📚 AQA AS Maths Unit 2 Jan 2019 Mark Scheme Guide | AQA AS数学第二单元2019年1月评分方案指南
The January 2019 AQA AS Mathematics Unit 2 paper, often referred to as Pure Core 2, tested a range of pure mathematics skills. The mark scheme reveals exactly how examiners awarded method marks, accuracy marks and independent marks across algebra, sequences, trigonometry, exponentials, differentiation and integration. This guide breaks down the key patterns and common pitfalls so you can use the mark scheme effectively in your revision.
2019年1月AQA AS数学第二单元试卷，通常称为纯数核心2，考查了一系列纯数学技能。评分方案准确展示了考官如何在代数、数列、三角函数、指数、微分和积分等题目中授予方法分、准确分和独立分。本指南将详细拆解这些关键规律和常见失分点，帮助你在复习中高效利用评分方案。
1. Overview of the Mark Scheme | 评分方案总览
The January 2019 mark scheme followed the standard AQA Pure Core 2 structure. Each question was broken into parts, with marks labelled M for method, A for accuracy and B for independent marks. The paper typically contained questions worth between 4 and 9 marks, with the final parts often testing connected skills such as finding a tangent after differentiation.
2019年1月的评分方案遵循AQA纯数核心2的标准结构。每道题被分成若干小问，分值标签包括M（方法分）、A（准确分）和B（独立分）。试卷中的题目通常在4到9分之间，最后一小问往往考查连贯技能，例如先求导再求切线。
| Topic | Typical Marks | Common Skills Tested |
|---|---|---|
| Algebra and Functions | 6-8 | Factor theorem, remainder theorem |
| Sequences and Series | 6-8 | Arithmetic and geometric formulas |
| Trigonometry | 8-10 | Radians, identities, solving equations |
| Exponentials and Logarithms | 6-8 | Log laws, modelling curves |
| Differentiation | 10-12 | Tangents, normals, stationary points |
| Integration | 10-12 | Definite integrals, area under a curve |
2. Algebra and Functions: Factor and Remainder Theorems | 代数与函数：因式定理与余式定理
st the method mark because they did not clearly show the substitution step, even if they wrote the correct result.
2019年1月的一道典型题目要求考生证明 (x - a) 是三次多项式的因式。评分方案对将正确值代入 f(x) 给 M1，对得到 f(a) = 0 给 A1。许多考生因为没有清晰展示代入步骤而失去方法分，即使他们写出了正确结果。
The remainder theorem also appeared in the paper. When finding the remainder after dividing by a linear factor, candidates needed to evaluate the polynomial at the zero of the divisor. The mark scheme often gave B1 for the correct remainder and M1 for a valid method, even if the final sign was wrong due to a minor slip.
余式定理也在试卷中出现。在求除以线性因式所得余数时，考生需要在除式的零点处计算多项式的值。评分方案通常对正确余数给 B1，对有效方法给 M1，即使因小失误导致最终符号错误仍可获得方法分。
f(x) = 2x³ - 5x² + 3x - 6; show (x - 2) is a factor
-4. This should equal 0 if the factor is valid, so checking your arithmetic is essential.
解题需要计算 f(2) = 2(2)³ - 5(2)² + 3(2) - 6 = 16 - 20 + 6 - 6 = -4。如果因式正确，该值应等于0，因此检查算术至关重要。
3. Coordinate Geometry: Circles and Tangents | 坐标几何：圆与切线
One question on circles required finding the centre and radius from an equation written in expanded form. The mark scheme rewarded completing the square correctly. A common error was forgetting to add the squared constants to both sides of the equation, which led to an incorrect radius.
有一道圆的问题要求从展开形式方程中找出圆心和半径。评分方案对正确配方给予分数。常见错误是忘记将平方常数加到等号两边，导致半径计算错误。
Another part asked for the equation of a tangent to a circle at a given point. The mark scheme gave M1 for finding the gradient of the radius, M1 for using the negative reciprocal to obtain the tangent gradient, and A1 for the final tangent equation in the required form. Marks were often lost by mixing up the gradient of the radius with the gradient of the tangent.
另一部分要求求圆上给定点处的切线方程。评分方案对求半径斜率给 M1，对使用负倒数得到切线斜率给 M1，对最终符合要求形式的切线方程给 A1。许多考生因混淆半径斜率和切线斜率而失分。
, the radius gradient is (3 - (-1))/(5 - 2) = 4/3. The tangent gradient is therefore -3/4.
例如，若圆心为 (2, -1)，圆上一点为 P(5, 3)，则半径斜率为 (3 - (-1))/(5 - 2) = 4/3。因此切线斜率为 -3/4。
4. Sequences and Series: Arithmetic and Geometric Progressions | 数列与级数：等差与等比数列
rect substitution. Errors often came from confusing n and n - 1 inside the formula, or from using the nth term formula a + (n - 1)d when the sum was required.
等差数列问题通常要求前 n 项和。评分方案对正确代入 Sₙ = n/2[2a + (n - 1)d] 给予 M1。常见错误包括混淆公式中的 n 和 n - 1，或在要求求和时误用了第 n 项公式 a + (n - 1)d。
Geometric series questions tested the sum to infinity. The mark scheme required the condition |r| < 1 to be stated or implied. Candidates who attempted to use the finite sum formula without checking convergence lost accuracy marks, even if the subsequent calculation was correct.
等比数列问题考查无穷级数和。评分方案要求考生陈述或隐含条件 |r| < 1。未检查收敛性就尝试使用有限和公式的考生会失去准确分，即使后续计算正确。
atio r, the sum to infinity is S∞ = a/(1 - r). This formula is only valid when -1 < r < 1.
对于首项为 a、公比为 r 的等比数列，无穷级数和为 S∞ = a/(1 - r)。该公式仅在 -1 < r < 1 时有效。
5. Trigonometry: Radians, Identities and Equations | 三角函数：弧度、恒等式与方程
Trigonometry questions often involved solving equations in radians. The mark scheme awarded B1 for the first correct solution, M1 for finding related angles using the CAST diagram or the graph of the function, and A1 for all solutions in the given interval. A major error was working in degrees instead of radians, which often gave no marks because the answers were outside the expected interval.
三角题通常涉及以弧度求解方程。评分方案对第一个正确解给 B1，对使用 CAST 图或函数图像求相关角给 M1，对区间内所有正确解给 A1。使用度数而非弧度计算是主要错误，因为所得答案往往不在预期区间内。
Identity questions such as using sin²θ + cos²θ ≡ 1 required careful algebraic manipulation. The mark scheme typically gave M1 for replacing one trigonometric function with another, and A1 for reaching the required simplified form. Candidates who wrote the identity incorrectly at the start lost both method and accuracy marks.
恒等式题如使用 sin²θ + cos²θ ≡ 1 需要仔细的代数变形。评分方案通常对将一种三角函数替换为另一种给 M1，对得到所需简化形式给 A1。一开始就写错恒等式的考生会同时失去方法分和准确分。
h 1 - cos²θ to get 2(1 - cos²θ) - cosθ - 1 = 0, which simplifies to 2cos²θ + cosθ - 1 = 0.
例如，要求解 2sin²θ - cosθ - 1 = 0，首先用 1 - cos²θ 替换 sin²θ，得到 2(1 - cos²θ) - cosθ - 1 = 0，化简后为 2cos²θ + cosθ - 1 = 0。
6. Exponentials and Logarithms | 指数与对数
A common question asked to solve an equation of the form 2ˣ eme gave M1 for applying logs to both sides, M1 for using the power rule to bring down the exponent, and A1 for the correct answer to three significant figures. Some candidates lost the accuracy mark by not giving the answer to the required precision.
常见题目要求通过取对数求解 2ˣ = 7 形式的方程。评分方案对两边取对数给 M1，对使用幂法则降下指数给 M1，对保留三位有效数字的正确结果给 A1。一些考生因为没有按要求的精度给出答案而失去准确分。
Modelling questions with exponential growth or decay required forming an equation from context. The mark scheme often awarded B1 for the initial value, M1 for the correct base or growth factor, and A1 for the final model. Interpreting the constant e or the base correctly was essential.
指数增长或衰减的应用题需要根据背景建立方程。评分方案通常对初始值给 B1，对正确底数或增长因子给 M1，对最终模型给 A1。正确理解常数 e 或底数的含义至关重要。
For example, if a population doubles every 5 years, the model can be written as P = P₀ × 2t/5, where P₀ is the initial population and t is time in years.
例如，若人口每5年翻倍，模型可写成 P = P₀ × 2t/5，其中 P₀ 为初始人口，t 为以年为单位的时间。
7. Differentiation: Tangents, Normals and Stationary Points | 微分：切线、法线与驻点
Differentiation questions tested polynomials and simple functions. The mark scheme gave M1 for reducing the power by one and A1 for each correct term. Simplifying before differentiating was often necessary to avoid errors, especially when dealing with fractional or negative powers.
微分题考查多项式及简单函数。评分方案对幂减少 1 给 M1，对每个正确项给 A1。求导前先化简通常可以避免错误，尤其是在处理分数幂或负幂时。
), and A1 for the correct equation. A normal line requires taking the negative reciprocal of the tangent gradient; forgetting this was a common mistake.
切线或法线题要求先求某点的梯度，再求直线方程。评分方案对将 x 代入 dy/dx 给 M1，对使用 y - y₁ = m(x - x₁) 给 M1，对正确方程给 A1。法线需要取切线梯度的负倒数；忘记这一点是常见错误。
, and A1 for using the second derivative or a sign change to determine maximum or minimum.
驻点题要求给出拐点的坐标并判断其性质。评分方案对令 dy/dx = 0 给 M1，对解出 x 给 M1，对使用二阶导数或符号变化判断最大值或最小值给 A1。
8. Integration: Definite Integrals and Area Under a Curve | 积分：定积分与曲线下面积
Indefinite integration questions were marked for each term. The mark scheme gave M1 for raising the power and dividing by the new power, and A1 for each correct term plus the constant of integration. Forgetting the constant of integration only lost a single accuracy mark, but it was still avoidable.
不定积分题按项给分。评分方案对幂加 1 并除以新幂给 M1，对每个正确项及积分常数给 A1。忘记积分常数只失去一个准确分，但这是可以避免的。
Definite integration and area under curves required substituting limits correctly. The mark scheme often gave M1 for integrating, M1 for substituting limits, and A1 for the exact area. Forgetting to subtract the lower limit value was a common error. Some questions asked for the area between a curve and the x-axis over an interval where the curve crosses the axis, requiring a split of the integral.
定积分及曲线下面积要求正确代入上下限。评分方案通常对求积分给 M1，对代入上下限给 M1，对精确面积给 A1。忘记减下限值是常见错误。有些题要求曲线与 x 轴之间在跨轴区间上的面积，需要将积分分段处理。
to x = 2, integrate to get x³ + 2x, then evaluate [2³ + 2(2)] - [0³ + 2(0)] = 8 + 4 = 12.
例如，要求 y = 3x² + 2 从 x = 0 到 x = 2 下的面积，先积分得到 x³ + 2x，再计算 [2³ + 2(2)] - [0³ + 2(0)] = 8 + 4 = 12。
9. Common Pitfalls from the January
Published by TutorHao | AS 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