📚 Review Set 10A – Non-Calculator Skills for IB Mathematics | IB 数学非计算器复习集 10A
The ability to solve problems without a calculator is a cornerstone of IB Mathematics, especially in Paper 1 for both Analysis & Approaches and Applications & Interpretation courses. Review set 10A has been carefully designed to strengthen your mental arithmetic, exact value reasoning, and algebraic fluency - all essential for top performance in non-calculator assessments. In this article, we break down the key concepts and strategies embedded in the set, so you can approach each question with confidence.
不依赖计算器解决问题的能力是 IB 数学的基石，尤其在分析与方法（AA）和应用与解释（AI）课程的试卷一当中。复习题集 10A 旨在强化心算、精确值推理和代数运算的熟练度——这些都是在非计算器考试中取得高分的关键。本文将逐一拆解题集中蕴含的核心概念与策略，帮助你从容应对每一道题目。
1. Why Non-Calculator Skills Decide Your Grade | 为什么非计算器技能决定你的分数
Paper 1 is not simply Paper 2 with the calculator taken away. It tests a different layer of understanding. When you cannot press a button, you are forced to recognise structure: a difference of two squares, a hidden common factor, a standard angle, or a logarithm that collapses into a single integer.
试卷一并不是"把计算器拿走的试卷二"。它考查的是另一个层面的理解。当你无法按键时，你被迫去识别结构：平方差、隐藏的公因式、标准角，或者一个能化简为整数结果的对数。
In Review set 10A the questions consistently reward students who work with exact forms - fractions, surds, π and logarithms - rather than decimal approximations. An answer of 0.7071 will lose marks where (√2)/2 is required, even if the value is numerically correct.
复习题集 10A 中的题目一贯奖励那些使用精确形式——分数、根式、π 和对数——而非小数近似值的学生。在要求 (√2)/2 的地方写出 0.7071 会被扣分，即使数值本身是正确的。
The three pillars the set is built on are:
本习题集建立在三大支柱之上：
- Mental arithmetic - fast, reliable computation with fractions, percentages and signed numbers.
心算——对分数、百分数和带符号数的快速且可靠的计算。 - Exact value reasoning - knowing when and how to leave an answer in surd, logarithmic or trigonometric form.
精确值推理——知道何时以及如何把答案保留为根式、对数或三角函数形式。 - Algebraic fluency - expanding, factorising, rearranging and substituting without hesitation.
代数运算的熟练度——毫不犹豫地展开、因式分解、移项与代入。
2. Understanding the Non-Calculator Format | 理解非计算器形式
IB Mathematics Paper 1 explicitly forbids any calculator. This means that all working, from simple arithmetic to complex calculus, must be done by hand. Questions are designed to test your understanding of exact forms: fractions, surds, π and logarithms must be left in their simplest exact representation. For example, an answer like √2 or ln 3 is expected, rather than a decimal approximation.
IB 数学试卷一明确禁止使用任何计算器。这意味着从简单运算到复杂微积分的所有步骤都要手算完成。试题旨在考查你对精确形式的理解：分数、根式、π 和对数必须保留最简精确表达。比如，答案应为 √2 或 ln 3，而不是小数近似值。
In Review set 10A, you will see questions that prompt you to rationalise denominators, combine logarithmic expressions, and evaluate trigonometric functions at standard angles. The non-calculator constraint encourages a deeper grasp of structure rather than rote button-pressing.
在复习题集 10A 中，你会遇到要求分母有理化、合并对数表达式以及计算标准角三角函数值的题目。非计算器的限制促使你更深入地把握数学结构，而非机械按键。
The table below summarises how the two courses differ in what they emphasise under non-calculator conditions.
下表总结了两个课程在非计算器条件下各自的侧重点。
| Aspect | 方面 | AA (Analysis & Approaches) | 分析与方法 | AI (Applications & Interpretation) | 应用与解释 |
| Core demand | 核心要求 | Symbolic manipulation, proof, exact calculus | 符号运算、证明、精确微积分 | Modelling arithmetic, financial and statistical formulas | 建模运算、金融与统计公式 |
| Typical exact form | 典型精确形式 | Surds, ln, inverse trig | 根式、ln、反三角 | Fractions, ratios, percentages, compound interest expressions | 分数、比例、百分数、复利表达式 |
| Skill stressed | 强调技能 | Algebraic fluency and rigour | 代数熟练度与严谨性 | Efficient hand computation and estimation | 高效手算与估算 |
3. Exact Values in Trigonometry | 三角函数的精确值
Memorising the exact values of sine, cosine and tangent at the standard angles is unavoidable. These values appear in identity questions, in solving trigonometric equations, and in calculus problems where an exact gradient or area is required.
背诵标准角的正弦、余弦和正切精确值是不可避免的。这些数值会出现在恒等式题目、解三角方程，以及需要精确斜率或面积的微积分问题中。
Standard angles and exact values:
标准角与精确值：
| θ | 0° | 30° | 45° | 60° | 90° |
| sin θ | 0 | ½ | (√2)/2 | (√3)/2 | 1 |
| cos θ | 1 | (√3)/2 | (√2)/2 | ½ | 0 |
| tan θ | 0 | (√3)/3 | 1 | √3 | undefined | 无定义 |
The fastest way to remember the sine row is the pattern √0/2, √1/2, √2/2, √3/2, √4 w is the same list read backwards. Tangent follows immediately from tan θ = sin θ ÷ cos θ.
记住正弦行最快的方法是模式 √0/2、√1/2、√2/2、√3/2、√4/2。余弦行就是把这张表倒着读。正切则由 tan θ = sin θ ÷ cos θ 立即得到。
sin²θ + cos²θ = 1 1 + tan²θ = sec²θ tan θ = sin θ ÷ cos θ
her with reference angles: 150° has reference angle 30°, so sin 150° = ½ and cos 150° = -(√3)/2.
对于第一象限以外的角，使用 ASTC 符号法则配合参考角：150° 的参考角是 30°，因此 sin 150° = ½，cos 150° = -(√3)/2。
4. Surds, Rationalising Denominators and Index Laws | 根式、分母有理化与指数法则
Surd manipulation is examined constantly in Paper 1. The examiner wants to see that you can simplify, expand and rationalise without a decimal check.
根式运算在试卷一中频繁出现。考官希望看到你能在不借助小数验证的情况下化简、展开和有理化。
Key rules to have at your fingertips:
必须烂熟于心的关键法则：
- √a × √b = √ab, and √a ÷ √b = √a ÷ b.
√a × √b = √ab，且 √a ÷ √b = √a ÷ b。 - √a²b = a√b for a > 0 - always extract the largest square factor.
当 a > 0 时 √a²b = a√b——永远先提取最大的平方因子。 - To rationalise 1 ÷ √a, multiply numerator and denominator by √a.
要化简 1 ÷ √a，分子分母同乘 √a。 - To rationalise 1 ÷ (a + √b), multiply by the conjugate (a - √b).
要化简 1 ÷ (a + √b)，乘以共轭式 (a - √b)。
Worked example: simplify 6 ÷ (√5 - 1). Multiply top and bottom by (√5 + 1) to get 6(√5 + 1) ÷ (5 - 1) = 6(√5 + 1) ÷ 4 = 3(√5 + 1) ÷ 2.
例题：化简 6 ÷ (√5 - 1)。上下同乘 (√5 + 1)，得 6(√5 + 1) ÷ (5 - 1) = 6(√5 + 1) ÷ 4 = 3(√5 + 1) ÷ 2。
Index laws run in parallel and must be exact:
指数法则与之并行，同样必须精确：
= aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ a⁰ = 1 a⁻ⁿ = 1 ÷ aⁿ a1/n = ⁿ√a
Combine these with fractional indices: 272/3 = (³√27 s (4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 125, 128, 243) turns these into one-line answers.
把这些与分数指数结合：272/3 = (³√27)² = 3² = 9。熟记完全幂（4、8、9、16、25、27、32、36、49、64、81、100、125、128、243）能让这类题目一步得出答案。
5. Logarithms and Exponent Laws Without a Calculator | 对数与指数法则
Logarithm questions in Review set 10A almost always reduce to a small set of laws. If you can move confidently between exponential and logarithmic form, most questions collapse.
复习题集 10A 中的对数题几乎总能归结为一小组法则。如果你能在指数形式与对数形式之间自如转换，多数题目都会迎刃而解。
The laws, stated once and for all:
以下法则需要一次性彻底掌握：
p; logₐ(m ÷ n) = logₐm - logₐn logₐ(mᵏ) = k logₐm logₐa = 1 logₐ1 = 0
cal non-calculator tasks include combining log₂6 + log₂4 - log₂3 into log₂8 = 3, solving 2ˣ = 32 by writing 32 = 2⁵, and changing base with logₐb = log_c b ÷ log_c a.
典型的非计算器任务包括：把 log₂6 + log₂4 - log₂3 合并为 log₂8 = 3；通过把 32 写成 2⁵ 来解 2ˣ = 32；以及用 logₐb = log_c b ÷ log_c a 进行换底。
e = 1, together with e⁰ = 1. When a question asks for ln 5, the expected answer is simply "ln 5" - do not convert to 1.609.
两个值得记住的精确值是 ln 1 = 0 与 ln e = 1，以及 e⁰ = 1。当题目要求 ln 5 时，期望答案就是 "ln 5"——不要写成 1.609。
A useful check within non-calculator work: if you obtain a logarithmic answer that is not one of the "nice" values (0, 1, 2, 3 or a simple fraction), look again - the question is almost certainly designed to simplify exactly.
非计算器作业中的一个有用自查：如果你得到的对数答案不是那些"漂亮"数值（0、1、2、3 或简单分数），请回头检查——题目几乎肯定被设计为能精确化简。
6. Algebraic Fluency: Expansion, Factorisation and Completing the Square | 代数熟练度：展开、因式分解与配方
Algebraic fluency is the single strongest predictor of a high Paper 1 score. It is not a topic; it is the machinery you use in every topic.
代数熟练度是试卷一高分的唯一最强预测指标。它不是一个知识点，而是你在每个知识点中都要使用的机器。
Three patterns appear again and again in the set:
在本习题集中反复出现三种模式：
- Difference of two squares: a² - b² = (a - b)(a + b).
平方差：a² - b² = (a - b)(a + b)。 - Perfect square trinomials: a² ± 2ab + b² = (a ± b)².
完全平方式：a² ± 2ab + b² = (a ± b)²。 - Completing the square: x² + bx + c = (x + b ÷ 2)² + c - b² ÷ 4.
配方：x² + bx + c = (x + b ÷ 2)² + c - b² ÷ 4。
The quadratic formula should also be known by heart and applied with exact surds:
求根公式同样应牢记于心，并用精确根式表示结果：
x = (-b ± √b² - 4ac) ÷ (2a)
Completing the square gives you the vertex (-b ÷ 2, c - b² ÷ 4) immediately, which is why examiners prefer it over the formula when a question asks for the minimum value of a quadratic.
配方能立刻给出顶点 (-b ÷ 2, c - b² ÷ 4)，这就是当题目要求二次函数最小值时，考官更青睐配方而非公式的原因。
nomials by inspection, and using the factor theorem: if f(a) = 0 then (x - a) is a factor.
此外还要练习分组分解、通过观察法进行多项式除法，以及使用因式定理：若 f(a) = 0，则 (x - a) 是公因式。
7. Fractions, Ratios and Percentage Reasoning by Hand | 分数、比例与百分数的心算推理
For AI students in particular, Paper 1 is full of financial and proportional reasoning that must be done exactly. Compound interest, depreciation and unit conversions all appear without a calculator.
尤其对 AI 学生来说，试卷一充满必须精确完成的金融与比例推理。复利、折旧和单位换算都会在无计算器情况下出现。
The compound interest formula should be quoted and then evaluated by hand:
复利公式需要背出，然后手算求值：
FV = PV × (1 + r ÷ 100 ÷ n)nt
1.05² = 2000 × 1.1025 = 2205. The arithmetic 1.05² is done mentally: 1.05 × 1.05 = 1.1025.
若题目给出 PV = 2000、r = 5、n = 1、t = 2，则 FV = 2000 × 1.05² = 2000 × 1.1025 = 2205。1.05² 的心算过程是：1.05 × 1.05 = 1.1025。
Handy percentage tools to internalise:
需要内化的百分数技巧：
- 10% is a decimal shift; 5% is half of that; 1% is a shift of two places.
10% 是小数点左移一位；5% 是它的一半；1% 是左移两位。 - To increase by 12%, multiply by 1.12; to decrease by 12%, multiply by 0.88.
增加 12% 就乘以 1.12；减少 12% 就乘以 0.88。 - Reverse percentage: if a price after a 20% discount is 64, the original was 64 ÷ 0.8 = 80.
反推百分数：若打折 20% 后价格为 64，则原价为 64 ÷ 0.8 = 80。
Ratio work is equally exact: a ratio of 3 : 5 : 7 over a total of 450 gives parts of 30, so the shares are 90, 150 and 210.
比例题同样要求精确：总量 450 按 3 : 5 : 7 分配，每一份为 30，因此各份额是 90、150 和 210。
8. Sequences, Series and Binomial Expansion | 数列、级数与二项展开
Sequences and series questions in a non-calculator paper are usually chosen so that the algebra stays clean. Learn the formulas, then apply them to small numbers.
非计算器试卷中的数列与级数题通常被设计成代数保持整洁。记住公式，然后代入较小数字运算。
The formulas you need:
你需要掌握的公式：
uₙ = a + (n - 1)d Sₙ = n ÷ 2 × [2a + (n - 1)d] = n ÷ 2 × (a + l)
uₙ = arⁿ⁻¹ Sₙ = a(1 - rⁿ) ÷ (1 - r) S∞ = a ÷ (1 - r) for |r| < 1
, ... Here a = 3, d = 4, so S₂₀ = 20 ÷ 2 × [2(3) + 19(4)] = 10 × (6 + 76) = 10 × 82 = 820.
典型题目：求 3, 7, 11, ... 的前 20 项之和。此处 a = 3，d = 4，故 S₂₀ = 20 ÷ 2 × [2(3) + 19(4)] = 10 × (6 + 76) = 10 × 82 = 820。
Binomial expansion reappears constantly, and Paper 1 usually asks only for a specific term. Pascal's triangle up to the fifth row is enough for most questions:
二项展开频繁出现，而试卷一通常只要求某一特定项。多数题目只需帕斯卡三角到第五行：
(a + b)ⁿ = aⁿ + n·aⁿ⁻¹b + n(n-1) ÷ 2 · aⁿ⁻²b² + ... + bⁿ
Use "n choose r" = n! ÷ (r!(n - r)!) to identify the coefficient of a single term rather than expanding everything.
使用组合数 "n 选 r" = n! ÷ (r!(n - r)!) 来直接确定某一项的系数，而不是把全部展开。
9. Calculus Without a Calculator: Differentiation and Integration | 非计算器微积分：微分与积分
Calculus in Paper 1 is always set with manageable numbers. The examiner wants to see correct differentiation rules and exact evaluation, not decimal output.
试卷一中的微积分题目总被设置成数字可处理。考官想看到的是正确的求导法则与精确求值，而不是小数输出。
The core rules:
核心法则：
d ÷ dx (xⁿ) = nxⁿ⁻¹ ∫ xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + C, n ≠ -1
dx = 12x³ - 2 = 12 - 2 = 10, and ∫ from 1 to 2 of (3x²) dx = [x³] from 1 to 2 = 8 - 1 = 7.
因此 y = 3x⁴ - 2x 在 x = 1 处的斜率为 dy ÷ dx = 12x³ - 2 = 12 - 2 = 10；而从 1 到 2 对 (3x²) 求积分等于 [x³] 在 1 到 2 的取值 = 8 - 1 = 7。
Other non-calculator calculus demands include finding stationary points by solving f' vative to classify them, and evaluating definite integrals that produce exact surds or fractions.
其他非计算器微积分要求包括：通过精确求解 f'(x) = 0 找出驻点、用二阶导数判别其性质，以及计算能得到精确根式或分数的定积分。
One recurring trap is forgetting the constant of integration C in indefinite integrals, and another is integrating x⁻¹ as x⁰ ÷ 0. The correct result is ln |x| + C.
一个反复出现的陷阱是在不定积分中忘记积分常数 C；另一个是把 x⁻¹ 按 x⁰ ÷ 0 处理。正确结果是 ln |x| + C。
10. Geometry, Vectors and Coordinate Work | 几何、向量与坐标
Coordinate geometry in Paper 1 rewards exact manipulation of gradients, distances and midpoints. Keep everything in surd or fraction form.
试卷一中的坐标几何奖励对斜率、距离和中点的精确运算。请把所有结果保留为根式或分数形式。
The formulas to recall:
需要回忆的公式：
)²] M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2) m = (y₂ - y₁) ÷ (x₂ - x₁)
isfy m₁ × m₂ = -1, and the distance from a point to a line can be handled with the formula |ax₀ + by₀ + c| ÷ √a² + b².
垂直直线满足 m₁ × m₂ = -1；点到直线的距离可用公式 |ax₀ + by₀ + c| ÷ √a² + b²。
roducts are computed as a₁b₁ + a₂b₂ + a₃b₃, and the angle follows from cos θ = (a · b) ÷ (|a||b|).
无计算器的向量题主要考验条理：i、j、k 分量逐项相加，数量积按 a₁b₁ + a₂b₂ + a₃b₃ 计算，夹角由 cos θ = (a · b) ÷ (|a||b|) 得出。
Remember also the standard exact results: the area of an equilateral triangle with side s is (√3 ÷ 4)s², and the diagonal of a unit square is √2.
同时记住标准精确结果：边长为 s 的等边三角形面积是 (√3 ÷ 4)s²，单位正方形的对角线是 √2。
11. Common Traps and Marking Pitfalls | 常见陷阱与评分要点
Most lost marks in a non-calculator paper come from a small set of repeated errors. Knowing them in advance is worth several points.
非计算器试卷中丢失的分数大多来自一小撮反复出现的错误。提前了解这些错误，价值相当于好几分。
- Leaving a decimal when an exact form is required
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