📚 AS Maths Unit 1 Mark Scheme (Jan 2021) – Concept Breakdown | AS数学单元1评分标准(2021年1月)知识点详解
This guide unpacks the key topics tested in the Edexcel AS Mathematics Unit 1 (Pure Mathematics) examination from January 2021. Using the official mark scheme, we highlight how marks are awarded for method (M), accuracy (A), and final answers, and we address common mistakes so you can refine your exam technique.
本指南深入剖析2021年1月爱德思AS数学单元1（纯数学）考试的核心知识点。借助官方评分标准，我们揭示方法分(M)、准确分(A)和最终答案的给分机制，并指出常见错误，帮助考生优化应试技巧。
1. Simplifying Surds and Indices | 根式与指数的化简
In Question 1, candidates were required to simplify an expression such as √48 + √27 – √75. The mark scheme awards an M1 mark for expressing each surd in the form k√3, for example √48 = 4√3, √27 = 3√3, √75 = 5√3.
在第一题中，考生需化简如√48 + √27 – √75的表达式。评分标准规定，若能将每个根式化为k√3的形式（如√48 = 4√3，√27 = 3√3，√75 = 5√3），即可获得M1方法分。
An A1 mark is then given for the final simplified answer, in this case 2√3. A common error is to incorrectly simplify √75 as 25√3 or to treat the sum as √120, losing both marks.
随后的A1分则授予最终化简结果2√3。常见错误包括将√75误算为25√3，或将原式直接相加得√120，导致两分全失。
Similarly, index laws were tested with expressions like (2x²)³ × 3x⁻⁴. The M1 requires a correct expansion of the power and the multiplication, leading to 8x⁶ × 3x⁻⁴ = 24x² for the final A1.
同样，指数运算考查了如(2x²)³ × 3x⁻⁴的表达式。方法分要求正确展开幂和乘法，最终得到8x⁶ × 3x⁻⁴ = 24x²，以获得A1准确分。
2. Factorising Quadratics and Cubic Expressions | 二次与三次式的因式分解
Question 2 involved solving a quadratic equation by factorisation, e.g. 2x² – 5x – 3 = 0. The mark scheme gives M1 for finding the correct factor pair (2x + 1)(x – 3), and A1 for both solutions x = –½ and x = 3.
第二题要求通过因式分解解二次方程，如2x² – 5x – 3 = 0。评分标准对正确找出因式对(2x + 1)(x – 3)给予M1分，对两个解x = –½和x = 3给予A1分。
If a candidate only wrote one solution, or incorrectly wrote x = ½, marks were withheld. The mark scheme also required the final answer to be presented as exact values, not rounded decimals.
如果考生只写了一个解，或误写成x = ½，均不得分。评分标准还要求最终答案以精确值给出，而非近似小数。
Later, a cubic equation x³ – 4x² + x + 6 = 0 appeared, where the factor theorem was needed. Substituting x = –1 yields 0, so (x + 1) is a factor, earning an M1. Long division or synthetic division then gives a quadratic factor, leading to A1 for all three solutions.
之后出现三次方程x³ – 4x² + x + 6 = 0，需使用因式定理。代入x = –1得0，因此(x + 1)是一个因式，获得M1分。通过长除法或综合除法得到二次因式，最终求出全部三个解可得A1分。
3. Completing the Square and Quadratic Graphs | 配方法与二次图像
The mark scheme for a completing-the-square question, such as rewriting 2x² – 8x + 5 in the form a(x + b)² + c, awards M1 for taking out the factor of 2 and correctly halving the coefficient of x. This yields 2(x – 2)² – 3.
对于配方法题目，如将2x² – 8x + 5改写为a(x + b)² + c的形式，评分标准对提取公因式2并正确取x系数的一半给予M1分。由此得到2(x – 2)² – 3。
The A1 mark is for the exact completed square form. A subsequent part asking for the coordinates of the vertex then gains B1 for (2, –3) provided the completed square is correct. A frequent error is forgetting to multiply the constant term when expanding back, leading to a sign mistake.
A1分则要求精确的完全平方形式。后续求顶点坐标的小问，若配方式正确，可得B1分，顶点为(2, –3)。常见错误是忘记在展开时乘以常数项，导致符号错误。
In the mark scheme, sketching the quadratic required the vertex, y-intercept (0, 5) and the correct shape to be shown, each earning a B mark.
在绘制二次图像时，评分标准要求标出顶点、y轴截距(0, 5)并画出正确开口方向，每项各得一个B分。
4. Equations of Lines and Circles | 直线与圆的方程
A typical coordinate geometry question gave two points A(1, 2) and B(5, 8) and asked for the equation of the perpendicular bisector. The mark scheme awards M1 for finding the midpoint (3, 5) and M1 for the gradient of AB (3/2), then the negative reciprocal –2/3 for the perpendicular gradient.
典型的坐标几何题给出两点A(1, 2)和B(5, 8)，要求中垂线方程。评分标准对中点(3, 5)给M1，对AB的斜率3/2给M1，再取其负倒数–2/3作为垂直斜率。
Using the point-slope form, y – 5 = –2/3(x – 3), earns an M1, and the final A1 is for the simplified equation, e.g. 2x + 3y = 23. Many candidates lost the final A1 by not simplifying to integer coefficients.
使用点斜式y – 5 = –2/3(x – 3)得M1，最终简化为2x + 3y = 23得A1。许多考生因未化简为整数系数而丢掉了最后的准确分。
Circle questions required finding the centre and radius from an equation like x² + y² – 6x + 2y = 15. Completing the square gives (x – 3)² + (y + 1)² = 25, centre (3, –1), radius 5. Each component is marked with B1 or M1A1.
圆方程题要求从x² + y² – 6x + 2y = 15中求出圆心和半径。配方得(x – 3)² + (y + 1)² = 25，圆心(3, –1)，半径5。各部分对应B1或M1A1。
5. Solving Simultaneous Equations | 解联立方程组
The paper included a linear–quadratic system: y = 2x + 1 and x² + y² = 10. The mark scheme awards M1 for substituting the linear equation into the circle equation to obtain an equation in one variable.
试卷中有一元二次联立方程组：y = 2x + 1和x² + y² = 10。评分标准对将直线方程代入圆方程得到一元方程给予M1分。
This yields x² + (2x + 1)² = 10 → 5x² + 4x – 9 = 0. Factorising gives (5x + 9)(x – 1) = 0, so x = –9/5 and x = 1, each A1. The corresponding y values are then found, and full A1 marks require all four coordinates presented clearly.
代入后得到x² + (2x + 1)² = 10 → 5x² + 4x – 9 = 0。因式分解得(5x + 9)(x – 1) = 0，因此x = –9/5和x = 1，各得A1。随后求出对应的y值，要拿到全部分数需清晰列出全部四组坐标。
A common error was to stop after finding x values, losing the final A marks. The mark scheme explicitly states that both solution pairs must be given.
常见错误是求出x值后就停笔，导致丢掉后续A分。评分标准明确指出必须给出两组解对。
6. Differentiation: Tangents and Normals | 微分：切线与法线
The derivative of a polynomial like y = x³ – 3x² + 5 was required. The mark scheme awards M1 for correct application of the power rule, and A1 for dy/dx = 3x² – 6x.
题目要求对多项式y = x³ – 3x² + 5求导。评分标准对正确使用幂函数求导法则给M1，对结果dy/dx = 3x² – 6x给A1。
To find the equation of the tangent at a point (2, 1), candidates first evaluate the gradient m = 3(2)² – 6(2) = 0. This is a horizontal tangent, so the equation is y = 1. M1 is for substituting x = 2 into the derivative, and A1 for the correct tangent equation.
求点(2, 1)处的切线方程，需先计算梯度m = 3(2)² – 6(2) = 0。这是一条水平切线，故方程为y = 1。将x = 2代入导函数给M1，正确切线方程得A1。
When a normal was required, the negative reciprocal gradient had to be used. A common slip was forgetting to take the reciprocal, or using the wrong sign. The mark scheme then awards a final A1 for the normal equation in the required form ax + by + c = 0.
若求法线，需使用负倒数斜率。常见失误是忘记取倒数，或符号弄错。评分标准最后对形如ax + by + c = 0的法线方程给出A1分。
7. Integration and Area Under Curves | 积分与曲线下方面积
An indefinite integral such as ∫(6x² – 4x + 3) dx appeared early in the paper. The mark scheme awards M1 for integrating term by term with the correct increase in power, and A1 for 2x³ – 2x² + 3x + c. The constant of integration is part of the A1 mark.
试卷前部出现了不定积分∫(6x² – 4x + 3) dx。评分标准对逐项积分且幂次正确增加给M1，对结果2x³ – 2x² + 3x + c给A1。积分常数是A1分的一部分。
For a definite integral, such as finding the area under y = 4x – x² between x = 0 and x = 3, the mark scheme requires setting up the integral ∫₀³ (4x – x²) dx. The M1 is for the correct integration to [2x² – x³/3]₀³. Substituting the limits gives (18 – 9) – (0) = 9, earning the final A1.
对于定积分，如求y = 4x – x²在x = 0到x = 3之间的面积，评分标准要求列出积分∫₀³ (4x – x²) dx。正确积分得[2x² – x³/3]₀³给M1。代入上下限得(18 – 9) – (0) = 9，最终得A1。
Errors in handling negative signs when substituting the lower limit were penalised. The mark scheme explicitly shows the subtraction step as part of the A1 requirement.
代入下限时负号处理错误会被扣分。评分标准明确将减法步骤作为A1分的一部分。
8. Exponential and Logarithmic Equations | 指数与对数方程
Solving equations like 2e²ˣ = 5 required a sequence of method marks. Taking natural logs of both sides earns an M1: ln(e²ˣ) = ln(2.5), simplifying to 2x = ln(2.5), then x = ln(2.5)/2. The final A1 is for the exact logarithmic form or a correctly rounded decimal if specified.
解形如2e²ˣ = 5的方程需要一系列方法分。两边取自然对数得M1：ln(e²ˣ) = ln(2.5)，化简得2x = ln(2.5)，然后x = ln(2.5)/2。最终A1分要求保留精确对数形式，或按要求给出正确的小数近似。
An equation involving log₂(x + 1) – log₂(x) = 3 was also tested. The first M1 is for combining the logs: log₂((x + 1)/x) = 3. Then rewriting as an exponential equation (x + 1)/x = 2³ = 8 earns the second M1. Solving to x = 1/7 yields the A1.
试卷还考查了方程log₂(x + 1) – log₂(x) = 3。第一个M1分是合并对数：log₂((x + 1)/x) = 3。然后改写为指数形式(x + 1)/x = 2³ = 8得第二个M1。解得x = 1/7给A1。
Candidates who omitted the domain check, or gave an extraneous solution, were not penalised unless the question specifically asked for justification, but the mark scheme shows the importance of verifying solutions in logarithmic equations.
省略定义域检查或给出增根的考生，除非题目明确要求说明理由，一般不会扣分，但评分标准显示了验证对数方程解的重要性。
9. Graph Transformations and Sketching | 图像变换与草图绘制
A function f(x) = √x underwent two transformations: first a translation by vector (3, 0), then a stretch parallel to the y-axis with scale factor 2. The mark scheme requires the final equation y = 2√(x – 3). M1 is given for each transformation correctly applied, and A1 for the fully correct expression.
函数f(x) = √x经过了两次变换：先沿x轴正方向平移3个单位，再沿y轴方向拉伸为原来的2倍。评分标准要求最终方程为y = 2√(x – 3)。每正确应用一个变换给一个M1，表达完全正确给A1。
Common mistakes included writing y = √(2x – 3) or y = 2√x – 3, which receive no marks because the order of operations and the horizontal translation were applied incorrectly.
常见错误包括写成y = √(2x – 3)或y = 2√x – 3，这些完全不得分，因为运算顺序和水平平移应用错误。
When sketching the transformed graph, the starting point and general shape had to be indicated. A mark was available for the new starting point (3, 0) and the correct direction of the stretch.
绘制变换后的图像时，必须标出起点和大致形状。新起点(3, 0)和正确的拉伸方向各可获得一个分数。
10. Proof and Mathematical Reasoning | 证明与数学推理
One question asked to prove that the sum of three consecutive integers is divisible by 3. The mark scheme awards M1 for setting up algebraic expressions: n, n+1, n+2. Their sum is 3n+3 = 3(n+1), which is clearly a multiple of 3, earning A1 for the conclusion with reasoning.
有一题要求证明三个连续整数之和可被3整除。评分标准对设立代数式n, n+1, n+2给M1。其和为3n+3 = 3(n+1)，显然是3的倍数，给出结论和推理可得A1。
A proof by exhaustion or by counterexample for another statement (e.g., “all prime numbers are odd”) required checking n=2, which is prime and even, thus disproving the statement. The mark scheme gives M1 for selecting a relevant counterexample and A1 for a clear disproof.
对于另一陈述（如“所有质数都是奇数”）的穷举或反证法，需要检验n=2，该数是质数且为偶数，从而推翻命题。评分标准对选取恰当反例给M1，对清晰驳斥给A1。
Many candidates lost marks by not explicitly stating the conclusion that the statement is false, or by giving an insufficient counterexample. The mark scheme insists on a complete logical argument.
许多考生因未明确写出命题为假的结论，或给出的反例不充分而失分。评分标准特别强调完整的逻辑论证过程。
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