📚 AS Maths Unit 2 (Jan 2021) Common Mistakes Summary | AS数学第二单元（2021年1月）易错点总结
The January 2021 AS Mathematics Unit 2 paper tested a wide range of pure mathematics skills, including functions, logarithms, trigonometry, differentiation and integration. Despite the straightforward structure of many questions, examiners reported recurring errors that prevented candidates from securing top marks. This article identifies the ten most common mistakes and explains how to correct them, helping you refine your technique for future assessments.
2021年1月的AS数学第二单元试卷考查了函数、对数、三角学、微分与积分等众多纯数技能。尽管许多题目结构清晰，但阅卷官反馈学生反复出现一些本可避免的错误，导致丢分。本文总结十大高频易错点并给出纠正方法，帮助你优化应试技巧。
1. Misreading Domain and Range in Functions | 函数定义域与值域的误读
When forming composite functions such as f(g(x)), candidates often forget that the output of g(x) must lie within the domain of f. For example, if f(x) = √(x – 2) and g(x) = 1/x, the composite f(g(x)) = √(1/x – 2) requires 1/x – 2 ≥ 0, leading to 0 < x ≤ 1/2. Many students only wrote x ≠ 0 and lost marks for incomplete domain reasoning.
在构造复合函数如 f(g(x)) 时，考生常忽略 g(x) 的输出必须落在 f 的定义域内。例如，若 f(x) = √(x – 2) 且 g(x) = 1/x，则复合函数 f(g(x)) = √(1/x – 2) 要求 1/x – 2 ≥ 0，解得 0 < x ≤ 1/2。不少学生只写了 x ≠ 0，未能完整分析定义域而失分。
Domain of f(g(x)): 1/x – 2 ≥ 0 ⇒ 0 < x ≤ 1/2
2. Incorrect Use of Logarithm Laws | 对数运算律的错误使用
A common error is to treat log(a – b) as log a – log b, or to mistakenly believe that log a / log b equals log(a/b). In the equation log₂(x + 1) – log₂(x – 1) = 3, the correct first step is to combine the logs: log₂((x + 1)/(x – 1)) = 3. Students who wrote log₂(x + 1) / log₂(x – 1) = 3 applied a non-existent rule and reached an unsolvable form.
常见错误是将 log(a – b) 当作 log a – log b，或误以为 log a / log b 等于 log(a/b)。在方程 log₂(x + 1) – log₂(x – 1) = 3 中，正确的第一步应合并为 log₂((x + 1)/(x – 1)) = 3。有学生直接写成 log₂(x + 1) / log₂(x – 1) = 3，使用了一条并不存在的运算律，导致方程无法求解。
Correct: log₂((x + 1)/(x – 1)) = 3 ⇒ (x + 1)/(x – 1) = 2³ = 8
3. Sketching Trig Graphs Without Key Features | 绘制三角图像时缺失关键特征
Questions that ask for a sketch of y = a cos(bθ + c) + d or y = a sin(bθ + c) + d require clear labelling of amplitude, period, vertical shift and the positions of maximum and minimum points. A typical mistake is to mark the maximum value as |a| but forget to add d, or to mislabel the period as 360°/b for degrees without adjusting for the coefficient inside the bracket. For y = 2 cos(3θ – 60°) + 1, the amplitude is 2, the period is 120°, the graph is shifted up by 1 and the first maximum occurs where 3θ – 60° = 0° ⇒ θ = 20°.
要求绘制 y = a cos(bθ + c) + d 或 y = a sin(bθ + c) + d 的图像时，必须清晰标出振幅、周期、垂直平移以及最大值和最小值的坐标。典型错误是只标记振幅为 |a| 却忘记加上 d，或者错把周期写成 360°/b 但忽略了括号内系数的影响。对于 y = 2 cos(3θ – 60°) + 1，振幅为2，周期为120°，图像上移1个单位，第一个最大值出现在 3θ – 60° = 0° ⇒ θ = 20° 处。
4. Errors in Binomial Expansion Validity | 二项展开有效性的错误
When expanding expressions like (1 + 4x)^(3/2) using the binomial theorem, students often calculate coefficients correctly but neglect to state the range of validity |4x| < 1, i.e. |x| < 1/4. Some also forget that the expansion is valid only for |x| < 1/4, and they substitute x = 0.5 into the series, yielding an inaccurate approximation. Another slip is misapplying the formula for the coefficient of x², forgetting to multiply by the correct factorial denominator.
用二项式定理展开如 (1 + 4x)^(3/2) 时，学生常能正确计算系数，却忘记注明有效性范围 |4x| < 1，即 |x| < 1/4。有些人甚至忽略展开只对 |x| < 1/4 有效，直接将 x = 0.5 代入级数，得到不精确的近似值。另一个常见失误是错误使用 x² 系数公式，忘记除以相应的阶乘分母。
(1 + 4x)^(3/2) = 1 + 6x + (3/2)(1/2)(4x)²/2! + … valid for |x| < 1/4
5. Solving Trigonometric Equations in a Given Interval | 在给定区间解三角方程时漏解
When solving 2 sin²θ – sin θ – 1 = 0 for 0° ≤ θ ≤ 360°, pupils often factorise correctly to (2 sin θ + 1)(sin θ – 1) = 0, giving sin θ = -1/2 and sin θ = 1. The error arises when they solve sin θ = 1 and only give θ = 90°, forgetting that 90° is indeed the only solution in this interval for that branch, yet they sometimes also discard sin θ = -1/2 solutions like θ = 210°, 330° due to misreading CAST or graphs. Always list all values using the periodicity of sine.
解方程 2 sin²θ – sin θ – 1 = 0，区间 0° ≤ θ ≤ 360°，学生通常能正确因式分解为 (2 sin θ + 1)(sin θ – 1) = 0，得到 sin θ = -1/2 和 sin θ = 1。错误发生在处理 sin θ = 1 时只给出 θ = 90°，然后误以为该支只有一个解，或者在求解 sin θ = -1/2 时漏掉了 210° 或 330°，原因是未能正确运用 CAST 图或正弦的周期性。务必列出所有满足条件的解。
sin θ = 1 ⇒ θ = 90°; sin θ = -1/2 ⇒ θ = 210°, 330°
6. Differentiation of Composite Functions | 复合函数求导的链式法则应用错误
The chain rule is frequently misapplied when differentiating powers of a bracket. For y = (3x² – 5)⁴, the outer derivative is 4(3x² – 5)³ and the inner derivative is 6x, giving dy/dx = 4(3x² – 5)³ × 6x. A common mistake is to forget the factor 6x entirely, writing only 4(3x² – 5)³, or to incorrectly multiply the constants inside the bracket. Another pitfall is failing to recognise that expressions like √(2x + 1) should first be written as (2x + 1)^(1/2) before differentiating.
在求导带幂次的括号时，链式法则常被误用。对 y = (3x² – 5)⁴，外部导数为 4(3x² – 5)³，内部导数为 6x，因此 dy/dx = 4(3x² – 5)³ × 6x。常见错误是完全遗忘乘以内层导数 6x，只写下 4(3x² – 5)³，或者错误地处理括号内的常数系数。另一个容易失分之处是忘记先将形如 √(2x + 1) 的表达式写成 (2x + 1)^(1/2) 再利用链式法则求导。
y = (3x² – 5)⁴ ⇒ dy/dx = 4(3x² – 5)³ · 6x
7. Integration Involving Substitution Limits | 换元积分法定限错误
When a definite integral is evaluated by substitution, the limits of integration must be converted to the new variable. For ∫₀¹ 2x/(x² + 1) dx, using u = x² + 1 gives du = 2x dx, and x = 0 → u = 1, x = 1 → u = 2. The integral becomes ∫₁² 1/u du = [ln|u|]₁² = ln 2. Many candidates forget to change limits, leaving the integral as ∫₀¹ and wrongly applying the antiderivative in terms of u with x‑limits, which yields nonsense or the wrong answer. Always rewrite limits clearly on substitution.
用换元法计算定积分时，必须将积分上下限转换为新变量的对应值。例如 ∫₀¹ 2x/(x² + 1) dx，令 u = x² + 1，则 du = 2x dx，且 x = 0 → u = 1，x = 1 → u = 2。积分变为 ∫₁² 1/u du = [ln|u|]₁² = ln 2。不少考生忘记转换上下限，仍保留 ∫₀¹，然后对 u 使用原函数却套用 x 的界限，导致结果无意义或错误。每次换元都应明确标出新限。
∫₀¹ 2x/(x² + 1) dx = ∫₁² 1/u du = ln 2
8. Handling Proof by Deduction | 演绎推理证明的逻辑缺失
Proof questions, such as ‘prove that for any integer n, n² – n is even’, demand a logical chain of reasoning. The correct approach is to factorise as n(n – 1), note that the product of two consecutive integers is always even, and hence conclude. A weak answer states ‘n² – n is even because n² and n have the same parity’ without further justification, which does not earn full marks. Always present clear algebraic steps and articulate why the conclusion follows.
证明题如“证明对任意整数 n，n² – n 为偶数”，需要展示清晰的逻辑链。正确做法是因式分解为 n(n – 1)，注意到两个连续整数的积恒为偶数，然后得出结论。低分答案往往只说“n² – n 是偶数因为 n² 与 n 同奇偶”，却没有进一步论证，无法获得全部分数。务必写出明确的代数步骤，并解释结论为何成立。
n² – n = n(n – 1) → product of consecutive integers → always even
9. Misinterpreting the Meaning of ‘Increasing’ Functions | 对“递增函数”含义的误解
The concept of an increasing function is often confused with having a positive derivative at every point. While a strictly increasing function has f'(x) > 0 for all x in an interval, a function can still be increasing if f'(x) ≥ 0 and the derivative is zero only at isolated points. In interval questions, students sometimes incorrectly exclude endpoints where f'(x) = 0, or they write x > a instead of x ≥ a. Always check the definition used in the specification: a function is increasing on an interval if for any x₁ < x₂, f(x₁) ≤ f(x₂).
递增函数的概念常与导数处处为正相混淆。严格递增函数满足区间内 f'(x) > 0，然而若 f'(x) ≥ 0 且只在孤立点处等于零，函数仍可视为递增。在求递增区间的问题中，学生有时错误地排除导数等于零的端点，或写成 x > a 而非 x ≥ a。建议始终对照考纲定义：若对区间内任意 x₁ < x₂ 均有 f(x₁) ≤ f(x₂)，则函数在该区间上递增。
10. Algebraic Manipulation in Circle Geometry | 圆几何中的代数变形错误
Finding the centre and radius of a circle from x² + y² – 4x + 6y – 12 = 0 requires completing the square: (x – 2)² – 4 + (y + 3)² – 9 – 12 = 0 → (x – 2)² + (y + 3)² = 25. A frequent slip is to mishandle the constant terms, writing +12 instead of -12 after moving the constant, or making sign errors when halving the coefficients of x and y. Another error occurs when finding the equation of a tangent or chord: students substitute correctly but then expand brackets incorrectly, losing the simplicity of the gradient relationship.
由 x² + y² – 4x + 6y – 12 = 0 求圆心和半径需配方：(x – 2)² – 4 + (y + 3)² – 9 – 12 = 0 → (x – 2)² + (y + 3)² = 25。常见失误是处理常数项时出错，移项后误写成 +12 而非 -12，或在取 x、y 系数一半时出现正负号错误。另一个错误发生在求切线或弦的方程时：学生代入正确但展开括号时计算出错，破坏了斜率的简洁关系。
(x – 2)² + (y + 3)² = 5² ⇒ centre (2, -3), radius 5
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