📚 IB & AQA Maths: Common Pitfalls Explained | IB AQA 数学：易错题精讲
Mathematical mistakes in IB and AQA examinations rarely come from a lack of ability. They almost always come from a small number of repeated conceptual slips that are easy to fix once they are named. This article walks through twelve of the most expensive pitfalls in the two specifications - illegal cancelling, lost trigonometric solutions, hidden quadratics, logarithm domains, differentiation and integration traps, vectors, complex numbers, probability, functions, sequences and exam technique - and shows the correct method for each one.
在 IB 与 AQA 数学考试中，失分往往并非因为能力不足，而是源于少数几个反复出现的概念性疏漏。只要把这些疏漏点一一说清楚，修正起来其实很快。本文逐一梳理两个考纲中最"昂贵"的十二个易错点——非法约分、三角方程漏解、隐藏二次方程、对数定义域、求导与积分的陷阱、向量、复数、概率、函数、数列以及应试技巧——并给出每个知识点的正确解法。
1. Cancelling Without Factoring | 忘记因式分解直接约分
The classic question asks you to simplify the expression (x² − 4)/(x − 2). It appears in both IB and AQA papers, sometimes as a warm-up, sometimes as the first line of a larger proof.
经典题目要求化简表达式 (x² − 4)/(x − 2)。它在 IB 和 AQA 试卷中都出现过，有时作为热身题，有时作为一道大题的第一步。
merator can be split as x² divided by x minus 4 divided by 2, which is not a legal operation. You cannot divide a sum term by term like that: (a + b)/c is not equal to a/c + b/a in general.
常见错误是直接约去 x 项，写成 (x² − 4)/(x − 2) = x − 2。这种做法把分子当作 x² 除以 x 再减去 4 除以 2，但这是不合法的：不能把和式的各项分别相除，(a + b)/c 一般不等于 a/c + b/c 的随意拆分。
The fraction then becomes (x − 2)(x + 2)/(x − 2). Provided x ≠ 2, the common factor (x − 2) cancels to leave x + 2.
正确做法是先因式分解。分子是平方差：x² − 4 = (x − 2)(x + 2)。于是分式变为 (x − 2)(x + 2)/(x − 2)。只要 x ≠ 2，即可约去公因式 (x − 2)，结果为 x + 2。
x² − 4 = (x − 2)(x + 2), x ≠ 2 ⇒ (x² − 4)/(x − 2) = x + 2
Two marks are typically at stake: one for the factorisation and one for the restriction. Always state x ≠ 2, because the original expression is undefined at that value, even though the simplified form is perfectly well defined there.
通常有两分在这里：一分给因式分解，一分给限制条件。务必写明 x ≠ 2，因为原式在该点无定义，尽管化简后的式子在 x = 2 处取值正常。
ntical expressions must agree at every value in the shared domain, so testing one value catches most cancelling errors immediately.
一个实用习惯是代入数字检验。取 x = 3：原式得 (9 − 4)/(3 − 2) = 5，化简式得 3 + 2 = 5。恒等的表达式在公共定义域内每个取值上都必须相等，因此随手代入一个数就能立刻发现大多数错误约分。
2. Losing Solutions in Trigonometric Equations | 解三角方程漏解
Problem: solve sin 2x = 1/2 for 0 ≤ x ≤ 2π.
问题：在 0 ≤ x ≤ 2π 范围内解方程 sin 2x = 1/2。
θ = 1/2 has infinitely many solutions, and the interval for x corresponds to an interval for 2x that is twice as wide, allowing two full revolutions.
陷阱在于只写出 2x = π/6 与 2x = 5π/6，除以 2 得到 x = π/12 和 x = 5π/12 就停笔。由于正弦函数周期为 2π，方程 sin θ = 1/2 有无穷多解；而 x 的区间宽度相对于 2x 的区间是被放大两倍的，因此可以转满两整圈。
boundary of the x-interval into a boundary for 2x. Since 0 ≤ x ≤ 2π, we have 0 ≤ 2x ≤ 4π.
正确方法：先写出角 2x 的通解，再把 x 的区间端点转换成 2x 的区间端点。由 0 ≤ x ≤ 2π 得 0 ≤ 2x ≤ 4π。
2x = π/6 + 2kπ or 2x = 5π/6 + 2kπ, k ∈ ℤ
12 + kπ. Now select the integers k that place x inside [0, 2π]. Remember that 2π = 24π/12.
两边除以 2 得 x = π/12 + kπ 或 x = 5π/12 + kπ。再选出使 x 落在 [0, 2π] 内的整数 k。注意 2π = 24π/12。
| k | x = π/12 + kπ | x = 5π/12 + kπ |
| 0 | π/12 ≈ 0.26 | 5π/12 ≈ 1.31 |
| 1 | 13π/12 ≈ 3.40 | 17π/12 ≈ 4.45 |
| 2 | 25π/12 > 2π ✗ | 29π/12 > 2π ✗ |
, rewrite the range as 0 ≤ θ ≤ 4π, solve sin θ = 1/2 completely within that range, and only at the very end convert back to x. This makes the extra revolution obvious rather than hidden.
因此完整解集为 x = π/12、5π/12、13π/12、17π/12，共四个解，而不是两个。另一种更稳妥的方法是换元：令 θ = 2x，把范围改写为 0 ≤ θ ≤ 4π，在该区间内完整解出 sin θ = 1/2，最后才换回 x。这样多出的一整圈就一目了然，不会藏起来。
Two related rules are worth memorising. Equations in sin θ or cos θ have period 2π, so add 2kπ. Equations in tan θ have period π, so add kπ. When the argument is doubled or halved, scale the interval first, solve second, and convert back last.
还有两条相关规则值得记住：含 sin θ 或 cos θ 的方程周期为 2π，加 2kπ；含 tan θ 的方程周期为 π，加 kπ。当角度被乘以 2 或除以 2 时，应先把区间按比例
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