📚 A-Level Maths Unit 5 Mark Scheme Jan21: Key Topic Breakdown | A-Level数学单元5 2021年1月评分标准核心知识点精讲
This article unpacks the essential techniques assessed in the January 2021 Unit 5 mark scheme. By examining the typical allocation of method (M), accuracy (A) and independent (B) marks, you will learn how to structure solutions and secure every available point. The topics covered span advanced integration, differential equations, numerical methods and connected rates of change – all core to the pure mathematics component of A-Level Maths.
本文深入解析2021年1月单元5评分标准中考察的核心方法。通过分析典型的方法分(M)、答案分(A)和独立分(B)的配置，你将学会如何规范地构建解题步骤并拿到每一分。所涵盖的主题包括进阶积分法、微分方程、数值方法以及关联变化率，这些都是A-Level数学纯数部分的重中之重。
1. Mark Scheme Structure and Key Command Words | 评分方案结构与指令词解析
The Unit 5 mark scheme typically awards M1 for a correctly stated method, A1 for an accurate intermediate or final result, and B1 for independent knowledge such as a formula or definition. Understanding this helps you present your working logically so that method marks are visible even if a numerical slip occurs later.
单元5评分标准通常对正确陈述的方法给予M1，对精确的中间或最终结果给予A1，对公式或定义等独立知识点给予B1。理解这一点有助于你有条理地展示解题过程，即便后面出现计算错误，方法分依然可见。
Common command words include ‘Find’, ‘Express’, ‘Determine’, ‘Show that’ and ‘Hence’. When a question says ‘Show that’, a given answer is provided; you must demonstrate a valid chain of reasoning without skipping steps. Marks are often awarded for setting up an integral, choosing the correct substitution, or manipulating an equation into standard form.
常见的指令词包括 ‘Find’、’Express’、’Determine’、’Show that’ 和 ‘Hence’。当问题要求 ‘Show that’ 时，通常会给出答案；你必须展示完整的推理链，不能跳步。评分点往往分配给建立积分式、选择正确替换或把方程化为标准形式等关键步骤。
2. Integration by Substitution | 换元积分法
Substitution is essential when the integrand contains a function and its derivative (or a linear variation of it). For the integral ∫ x√(2x+1) dx, the mark scheme credits M1 for setting u = 2x+1, then finding dx = du/2 and x = (u-1)/2. The integral transforms to ∫ (u-1)/2 · √u · (du/2) which simplifies to ¼ ∫ (u3/2 – u1/2) du.
当被积函数包含一个函数及其导数（或其线性变形）时，换元法必不可少。对于积分 ∫ x√(2x+1) dx，评分标准会因设 u = 2x+1 而给出 M1，随后求出 dx = du/2 和 x = (u-1)/2。积分转化为 ¼ ∫ (u3/2 – u1/2) du。
For definite integrals, remember to change the limits: if x = 0, u = 1; if x = 4, u = 9. The A1 mark depends on correct evaluation of the simplified power functions and substituting the new limits accurately. A common error is to forget the constant multiplier from dx, leading to a missing factor of ½ or ¼.
对于定积分，切记变换积分限：若 x = 0，则 u = 1；若 x = 4，则 u = 9。A1分取决于正确计算简化后的幂函数并准确代入新限。常见错误是忘记 dx 带来的常数相乘因子，导致遗漏 ½ 或 ¼。
3. Integration by Parts | 分部积分法
The formula ∫ u dv = uv – ∫ v du is tested with products of algebraic and exponential/trigonometric functions. For ∫ x eˣ dx, the mark scheme expects you to choose u = x so that du = dx, and dv = eˣ dx giving v = eˣ. This awards M1. The result x eˣ – ∫ eˣ dx = x eˣ – eˣ + C secures the A1.
公式 ∫ u dv = uv – ∫ v du 常用于代数式与指数/三角函数的乘积。对于 ∫ x eˣ dx，评分标准期望你选取 u = x 使得 du = dx，而 dv = eˣ dx 得到 v = eˣ。这一设置给出 M1。结果 x eˣ – ∫ eˣ dx = x eˣ – eˣ + C 锁定 A1。
When the integrand is a product of a polynomial and a logarithm, such as ∫ ln x dx, set u = ln x and dv = dx. Then v = x and du = (1/x) dx. The integral becomes x ln x – ∫ x·(1/x) dx = x ln x – x + C. Many mark schemes award B1 for correctly identifying the ‘parts’ arrangement and A1 for the final simplified form.
当被积函数为多项式与对数的乘积，如 ∫ ln x dx，设 u = ln x，dv = dx。则 v = x，du = (1/x) dx。积分变为 x ln x – ∫ x·(1/x) dx = x ln x – x + C。许多评分标准会因正确确定“分部”而给 B1，并为最终简化形式给 A1。
Repeated integration by parts may be required for ∫ x² sin x dx. Here the M-mark scheme follows the tabular or repeated application and rewards consistent sign management.
对于 ∫ x² sin x dx，可能需要重复分部积分。此时 M 分会根据表格法或反复运算给予，并奖励符号处理的一致性。
4. Integration Using Partial Fractions | 部分分式积分法
Rational functions where the denominator factorises are integrated by splitting into partial fractions. Consider ∫ (2x+3)/(x² + x – 2) dx. The denominator (x+2)(x-1) leads to the decomposition A/(x+2) + B/(x-1). Solving 2x+3 = A(x-1) + B(x+2) gives A = 1/3, B = 5/3. Method marks are given for setting up the identity correctly.
分母可分解因式的有理函数可通过拆成部分分式进行积分。以 ∫ (2x+3)/(x² + x – 2) dx 为例，分母 (x+2)(x-1) 可分解为 A/(x+2) + B/(x-1)。由 2x+3 = A(x-1) + B(x+2) 解得 A = 1/3, B = 5/3。正确设立恒等式可获得方法分。
The integral becomes (1/3) ∫ 1/(x+2) dx + (5/3) ∫ 1/(x-1) dx = (1/3) ln|x+2| + (5/3) ln|x-1| + C. Marks for accuracy are awarded for the logarithmic integration and for combining constants correctly. If the question asks for a definite integral, absolute value signs are often omitted but must be considered for C marks.
积分变为 (1/3) ∫ 1/(x+2) dx + (5/3) ∫ 1/(x-1) dx = (1/3) ln|x+2| + (5/3) ln|x-1| + C。准确分会给在对数积分和正确合并常数上。如果问题要求计算定积分，绝对值符号常被省略，但评判时会关注。
5. Differential Equations: Separation of Variables | 微分方程：变量分离法
First-order differential equations of the form dy/dx = f(x)g(y) are solved by separating variables. For dy/dx = xy, rewrite as (1/y) dy = x dx. Integrating both sides gives ln|y| = ½ x² + C, hence y = A e^(½ x²). The mark scheme assigns M1 for separating correctly, M1 for integrating both sides, and A1 for solving explicitly for y, including the constant.
形如 dy/dx = f(x)g(y) 的一阶微分方程通过变量分离求解。对于 dy/dx = xy，改写为 (1/y) dy = x dx。两边积分得 ln|y| = ½ x² + C，因而 y = A e^(½ x²)。评分标准给分离正确 M1，两边积分 M1，并给出显式解（含常数）A1。
When an initial condition is given, e.g. y(0) = 2, substitute to find A = 2. This step often earns a B1 mark. A common pitfall is forgetting to include the constant of integration immediately and losing the A mark for the general solution.
当给出初始条件，如 y(0) = 2 时，代入可求得 A = 2。此步骤常获 B1 分。常见的陷阱是未能立即包含积分常数，从而失去通解的 A 分。
6. Numerical Methods: The Trapezium Rule | 数值方法：梯形法则
The trapezium rule approximates ∫ₐᵇ y dx ≈ h/2 [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)], where h = (b-a)/n. In the Jan21 Unit 5 mark scheme, a typical question provides a table of values and asks for an approximation to a definite integral. Method marks hinge on reading ordinates correctly, identifying h, and applying the formula structure.
梯形法则近似计算 ∫ₐᵇ y dx ≈ h/2 [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]，其中 h = (b-a)/n。在2021年1月单元5评分标准中，典型题目会提供数值表格并要求近似求积。方法分取决于正确读取纵坐标、确定 h 并应用公式结构。
For a table with x = 0, 0.5, 1.0, 1.5, 2.0 and corresponding y-values, h = 0.5. The mark scheme awards M1 for h/2 × [first + last + 2×(sum of the rest)]. Any arithmetic slip only costs an A1, while using a wrong h loses the M1.
对于 x = 0, 0.5, 1.0, 1.5, 2.0 及其对应的 y 值表格，h = 0.5。评分标准对 h/2 × [首项+末项+2×(其余总和)] 给予 M1。任何算术错误只会丢掉 A1，而用错 h 则会失去 M1。
Students often ask whether to use more strips for better accuracy. The mark scheme may include a ‘comment on accuracy’ part, where B1 is given for stating that the trapezium rule overestimates or underestimates depending on the curve’s concavity. Know that for convex curves the rule overestimates, for concave it underestimates.
学生常问是否需要用更多等分提高精度。评分标准有时包含“评论精度”的部分，会根据曲线凹凸性说明梯形法高估或低估而给 B1。记住凸曲线则高估，凹曲线则低估。
7. Parametric Differentiation and Integration | 参数方程求导与积分
When a curve is given by x = f(t), y = g(t), the gradient is dy/dx = (dy/dt) / (dx/dt). For x = t², y = 2t, we have dx/dt = 2t, dy/dt = 2, so dy/dx = 2/(2t) = 1/t. The mark scheme gives M1 for differentiating both, and A1 for the simplified fractional form.
当曲线由 x = f(t), y = g(t) 给出时，斜率为 dy/dx = (dy/dt) / (dx/dt)。对于 x = t², y = 2t，有 dx/dt = 2t, dy/dt = 2, 故 dy/dx = 1/t。评分标准对两者求导给 M1，对化简分式给 A1。
To find the area under a parametric curve, use ∫ y dx = ∫ y (dx/dt) dt. Limits of t correspond to the x-limits. In the mark scheme, setting up the integral correctly with the substituted dx/dt earns M1. For example, with x = 2 sin θ, y = cos 2θ, the integral for area becomes ∫ cos 2θ · (2 cos θ) dθ.
求参数曲线下的面积，使用 ∫ y dx = ∫ y (dx/dt) dt，t 的上下限对应 x 的上下限。评分标准中，正确建立带 dx/dt 的积分式可得 M1。例如 x = 2 sin θ, y = cos 2θ，面积积分即为 ∫ cos 2θ · (2 cos θ) dθ。
8. Connected Rates of Change | 关联变化率
Problems involving two or more rates related by a geometric equation require the chain rule. A sphere’s volume V = ⁴⁄₃ π r³. If dr/dt is given, dV/dt = dV/dr × dr/dt = 4π r² (dr/dt). The Jan21 scheme awards M1 for a correct chain rule statement and A1 for correct substitution and simplification.
涉及两个或多个变化率且由几何方程关联的问题需要链式法则。球的体积 V = ⁴⁄₃ π r³。若给出 dr/dt，则 dV/dt = dV/dr × dr/dt = 4π r² (dr/dt)。2021年1月评分方案对正确链式法则陈述给 M1，对正确代入和化简给 A1。
For a cone with height h and radius r where h = 3r, volume V = ⅓ π r² h = π r³. The rate of change of volume when r = 5 cm and dr/dt = 0.2 cm/s is dV/dt = 3π r² (0.2) = 15π cm³/s. Marks are often lost when students fail to differentiate the volume expression correctly with respect to r.
对于一个高 h、半径 r 且满足 h = 3r 的圆锥，体积 V = ⅓ π r² h = π r³。当 r = 5 cm, dr/dt = 0.2 cm/s 时，体积变化率为 dV/dt = 3π r² (0.2) = 15π cm³/s。学生常因未能正确对 r 求导而失分。
9. Implicit Differentiation | 隐函数求导
When y is defined implicitly, we differentiate term-by-term with respect to x, treating y as a function of x. For x² + y² = 25, differentiating gives 2x + 2y (dy/dx) = 0, so dy/dx = -x/y. The mark scheme gives M1 for applying the chain rule to the y² term and A1 for the final expression.
当 y 由隐式定义时，逐项对 x 求导，将 y 视为 x 的函数。对于 x² + y² = 25，求导得 2x + 2y (dy/dx) = 0，故 dy/dx = -x/y。评分标准对 y² 项使用链式法则给 M1，对最终表达式给 A1。
If a product xy appears, use the product rule: d(xy)/dx = y + x(dy/dx). The Jan21 mark scheme often includes implicit differentiation in the context of finding tangents or normals to a curve. After finding dy/dx, substituting the point coordinates secures the A1 for the gradient.
若出现乘积 xy，用乘法法则：d(xy)/dx = y + x(dy/dx)。2021年1月评分方案常在求切线或法线时考察隐函数求导。求出 dy/dx 后，代入点坐标可得斜率 A1。
10. Proof and Mathematical Argument | 证明与数学论证
Proof questions in Unit 5 may require you to verify a reduction formula or show a trigonometric identity. For example, to prove ∫ sinⁿ x dx = -1/n sinⁿ⁻¹ x cos x + (n-1)/n ∫ sinⁿ⁻² x dx, the mark scheme rewards M1 for using integration by parts with u = sinⁿ⁻¹ x and dv = sin x dx. Subsequent manipulation using sin² x = 1 – cos² x gains further M1 and A1 for the algebraic rearrangement.
单元5中的证明题可能要求验证递推公式或证明三角恒等式。例如，要证明 ∫ sinⁿ x dx = -1/n sinⁿ⁻¹ x cos x + (n-1)/n ∫ sinⁿ⁻² x dx，评分标准因设 u = sinⁿ⁻¹ x, dv = sin x dx 进行分部积分而给 M1。随后利用 sin² x = 1 – cos² x 进行代数变形再获 M1 和 A1。
Examiners look for clear logical flow: stating the method, showing the differentiation steps, and simplifying to the required form. B1 marks are often independent and can be gained by quoting a relevant identity, even if the integration falters.
考官看重清晰的逻辑流程：陈述方法、展示求导步骤并化为所需形式。B1 分往往是独立的，即便积分受阻，引用相关恒等式也能获得。
11. Tips for Maximising Marks from the Mark Scheme | 从评分标准中榨取最高分的技巧
Always write down the formula you intend to use before substituting numbers. For example, write ‘Trapezium rule: Area ≈ h/2 […]’ – this alone can secure a B1 mark. Similarly, label your substitution ‘Let u = …’ and show du/dx even if it’s trivial. This generates evidence for M marks.
始终先写下你打算使用的公式再代入数字。例如，写出“梯形法则：面积 ≈ h/2 […]”——这可能本身就锁定了 B1 分。同样，标明你的替换“Let u = …”并写出 du/dx，即使很简单，这为 M 分提供了证据。
When solving differential equations, write ‘Separate variables: 1/g(y) dy = f(x) dx’ as a separate line. The mark scheme explicitly awards M1 for this line. Do not jump straight to the integrated form, as you could accidentally miss the method mark.
解微分方程时，单独写一行“分离变量：1/g(y) dy = f(x) dx”。评分标准明确为此行给出 M1。不要直接跳到积分后的形式，否则可能无意间丢掉方法分。
Finally, check that your final answer matches the required form. If the question asks for y = f(x), do not leave it as ln|y| = …; continue to exponentials. The A1 mark often requires the answer in a specific, simplified form.
最后，检查最终答案是否符合要求的形式。如果题目要求 y = f(x)，不要停留在 ln|y| = …；继续推导至指数形式。A1 分通常要求答案为指定且简化的形式。
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