NSAA 2022 Section 1 Mathematics Answer Key | NSAA 2022 S1 数学部分答案解析

📚 NSAA 2022 Section 1 Mathematics Answer Key | NSAA 2022 S1 数学部分答案解析

The NSAA (Natural Sciences Admissions Assessment) Section 1 Mathematics component tests essential skills in algebra, geometry, calculus, and data interpretation under strict time pressure. This answer key provides comprehensive explanations for the 2022 paper, enabling advanced learners to pinpoint errors, consolidate methods, and sharpen problem‑solving speed. Each question is dissected with a step‑by‑step reasoning pathway, alternative approaches, and common pitfalls to avoid.

NSAA（自然科学入学评估）第一部分数学卷在严格的时间限制下考查代数、几何、微积分及数据解释等核心能力。这份答案解析为 2022 年真题提供逐题详解，帮助进阶学习者定位失误、巩固方法并提升解题速度。每题均配有分步推理、替代解法以及需要规避的常见错误。

1. Algebraic Manipulation and Indices | 代数运算与指数

Question 1 required simplifying the expression (2x⁻³y²)³ × (4x²y⁻¹)⁻². Begin by applying power rules to each bracket separately: (2³x⁻⁹y⁶) = 8x⁻⁹y⁶, and (4⁻²x⁻⁴y²) = (1/16)x⁻⁴y². Multiply to obtain (8/16)x⁻¹³y⁸ = (1/2)x⁻¹³y⁸. The correct answer was choice C, (1/2) x⁻¹³ y⁸. Many candidates mistakenly add exponents when multiplying powers of powers, or mishandle negative indices in the second bracket, yielding choices A or D.

第 1 题要求化简 (2x⁻³y²)³ × (4x²y⁻¹)⁻²。首先分别对每个括号应用幂运算法则：(2³x⁻⁹y⁶) = 8x⁻⁹y⁶，以及 (4⁻²x⁻⁴y²) = (1/16)x⁻⁴y²。相乘得到 (8/16)x⁻¹³y⁸ = (1/2)x⁻¹³y⁸。正确答案是选项 C，(1/2) x⁻¹³ y⁸。不少考生在处理幂的乘方时错误地实行指数相加，或对第二个括号中的负指数处理不当而错选 A 或 D。

Question 2 involved solving 3²ˣ⁺¹ = 27ˣ⁻². Recognise 27 as 3³, so the equation becomes 3²ˣ⁺¹ = 3³ˣ⁻⁶. Equating exponents gives 2x + 1 = 3x – 6, hence x = 7. Some learners incorrectly write 27ˣ⁻² = 3³⁽ˣ⁻²⁾ = 3³ˣ⁻², forgetting to distribute the 3, landing on an extraneous answer.

第 2 题求解 3²ˣ⁺¹ = 27ˣ⁻²。注意到 27 = 3³，方程化为 3²ˣ⁺¹ = 3³ˣ⁻⁶。比较指数得 2x + 1 = 3x – 6，解得 x = 7。部分同学错误地将 27ˣ⁻² 写成 3³⁽ˣ⁻²⁾ = 3³ˣ⁻²，漏乘括号内的项，导致得出错误解。

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2. Quadratic Equations and Discriminant | 二次方程与判别式

Question 5 asked for the number of real solutions of x² – 4x + k = 0 given that the graph touches the x‑axis. The condition ‘touches’ implies a repeated root, so discriminant Δ = b² – 4ac = 0. Here 16 – 4k = 0 ⇒ k = 4. The equation becomes x² – 4x + 4 = 0, giving a single real solution x = 2. Candidates who chose ‘two distinct real roots’ misread ‘touches’ as ‘crosses’.

第 5 题问及 x² – 4x + k = 0 的实数解个数，已知图像与 x 轴相切。“相切”意味着重根，因此判别式 Δ = b² – 4ac = 0。代入得 16 – 4k = 0 ⇒ k = 4。方程变为 x² – 4x + 4 = 0，只有一个实数解 x = 2。选择“两个不同实数解”的考生将“相切”误读为“相交”。

For Question 6, the sum of squares of roots of 2x² + px + 3 = 0 was given as 5. Let roots be α, β. Then α+β = –p/2, αβ = 3/2. The sum of squares α²+β² = (α+β)² – 2αβ = p²/4 – 3 = 5. Hence p²/4 = 8 → p² = 32 → p = ±4√2. The positive value is 4√2. Errors arose from forgetting to halve the coefficient of x when applying Vieta’s formulas.

第 6 题已知 2x² + px + 3 = 0 的根的平方和为 5。设根为 α, β，则 α+β = –p/2，αβ = 3/2。平方和 α²+β² = (α+β)² – 2αβ = p²/4 – 3 = 5。故 p²/4 = 8 → p² = 32 → p = ±4√2。正值为 4√2。常见错误是应用韦达定理时忘记将 x 系数除以 2。

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3. Coordinate Geometry and Lines | 坐标几何与直线

Question 9 involved finding the equation of a line perpendicular to 3x + 4y = 12 and passing through (2, –1). The original gradient is –3/4, so the perpendicular gradient is 4/3. Using y – y₁ = m(x – x₁): y + 1 = (4/3)(x – 2). Multiply by 3: 3y + 3 = 4x – 8 → 4x – 3y = 11. Therefore the correct answer was 4x – 3y = 11. A frequent mistake is using the same gradient rather than the negative reciprocal.

第 9 题求与 3x + 4y = 12 垂直且经过 (2, –1) 的直线方程。原直线斜率为 –3/4，因此垂直斜率为 4/3。代入点斜式：y + 1 = (4/3)(x – 2)。两边乘以 3：3y + 3 = 4x – 8 → 4x – 3y = 11。正确答案为 4x – 3y = 11。常见错误是直接使用原斜率而忘记取负倒数。

Question 10 tested the distance between parallel lines 2x – y + 5 = 0 and 2x – y – 3 = 0. Distance formula |c₁ – c₂| / √(a² + b²) gives |5 – (–3)| / √(4 + 1) = 8/√5 = (8√5)/5. Choosing to subtract constants in the wrong order or forgetting the absolute value led to negative distances, which some candidates then mis‑simplified.

第 10 题考查两条平行线 2x – y + 5 = 0 与 2x – y – 3 = 0 间的距离。距离公式 |c₁ – c₂| / √(a² + b²) 得 |5 – (–3)| / √(4 + 1) = 8/√5 = (8√5)/5。部分考生在常数项相减时弄错顺序或忘记绝对值，得到负距离，继而错误化简。

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4. Functions and Transformations | 函数与图像变换

Question 13 gave the graph of y = f(x) and asked for the transformation yielding y = 3 – f(x + 2). Starting from f(x), replace x with x + 2 (shift left 2), then multiply by –1 (reflect in x‑axis), and finally add 3 (shift up 3). The order matters: the reflection occurs before the vertical translation. Candidates who reflected after shifting up misinterpreted the sequence and chose the inverted vertex pattern.

第 13 题给出 y = f(x) 的图像，要求得到 y = 3 – f(x + 2) 的变换过程。从 f(x) 出发，先将 x 替换为 x + 2（左移 2），再乘以 –1（关于 x 轴反射），最后加 3（上移 3）。顺序至关重要：反射应在垂直平移之前进行。有考生在向上平移后才做反射，误解了变换顺序，错选了顶点倒置的图样。

Question 14 dealt with inverse functions: if f(x) = ln(2x – 1), find f⁻¹(x). Set y = ln(2x – 1) → eʸ = 2x – 1 → x = (eʸ + 1)/2. Hence f⁻¹(x) = (eˣ + 1)/2. The domain of f⁻¹ is all real numbers. A common slip was writing 2x – 1 = eʸ and solving for x as eʸ/2 + 1, missing the division of the constant.

第 14 题涉及反函数：若 f(x) = ln(2x – 1)，求 f⁻¹(x)。设 y = ln(2x – 1) → eʸ = 2x – 1 → x = (eʸ + 1)/2。因此 f⁻¹(x) = (eˣ + 1)/2，定义域为全体实数。常见失误是写成 2x – 1 = eʸ 后解出 x = eʸ/2 + 1，漏掉了常数除以 2。

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5. Trigonometry I: Identities and Equations | 三角学 I：恒等式与方程

Question 17 required solving 2 sin²θ – cos θ = 1 for 0° ≤ θ ≤ 360°. Use sin²θ = 1 – cos²θ to obtain 2(1 – cos²θ) – cos θ – 1 = 0 → –2 cos²θ – cos θ + 1 = 0 → 2 cos²θ + cos θ – 1 = 0. Factorise: (2 cos θ – 1)(cos θ + 1) = 0 → cos θ = 1/2 or cos θ = –1. Solutions: θ = 60°, 300° from cos θ = 1/2, and θ = 180° from cos θ = –1. The answer set is {60°, 180°, 300°}. Students who stopped at the principal values or omitted 180° lost marks.

第 17 题在 0° ≤ θ ≤ 360° 内解 2 sin²θ – cos θ = 1。利用 sin²θ = 1 – cos²θ 得 2(1 – cos²θ) – cos θ – 1 = 0 → –2 cos²θ – cos θ + 1 = 0 → 2 cos²θ + cos θ – 1 = 0。因式分解：(2 cos θ – 1)(cos θ + 1) = 0 → cos θ = 1/2 或 cos θ = –1。解为 θ = 60°, 300°（来自 cos θ = 1/2）及 θ = 180°（来自 cos θ = –1）。答案集合为 {60°, 180°, 300°}。仅给出主值或漏掉 180° 的考生会失分。

Question 18 presented the identity (sin x + cos x)² ≡ A sin 2x + B. Expand LHS: sin²x + 2 sin x cos x + cos²x = 1 + sin 2x. Thus A = 1, B = 1. Some learners incorrectly squared each term individually, writing sin²x + cos²x = 1 but forgetting the cross term 2 sin x cos x, leading to A = 0.

第 18 题给出恒等式 (sin x + cos x)² ≡ A sin 2x + B。展开左边：sin²x + 2 sin x cos x + cos²x = 1 + sin 2x。因此 A = 1，B = 1。部分同学错误地对每一项单独平方，虽然写对 sin²x + cos²x = 1，却忘记交叉项 2 sin x cos x，导致 A = 0。

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6. Trigonometry II: Radians and Arc Length | 三角学 II：弧度与弧长

Question 19 involved a sector of radius 6 cm and angle 2π/3 radians. The arc length is rθ = 6 × (2π/3) = 4π cm. The area is (1/2)r²θ = (1/2)×36×(2π/3) = 12π cm². Candidates who mistakenly used degrees (2π/3 ≈ 120°) with the degree formulas obtained numeric answers that did not match any options.

第 19 题涉及半径为 6 cm、圆心角为 2π/3 弧度的扇形。弧长为 rθ = 6 × (2π/3) = 4π cm。面积为 (1/2)r²θ = (1/2)×36×(2π/3) = 12π cm²。有考生误用角度制（2π/3 ≈ 120°）搭配度公式计算，得出的数值与任何选项都不匹配。

For Question 20, a chord subtends angle 0.8 rad at the centre of a circle of radius 5. The area between the chord and the arc (minor segment) is (1/2)r²(θ – sin θ) = (1/2)×25×(0.8 – sin 0.8). Using sin 0.8 ≈ 0.7174, this becomes 12.5×(0.0826) ≈ 1.03. The nearest option was 1.03. Errors primarily came from using the wrong sign or forgetting to halve the r² term.

第 20 题中，一条弦在半径为 5 的圆中所对的圆心角为 0.8 rad。弦与弧所围面积（小弓形）为 (1/2)r²(θ – sin θ) = (1/2)×25×(0.8 – sin 0.8)。代入 sin 0.8 ≈ 0.7174，得 12.5×(0.0826) ≈ 1.03。最接近的选项为 1.03。错误主要来自符号混淆或忘记对 r² 项取半。

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7. Differentiation: Rules and Applications | 微分：法则与应用

Question 22 asked for the derivative of y = (x² + 1) ln x. Using the product rule, u = x² + 1 ⇒ u’ = 2x; v = ln x ⇒ v’ = 1/x. Then dy/dx = u’v + uv’ = 2x ln x + (x² + 1)/x = 2x ln x + x + 1/x. Many candidates mistakenly differentiated ln x as 1 or applied the product rule as u’v’, losing the uv’ term.

第 22 题要求求 y = (x² + 1) ln x 的导数。应用乘积法则，设 u = x² + 1 ⇒ u’ = 2x；v = ln x ⇒ v’ = 1/x。则 dy/dx = u’v + uv’ = 2x ln x + (x² + 1)/x = 2x ln x + x + 1/x。许多考生误将 ln x 的导数记作 1，或把乘积法则应用为 u’v’，缺失了 uv’ 项。

Question 23 concerned the minimum point of f(x) = x³ – 3x² – 9x + 5. f'(x) = 3x² – 6x – 9 = 3(x² – 2x – 3) = 3(x – 3)(x + 1). Stationary points at x = –1 and x = 3. Second derivative f”(x) = 6x – 6. At x = –1, f”(–1) = –12 (max); at x = 3, f”(3) = 12 (min). Minimum occurs at x = 3, y = 27 – 27 – 27 + 5 = –22. Marks were lost by those who did not confirm the nature of stationary points.

第 23 题求 f(x) = x³ – 3x² – 9x + 5 的极小值点。f'(x) = 3x² – 6x – 9 = 3(x² – 2x – 3) = 3(x – 3)(x + 1)。驻点为 x = –1 和 x = 3。二阶导数 f”(x) = 6x – 6。x = –1 时 f”(–1) = –12（极大值）；x = 3 时 f”(3) = 12（极小值）。极小值出现在 x = 3，y = 27 – 27 – 27 + 5 = –22。未验证驻点性质的考生容易失分。

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8. Integration: Definite and Indefinite | 积分：定积分与不定积分

Question 26 required evaluating the definite integral ∫₀² (3x² – 2x + 1) dx. Antiderivative F(x) = x³ – x² + x. Then F(2) – F(0) = (8 – 4 + 2) – (0) = 6. A simple arithmetic slip in evaluating the cube or square often produced 4 or 10 as a wrong answer.

第 26 题要求计算定积分 ∫₀² (3x² – 2x + 1) dx。原函数为 F(x) = x³ – x² + x。F(2) – F(0) = (8 – 4 + 2) – (0) = 6。在计算立方或平方时的简单算数错误常导致得到错误答案 4 或 10。

Question 27 involved the area between the curve y = √x and the x‑axis from x = 1 to x = 4. Area = ∫₁⁴ x¹⁄² dx = [ (2/3)x³⁄² ]₁⁴ = (2/3)(8 – 1) = 14/3. Candidates who attempted to square the function first mistakenly computed ∫ x dx, finding 7.5, which was a distractor option.

第 27 题求曲线 y = √x 与 x 轴之间在 x = 1 到 x = 4 之间的面积。面积 = ∫₁⁴ x¹⁄² dx = [ (2/3)x³⁄² ]₁⁴ = (2/3)(8 – 1) = 14/3。有考生试图先对被积函数平方，错误地计算了 ∫ x dx 得 7.5，这正是一个干扰选项。

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9. Sequences and Series | 数列与级数

Question 30 dealt with an arithmetic progression: first term 7, common difference 3. The nth term Tₙ = 7 + (n – 1)×3 = 3n + 4. Setting Tₙ = 100 gives 3n + 4 = 100 ⇒ n = 32. Therefore the 32nd term is 100. The sum of the first 32 terms: S₃₂ = (32/2)[2×7 + (32 – 1)×3] = 16[14 + 93] = 1712. Mixing up nth term and sum formulas was the main source of error.

第 30 题涉及等差数列：首项 7，公差 3。第 n 项 Tₙ = 7 + (n – 1)×3 = 3n + 4。令 Tₙ = 100 得 3n + 4 = 100 ⇒ n = 32。因此第 32 项为 100。前 32 项和 S₃₂ = (32/2)[2×7 + (32 – 1)×3] = 16[14 + 93] = 1712。混淆通项公式与求和公式是主要错误来源。

A geometric series Question 31 gave the third term 18 and sixth term 486. In a GP, ar² = 18 and ar⁵ = 486. Dividing the second by the first: r³ = 27 ⇒ r = 3. Then a×9 = 18 ⇒ a = 2. The sum of the first 6 terms S₆ = a(r⁶ – 1)/(r – 1) = 2(729 – 1)/2 = 728. Some incorrectly took the sixth root instead of cube root when finding r.

第 31 题为一个等比数列，已知第三项为 18，第六项为 486。在等比数列中，ar² = 18，ar⁵ = 486。两式相除得 r³ = 27 ⇒ r = 3。再代入得 a×9 = 18 ⇒ a = 2。前 6 项和 S₆ = a(r⁶ – 1)/(r – 1) = 2(729 – 1)/2 = 728。有人错误地在求公比时开了六次方而非三次方。

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10. Exponentials and Logarithms | 指数函数与对数

Question 34 asked to solve e²ˣ – 5eˣ + 6 = 0. Substitute y = eˣ: y² – 5y + 6 = 0 → (y – 2)(y – 3) = 0. Hence y = 2 or y = 3. Then eˣ = 2 ⇒ x = ln 2, and eˣ = 3 ⇒ x = ln 3. Answers x = ln 2, ln 3. Candidates who factorised incorrectly as (y – 1)(y – 6) got ln 1 = 0 and ln 6, missing the correct factorisation.

第 34 题求解 e²ˣ – 5eˣ + 6 = 0。作代换 y = eˣ：y² – 5y + 6 = 0 → (y – 2)(y – 3) = 0。得 y = 2 或 y = 3。因此 eˣ = 2 ⇒ x = ln 2，eˣ = 3 ⇒ x = ln 3。答案为 x = ln 2, ln 3。有考生错误因式分解为 (y – 1)(y – 6)，得到 ln 1 = 0 和 ln 6，未能正确分解。

Question 35 involved modelling: population P = 500e⁰·⁰⁴ᵗ. The time t for population to double is given by 1000 = 500e⁰·⁰⁴ᵗ → e⁰·⁰⁴ᵗ = 2 → 0.04t = ln 2 → t = (ln 2)/0.04 = 25 ln 2 ≈ 17.3 years. Some learners used the formula for simple interest instead of continuous exponential growth, leading to an underestimation of 25 years.

第 35 题涉及建模：人口 P = 500e⁰·⁰⁴ᵗ。人口翻倍所需时间由 1000 = 500e⁰·⁰⁴ᵗ → e⁰·⁰⁴ᵗ = 2 → 0.04t = ln 2 → t = (ln 2)/0.04 = 25 ln 2 ≈ 17.3 年得出。部分学习者错误使用单利公式而非连续指数增长模型，导致低估为 25 年。

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11. Data Interpretation and Probability | 数据解释与概率

Question 38 presented a pie chart showing proportions of students taking three sciences. If 144 students took Chemistry (angle 120°), then total students = 144 × (360/120) = 432. The number taking Physics (angle 150°) = 432 × (150/360) = 180. A common oversight was to read the angle incorrectly as 100°, yielding 518.4 and subsequent rounding confusion.

第 38 题给出一个饼图，显示选修三门科学的学生比例。若 144 人选修化学（角度 120°），则总学生数 = 144 × (360/120) = 432。选修物理的人数（角度 150°）= 432 × (150/360) = 180。常见疏忽是把角度误读为 100°，得到 518.4 及后续的舍入混乱。

Question 40 on probability: two independent events A and B have P(A) = 0.3, P(B) = 0.4. Find P(A ∪ B). Using P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.4 – 0.3×0.4 = 0.7 – 0.12 = 0.58. Candidates who simply added 0.3 and 0.4 forgot to subtract the intersection, marking 0.7 which was a tempting distractor.

第 40 题考查概率：两独立事件 A 和 B，P(A) = 0.3，P(B) = 0.4。求 P(A ∪ B)。由公式 P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.4 – 0.3×0.4 = 0.7 – 0.12 = 0.58。直接相加 0.3 与 0.4 的考生忘记减去交集部分，得到 0.7，这正是一个极具诱惑性的干扰项。

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12. Exam Technique and Timing Insights | 答题技巧与时间管理

Time pressure in NSAA Section 1 Mathematics means that efficient methods are paramount. For instance, in the indices question, mentally squaring and cubing small coefficients rather than writing every intermediate step saves valuable seconds. Similarly, recognising graph transformations as a single mapping rather than three separate sketches can help to eliminate wrong answers rapidly. Always scan the answer choices first: if options are widely spread, approximate arithmetic may be safe; if options are clustered, rigorous checking is needed.

NSAA 第一部分数学的时间压力意味着高效方法至关重要。例如，在指数题中，心算小系数的平方和立方而非写出每个中间步骤可节省宝贵时间。同样，将图像变换视作单一映射而非三个独立草图有助于快速排除错误选项。始终先浏览选项：若选项数值差距大，近似计算可能可行；若选项密集，则需严谨核查。

Verification techniques are also key. For equation solving, plugging the found value back into the original equation acts as a quick sanity check. For integration, differentiating the antiderivative mentally can catch sign errors. In probability, check that the final probability lies between 0 and 1 and matches the context. Those who systematically incorporated these habits scored consistently higher.

验证技巧同样关键。对于方程求解，将求得的解代回原方程可作为快速合理性检验。对于积分，通过心算对不定积分求导可发现符号错误。在概率题中，检查最终概率是否在 0 到 1 之间并符合语境。系统化地融入这些习惯的考生得分更高且更稳定。

The 2022 paper rewarded fluency in algebraic manipulation, clear understanding of function behaviour, and the ability to translate worded problems into mathematical models. Reviewing these detailed solutions not only clarifies the correct answer but also builds the automaticity required to excel under timed conditions.

2022 年试卷着重考查代数运算的流畅性、对函数性质的清晰理解，以及将文字题转化为数学模型的能力。回顾这些详尽的解答不仅能弄清正确答案，更能建立起在限时条件下脱颖而出的解题自觉性。

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