📚 Review Set 25A - Non-Calculator | 复习集 25A - 非计算器
In IB Mathematics, the non-calculator paper challenges students to demonstrate deep understanding, quick mental arithmetic, and a solid grasp of algebraic structure. Review Set 25A brings together the essential skills you need to solve problems accurately without relying on technology. This article walks you through key topic areas, provides example problems and solutions, and reinforces the reasoning that leads to success when every step must be performed by hand.
在 IB 数学中，非计算器试卷考验的是学生对概念的深入理解、快速心算能力和扎实的代数基本功。复习集 25A 汇集了在不依赖技术工具的情况下准确解题所必需的核心技能。本文带你回顾重点主题，给出示例题目与解答，并强化手工完成每一步所需的推理过程，帮助你在非计算器考试中稳操胜券。
1. Why Non-Calculator Skills Matter | 为何非计算器技能重要
The non-calculator section is not about tedious arithmetic; it tests your ability to recognise patterns, simplify expressions before evaluating, and use exact values. Examiners design questions where mental shortcuts and algebraic insight give you the answer faster than a calculator ever could. Developing this mindset early makes the entire IB course more manageable.
非计算器部分并非考查繁琐的运算，而是检验你识别模式、先化简再求值以及运用精确值的能力。出题者设计的题目往往通过心算技巧和代数洞察就能比用计算器更快地得出答案。尽早培养这种思维模式会使整个 IB 课程学起来更轻松。
, knowing that log₂8 = 3 because 2³ = 8 lets you avoid unnecessary steps. Recognizing that √48 simplifies to 4√3 turns a messy-looking expression into a manageable one. All these habits save time and reduce errors.
题目常涉及分数、根式、对数以及具有标准精确值的三角比。例如，若知道 log₂8 = 3，因为 2³ = 8，就可省去不必要的步骤。意识到 √48 可化简为 4√3，能将看似凌乱的表达式变得易于处理。这些习惯都能节约时间并减少错误。
2. Algebraic Techniques: Perfecting the Basics | 代数技巧：夯实基础
Strong algebra is the backbone of every non-calculator solution. You must be comfortable expanding brackets, factoring quadratics, simplifying rational expressions, and manipulating indices and surds. For example, simplify (x² - 9) / (x² + x - 6). Factoring reveals (x - 3)(x + 3) in the numerator and (x + 3)(x - 2) in the denominator, giving (x - 3)/(x - 2) provided x ≠ -3, x ≠ 2. This simplification is impossible to see if you reach for a calculator too soon.
扎实的代数功底是所有非计算器解题的支柱。你必须熟练展开括号、因式分解二次式、化简有理表达式以及处理指数和根式。例如，化简 (x² - 9) / (x² + x - 6)。分子分解为 (x - 3)(x + 3)，分母为 (x + 3)(x - 2)，结果得到 (x - 3)/(x - 2)，前提是 x ≠ -3, x ≠ 2。如果过早依赖计算器，这类化简便无从察觉。
Index laws also appear frequently. Solve 22x+1 = 8x-1 to x = 4. No calculator needed. Surd questions, like rationalising the denominator of 1/(√5 - 2), require multiplying by the conjugate. The answer, √5 + 2, emerges neatly through algebraic manipulation.
指数运算律也频繁出现。解方程 22x+1 = 8x-1。将 8 写成 2³，再让指数相等：2x+1 = 3(x-1)，得到 x = 4，完全无需计算器。根式题目，例如将 1/(√5 - 2) 的分母有理化，只需乘以共轭式，结果 √5 + 2 便通过代数处理清晰地呈现出来。
3. Equation Solving: Quadratic, Exponential, and More | 解方程：二次方程、指数方程等
thm properties. For a quadratic like 2x² - 3x - 5 = 0, use the quadratic formula: x = [3 ± √9 + 40 ou must be able to handle fractions and square roots confidently without a decimal approximation.
非计算器的方程求解依赖精确方法：因式分解、配方法以及运用对数性质。对于二次方程 2x² - 3x - 5 = 0，使用求根公式：x = [3 ± √9 + 40] / 4 = [3 ± 7] / 4，得到 x = 5/2 或 x = -1。你必须能自信地处理分数和平方根，而不依赖小数近似。
Exponential equations like e2x 1) = 3 becomes log₂((x+1)(x-1)) = 3, then (x²-1) = 8, x = ±3, but reject -3 due to domain.
指数方程如 e2x = 5 可通过取自然对数求解：2x = ln 5，故 x = (ln 5)/2，答案保留精确对数形式。类似地，对数方程需要合并对数：log₂(x+1) + log₂(x-1) = 3 化为 log₂((x+1)(x-1)) = 3，进而 (x²-1) = 8，x = ±3，但要根据定义域舍去 -3。
4. Functions and Transformations | 函数与变换
on notation and transformations without graph-plotting technology is crucial. Given f(x) = √x, sketch g(x) = 2√x-3 d to determine the new domain x ≥ 3 and range y ≥ 1 by reasoning, not by viewing a screen.
在不借助绘图技术的情况下理解函数符号与变换至关重要。已知 f(x) = √x，画出 g(x) = 2√x-3 + 1 的草图。变换包括向右平移 3 个单位、纵向拉伸为 2 倍以及向上平移 1 个单位。你需要通过推理得出新的定义域 x ≥ 3 和值域 y ≥ 1，而不是靠屏幕显示。
= -3x - 1 → y(x - 2) = -3x - 1 → f⁻¹(x) = (-3x - 1)/(x - 2). The algebra must be flawless.
复合函数和反函数经常出现在非计算器题目中。若 f(x) = (2x - 1)/(x + 3)，求 f⁻¹(x)。交换 x 和 y，然后求解： ) → xy + 3x = 2y - 1 → xy - 2y = -3x - 1 → y(x - 2) = -3x - 1 → f⁻¹(x) = (-3x - 1)/(x - 2)。代数变形必须完美无瑕。
5. Trigonometry: Exact Values and Identities | 三角学：准确值与恒等式
150° = sin 30° = 1/2. For cos(2π/3), reference π/3, cosine negative, so cos(2π/3) = -1/2.
不借助计算器的三角学依赖于熟记 0°、30°、45°、60°、90° 及其弧度对应的精确值，并利用单位圆判断符号。计算 sin 150°：参考角为 30°，正弦在第二象限为正，因此 sin 150° = sin 30° = 1/2。对于 cos(2π/3)，参考角 π/3，余弦为负，故 cos(2π/3) = -1/2。
θ = 1/2 or cos θ = -1, giving θ = π/3, 5π/3, π. Exact reasoning replaces decimal guessing.
三角恒等式用于化简和解方程。在 0 ≤ θ ≤ 2π 内解 2 sin²θ - cos θ - 1 = 0。用 1 - cos²θ 替换 sin²θ： θ - 1 = 0 → -2cos²θ - cos θ + 1 = 0 → 2cos²θ + cos θ - 1 = 0 → (2cos θ - 1)(cos θ + 1) = 0。从而 cos θ = 1/2 或 cos θ = -1，得到 θ = π/3, 5π/3, π。精确推理取代了小数猜测。
6. Sequences and Series: Patterns Without Technology | 数列与级数：无需技术找规律
2 (7 + 64) = 10 × 71 = 710. All calculations are integer-based and easy to check mentally.
等差数列和等比数列依靠公式，而不是反复累加。对于首项 u₁ = 7，公差 d = 3 的等差数列，求第 20 项：u₂₀ = 7 + (19)(3) = 64。前 20 项之和为 S₂₀ = 20/2 (7 + 64) = 10 × 71 = 710。所有计算都是整数运算，便于心算检查。
0/243. Maintaining fractional form demonstrates precision that a calculator might obscure.
首项为 5、公比为 2/3 的等比级数可以精确求值。无穷项之和为 S∞ = 5 / (1 - 2/3) = 5 / (1/3) = 15。若求第 6 项，u₆ = 5 × (2/3)⁵。答案保留分数形式：5 × 32/243 = 160/243。保持分数形式体现了计算器可能掩盖的精确性。
7. Differential Calculus: First Principles and Rules | 微分：第一原理与规则
rces why the power rule works. For f(x) = 3x⁴ - 2/x + 5, rewrite as 3x⁴ - 2x⁻¹ + 5, then f'(x) = 12x³ + 2x⁻².
不借助计算器的微分考查你对极限定义和求导法则的掌握。用第一原理对 f(x) = x² 求导。当 h → 0 时，极限 ( (x+h)² - x² ) / h 给出 2x。这种概念性方法强化了理解幂法则的来源。对于 f(x) = 3x⁴ - 2/x + 5，改写为 3x⁴ - 2x⁻¹ + 5，然后 f'(x) = 12x³ + 2x⁻²。
etting the derivative to zero to find stationary points often leads to factorised equations you can solve without approximation.
链式法则、乘法法则和除法法则必须手工运用。对 y = (2x-1)³ sin x 求导。使用乘法法则：dy/dx = 3(2x-1)²·2·sin x + (2x-1)³ cos x = 6(2x-1)² sin x + (2x-1)³ cos x。令导数为零求驻点时，往往得到可因式分解的方程，无需近似值便能求解。
8. Integral Calculus: Definite and Indefinite | 积分：定积分与不定积分
- 3) dx, find the antiderivative x² - 3x, then F(3) - F(1) = (9-9) - (1-3) = 0 - (-2) = 2.
非计算器背景下的积分要求识别逆导数并运用基本定理。计算 ∫(4x³ - 6x + 1) dx。逐项积分：4x⁴/4 - 6x²/2 + x + C = x⁴ - 3x² + x + C。对于定积分如 ∫₁³ (2x - 3) dx，求出原函数 x² - 3x，然后 F(3) - F(1) = (9-9) - (1-3) = 0 - (-2) = 2。
Substitution and area under curves appear as well. Evaluate ∫₀² x √x²+1 Let u = x²+1, du = 2x dx → x dx = du/2. Limits: x=0, u=1; x=2, u=5. Integral becomes ½ ∫₁⁵ √u du = ½ × (2/3) u³/² |₁⁵ = ⅓ (5³/² - 1³/²) = ⅓(5√5 - 1). Exact surd form is expected.
换元积分和曲线下面积也会出现。计算 ∫₀² x √x²+1 dx。令 u = x²+1，du = 2x dx → x dx = du/2。积分限：x=0 时 u=1；x=2 时 u=5。积分变为 ½ ∫₁⁵ √u du = ½ × (2/3) u³/² |₁⁵ = ⅓ (5³/² - 1³/²) = ⅓(5√5 - 1)。精确的根式形式是预期答案。
9. Probability and Combinatorics: Logical Counting | 概率与组合：逻辑计数
tations, and combinations systematically. How many ways can 5 books be arranged on a shelf? 5! = 120. If three specific books must be together, treat them as one block: 3! × 3! ting without a calculator demands clear organisation to avoid double-counting or omission.
非计算器的概率题考查你系统处理阶乘、排列与组合的能力。5 本书在书架上排列有多少种方式？5! = 120。若某三本书必须放在一起，可将它们视为一个整体：3! × 3! = 6 × 6 = 36。不使用计算器进行计数要求思路清晰，以避免重复或遗漏。
/15. Always simplify fractions to their lowest terms. Expect answers like 2/15 rather than approximate decimals.
条件概率可通过画树形图和约分来解决。一个袋子装有 4 粒红弹珠和 6 粒蓝弹珠，不放回地抽取两粒。两粒都是红色的概率为：(4/10) × (3/9) = 12/90 = 2/15。务必把分数化为最简形式。预期答案如 2/15，而非近似小数。
10. Geometry and Vectors: Coordinate-Free Thinking | 几何与向量：无坐标系思维
s are perpendicular (90°). Recognising special results like orthogonal vectors saves time.
非计算器的几何题常涉及角度性质、圆定理和向量模的计算。求向量 a = 2i + j 与 b = i - 2j 之间的夹角。使用点积：a·b = 2×1 + 1×(-2) = 0，因此两向量垂直（90°）。快速识别正交等特殊结果可以节省时间。
ntre (-1,2), radius 3. Tangent gradients and chord lengths follow algebraically. Vector line intersections require solving parametric equations exactly, yielding neat fractions or surds.
圆的方程如 x² + y² + 2x - 4y - 4 = 0 必须通过配方法重排：(x+1)² + (y-2)² = 9。圆心 (-1,2)，半径 3。切线斜率和弦长通过代数推导得出。向量直线交点需要精确求解参数方程，得到整洁的分数或根式。
11. Exam Strategies: Time Management and Checking | 考试策略：时间管理与检查
Approach non-calculator questions by first identifying what exact form the answer should take. Read the question: 'Give your answer in the form a + b√c' or 'leaving your answer in terms of π'. This tells you to avoid decimals entirely. Structure your working stepwise; marks are awarded for method even if a sign error occurs later.
应对非计算器题目时，首先确定答案应为何种精确形式。仔细读题：“将答案表示为 a + b√c 的形式”或“答案保留 π”。这提示你彻底避免小数。将解题过程步骤化；即使后面出现符号错误，方法正确也能得分。
This mental check helps catch algebra slips. When simplifying surds, verify by squaring: (√3 + 1)² = 4 + 2√3. Such habits build confidence and accuracy under time pressure.
使用估算和合理性检查。解方程前，先预测近似值：若 log₂(x-1) = 5，则 x-1 ≈ 32，x ≈ 33。这种心算核对有助于发现代数错误。化简根式时，通过平方验证：(√3 + 1)² = 4 + 2√3。这些习惯能在时间压力下增强信心与准确性。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
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