📚 A-Level Maths MA03 Unit P2: Example Responses and Question Analysis | A-Level数学 MA03 单元 P2 题型解析与范例解答
In the A-Level Mathematics specification, the MA03 Unit P2 (Pure Mathematics 2) paper assesses core algebraic, trigonometric, exponential and calculus skills through structured and unstructured questions. This article breaks down typical question types, provides step-by-step model solutions and highlights key marking points to help you maximise your score.
在A-Level数学考试中，MA03单元P2（纯数学2）试卷通过结构化和非结构化题目考查代数、三角、指数、对数及微积分等核心技能。本文解析常见题型，逐步展示范例解答，并强调评分要点，助你斩获高分。
1. Polynomial Division and Factor Theorem | 多项式除法与因式定理
) = 2x³ − 3x² − 3x + 2, show that (x − 2) is a factor and hence factorise f(x) completely.
示例：已知 f(x) = 2x³ − 3x² − 3x + 2，证明 (x − 2) 是其因式，并将 f(x) 完全分解。
so (x − 2) is a factor. Then divide f(x) by (x − 2) using algebraic long division or synthetic division. The quotient is 2x² + x − 1. Factorising the quadratic gives (2x − 1)(x + 1). Thus the full factorisation is (x − 2)(2x − 1)(x + 1).
应用因式定理：计算 f(2) = 2(8) − 3(4) − 3(2) + 2 = 16 − 12 − 6 + 2 = 0，因此 (x − 2) 是因式。然后用长除法或综合除法将 f(x) 除以 (x − 2)，商为 2x² + x − 1。对该二次式分解得 (2x − 1)(x + 1)。因此完整分解为 (x − 2)(2x − 1)(x + 1)。
Key marking points: ng division, and fully factorising the quadratic. Missing the final factorisation loses accuracy marks.
评分关键点：展示 f(2)=0，正确完成除法运算，并完全分解二次式。若遗漏最终分解，将损失准确性分数。
2. Binomial Expansion and Range of Validity | 二项展开与有效范围
Example: Expand (1 + 3x)⁻² in ascending powers of x up to the term in x³, stating the values of x for which the expansion is valid.
示例：将 (1 + 3x)⁻² 按 x 的升幂展开至 x³ 项，并说明展开有效的 x 的取值范围。
Recall the binomial series for any rational n: (1 + u)ⁿ = 1 + n u + n(n−1)/2! u² + n(n−1)(n−2)/3! u³ + ... Here u = 3x and n = −2. Substituting: 1 + (−2)(3x) + (−2)(−3)/2! (3x)² + (−2)(−3)(−4)/3! )/6 = −4, times (3x)³ = −4 × 27x³ = −108x³. Hence the expansion is 1 − 6x + 27x² − 108x³ + …
对任意有理数 n 的二项级数公式为 (1 + u)ⁿ = 1 + n u + n(n−1)/2! u² + n(n−1)(n−2)/3! u³ + ... 此处 u = 3x，n = −2。逐项代入并化简：1 − 6x + 27x²，x³ 项的系数为 (−2)(−3)(−4)/6 = −4，乘以 (3x)³ = 27x³，得 −108x³。故展开式为 1 − 6x + 27x² − 108x³ + …
The expansion is valid when |3x| < 1, i.e. |x| < 1/3. Some exam boards may accept −1/3 < x < 1/3. Always include the range statement.
展开式在 |3x| < 1 时有效，即 |x| < 1/3。某些考试局也接受 −1/3 < x < 1/3。务必明确写出取值范围。
3. Exponential and Logarithmic Equations | 指数与对数方程
Example: Solve 2e3x iving your answer in exact logarithmic form and as a decimal to three significant figures.
示例：解方程 2e3x = 5，给出精确对数形式答案及保留三位有效数字的小数近似值。
Divide both sides by 2: e3x = 5/2. Take natural logarithms: ln(e3x ) ln(5/2) is the exact form. Using a calculator, ln(2.5) ≈ 0.91629, so x ≈ 0.305 (3 s.f.).
两边同除以 2 得 e3x = 5/2。取自然对数：ln(e3x) = ln(5/2) → 3x = ln(5/2)。因此精确解为 x = (1/3) ln(5/2)。用计算器求得 ln(2.5) ≈ 0.91629，故 x ≈ 0.305（保留三位有效数字）。
Key marking points: correct manipulation to isolate the exponential, taking logs, and handling the coefficient. The exact form must contain the logarithm; the decimal is often required but is not the primary answer.
评分关键点：正确移项分离指数，取对数，并处理系数。精确解必须含对数表达式；小数值常作为额外要求，但不是主要得分点。
4. Trigonometric Equations | 三角方程
Example: Solve 2 sin θ = 1 for 0 ≤ θ ≤ 2π, giving your answers in terms of π.
示例：在区间 0 ≤ θ ≤ 2π 内解 2 sin θ = 1，答案用 π 表示。
t, the other solution is θ = π − π/6 = 5π/6. Therefore the solutions are θ = π/6 and 5π/6.
首先化得 sin θ = 1/2。基本解为 θ = π/6。由于正弦在第二象限也为正，另一解为 θ = π − π/6 = 5π/6。因此解为 θ = π/6 和 5π/6。
A clear sketch of the sine graph or the CAST diagram helps justify the second solution and avoid losing marks for missing values.
清晰画出正弦图像或使用 CAST 图有助于说明第二个解，避免因漏解而失分。
5. Differentiation Using Chain, Product and Quotient Rules | 链式、乘积与商法则求导
Example: Differentiate y = (x² + 1)³ sin(2x) with respect to x.
示例：对 y = (x² + 1)³ sin(2x) 关于 x 求导。
Let u = (x² + 1)³ and v = sin(2x). Then u' = 3(x²
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图