📚 Non-calculator Questions in IB Mathematics | IB数学非计算器问题
The non-calculator paper in IB Mathematics challenges you to rely on algebraic insight, exact values, and logical reasoning rather than pressing buttons. Mastering this paper means cultivating fluency with symbolic manipulation, memorised exact trigonometric values, and careful mental arithmetic.
IB数学的非计算器试卷要求你依赖代数洞察、精确值和逻辑推理，而不是按键操作。要驾驭这份试卷，你必须熟练运用符号操作、牢记精确三角值，并保持细致的心算习惯。
1. The Nature of Non-calculator Exams | 非计算器考试的本质
In IB Mathematics: Analysis and Approaches (AA), Paper 1 is completely calculator‑free, while Applications and Interpretation (AI) Paper 1 includes short‑response questions where no calculator is allowed. The examiner expects answers as simplified radicals, fractions, or exact logarithmic expressions.
在IB数学分析与方法(AA)中，试卷一完全禁用计算器；应用与解释(AI)试卷一中的简答题也不得使用计算器。阅卷人期望你给出化简后的根式、分数或精确的对数表达式。
Working without a calculator trains you to recognise algebraic patterns. For example, you learn to spot a difference of squares or a perfect cubic factorisation instantly, skills that are often hidden when a machine does the factoring for you.
不用计算器能够训练你识别代数模式。例如，你会立即辨识平方差公式或完全立方的因式分解，而这些技巧在使用机器代为分解时常常会被掩盖。
2. Algebraic Manipulation Mastery | 精通代数运算
Expanding, factorising, and simplifying are the backbone of non‑calculator success. You must be able to rewrite x² − 5x + 6 as (x − 2)(x − 3) in seconds, or expand (2x − 1)³ without misplacing a coefficient.
展开、因式分解和化简是非计算器考试成功的基石。你必须能在几秒内将 x² − 5x + 6 写成 (x − 2)(x − 3)，或在展开 (2x − 1)³ 时准确无误地处理每一项系数。
Long division of polynomials appears frequently when finding oblique asymptotes or factorising a cubic whose root you have identified. Practise setting out the division carefully so the working remains clear and you avoid sign errors.
多项式长除在寻找斜渐近线或因式分解已知一根的三次多项式时经常出现。练习工整地写出除法竖式，以确保过程清晰并避免符号错误。
Rationalising denominators such as 1/(√3 − 1) requires multiplying numerator and denominator by the conjugate √3 + 1. Always present the final answer with a rational denominator unless the question specifies otherwise.
分母有理化，例如 1/(√3 − 1)，要求分子分母同乘共轭根式 √3 + 1。除非题目另有说明，最终答案应始终保留有理化分母的形式。
3. Exponents and Logarithms Without a Calculator | 不用计算器处理指数与对数
With no calculator, you evaluate expressions like log₂8 by asking "to what power must 2 be raised to obtain 8?" The answer 3 is instant. Similarly, log₅(1/25) = −2 because 5⁻² = 1/25.
没有计算器时，你通过思考“2的几次方等于8”来计算 log₂8，答案 3 瞬间得出。同理，log₅(1/25) = −2，因为 5⁻² = 1/25。
Laws of logarithms become your most reliable tool: combine logₐ(x) + logₐ(y) into logₐ(xy), or expand logₐ(x³/y) into 3 logₐx − logₐy. These rules let you solve exponential equations neatly.
对数运算律是你最可靠的工具：将 logₐ(x) + logₐ(y) 合并为 logₐ(xy)，或将 logₐ(x³/y) 展开为 3 logₐx − logₐy。利用这些规则可以干净利落地解出指数方程。
When simplifying expressions like 82/3 , and 2² = 4. Avoid reaching for a calculator; the mental steps sharpen your number sense.
化简 82/3 这类表达式时，先想8的立方根再平方：8的立方根是2，2² = 4。尽量避免使用计算器；这些心算步骤能提升你对数字的感觉。
4. Trigonometric Exact Values and Equations | 三角函数精确值与方程
Non‑calculator trigonometry relies heavily on known exact values: sin(π/6) = 1/2, cos(π/4) = √2/2, tan(π/3) = √3. Drawing the 30°‑60°‑90° and 45°‑45°‑90° triangles helps you recall these values instantly.
非计算器三角部分严重依赖已知的精确值：sin(π/6) = 1/2，cos(π/4) = √2/2，tan(π/3) = √3。画出30°‑60°‑90°和45°‑45°‑90°三角形能帮助你立即回忆起这些值。
act values to angles greater than π/2. For instance, sin(2π/3) = sin(π − π/3) = sin(π/3) = √3/2, without a single button press.
单位圆的图像化展示能让你将精确值推广到大于 π/2 的角度。例如，sin(2π/3) = sin(π − π/3) = sin(π/3) = √3/2，完全无需按键。
= 0, treat it as a quadratic in sinx, then use exact values to find x in a given interval.
必须熟练应用三角恒等式，如 sin²θ + cos²θ = 1 或 tanθ = sinθ/cosθ。在求解 2 sin²x − sinx − 1 = 0 时，将其视为 sinx 的二次方程，再利用精确值求出给定区间内的 x。
5. Functions and Transformations by Hand | 手绘函数与变换
rent function y = 1/x. Marks are awarded for correctly labelled intercepts and asymptotes.
手绘 y = 1/(x + 2) − 3 的函数图像时，你需要通过解读从母函数 y = 1/x 的平移变换，确定垂直渐近线 x = −2 和水平渐近线 y = −3。正确标注截距和渐近线才能得分。
ompute g(2) = 5, then f(5) = 25 − 15 = 10. Show each step so partial credit can be earned.
复合函数需要仔细代入。计算 f(g(2))，已知 f(x) = x² − 3x，g(x) = 2x + 1，先算出 g(2) = 5，再求 f(5) = 25 − 15 = 10。应展示每一步，以便获得步骤分。
then solve for y algebraically. The result must be expressed as f⁻¹(x) with its domain stated clearly.
反函数也经常考查：在 y = (2x − 1)/(x + 3) 中交换 x 和 y，然后通过代数操作解出 y。结果必须表示为 f⁻¹(x)，并明确给出定义域。
6. Calculus: Differentiation and Integration Pencil‑and‑paper | 微积分：纸笔求导与积分
− 1)⁵, multiply by the derivative of the inside: dy/dx = 5(3x² − 1)⁴ · 6x = 30x(3x² − 1)⁴.
不使用计算器时，求导依赖链式法则、乘积法则和商法则的准确应用。对 y = (3x² − 1)⁵ 求导，需要乘上内层函数的导数：dy/dx = 5(3x² − 1)⁴ · 6x = 30x(3x² − 1)⁴。
x²]₁² = (16 − 4) − (1 − 1) = 12. Leave answers as rational numbers or simplified radicals.
积分题常要求计算精确面积。计算 ∫₁² (4x³ − 2x) dx 时，将幂次加一后除以新幂次：[x⁴ − x²]₁² = (16 − 4) − (1 − 1) = 12。将答案保留为有理数或最简根式。
Recognising patterns for reverse chain rule (u‑substitution) is essential. In ∫ 2x √x² + 1 dx, let u = x² + 1, du = 2x dx, so the integral becomes ∫ √u du = (2/3) u3/2 + C = (2/3)(x² + 1)3/2 + C.
识别逆链式法则（u-代换）的模式至关重要。在 ∫ 2x √x² + 1 dx 中，令 u = x² + 1，du = 2x dx，于是积分变为 ∫ √u du = (2/3) u3/2 + C = (2/3)(x² + 1)3/2 + C。
7. Sequences and Series: Summation and Patterns | 数列与级数：求和与模式
Arithmetic and geometric sequences are core non‑ 3 = 3n + 4. The sum of the first 10 terms uses S₁₀ = (10/2)(2×7 + 9×3) = 5(14 + 27) = 205.
等差与等比数列是非计算器的核心主题。对于首项 u₁ = 7、公差 d = 3 的等差数列，通项公式为 uₙ = 7 + (n − 1)·3 = 3n + 4。前10项的和用 S₁₀ = (10/2)(2×7 + 9×3) = 5(14 + 27) = 205。
(1 − 1/64)/(1/2) = 8(63/64) = 63/8. Simplify completely and leave as an improper fraction.
对于等比级数，公式 Sₙ = a(1 − rⁿ)/(1 − r) 常以精确分数形式考查。若 r = 1/2，a = 4，n = 6，则 S₆ = 4(1 − (1/2)⁶)/(1 − 1/2) = 4(1 − 1/64)/(1/2) = 8(63/64) = 63/8。需彻底化简并保留为假分数形式。
Sigma notation requires careful index manipulation. To evaluate Σk=3⁷ (2k − 1), either list terms (5 + 7 + 9 + 11 + 13 = 45) or shift the index: Σk=1⁷ (2k − 1) − Σk=1² (2k − 1) = 49 − 4 = 45.
求和符号要求你小心处理指标。计算 Σk=3⁷ (2k − 1) 时，可以列出各项 (5 + 7 + 9 + 11 + 13 = 45)，或平移指标：Σk=1⁷ (2k − 1) − Σk=1² (2k − 1) = 49 − 4 = 45。
8. Probability and Counting Principles | 概率与计数原理
ions appear without a calculator, so you must manually compute expressions like C(8,3) = 8!/(3!5!) = (8×7×6)/(3×2×1) = 56. Cancel common factors before multiplying to keep numbers small.
组合与排列经常以非计算器形式出现，因此你必须手动计算诸如 C(8,3) = 8!/(3!5!) = (8×7×6)/(3×2×1) = 56 的表达式。先约去公因子再相乘，以保持数字较小。
The binomial distribution probability P(X = k) = C(n,k) pk (1−p)n−k lculate C(5,2) = 10, (1/3)² = 1/9, (2/3)³ = 8/27, then multiply: 10 × 1/9 × 8/27 = 80/243.
二项分布概率 P(X = k) = C(n,k) pk (1−p)n−k 经常以分数形式考查。若 n = 5，p = 1/3，k = 2，则计算 C(5,2) = 10，(1/3)² = 1/9，(2/3)³ = 8/27，然后相乘：10 × 1/9 × 8/27 = 80/243。
Probability trees remain indispensable: the branches must carry exact fractions, and the final answer is obtained by multiplying along paths and adding disjoint events. Never convert fractions to decimals; leave everything exact.
概率树形图仍不可或缺：树的分支应标注精确分数，最终答案通过路径上的乘积及互斥事件的加和得到。绝不将分数化为小数；保持所有结果精确。
9. Vectors and Geometry Proofs | 向量与几何证明
4) + (−1)×2 + 3×1 = −8 −2 + 3 = −7. Used to find angles, the cosine formula cosθ = (a·b)/(|a||b|) must yield an exact value.
向量计算要求精确的算术。对于向量 a = (2, −1, 3) 和 b = (−4, 2, 1)，点积 a·b = 2×(−4) + (−1)×2 + 3×1 = −8 −2 + 3 = −7。在计算角度时，余弦公式 cosθ = (a·b)/(|a||b|) 必须给出精确值。
, −1) + t(3, −1, 4). To test whether a point lies on the line, solve three equations simultaneously for t — without a calculator.
求三维空间直线方程：给定点 A(1, 2, −1) 和方向向量 d = (3, −1, 4)，直线为 r = (1, 2, −1) + t(3, −1, 4)。要检验某点是否在直线上，需同时解出三个关于 t 的方程——不用计算器。
y two vectors is half the magnitude of their cross product, leaving results as √ integers.
向量叉积（高等级）需要计算三阶行列式。由两个向量构成的三角形面积是叉积模长的一半，结果应保留为根号形式。
10. Solving Equations and Inequalities | 解方程与不等式
adratic equations are solved by factorising first, then by the quadratic formula x = [−b ± √b² − 4ac e discriminant Δ = b² − 4ac tells you about the nature of the roots without solving fully.
二次方程优先尝试因式分解；若不明显，则使用求根公式 x = [−b ± √b² − 4ac]/(2a)。判别式 Δ = b² − 4ac 无需完全解出即可告诉你根的性质。
Higher‑ zero, then factorise step by step. All coefficients stay integers, so the arithmetic is manageable by hand.
高次多项式通常利用余式定理和因式定理求解。依次检验 x = 1, −1, 2, −2，直至余式为零，再逐步因式分解。所有系数保持整数，因此手工算术完全可行。
to one side, obtaining a single fraction, and building a sign table. Critical values come from setting numerator and denominator to zero — simple integer arithmetic.
像 (2x − 3)/(x + 1) ≥ 1 这样的有理不等式，通过将各项移项、化为单一分式、建立符号表来求解。临界值来自令分子和分母为零的方程——只是简单的整数运算。
11. Exam Technique and Time Management | 考试技巧与时间管理
Start by scanning the whole paper and tackling the questions you are most confident about. This warm‑up builds momentum and ensures you collect marks quickly, leaving more time for demanding proof or multi‑step problems.
先浏览整份试卷，从最有把握的题目入手。这种热身能帮助你建立答题节奏，快速拿下分数，为要求较高的证明题或多步骤问题留出更多时间。
Show all working lines, even for simple substitutions. If a slip occurs, the examiner can follow the logical flow and award method marks. A correct answer with no working often earns only the answer mark.
展示所有计算步骤，即便是简单的代入。如果出现笔误，阅卷人也能顺着逻辑流程给予方法分。仅有正确答案而没有过程，往往只能得到答案分。
Double‑check exactness: radicals simplified, denominators rationalised, fractions reduced, and integrals given with ‘+C’. These small points add up and demonstrate the precision expected in a non‑calculator setting.
反复检查答案的精确性：根式已化简、分母已有理化、分数已约分、不定积分带有“+C”。这些细节累积起来，能体现非计算器环境下所需的严谨性。
12. Practice with Purpose | 有针对性的练习
Consistent practice on past Paper 1 questions builds speed with algebraic manipulation. Set a timer and work under exam conditions, resisting any temptation to look up a value on a phone.
持之以恒地练习历年试卷一，能提升代数运算速度。设定计时器，在模拟考试环境下做题，克制用手机查询数值的冲动。
Create a personal ‘non‑calculator toolkit’ summary card: exact trig values, log laws, derivative rules, and geometric formulas. Reviewing this card daily embeds the facts into long‑term memory, so they become automatic under pressure.
制作一张个人“非计算器工具包”摘要卡：精确三角值、对数法则、导数规则和几何公式。每日复习这张卡片，将知识嵌入长期记忆，使其在压力环境下也能自动浮现。
After each practice session, analyse every error. Did you mis‑remember an exact value? Lose a negative sign? Forgetting a constant of integration? Targeted correction turns weaknesses into strengths.
每次练习后，分析每一个错误。是记错了精确值？丢掉了负号？还是忘了积分常数？有针对性的纠正能把弱点转化为强项。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
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