📚 Mastering TMUA Pacing: 20 Application Questions in 75 Minutes | TMUA 七十五分钟内稳定处理 20 道应用题的节奏控制
The TMUA Paper 1 (Applications of Mathematical Knowledge) hands you twenty multiple-choice questions and seventy-five minutes, with no calculator and no formula booklet. Most candidates who underperform do not fail because they cannot do the mathematics; they fail because they spend nine minutes on question 7 and then meet question 13 with four minutes left. Pacing is not a soft skill bolted onto the maths – it is a scoring skill in its own right, and it can be trained with the same discipline as a trigonometric identity.
TMUA 的 Paper 1（Applications of Mathematical Knowledge，数学知识应用）给你 20 道选择题和 75 分钟，不允许使用计算器，也没有公式表。大多数发挥失常的考生并不是不会做题，而是把 9 分钟耗在了第 7 题上，然后在还剩 4 分钟时遇到第 13 题。节奏控制不是附加在数学能力之外的”软技能”，它本身就是一项得分技能，并且可以像训练三角恒等式一样被系统训练。
1. What Paper 1 Actually Asks You to Do | Paper 1 究竟要求你做什么
Paper 1 is drawn from the core of A-level or IB Higher Level mathematics: algebra and functions, sequences and series, coordinate geometry, trigonometry, exponentials and logarithms, differentiation, integration, graph sketching, and elementary probability and statistics. Nothing on the paper requires knowledge outside that syllabus. What it does require is method selection under time pressure – recognising, within a few seconds, which standard technique a question has been dressed up to hide.
Paper 1 的考点全部来自 A-level 或 IB HL 数学的核心内容：代数与函数、数列与级数、解析几何、三角、指数与对数、微分、积分、函数图像，以及基础概率统计。试卷不考查超纲知识，它真正考查的是在时间压力下的方法选择能力——在几秒之内识别出这道题”伪装”的是哪一个标准技巧。
Paper 2 (Mathematical Reasoning) is a different exam: logic, proof, necessary and sufficient conditions, and error-spotting. Its rhythm is different because its questions are often fast to read but slow to verify. Everything in this article concerns Paper 1, where the bottleneck is calculation and decision time.
Paper 2（Mathematical Reasoning，数学推理）是另一套卷子：逻辑、证明、充分必要条件、找错。它的节奏不同——题目往往读得快但验证慢。本文讨论的全部是 Paper 1，其瓶颈在于计算与决策时间。
2. The Arithmetic of 75 Minutes | 75 分钟的算术
Start with the honest numbers. Seventy-five minutes is 4500 seconds; divided by twenty questions that is 225 seconds, or 3 minutes 45 seconds per question. Allow roughly 15 seconds per question for reading the stem, scanning the five options and transferring an answer, and the effective working time drops to about 3 minutes 30 seconds. That is your true budget – and it is an average, not a plan.
先算清最基本的账：75 分钟 = 4500 秒，除以 20 道题，平均每题 225 秒，即 3 分 45 秒。若每题再扣除约 15 秒用于读题、扫选项和填涂答案，实际可用的解题时间约为 3 分 30 秒。这是你的真实预算，但它是平均值，不是方案。
75 min × 60 s ÷ 20 questions = 225 s = 3 min 45 s per question
Two consequences follow immediately. First, a uniform 3-minute-45-second rule is a trap: some questions genuinely take 45 seconds, and others genuinely take 8 minutes. Second, since TMUA awards marks for correct answers only and applies no penalty for wrong ones, an unanswered question is a guaranteed zero while a guessed one has a real, positive expected value. You must never walk out with a blank.
由此立刻得到两个结论。第一，”每题平均 3 分 45 秒”是一个陷阱：有些题真的只要 45 秒，有些题真的需要 8 分钟。第二，TMUA 只对正确答案给分，答错不扣分，所以空题等于必得 0 分，而猜一个选项的期望值是正的。绝对不要交白卷式的空题。
3. The Three-Pass System | 三轮答题法
The single most reliable structure for Paper 1 is a three-pass sweep with a fixed checkpoint clock, followed by a short verification window. In Pass 1 you harvest everything cheap; in Pass 2 you work the medium questions with a real time allowance; in Pass 3 you attack the stubborn ones with elimination and educated guessing. The critical rule is that no question may consume more than 90 seconds in Pass 1.
Paper 1 最可靠的结构是”三轮扫描 + 固定检查点时钟”，最后加一个简短的核对窗口。第一轮收割所有便宜分；第二轮用充足的时间预算处理中等题；第三轮用排除法和合理猜测强攻顽固题。最关键的一条规则是：第一轮中任何一道题都不得超过 90 秒。
- Pass 1 (0-30 min): Answer every question you can finish in under 90 seconds. Do not verify. Flag everything else and move on. Target: 11-13 answers banked. | 第一轮（0-30 分钟）：凡 90 秒内能完成的题立刻作答，不要验算，其余全部标记后跳过。目标：已确定 11-13 题。
- Pass 2 (30-58 min): Return to flagged questions. These now deserve 3 to 4 minutes each, because the easy material has been cleared. Target: 5-6 more answers. | 第二轮（30-58 分钟）：回到标记题。由于简单题已清空，这些题现在值得每道 3-4 分钟。目标：再拿下 5-6 题。
- Pass 3 (58-70 min): For anything still unsolved, use structural elimination to cut the five options to two, then choose. Never leave the last stage empty. | 第三轮（58-70 分钟）：对仍未解决的题，用结构性排除把 5 个选项缩到 2 个再选。最后阶段绝不能留空。
- Pass 4 (70-75 min): Verification. Confirm no blanks, confirm answers were transferred correctly, and re-read only the questions you flagged as “high risk”. | 第四轮（70-75 分钟）：核对。确认没有空题、答案填涂正确，只重读你标记为”高风险”的题目。
4. Question Tiering: Cost Against Value | 题目分级：认知成本与分值
Every TMUA question is worth exactly one mark, but the questions do not cost the same. Treat each question as an investment with a known return – one mark – and an unknown price in seconds. Your job is to buy the cheapest marks first. In practice, questions fall into three cost tiers, and the distribution is roughly stable across past papers.
TMUA 每题分值完全相同（1 分），但”成本”天差地别。把每道题看成一笔投资：回报已知（1 分），价格未知（秒数）。你的任务就是先买最便宜的分数。实际做题时，题目大致分为三个成本层级，这个分布在不同年份的真题中相当稳定。
| Tier 层级 | Typical cost 典型耗时 | Expected count 预期数量 | Total time 时间合计 |
|---|---|---|---|
| A – routine 常规 | 45-90 s | 9-11 | ~15 min |
| B – structured 结构化 | 2-4 min | 6-8 | ~24 min |
| C – multi-stage 多阶段 | 5-7 min | 2-4 | ~17 min |
The arithmetic of tiering is what makes the strategy work. In Tier A you buy a mark for roughly 80 seconds; in Tier C you buy the same mark for 360 seconds or more. Four and a half times the price, identical value. A candidate who spends nine minutes trapped in one Tier C question has effectively paid for one mark with the currency of two or three others.
分级的意义就在于这笔算术：A 级题用约 80 秒换 1 分，C 级题要用 360 秒以上换同样 1 分，价格差 4 倍以上，价值完全相同。一个在 C 级题上被困 9 分钟的考生，实际上是用两三分的机会成本去换 1 分。
5. The 30-Second Diagnostic | 30 秒诊断法
You cannot tier a question until you have read it, so the diagnostic has to be fast. Read the stem once, without writing, and ask a single question: how many separate steps separate the given data from the required answer? One step is Tier A, two or three steps is Tier B, and “I cannot yet see the path” is Tier C by definition. Then glance at the answer options for a second signal.
分级的前提是读题，所以诊断必须快。不落笔先读一遍题干，只问一个问题：从已知到所求之间隔着几步？一步是 A 级，两三步是 B 级，”我还看不到路径”按定义就是 C 级。然后扫一眼选项，寻找第二个信号。
- Tier A signals: single substitution; standard formula application; a graph read-off; direct differentiation or integration of a polynomial; simple ratio or percentage. | A 级信号：一步代入；套标准公式；读图取点；多项式直接求导或积分；简单比例或百分比。
- Tier B signals: one intermediate unknown to find first; a system of two equations; setting up an equation from a word problem; a definite integral requiring a substitution or expansion. | B 级信号：需要先求一个中间量；两个方程的方程组；由应用题列方程；需要换元或展开的定积分。
- Tier C signals: parameters such as a, k or θ that must be handled symbolically; absolute values or inequalities; “for all x” phrasing; nested fractions or surds; more than one constraint in a word problem; a function defined piecewise. | C 级信号：出现 a、k、θ 等参数需要符号处理；绝对值或不等式；”对所有 x”的表述；嵌套分式或根式；应用题含多个约束；分段函数。
The diagnostic should take no more than 30 seconds. If reading a question twice has still not revealed a method, that is not a failure of intelligence – it is information. The question is telling you it belongs in Pass 3, and you should believe it immediately.
整个诊断不应超过 30 秒。如果读了两遍仍看不出方法，这不是智力问题，而是信息：这道题在告诉你它属于第三轮，你应当立刻相信这个信号。
6. Triage Rules and the Sunk-Cost Trap | 止损规则与沉没成本陷阱
The hardest discipline in the exam is abandoning work you have already invested. Human beings are loss-averse: after three minutes of algebra you feel obliged to finish, even when the route is clearly collapsing. Set two hard checkpoints before the exam begins and obey them mechanically, without renegotiating.
考场上最难的自律，是放弃已经投入的工作。人是厌恶损失的：算了三分钟之后你会觉得”必须做完”，即使路径明显已经崩塌。解决办法是在考前就设定两个硬性检查点，考场上机械执行，不做二次谈判。
Checkpoint 1: 90 seconds – no visible path means move on. Checkpoint 2: 3 min 30 s – no answer in sight means guess, flag, leave.
- The 90-second rule: applies in Pass 1 only. It exists to protect the cheap marks behind the current question. | 90 秒规则：仅适用于第一轮，目的是保护后面那些便宜分。
- The 3 min 30 s rule: applies in Pass 2. If you are past this and have neither a value nor a two-option shortlist, bank a guess. | 3 分 30 秒规则：适用于第二轮。超过这个时间既没算出结果也没缩到两个选项，就先猜一个。
- The recovery rule: after abandoning a question, take one deliberate breath and read the next stem twice. Panic carried forward is more expensive than the question you just left. | 恢复规则：放弃一题后，有意识地呼吸一次，再把下一题读两遍。把慌乱带进下一题，比丢掉的那道题代价更大。
- The no-return rule: once a question is flagged and abandoned, it is answered only in Pass 3, never earlier. | 不回头规则：被标记放弃的题只在第三轮处理，绝不在中途回头看。
7. Topic-by-Topic Speed Skills | 分题型提速技巧
Speed is not generic; it is topic-specific. Each syllabus area has two or three standard manipulations that convert a two-minute derivation into a fifteen-second recognition. Learn these as reflexes, not as facts to be re-derived.
速度不是通用的，而是分专题的。每个板块都有两三个标准变形，能把两分钟的推导压缩成十五秒的识别。要把它们练成条件反射，而不是每次重新推导。
Algebra and functions 代数与函数
- Factor theorem: if p(a) = 0 then (x − a) is a factor. Test integer candidates before dividing. | 因式定理：若 p(a) = 0，则 (x − a) 是因子。先用整数试根，再做除法。
- Completing the square to read off vertex, range or minimum instantly. | 配方直接读出顶点、值域或最小值。
- Symmetric systems: swap to u = x + y, v = xy. | 对称方程组：令 u = x + y，v = xy。
(x + y)² + (x − y)² = 2(x² + y²) and (x + y)² − (x − y)² = 4xy
Sequences and series 数列与级数
- Arithmetic: nth term a + (n − 1)d; sum Sₙ = n/2 [2a + (n − 1)d]. | 等差：通项 a + (n − 1)d；前 n 项和 Sₙ = n/2 [2a + (n − 1)d]。
- Geometric: nth term arⁿ⁻¹; infinite sum a ÷ (1 − r) for |r| < 1. | 等比：通项 arⁿ⁻¹；|r| < 1 时无穷和为 a ÷ (1 − r)。
- Check convergence conditions before using an infinite sum – it is a classic distractor. | 使用无穷和之前先验证收敛条件，这是经典的干扰点。
Exponentials and logarithms 指数与对数
logₐ(xy) = logₐx + logₐy logₐ(x ÷ y) = logₐx − logₐy logₐ(xⁿ) = n logₐx logₐb = log b ÷ log a
- To solve aˣ = b, take logs immediately: x = log b ÷ log a. | 解 aˣ = b 时立刻取对数：x = log b ÷ log a。
- Recognise disguised quadratics such as e²ˣ − 3eˣ + 2 = 0 as (eˣ)² − 3eˣ + 2 = 0. | 识别伪二次型，如 e²ˣ − 3eˣ + 2 = 0 即 (eˣ)² − 3eˣ + 2 = 0。
Trigonometry 三角
sin²θ + cos²θ = 1 sin 2θ = 2 sin θ cos θ cos 2θ = 1 − 2sin²θ tan θ = sin θ ÷ cos θ
- Memorise exact values at 0, 30°, 45°, 60°, 90° in both degrees and radians. | 熟记 0°、30°、45°、60°、90° 的精确值，角度与弧度都要会。
- For “solve in the interval” questions, find the principal solution, then add or subtract the period before listing. | 区间解方程题：先求主解，再加减周期后再列答案。
Coordinate geometry 解析几何
d = √((x₂ − x₁)² + (y₂ − y₁)²) circle: (x − a)² + (y − b)² = r² perpendicular: m₁m₂ = −1
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