📚 AS Mathematics Unit 1 Mark Scheme June 2019: Question Types & Solutions Analysis | AS 数学单元一 2019年6月评分方案题型解析
The June 2019 AS Mathematics Unit 1 paper tests core pure mathematics skills, covering algebra, coordinate geometry, calculus, trigonometry, sequences and proof. The accompanying mark scheme reveals exactly how examiners award marks – for method, accuracy and communication. In this article, we break down each question type, highlight the key steps according to the mark scheme, and provide bilingual explanations to help students understand the reasoning and avoid common pitfalls.
2019年6月的AS数学单元一试卷考查了纯数学的核心技能，涵盖代数、坐标几何、微积分、三角学、数列与证明。配套的评分标准精确揭示了考官如何给分——方法分、准确性分和表达分。本文逐一解析各题型，根据评分方案指出关键步骤，并提供中英双语解释，帮助学生理清逻辑、避开常见失分点。
1. Overview of the Paper and Mark Scheme | 试卷与评分标准概览
The 2019 June Unit 1 paper typically contains around 11 to 14 questions, with a total of 100 marks. Questions range from short, single-step items to multi-part structured problems. The mark scheme shows that marks are allocated as M marks (method), A marks (accuracy) and B marks (independent/statement marks). Some final answers are awarded A1 marks; others require both M1 and A1. Understanding this structure helps students decide where to show full working.
2019年6月的单元一试卷通常包含11至14道题，总分100分。题目范围从简短的单一步骤题到多部分的综合题。评分方案显示，分数分配为M分（方法分）、A分（准确分）和B分（独立/结论分）。部分最终答案获得A1分；另一些则需要M1和A1。了解这种结构有助于学生确定在何处展示完整步骤。
- M marks are earned by demonstrating a correct mathematical method, even if a minor slip occurs later.
- M分通过展示正确的数学方法获得，即使后续出现小错误。
- A marks require a correct final answer, or an expression matching the printed answer in the mark scheme.
- A分要求最终答案正确，或表达式与评分方案中的答案一致。
- B marks are for key statements, such as stating the vertex of a quadratic or the centre of a circle, without requiring working.
- B分用于关键陈述，例如写出二次函数的顶点或圆的圆心，无需展示步骤。
Always show your working – a missing step can lose an M mark even if the final answer is correct. The examiner cannot award a method mark if no method is seen.
务必展示解题步骤——即便最终答案正确，缺少步骤也可能丢失M分。如果没有展示方法，考官无法给出方法分。
2. Algebra and Simplification | 代数与化简
− 12) / (x² − 4x + 4). The mark scheme awards M1 for factorising numerator and denominator, and A1 for the simplified expression. The denominator x² − 4x + 4 = (x − 2)²; the numerator 3(x² − 4) = 3(x − 2)(x + 2). Cancelling (x − 2) gives 3(x + 2)/(x − 2). Full marks require stating the restriction x ≠ 2 – missing this may lose the A1 mark if the question asks for the set of values.
典型的第一题涉及有理式的化简，通常通过因式分解并约去公因式完成。例如：化简 (3x² − 12) / (x² − 4x + 4)。评分方案对分子分母因式分解给予M1，对化简后的表达式给予A1。分母 x² − 4x + 4 = (x − 2)²；分子 3(x² − 4) = 3(x − 2)(x + 2)。约去 (x − 2) 得到 3(x + 2)/(x − 2)。如果要写出定义域，还需注明 x ≠ 2——如果问题要求写出数值集，遗漏此限制可能丢失A1分。
Another type is expanding brackets and collecting like terms: e.g., (2x + 1)(x − 3) − (x + 5)². The mark scheme gives M1 for expanding each bracket correctly, and A1 for the fully simplified quadratic. Any sign errors in collecting terms lead to loss of the A mark.
另一种题型是展开括号并合并同类项：例如 (2x + 1)(x − 3) − (x + 5)²。评分方案对每个括号的正确展开给M1，对完全化简的二次式给A1。合并同类项时的任何符号错误将导致失去A分。
3. Quadratic Functions and Completing the Square | 二次函数与配方法
e factor 2, and then for correctly completing the square inside the bracket. The solution: 2[x² − 6x] + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11. The final A1 is for the fully correct form. The coordinates of the vertex (3, −11) can then earn a B1 if requested.
配方法题型非常常见，例如将 2x² − 12x + 7 写成 a(x + b)² + c 的形式。评分方案对提取系数2给出M1，随后对括号内正确配方给予认可。解答：2[x² − 6x] + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11。最终A1是完整正确形式。若要求写出顶点坐标 (3, −11)，则可获得B1。
A tricky part can be when the coefficient of x² is negative. The mark scheme still expects the same logical steps. Students should check their expansion to verify the completed-square form.
难点可能在于 x² 系数为负的情况。评分方案仍期望相同的逻辑步骤。学生应通过展开验证配方结果是否正确。
4. Differentiation and Tangent Lines | 求导与切线方程
he tangent or normal. For instance: given y = x³ − 5x² + 3x − 2. The mark scheme: M1 for lowering the power on each term, A1 for the correct derivative dy/dx = 3x² − 10x + 3. Then, to find the gradient at x = 2, evaluate: 3(4) − 20 + 3 = −5. The tangent equation uses y − y₁ = m(x − x₁). M1 for correct substitution and A1 for y = −5x + 8 (if the point is (2, −2)).
求导题在2019年6月的试卷中常要求对多项式求 dy/dx，在某点计算梯度，然后求切线或法线方程。例如：已知 y = x³ − 5x² + 3x − 2。评分方案：对各项降幂给予M1，正确导数 dy/dx = 3x² − 10x + 3 给予A1。然后求 x = 2 处的梯度：3(4) − 20 + 3 = −5。切线方程使用 y − y₁ = m(x − x₁)。正确代入得M1，最终 y = −5x + 8（若点为(2, −2)）得A1。
Finding the normal requires using the negative reciprocal of the gradient. The mark scheme clearly states that failing to use the perpendicular gradient loses the M mark for the normal line.
求法线需要使用梯度的负倒数。评分方案明确指出：若未使用垂直梯度，将失去法线部分的M分。
| Step | Marks | Notes |
| Differentiate correctly | M1 A1 | At least one power reduced |
| Substitute x to find gradient | M1 | Numerical evaluation |
| Equation of tangent | M1 A1 | Correct form, simplified |
5. Integration and Area Under a Curve | 积分与曲线下方面积
scheme awards M1 for each term integrated correctly: e.g., ∫(4x³ − 2x + 3) dx = x⁴ − x² + 3x + C. The constant of integration is not always required for a definite integral; for an indefinite integral, missing '+ C' can cost the A1 mark.
积分题通常要求求多项式的原函数，或计算曲线与x轴之间的面积。评分方案对每一项正确积分给予M1：例如 ∫(4x³ − 2x + 3) dx = x⁴ − x² + 3x + C。对定积分，积分常数不总是必需；对不定积分，漏掉 '+ C' 可能导致丢失A1分。
For area problems, the mark scheme expects the student to integrate y with respect to x between the given limits, subtract the lower limit value from the upper limit value, and give the exact area. A common error is forgetting to apply limits correctly or mixing up signs when the curve is below the axis. If part of the curve lies below the x-axis, the area must be calculated separately or by integrating the absolute value; the mark scheme will specify.
对于面积问题，评分方案期望学生对y关于x在给定上下限之间积分，用上限值减去下限值，并给出准确面积。常见错误是未正确代入上下限，或当曲线在x轴下方时混淆符号。若部分曲线在x轴下，面积需分段计算或对绝对值积分；评分方案会明确说明。
Definite integral set-up: Area = ∫ₐᵇ f(x) dx. Evaluate F(b) − F(a), where F is the antiderivative.
定积分设置：面积 = ∫ₐᵇ f(x) dx。计算 F(b) − F(a)，其中F为原函数。
6. Trigonometric Equations | 三角函数方程
es of 2θ, and then adjusting for the given range. The principal value: 2θ = 30°, 150°, then add 360° to each: 2θ = 30°, 150°, 390°, 510°. Dividing by 2 gives θ = 15°, 75°, 195°, 255°. Each correct angle earns an A1; missing one solution loses marks. The mark scheme often requires the answer in degrees or radians as specified in the question.
标准的三角题要求解方程 sin 2θ = ½，其中 0° ≤ θ ≤ 360°。评分方案对求反正弦得到 2θ 的主值给予M1，然后调整到指定区间。主值：2θ = 30°, 150°，然后各自加 360°：2θ = 30°, 150°, 390°, 510°。除以2得 θ = 15°, 75°, 195°, 255°。每个正确角度可获得A1分；漏解会丢分。评分方案通常要求答案以度或弧度为单位，依题目要求。
method. The mark scheme expects M1 for making the substitution and solving for u, then converting back to θ. Students must adjust the range for u accordingly: if θ ∈ [0°, 360°], then u ∈ [−30°, 330°]. Missing the lower bound adjustment is a common pitfall.
对于 cos(θ − 30°) = 0.5 这类方程，使用替换 u = θ − 30° 十分有效。评分方案期望先进行替换并解出u，然后转换回θ。学生需相应调整u的范围：若 θ ∈ [0°, 360°]，则 u ∈ [−30°, 330°]。忽略下限调整是常见失误。
7. Exponentials and Logarithms | 指数与对数
ogarithms of both sides (any base) and using log properties. For example, x log 5 = log 30 ⇒ ark scheme may also accept natural logs: x = ln 30 / ln 5. Both methods are equally valid.
指数增长或衰变问题，如解 5ᵡ = 30，需要取对数。评分方案对两边取对数（任何底数）并运用对数性质给予M1。例如，x log 5 = log 30 ⇒ x = log 30 / log 5。最终A1要求正确的值，通常保留3位有效数字。评分方案也可能接受自然对数：x = ln 30 / ln 5。两种方法同等有效。
Another typical question involves simplifying an expression like 2 logₐ x + ½ logₐ y − logₐ z. The mark scheme gives M1 for using the power rule correctly, and A1 for a single logarithm. The answer: logₐ (x² √y / z). A further part might ask to evaluate the expression for given values. The process is step-by-step, and showing the combination of logs clearly helps secure both M and A marks.
另一典型题涉及化简表达式，如 2 logₐ x + ½ logₐ y − logₐ z。评分方案对正确使用幂律给予M1，对合并为一个对数给予A1。答案：logₐ (x² √y / z)。后续部分可能要求代入具体数值求值。逐步展示对数的合并有助于稳拿M分和A分。
8. Arithmetic Sequences | 等差数列
rms. The mark scheme awards B1 for the nth term formula uₙ = a + (n − 1)d, M1 for substituting correctly, and A1 for the 20th term = 7 + 19×3 = 64. For the sum, the formula Sₙ = n/2 [2a + (n − 1)d] or n/2 (a + l) can be used. Correct substitution and simplification lead to A1. Using the wrong formula loses the M mark.
2019年6月的试卷可能包含数列问题：已知首项 a = 7，公差 d = 3，求第20项和前50项之和。评分方案对通项公式 uₙ = a + (n − 1)d 给予B1，对正确代入给予M1，对第20项 = 7 + 19×3 = 64 给予A1。对于求和，可使用公式 Sₙ = n/2 [2a + (n − 1)d] 或 n/2 (a + l)。正确代入并化简获得A1。误用公式将失去M分。
A challenging variant includes finding the first term or common difference when given two conditions. This requires setting up simultaneous equations. The mark scheme allocates M1 for each correct equation and M1 for solving them. Full marks require both a and d correctly found.
更具挑战性的变体是根据两个条件求首项或公差。这需要建立联立方程组。评分方案对每个正确方程给予M1，对求解过程给予M1。完整得分需要正确求得a和d。
9. Polynomial Division and Factor Theorem | 多项式除法与因式定理
: M1 for evaluating f(2) = 8 − 16 + 2 + 6 = 0, and A1 for concluding (x − 2) is a factor. The next step is division by (x − 2) using synthetic division or long division. The quotient x² − 2x − 3 is then factorised into (x − 3)(x + 1). Full factorisation: (x − 2)(x − 3)(x + 1). Each step carries an M1 for method and A1 for accuracy.
因式分解三次多项式，例如 f(x) = x³ − 4x² + x + 6，通常先要求用因式定理证明 (x − 2) 是一个因式。评分方案：计算 f(2) = 8 − 16 + 2 + 6 = 0 给予M1，得到 (x − 2) 为因式给予A1。下一步是用综合除法或长除法除以 (x − 2)。商 x² − 2x − 3 再分解为 (x − 3)(x + 1)。完全分解结果为 (x − 2)(x − 3)(x + 1)。每个步骤都有方法分M1和准确性分A1。
Long division errors, such as misaligning terms, cause loss of both M and A marks. Showing the division process clearly, including the subtraction lines, helps examiners award method marks even if a sign error is made.
长除法的错误，例如项不对齐，会导致同时失去M和A分。清晰地展示除法过程，包括每步减法，有助于考官即便在符号错误时也能给方法分。
10. Equation of a Circle | 圆的方程
identifying the centre (1, −4) and radius 5. A subsequent part might ask to find the distance between the centre and a point P(7, 2) and determine whether P lies inside, on or outside the circle. Using the distance formula: √[(7 − 1)² + (2 + 4)²] = √36 + 36 = √72 = 6√2 5, so P is outside. M1 is for the distance calculation and A1 for the correct conclusion.
坐标几何题常涉及圆的方程，例如 (x − 1)² + (y + 4)² = 25。评分方案对正确写出圆心 (1, −4) 和半径 5 给予B1。后续部分可能要求求圆心到点 P(7, 2) 的距离，并判断P在圆内、圆上还是圆外。使用距离公式：√[(7 − 1)² + (2 + 4)²] = √36 + 36 = √72 = 6√2 ≈ 8.49 > 5，因此P在圆外。距离计算获得M1，正确结论获得A1。
Finding the equation of a tangent to a circle at a given point uses the perpendicularity of radius and tangent. The gradient of the radius CP is mr; the tangent gradient mt = −1/mr. M1 for finding mr, M1 for using perpendicular gradient, A1 for the tangent equation. Careful algebraic manipulation is required.
求圆上一点处的切线方程需要利用半径与切线垂直的性质。半径CP的梯度为 mr；切线梯度 mt = −1/mr。求 mr 获M1，使用垂直梯度获M1，切线方程获A1。需仔细进行代数运算。
11. Proof and Mathematical Reasoning | 证明与数学推理
bers be 2n + 1 and 2n + 3. Sum of squares: (2n + 1)² + (2n + 3)² = 4n² + 4n + 1 + 4n² + 12n + 9 = 8n² + 16n + 10 = 2(4n² + 8n + 5). The expression inside is an integer, so the sum is even. Since the bracket is odd + odd + odd = odd ×, the factor 2 cannot absorb a 4, so the sum is not a multiple of 4. The mark scheme awards M1 for algebraic expressions of the odds, M1 for correct expansion, A1 for the final factorised form and correct conclusion.
证明题时常出现，例如证明两个连续奇数的平方和是偶数但不是4的倍数。设奇数为 2n + 1 和 2n + 3。平方和：(2n + 1)² + (2n + 3)² = 4n² + 4n + 1 + 4n² + 12n + 9 = 8n² + 16n + 10 = 2(4n² + 8n + 5)。括号内为整数，因此和为偶数。由于括号内奇数+奇数+奇数仍为奇数，因子2不能吸收一个4，因此和不是4的倍数。评分方案对奇数的代数表达给予M1，正确展开给予M1，最终因式分解形式和正确结论给予A1。
Proving an identity or a trigonometric identity also requires transforming one side into the other. Marks are given for each valid step; the final A1 is for the completion of the proof. Insufficient justification or skipping steps can lose these marks.
证明恒等式或三角恒等式也需要将一边变换为另一边。每个有效步骤都有分；最终A1是证明完成。理由不充分或跳步可能丢分。
12. Common Mistakes and Mark Scheme Tips | 常见失分点与评分标准提示
Based on the June 2019 mark scheme, several patterns of errors are penalised. The most frequent include: omission of the '+ C' in indefinite integration; forgetting to consider both positive and negative roots when solving equations involving squares; misreading the range for trigonometric solutions; failing to rationalise denominators when explicitly asked; and not providing exact answers when specified. The mark scheme often applies a 'deduct once' rule – a mistake carried forward may not be penalised again if the method there is sound, but check the specifics.
基于2019年6月的评分方案，几种错误模式被反复扣分。最常见的有：不定积分中遗漏 '+ C'；解平方方程时忘记考虑正负根；误读三角函数的求解区间；题目明确要求时分母未有理化；指定保留精确值时给出近似值。评分方案常采用“只扣一次”原则——如果后续方法正确，前一步的错可能不再重复扣分，但需查看具体细则。
- Always read the question: 'give your answer in exact form' means keep surds, π, and fractions – do not give a decimal.
- 仔细审题：“请给出精确值”意味着保留根号、π 和分数——不要写小数。
- Show substitution steps clearly in calculus and sequence problems to gain method marks.
- 在微积分和数列问题中清晰展示代入步骤，以获取方法分。
- When using the quadratic formula, write it out first, then substitute values. This guarantees M1 even if arithmetic slips.
- 使用二次公式时，先写出公式，再代入数值。即使计算有误，也能确保获得M1。
- For proofs, state the conclusion explicitly – a proof is not complete without a closing statement.
- 证明题要明确写出结论——没有结束语证明不算完整。
Familiarity with the mark scheme's language is invaluable. Phrases such as 'allow for equivalent form', 'condone omission of brackets if recovered' and 'award for correct unsimplified expression' indicate where leniency is applied. Use past papers alongside the mark scheme to internalise these patterns and boost your confidence.
熟悉评分标准的用语极具价值。诸如“允许等价形式”、“若后续恢复则宽容括号遗漏”、“正确但未化简的表达式给分”等表述，指明了宽容之处。利用真题配合评分方案，将这种模式内化，可以极大提升自信。
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