📚 AS Maths Unit 1 Report on Examination (Jan 20): Common Pitfalls | AS数学单元1考试报告（2020年1月）易错点总结
The January 2020 AS Mathematics Unit 1 examination tested core topics in pure mathematics, including algebra, coordinate geometry, differentiation, integration, trigonometry, and proof. Examiner reports highlighted recurring mistakes that prevented many candidates from achieving top marks. This article summarises those common pitfalls and provides targeted advice to help you avoid them in your revision and future exams.
2020年1月的AS数学单元1考试涵盖了纯数学的核心主题，包括代数、坐标几何、微分、积分、三角学和证明。考官报告指出了许多考生反复出现的错误，这些错误使他们无法获得高分。本文总结了这些常见易错点，并提供针对性建议，帮助你在复习和未来考试中避免这些问题。
1. Algebraic Simplification Errors | 代数化简错误
to simplify fractions or surds fully. For example, when simplifying an expression like (x² – 4)/(x – ncorrectly without factorising first. Examiners insisted on showing the factorised form (x – 2)(x + 2)/(x – x + 2, provided x ≠ 2. Omitting the condition x ≠ 2 was also penalised in some questions.
许多考生因没有完全化简分式或根式而失分。例如，在化简 (x² – 4)/(x – 2) 这样的表达式时，考生常常没有先因式分解就直接错误地约分。考官坚持要求先写出因式分解形式 (x – 2)(x + 2)/(x – 2) = x + 2，并注明 x ≠ 2。在某些题目中，遗漏 x ≠ 2 的条件也会被扣分。
When dealing with surds, errors like √a + b = √a + √b were surprisingly common. Remember that √a + b cannot be split in this way. Rationalising denominators also caused trouble; for instance, to simplify 1/(√3 – 1), multiply numerator and denominator by (√3 + 1) to obtain (√3 + 1)/2. Not expanding the denominator correctly, or failing to simplify the final answer, led to unnecessary mark loss.
在处理根式时，√a + b = √a + √b 这类错误出人意料地普遍。请记住，√a + b 不能这样拆分。分母有理化也带来了麻烦；例如，要化简 1/(√3 – 1)，应将分子和分母同乘以 (√3 + 1)，得到 (√3 + 1)/2。如果分母展开不正确，或最终答案未化简，都会造成不必要的失分。
2. Solving Equations: Missing Solutions and Incorrect Methods | 解方程：漏解和方法错误
A significant number of candidates squared both sides of an equation without considering the implications. For example, solving √x + 3 = x – 3 by squaring gives x + 3 = (x – 3)², which simplifies to x² – ed to check for extraneous roots. Substituting x = 1 back into the original equation gives √4 = 1 – which is false. Only x = 6 is valid. Always verify solutions when squaring has been used.
相当多的考生在方程两边平方时没有考虑其影响。例如，解 √x + 3 = x – 3 时，通过平方得到 x + 3 = (x – 3)²，进而化简为 x² – 7x + 6 = 0。许多人随后正确地找到了 x = 1 和 x = 6，但没有检查增根。将 x = 1 代回原方程得 √4 = 1 – 3，这是不成立的。只有 x = 6 是有效解。每当使用平方法时，一定要验证所得解。
ded by a variable instead of factorising, thereby losing a root. For instance, solving 2x² – 5x = 0 by dividing both sides by x gives 2x – 5 = 0, yielding only x = 2.5. The correct method is to factorise: x(2x – Never divide by an expression that could be zero unless you explicitly consider that case.
在一元二次方程中，考生经常用变量去除方程而使根丢失。例如，解 2x² – 5x = 0 时，两边同除以 x 得到 2x – 5 = 0，只得出 x = 2.5。正确的方法是因式分解：x(2x – 5) = 0，因此 x = 0 或 x = 2.5。除非明确考虑了可能为零的情况，否则绝不要除以一个可能为零的表达式。
3. Functions and Graph Transformations | 函数与图像变换
incorrectly described a horizontal stretch by factor 1/2 followed by a translation, but got the order or direction wrong. The correct sequence is to rewrite f(2x + 4) as f(2(x + 2)), which represents a translation of 2 units to the left, followed by a horizontal stretch by factor 1/2. Reversing the order changes the outcome.
图像变换是常见的混淆点。当要求描述从 y = f(x) 到 y = f(2x + 4) 的变换时，许多考生错误地描述为先水平拉伸到 1/2 倍再平移，但弄错了顺序或方向。正确的做法是将 f(2x + 4) 改写为 f(2(x + 2))，这表示先向左平移 2 个单位，然后进行水平拉伸，压缩到原来的 1/2 倍。颠倒顺序会改变结果。
Inverse functions were also mishandled. To find the inverse of f(x) = (2x + 1)/(x – then make algebraic slips when rearranging. A robust approach is to write y = (2x + 1)/(x – 3), then multiply both sides by (x – 3) to get y(x – domain and range of the inverse must be stated correctly; many forgot to give the domain.
反函数也常被错误处理。求 f(x) = (2x + 1)/(x – 3) 的反函数时，考生会交换 x 和 y，但之后整理时出现代数错误。可靠的方法是先写出 y = (2x + 1)/(x – 3)，两边乘以 (x – 3) 得 y(x – 3) = 2x + 1，展开并整理项，最终解出用 y 表示 x 的表达式。交换变量后，必须正确写出反函数的定义域和值域；许多人忘记了给出定义域。
4. Differentiation: Tangents, Normals, and Notation | 微分：切线、法线和符号表达
The differentiation questions were generally well attempted, but the main errors occurred when finding equations of tangents and normals. After calculating the derivative dy/dx at the given point to find the gradient of the tangent, candidates often used the same gradient for the normal, forgetting that the normal gradient is the negative reciprocal. For a tangent gradient m, the normal gradient is –1/m, provided m ≠ 0.
微分题的解答总体较好，但主要在求切线和法线方程时出现错误。在计算出给定点的导数 dy/dx 得到切线斜率后，考生经常用相同的斜率来求法线，忘记了法线斜率应为负倒数。若切线斜率为 m，法线斜率则为 –1/m（前提是 m ≠ 0）。
ect, but writing dy/dx (x² + 5) is meaningless unless y is defined. Similarly, confusing dy/dx with Δy/Δx in explanatory text lost marks for clarity. Practice using precise mathematical language.
考官特别指出了符号的误用。写成 d/dx (x² + 5) = 2x 是正确的，但如果写成 dy/dx (x² + 5) 则是无意义的，除非先定义了 y。类似地，在解释性文字中将 dy/dx 与 Δy/Δx 混淆会导致表达不清晰而被扣分。请练习使用精确的数学语言。
5. Integration and Area Calculations | 积分与面积计算
Indefinite integration was mostly fine, but candidates frequently forgot the constant of integration '+ c'. In a question that required finding the equation of a curve from its derivative, missing the '+ c' meant the subsequent substitution of coordinates could not yield the correct constant. Always write '+ c' after every indefinite integral unless the question involves limits.
不定积分大多做得不错，但考生经常忘记积分常数 '+ c'。在需要根据导数求曲线方程的题目中，遗漏 '+ c' 意味着后续代入坐标无法求出正确的常数。每次求不定积分时，都要记得加上 '+ c'，除非题目含有积分上下限。
Definite integration problems involving area between a curve and the x‑axis revealed a conceptual gap. Many candidates simply integrated the function between the given limits without first determining whether the curve crossed the x‑axis within the interval. If the function changes sign, the total area is the sum of the absolute values of the separate integrals. Submitting a negative total area was also penalised.
涉及曲线与 x 轴之间面积的定积分问题暴露了一个概念上的漏洞。许多考生只是简单地在给定区间内对函数积分，而没有先判断曲线在此区间内是否穿越 x 轴。如果函数变号，总面积应该是各段积分绝对值的总和。提交负的总面积也会被扣分。
6. Trigonometric Equations and Identities | 三角方程与恒等式
nometric identities were not always applied correctly. When solving something like 2 sin²θ – cos θ = 1, many attempted to replace sin²θ with 1 – cos²θ, which is correct, but then mis‑solved the resulting quadratic in cos θ. The key steps are: substitute to get 2(1 – cos²θ) – cos θ = 1, rearrange to 2 cos²θ + cos θ – s often made sign errors or forgot to consider all solutions within the required interval.
三角恒等式的应用并不总是正确的。当解像 2 sin²θ – cos θ = 1 这样的方程时，许多人试图用 1 – cos²θ 替换 sin²θ，这样做是正确的，但随后求解关于 cos θ 的二次方程时却出了错。关键步骤是：代换得到 2(1 – cos²θ) – cos θ = 1，整理得 2 cos²θ + cos θ – 1 = 0，然后解出 cos θ。考生常常出现符号错误，或忘记在所给区间内找出所有解。
Another common oversight was giving solutions in degrees when radians were required, or vice versa. The specification often states the mode, but if the question uses π, you must work in radians. Always check the domain and the mode specified in the question.
另一个常见疏忽是题目要求用弧度制时却给出了角度制的解，或反之。考试说明通常会指明模式，但如果题目中使用了 π，那么就必须使用弧度制。请务必检查题目中指定的定义域和角度模式。
7. Proof and Mathematical Reasoning | 证明与数学推理
enge candidates. When asked to prove that the square of any odd integer is odd, many wrote 'let the odd integer be 2n + 1' ich is correct, but then expanded (2n + 1)² = 4n² + 4n + 1 = 2(2n² + 2n) + 1 and concluded 'therefore odd' plete proof must include a statement that the expression inside the bracket is an integer.
证明题仍然挑战着考生。当要求证明任意奇数的平方仍为奇数时，许多考生设奇数为 2n + 1，这是正确的，但接着展开 (2n + 1)² = 4n² + 4n + 1 = 2(2n² + 2n) + 1，并得出结论“因此是奇数”。然而，有些人没有说明 2n² + 2n 是一个整数，这对于严密的证明至关重要。一个完整的证明必须包含括号内的表达式为整数的陈述。
es supplied an example instead of a general deductive argument, or gave an incorrect counterexample. For instance, to disprove 'n² + n + 1 is prime for all integers n' ives 21, not prime). A single valid counterexample suffices to disprove a universal claim.
在演绎证明和反例证伪的题目中，考生有时会用一个例子代替一般的演绎论证，或者给出了错误的反例。例如，要证伪“对任意整数 n，n² + n + 1 都是质数”这一命题，只需给出 n = 4（得出 21，非质数）即可。一个有效的反例就足以证伪全称命题。
8. Inequalities and Set Notation | 不等式与集合符号
Solving quadratic inequalities such as x² – the answer x ≥ 3 and x ≤ 2, which is correct, but many wrote it incorrectly using the word 'and' instead of 'or'. The correct set notation is {x : x ≤ 2} ∪ {x : x ≥ 3}. Using 'and' implies the intersection, which is empty. Small notational slips like this cost marks.
求解如 x² – 5x + 6 ≥ 0 的二次不等式时，答案 x ≥ 3 和 x ≤ 2 常常是对的，但许多人用了错误的连接词“且”而不是“或”。正确的集合表示是 {x : x ≤ 2} ∪ {x : x ≥ 3}。使用“且”意味着交集，而这是空集。这类小的符号错误也会被扣分。
le, while ≤ or ≥ requires a closed circle. Emphasising these details will help you present answers clearly and earn all the marks.
在数轴上表示解时，空心圈和实心圈偶尔会被用错。严格不等式（> 或 <）需要用空心圈，而 ≤ 或 ≥ 则需要实心圈。注意这些细节有助于清晰地呈现答案，拿到所有分数。
9. Coordinate Geometry: Lines and Circles | 坐标几何：直线与圆
used the gradient of the original line segment instead of the negative reciprocal. The perpendicular bisector must be perpendicular to the original line, so its gradient is the negative reciprocal of the segment's gradient. After that, use the midpoint in y – y₁ = m(x – x₁).
在一道要求找出中垂线方程的题中，许多考生正确地求出了两点间线段的中点，但随后却使用了原线段的斜率，而不是负倒数。中垂线必须垂直于原线段，因此其斜率应是原线段斜率的负倒数。然后，将中点代入 y – y₁ = m(x – x₁) 即可。
equently required completing the square to find centre and radius. When faced with x² + y² – 6x + 10y + 9 = 0, candidates would often mismanage the signs. The correct grouping is (x² – 6x) + (y² + 10y) = –9, then complete the square to get (x – 3)² – 9 + (y + 5)² – 25 = –9, leading to (x – 3)² + (y + 5)² = 25. The centre is (3, –5), radius 5. Slips in signs for the centre were extremely common.
圆的几何题经常要求通过配方法求圆心和半径。面对 x² + y² – 6x + 10y + 9 = 0 这样的方程时，考生的符号处理经常出错。正确的分组是 (x² – 6x) + (y² + 10y) = –9，然后配方得到 (x – 3)² – 9 + (y + 5)² – 25 = –9，从而得出 (x – 3)² + (y + 5)² = 25。圆心为 (3, –5)，半径为 5。圆心的符号错误极为常见。
10. Common Arithmetic and Calculator Mistakes | 常见算术和计算器错误
e often dropped, brackets misused, or BIDMAS/BODMAS rules ignored. For example, evaluating –3² was frequently interpreted as 9 instead of – emind yourself that powers apply before the negative sign unless parentheses group them: (–3)² = 9, but –3² = –9.
即使是简单的算术问题，在时间压力下也会出现错误。负号经常被遗漏、括号使用不当，或者忽略 BIDMAS/BODMAS 运算次序。例如，计算 –3² 经常被误解为 9 而不是 –9。请记住，如果没有括号括起，幂运算优先于负号：(–3)² = 9，但 –3² = –9。
Calculator misuse also appeared. Relying on the calculator to handle fractions or trigonometric evaluations without verifying the mode (degrees/radians) led to wrong answers. It is wise to do a rough mental estimate to check the reasonableness of calculator output.
计算器使用不当的问题也出现了。依赖计算器处理分数或三角计算，却没有核实模式（角度/弧度），会导致错误答案。明智的做法是先进行一次粗略的心算，以检验计算器输出的合理性。
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