90°. There is no ambiguous case because sin B = 1 gives a unique angle. Answer: D.

题目 5：三角形 ABC 中，角 A = 30°，边 a = 10，边 b = 20。运用正弦定理：sin B / 20 = sin 30° / 10。sin 30° = 0.5，故 sin B / 20 = 0.05，得 sin B = 1。因此 B = 90°。因 sin B = 1 解得唯一角度，不存在多解情况。答案为 D。

Question 6: Evaluate cos 150°. Using the unit circle, cos 150° = −cos 30° = −√3/2. Answer A.

题目 6：求 cos 150° 的值。借助单位圆，cos 150° = −cos 30° = −√3/2。答案为 A。

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4. Sequences and Series | 数列与级数 (Questions 7-8)

Question 7: An arithmetic sequence has nth term aₙ = 3n − 2. The sum of the first 10 terms is S₁₀ = 10/2 × (a₁ + a₁₀). Compute a₁ = 1, a₁₀ = 28, so S₁₀ = 5 × 29 = 145. Answer E.

题目 7：等差数列通项 aₙ = 3n − 2，求前 10 项和。S₁₀ = 10/2 × (a₁ + a₁₀)。a₁ = 1，a₁₀ = 28，故 S₁₀ = 5 × 29 = 145。答案为 E。

Question 8: A geometric progression begins 2, 6, 18, … with first term 2 and common ratio 3. The 6th term is ar⁵ = 2 × 3⁵ = 2 × 243 = 486. Answer B.

题目 8：等比数列首项 2、公比 3，前三项为 2, 6, 18。第六项为 ar⁵ = 2 × 3⁵ = 2 × 243 = 486。答案为 B。

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5. Calculus: Differentiation | 微积分：微分 (Questions 9-11)

Question 9: Differentiate y = e²ˣ cos x. Apply the product rule: dy/dx = 2e²ˣ cos x + e²ˣ (−sin x) = e²ˣ (2cos x − sin x). Answer C.

题目 9：求 y = e²ˣ cos x 的导数。使用乘积法则：dy/dx = 2e²ˣ cos x + e²ˣ (−sin x) = e²ˣ (2cos x − sin x)。答案为 C。

Question 10: Find the stationary points of f(x) = x³ − 3x. Set f'(x) = 3x² − 3 = 0 → x = ±1. The second derivative f”(x) = 6x gives f”(1) = 6 > 0 (minimum), f”(−1) = −6 < 0 (maximum). The maximum point is (−1, 2). Answer D refers to this maximum.

题目 10：求 f(x) = x³ − 3x 的驻点。令 f'(x) = 3x² − 3 = 0 → x = ±1。二阶导数 f”(x) = 6x：f”(1) = 6 > 0（极小值），f”(−1) = −6 < 0（极大值）。极大值点为 (−1, 2)，对应选项 D。

Question 11: Find the equation of the tangent to y = 1/x at x = 2. The derivative is y’ = −1/x², so at x = 2 the gradient is −1/4. The point is (2, 1/2). Using y − y₁ = m(x − x₁): y − 1/2 = −1/4 (x − 2). Simplify to y = −1/4 x + 1. Answer A.

题目 11：求曲线 y = 1/x 在 x = 2 处的切线方程。导数 y’ = −1/x²，故 x = 2 时斜率为 −1/4，点为 (2, 1/2)。点斜式：y − 1/2 = −1/4 (x − 2)，整理得 y = −1/4 x + 1。答案为 A。

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6. Calculus: Integration | 微积分：积分 (Questions 12-14)

Question 12: Evaluate the definite integral ∫₁³ (2x + 1) dx. The antiderivative is x² + x. Substituting limits gives (3²+3) − (1²+1) = (9+3) − (1+1) = 12 − 2 = 10. Answer E.

题目 12：计算定积分 ∫₁³ (2x + 1) dx。原函数为 x² + x，代入上下限得 (3²+3) − (1²+1) = 12 − 2 = 10。答案为 E。

Question 13: Find the indefinite integral ∫ (3x² + 2) dx. Integrating term by term: 3 * (x³/3) + 2x + C = x³ + 2x + C. Answer B.

题目 13：求不定积分 ∫ (3x² + 2) dx。逐项积分：3·(x³/3) + 2x + C = x³ + 2x + C。答案为 B。

Question 14: Find the area enclosed by y = sin(2x) and the x-axis from x = 0 to x = π/2. The integral is ∫₀^{π/2} sin(2x) dx = [−½ cos(2x)]₀^{π/2} = −½ cos π − (−½ cos 0) = −½(−1) + ½(1) = 1. Answer C.

题目 14：求 y = sin 2x 与 x 轴在 x = 0 到 x = π/2 之间围成的面积。积分 ∫₀^{π/2} sin 2x dx = [−½ cos 2x]₀^{π/2} = −½ cos π − (−½ cos 0) = ½ + ½ = 1。答案为 C。

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7. Probability and Statistics | 概率与统计 (Questions 15-16)

Question 15: A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Find the probability they are the same colour. Total ways to choose 2 from 5 is C(5,2) = 10. Favorable: both red C(3,2)=3, both blue C(2,2)=1, total 4. Probability = 4/10 = 2/5. Answer A.

题目 15：袋中有 3 红 2 蓝共 5 个球，不放回抽取两球。求两球同色的概率。总组合数 C(5,2) = 10；同色情况：两红 C(3,2)=3，两蓝 C(2,2)=1，共 4 种。概率 = 4/10 = 2/5。答案为 A。

Question 16: Find the variance of the data set {2, 4, 6, 8}. The mean is (2+4+6+8)/4 = 5. Variance = [(2−5)² + (4−5)² + (6−5)² + (8−5)²] / 4 = (9 + 1 + 1 + 9)/4 = 20/4 = 5. Answer D.

题目 16：求数据集 {2, 4, 6, 8} 的方差。均值为 5，方差 = [(2−5)²+(4−5)²+(6−5)²+(8−5)²] / 4 = 20/4 = 5。答案为 D。

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8. Vectors and Matrices | 向量与矩阵 (Questions 17-18)

Question 17: Given vectors a = (1, 2, 3) and b = (4, 5, 6), compute the dot product a · b = 1×4 + 2×5 + 3×6 = 4 + 10 + 18 = 32. Answer E.

题目 17：已知向量 a = (1, 2, 3)，b = (4, 5, 6)，求点积 a·b = 1×4 + 2×5 + 3×6 = 32。答案为 E。

Question 18: Find the inverse of the matrix A = [[2, 1], [3, 2]]. The determinant is 2×2 − 1×3 = 1. The inverse is (1/det) × [[2, −1], [−3, 2]] = [[2, −1], [−3, 2]]. Answer B.

题目 18：求矩阵 A = [[2, 1], [3, 2]] 的逆矩阵。行列式 det = 4 − 3 = 1，逆矩阵为 [[2, −1], [−3, 2]]。答案为 B。

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9. Complex Numbers | 复数 (Questions 19-20)

Question 19: Find the modulus of z = 3 + 4i. |z| = √(3² + 4²) = √25 = 5. Answer C.

题目 19：求复数 z = 3 + 4i 的模。|z| = √(3² + 4²) = 5。答案为 C。

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