📚 AS-Level Maths Unit 1 Mark Scheme Jan 19: Question Breakdown | AS数学：2019年1月Unit 1评分标准题型解析
Mastering AS Mathematics Unit 1 is as much about understanding how marks are awarded as it is about solving the problems. The January 2019 AQA 7356/1 mark scheme reveals a consistent pattern: method marks (M) are generously given for starting a correct process, accuracy marks (A) depend on exact values, and independent marks (B) are awarded for stating definitions or key results without working. This breakdown will walk you through the main question types from that paper, highlighting where examiners award and deduct marks. By internalising these patterns, you can avoid dropping simple marks and present your solutions in the most mark-friendly way.
掌握AS数学Unit 1不仅在于解题，更在于理解评分规则。2019年1月AQA 7356/1评分标准揭示了一个稳定模式：方法分(M)对于开启正确步骤给得很大方，准确度分(A)依赖于精确数值，而独立分(B)则奖励无需过程的陈述或关键结果。本文将带你拆解这套试卷中的主要题型，指出考官在哪里给分、在哪里扣分。一旦内化了这些规律，你就能避开简单丢分，用最适应评分标准的格式呈现解答。
1. Understanding the Mark Scheme: M, A, B Marks | 评分标准解读：M分、A分和B分
The Jan 19 Unit 1 mark scheme uses three fundamental mark types. An M mark is for a correct method applied to the candidate’s numbers; it does not require the final answer. An A mark is accuracy — the answer must be exact or rounded to a specified degree. A B mark is independent, given for stating a fact such as a derivative or an identity without any working. In many questions you will see ‘M1 A1’ or ‘B1’. Knowing this helps you attempt every part, even if you are unsure of the final answer: always show a valid method to grab the M mark.
2019年1月Unit 1评分标准使用了三种基本分数。M分是方法分，只要对考生的数值使用了正确方法即可获得，不要求最终答案。A分是准确性分——答案必须精确或按指定精度四舍五入。B分是独立分，奖励在没有计算过程的情况下陈述事实，比如导数或恒等式。很多题目中你会看到’M1 A1’或’B1’。了解这一点能帮助你尝试每一部分，即使你不确定最终答案：总是展示一个有效的方法来抓住M分。
| Mark Type | Meaning | Typical Command |
|---|---|---|
| M1 | Method mark | Attempt to integrate, set up equation, etc. |
| A1 | Accuracy mark | Correct answer following from M1 |
| B1 | Independent mark | State a fact or complete a statement |
2. Algebraic Simplification and Surds | 代数运算与根式化简
Early questions often test index laws and surd manipulation. For example, simplifying ( √8 + √2 )² may appear. The method involves expanding the bracket correctly: (a+b)² = a² + 2ab + b². A typical solution would show √8 = 2√2, so (2√2 + √2)² = (3√2)² = 18. The mark scheme awards M1 for the attempt to express √8 as 2√2 or for expanding properly, A1 for the final integer 18. Never skip the step of rewriting the surd — the M mark depends on seeing that transformation.
卷首题目常考察指数律和根式操作。比如可能出现化简 ( √8 + √2 )²。方法包括正确展开括号：(a+b)² = a² + 2ab + b²。典型解答会显示 √8 = 2√2，所以 (2√2 + √2)² = (3√2)² = 18。评分标准中，尝试将 √8 写成 2√2 或正确展开可获M1，最终整数18获A1。千万不要跳过重写根式的步骤——方法分就在那个变换中。
(2√2 + √2)² = (3√2)² = 9 × 2 = 18
3. Solving Quadratic Equations | 二次方程求解
Paper 1 frequently includes a quadratic, often disguised or part of a larger problem. Jan 19 examiners expected candidates to solve 2x² − 5x − 3 = 0. Method marks are awarded for factorisation into (2x+1)(x−3)=0 or correct substitution into the quadratic formula. The answers x = −1/2 and x = 3 each get A1. If you use the formula, write it out: x = [−b ± √(b² − 4ac)] / (2a). Substituting a=2, b=−5, c=−3 earns M1. Then computing the discriminant (49) and simplifying to the two roots gets the accuracy marks. Even a sign error can still earn M1, so always write down the formula.
试卷一经常包含二次方程，往往隐藏在更大的问题中。2019年1月考卷期望考生解 2x² − 5x − 3 = 0。方法分可以通过因式分解 (2x+1)(x−3)=0 或代入求根公式获得。答案 x = −1/2 和 x = 3 各得A1。如果使用公式，请写出：x = [−b ± √(b² − 4ac)] / (2a)。代入 a=2, b=−5, c=−3 可获M1。然后计算判别式(49)并化简到两个根获准确性分。即使符号错了仍可能得到M1，所以一定要写出公式。
x = [5 ± √(25 − 4(2)(−3))] / 4 = [5 ± √49] / 4 → x = 3, x = −½
4. Coordinate Geometry: Equations of Lines | 坐标几何：直线方程
Straight line questions test gradient, midpoint, and perpendicular lines. A Jan 19 task gave two points A(2, 1) and B(8, 7) and asked for the equation of line AB. The method: compute gradient m = (7−1)/(8−2) = 1, then use y − y₁ = m(x − x₁). Substituting A gives y − 1 = 1(x − 2) → y = x − 1. M1 for using the gradient formula, M1 for applying the point-slope form, A1 for the final equation in the requested form. If the question then asks for a perpendicular line through a point, find the negative reciprocal gradient (−1) and repeat. Setting out each step clearly earns all method marks even if a sign is flipped later.
直线题考察斜率、中点和垂直关系。2019年1月有道题给出点A(2, 1)和B(8, 7)，要求直线AB的方程。方法：计算斜率 m = (7−1)/(8−2) = 1，然后用 y − y₁ = m(x − x₁)。代入A得 y − 1 = 1(x − 2) → y = x − 1。使用斜率公式可得M1，应用点斜式得M1，按要求整理成最终形式得A1。如果题目接着要求过某点的垂线，求出负倒数斜率(−1)并重复过程。清晰展示每一步能拿到所有方法分，即便后边符号错了。
5. Circles and Tangents | 圆与切线
Circle geometry appears every year. A subtopic from Jan 19 is the tangent condition. The equation of a circle (x−2)² + (y+3)² = 25 and a line y = mx + c may be given. To show the line is a tangent, substitute the line into the circle, form a quadratic in x, and set the discriminant b² − 4ac = 0. M1 is for the substitution, M1 for simplifying to a quadratic, M1 for equating the discriminant to zero. Solving for m gives the gradient. Many candidates lose marks by not stating ‘discriminant = 0’ explicitly; the examiner needs to see the condition. Write: ‘For tangency, discriminant = 0’.
圆几何每年都考。2019年1月的一个子题是切线条件。可能给圆方程 (x−2)² + (y+3)² = 25 和一条直线 y = mx + c。要证明直线是切线，需将直线代入圆方程，形成关于x的二次式，并令判别式 b² − 4ac = 0。代入得M1，化简成二次式得M1，让判别式等于零再得M1。解出m就得到斜率。许多考生因不明确写出’判别式 = 0’而丢分；考官必须看到这个条件。写出：’For tangency, discriminant = 0’。
(x−2)² + (mx+c+3)² = 25 → (1+m²)x² + 2(m(c+3)−2)x + (c+3)² − 21 = 0
6. Binomial Expansion | 二项式展开
AS Unit 1 expects expansion of (a+b)ⁿ for integer n. Jan 2019 included (2+3x)⁵. The method uses nCr coefficients: ⁵C₀, ⁵C₁, ⁵C₂… Writing Pascal’s triangle or the formula earns M1. The expansion is: 32 + 5×16×(3x) + 10×8×(3x)² + 10×4×(3x)³ + 5×2×(3x)⁴ + (3x)⁵. Simplify to 32 + 240x + 720x² + 1080x³ + 810x⁴ + 243x⁵. M1 for using binomial coefficients, A1 for first three terms correct, A1 for the rest. If the question asks for the term independent of x, set the power of x to zero after expansion.
AS Unit 1要求展开(a+b)ⁿ，n为整数。2019年1月考了(2+3x)⁵。方法使用组合数：⁵C₀, ⁵C₁, ⁵C₂…写出杨辉三角或公式可得M1。展开为：32 + 5×16×(3x) + 10×8×(3x)² + 10×4×(3x)³ + 5×2×(3x)⁴ + (3x)⁵。化简得32 + 240x + 720x² + 1080x³ + 810x⁴ + 243x⁵。使用二项系数给M1，前三项正确给A1，其余给A1。若题目要求x无关的项，展开后令x的幂次为零。
7. Trigonometric Equations | 三角方程
A staple question: solve sinθ = 0.4 for 0° ≤ θ ≤ 360°. The mark scheme gives M1 for finding the principal value θ = sin⁻¹(0.4) ≈ 23.6°. Then M1 for using symmetry: second solution 180° − 23.6° = 156.4°. Both rounded to 1 decimal place are required for A1. If the equation is cos2θ = −0.5, the method is similar but you must adjust the range: 0° ≤ 2θ ≤ 720°. Principal value for cos⁻¹(−0.5) is 120°, then find all four solutions. Show the CAST diagram or wave sketch to earn method marks; simply writing final answers risks losing M marks.
必考题：在0° ≤ θ ≤ 360°解 sinθ = 0.4。评分标准给出M1，求主值 θ = sin⁻¹(0.4) ≈ 23.6°。然后M1，利用对称性：第二个解 180° − 23.6° = 156.4°。两者四舍五入到1位小数得A1。如果是 cos2θ = −0.5，方法类似但必须调整区间：0° ≤ 2θ ≤ 720°。cos⁻¹(−0.5)主值为120°，然后找出全部四个解。画出CAST图或波形草图以获取方法分；只写最终答案可能失掉M分。
sinθ = 0.4 → θ = 23.6°, 180°−23.6° = 156.4°
8. Exponential and Logarithmic Equations | 指数与对数方程
Logarithm problems appear regularly. Jan 19 had an equation like 2eˣ − 5 = 1. Solve: 2eˣ = 6 → eˣ = 3 → x = ln3. Method: M1 for isolating eˣ, A1 for exact ln3. Another style: log₂(x+1) − log₂x = 3. Combine logs: log₂[(x+1)/x] = 3 → (x+1)/x = 2³ = 8. M1 for using log law, M1 for converting to exponential, A1 for x = 1/7. Never attempt to solve such equations without stating the relevant log law or conversion; the method mark depends on that explicit step.
对数题经常出现。2019年1月有类似方程 2eˣ − 5 = 1。解：2eˣ = 6 → eˣ = 3 → x = ln3。方法：分离eˣ得M1，精确值ln3得A1。另一种风格：log₂(x+1) − log₂x = 3。合并对数：log₂[(x+1)/x] = 3 → (x+1)/x = 2³ = 8。运用对数律给M1，转化为指数形式给M1，x = 1/7得A1。千万不要在不解明显示相关对数律或转换的情况下就解方程；方法分就靠那一步。
9. Differentiation: First Principles and Rules | 求导：第一原理与法则
Differentiation is central. You may be asked for the derivative from first principles of f(x)=x². The limit definition: f ‘(x) = lim[h→0] ((x+h)² − x²)/h = lim (2xh + h²)/h = 2x. M1 for writing the limit expression, M1 for expanding, A1 for the correct limit. More often, you use standard rules: for y=4x³−1/x², rewrite as 4x³−x⁻², then dy/dx = 12x² + 2x⁻³. The method mark is for simplification to power form. Then find the equation of a tangent: at x=1, y=3, gradient=14, tangent: y−3=14(x−1). Substituting coordinates earns M1, correct derivative A1, final equation A1.
求导是核心内容。可能要求用第一原理求f(x)=x²的导数。极限定义：f ‘(x) = lim[h→0] ((x+h)² − x²)/h = lim (2xh + h²)/h = 2x。写出极限表达式给M1，展开得M1，正确极限得A1。更常见的是标准法则：对y=4x³−1/x²，改写为4x³−x⁻²，则dy/dx = 12x² + 2x⁻³。方法分在于化简成幂次形式。接着求切线方程：在x=1处，y=3，斜率=14，切线：y−3=14(x−1)。代入坐标得M1，导数正确得A1，最终方程得A1。
f ‘(x) = lim[h→0] ( (x+h)² − x² ) / h = lim (2xh + h²)/h = 2x
10. Integration and Area Under a Curve | 积分与曲线下面积
Integration appears both as reverse differentiation and area finding. Jan 19 had ∫ (8x³ − 2/√x) dx. First rewrite √x as x^½, so 2/√x = 2x⁻^½. Integrate term by term: ∫ 8x³ dx = 2x⁴, ∫ 2x⁻^½ dx = 4x^½. Don’t forget the constant +C: answer 2x⁴ + 4√x + C. M1 for raising power by 1 and dividing by new power, A1 for each term, B1 for +C. For area between curve y = 4 − x² and x-axis, set limits: solve 4−x²=0 → x=±2. Area = ∫[-2,2] (4−x²) dx = [4x − x³/3] from −2 to 2 = (8−8/3) − (−8+8/3) = 32/3. M1 for using correct limits, M1 for integration, A1 for exact fraction.
积分既作为逆运算又要求求面积。2019年1月考了 ∫ (8x³ − 2/√x) dx。首先将√x写成 x^½，所以 2/√x = 2x⁻^½。逐项积分：∫ 8x³ dx = 2x⁴，∫ 2x⁻^½ dx = 4x^½。不要忘记常数+C：答案为 2x⁴ + 4√x + C。幂次加1并除以新幂次得M1，每一项正确得A1，+C得B1。对于曲线 y = 4 − x² 与x轴之间的面积，定限：解4−x²=0 → x=±2。面积 = ∫[-2,2] (4−x²) dx = [4x − x³/3] 从−2到2 = (8−8/3) − (−8+8/3) = 32/3。使用正确上下限得M1，积分得M1，精确分数得A1。
11. Sequences and Series: Arithmetic Progressions | 数列与级数：等差数列
Arithmetic sequences frequently give easy marks if you learn the formulas. Jan 19 had an AP with first term a=5 and common difference d=2. Find the 20th term: u₂₀ = a + 19d = 5 + 38 = 43. M1 for using uₙ = a+(n−1)d. Then sum of first 30 terms: S₃₀ = n/2 [2a + (n-1)d] = 15 [10 + 29×2] = 15×68 = 1020. M1 for quoting sum formula, A1 for correct substitution, A1 for answer. If asked ‘which term equals 101?’, solve a+(n−1)d=101 → 5+2(n−1)=101 → n=49. Always write the formula before substituting to secure the method mark.
等差数列常常是送分题，只要记住公式。2019年1月考了一个AP，首项a=5，公差d=2。求第20项：u₂₀ = a + 19d = 5 + 38 = 43。运用uₙ = a+(n−1)d得M1。然后求前30项和：S₃₀ = n/2 [2a + (n-1)d] = 15 [10 + 29×2] = 15×68 = 1020。引用求和公式M1，代入正确A1，结果A1。如果问’第几项等于101？’，解 a+(n−1)d=101 → 5+2(n−1)=101 → n=49。务必先写出公式再代入，确保拿到方法分。
12. Common Mistakes and How to Maximise Marks | 常见失分点与提分策略
Examiners’ reports for Jan 19 Unit 1 highlight repeated errors. The biggest one is not showing working: a candidate writes only the final answer and loses all method marks if the answer is wrong. Always write the formula, the substitution, and one simplification step. Another is algebraic mishandling of surds or negative/fractional indices. Practice rewriting terms like 1/x² as x⁻² before differentiating. Also, misreading the domain in trigonometry leads to missing solutions; always expand the interval for double/triple angles. Finally, incomplete simplification — leaving an answer as 2/√8 instead of √2/2 — loses the accuracy mark. Adopt a checking routine: after solving, substitute your answer back mentally to verify the equation balances.
2019年1月Unit 1的考官报告指出了重复出现的错误。最大问题是不展示过程：考生只写最终答案，一旦答案错误，所有方法分全丢。务必写出公式、代入步骤和至少一步化简。另一个是根式或负指数、分数指数代数处理错误。练习在微分前将项如 1/x² 改写成 x⁻²。此外，误读三角函数的定义域导致漏解；涉及倍角时总是扩大区间。最后，未完成化简——把答案留成 2/√8 而不是 √2/2 ——会丢掉准确性分。养成检验习惯：解完后在脑中代回原式核验是否成立。
| Action | Mark Gain |
|---|---|
| Write formula before substituting | Secure M1 |
| Show surd simplification step | Avoid losing A1 |
| State discriminant = 0 for tangency | Gain method mark |
| Expand trig range for double angles | Prevent missing solutions |
Published by TutorHao | AS Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导