📚 Review Set 24A – Non-Calculator | IB数学非计算器复习集24A
Welcome to Review Set 24A, a comprehensive non-calculator workout designed specifically for IB Mathematics students. In both Analysis & Approaches and Applications & Interpretation, paper sections that forbid calculators demand fluency in exact algebraic manipulation, trigonometric values, logarithmic properties, and calculus shortcuts. This article revisits the most frequently tested topics, presents concise worked examples, and highlights common pitfalls. Use it as a self‑check before your next mock exam, ensuring every step is grounded in precise, calculator‑free reasoning.
欢迎来到复习集24A，这是一套专为IB数学学生精心设计的非计算器练习。在分析与方法和应用与解释课程中，禁止使用计算器的试卷部分要求你熟练掌握精确的代数运算、三角特殊值、对数性质以及微积分速算技巧。本文重温最高频的考点，给出精炼的例题并指出常见误区。在下次模拟考之前，用它进行一次自我检测，让你的每一步推理都不依赖计算器，做到准确无误。
1. Algebraic Manipulation & Factoring | 代数变形与因式分解
Non‑calculator algebra rewards pattern recognition. Start by simplifying rational expressions through factoring quadratics and cancelling common factors. For instance, consider (x² – 4) / (x² – x – 6). Factor numerator as (x–2)(x+2) and denominator as (x–3)(x+2). Provided x ≠ –2, the expression reduces to (x–2)/(x–3). Always state domain restrictions explicitly; marks are often lost by ignoring excluded values.
非计算器代数问题看重模式识别。从因式分解二次式、约去公因子入手，化简有理表达式。例如 (x² – 4) / (x² – x – 6)。将分子分解为 (x–2)(x+2)，分母分解为 (x–3)(x+2)。当 x ≠ –2 时，表达式化简为 (x–2)/(x–3)。务必明确写出定义域限制；忽略不允许的取值常会导致失分。
When dealing with higher‑degree polynomials, the factor theorem is your ally. If P(c)=0, then (x–c) is a factor. In a non‑ e the degree, then factorise the quotient. Mastery of completing the square is equally essential for deriving vertex form and solving quadratic equations with irrational roots given in exact surd form.
处理高次多项式时，因式定理是你的利器。若 P(c)=0，则 (x–c) 为一个因式。在无计算器情况下，试用 ±1, ±2, ±3 等明显的候选值。找到一个因式后，用多项式长除法或综合除法降低次数，再对商式进行因式分解。同样重要的是熟练掌握配方法，以便推导顶点式，并用精确根式形式表示无理根。
2. Exponents and Logarithms | 指数与对数
Exact manipulation of indices and logs is a non‑ ws aᵐ·aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, and a⁻¹ = 1/a must be second nature. Solve equations like 2²ˣ = 8ˣ⁻¹ by expressing both sides with base 2: 2²ˣ = (2³)ˣ⁻¹ → 2²ˣ = 2³ˣ⁻³, hence 2x = 3x–3, giving x=3. No logarithmic form needed, but check your answer.
精确处理指数与对数是不可或缺的技能。法则 aᵐ·aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ 以及 a⁻¹ = 1/a 必须熟练到如同本能。求解 2²ˣ = 8ˣ⁻¹ 时，将两边化为同底数 2：2²ˣ = (2³)ˣ⁻¹ → 2²ˣ = 2³ˣ⁻³，于是 2x = 3x–3，解得 x=3。整个过程无需对数形式，但要记得检验。
Logarithmic equations require careful domain checking. For log₂(x) + log₂(x–2) = 3, combine to log₂[x(x–2)] = 3, then rewrite as x(x–2) = 2³ = 8. This yields x² – 2x – 8 = 0, so (x– st be discarded because log arguments must be positive, leaving x=4. Memorising the change‑of‑ logₐb = log_c(b)/log_c(a), helps when dealing with unfamiliar bases in exact calculations.
对数方程需要仔细检查定义域。对于 log₂(x) + log₂(x–2) = 3，合并为 log₂[x(x–2)] = 3，再写成指数形式 x(x–2) = 2³ = 8，得到 x² – 2x – 8 = 0，分解得 (x–4)(x+2)=0。必须舍弃负数解，因为真数必须为正，最终 x=4。熟记换底公式 logₐb = log_c(b)/log_c(a)，有助于在精确计算中处理不熟悉的底数。
3. Functions and Inverse Functions | 函数与反函数
Finding an inverse function f⁻¹(x) on a non‑calculator paper tests your algebraic rearrangement. Given f(x)=√2x–3, with domain x ≥ 1.5, write y = √2x–3, swap x and y, and solve: x = √2y–3 ⇒ x² = 2y–3 ⇒ h domain x ≥ 0 to match the range of f. Always state the domain of the inverse explicitly.
在非计算器试卷上求反函数 f⁻¹(x)，考察的是你的代数变形能力。对于 f(x)=√2x–3，定义域 x ≥ 1.5，设 y = √2x–3，交换 x 和 y，再求解：x = √2y–3 ⇒ x² = 2y–3 ⇒ y = (x²+3)/2。反函数为 f⁻¹(x) = (x²+3)/2，定义域 x ≥ 0 以匹配原函数的值域。一定要明确给出反函数的定义域。
Composite functions, such as (f∘ ul attention to the order of operations and domain restrictions. If g(x) = ln x and f(x)=eˣ, then (f∘g)(x)=eln x=x for x>0. Watch out for ‘ unction’s range falls outside the outside function’s domain. Practise sketching key graphs—reciprocal, cubic, absolute‑value—to answer transformation questions without a calculator.
复合函数如 (f∘g)(x)=f(g(x))，需要格外注意运算顺序和定义域限制。若 g(x)=ln x，f(x)=eˣ，则 (f∘g)(x)=eln x=x，x>0。要当心“定义域缺口”——当内层函数的值域超出外层函数定义域时产生的限制。练习描绘倒数、三次、绝对值等关键函数图像，以便在不借助计算器的情况下回答图像变换问题。
4. Trigonometry: Exact Values and Identities | 三角学：精确值与恒等式
Non‑ 0°, 30°, 45°, 60°, 90° and their radian equivalents. Use the special triangles (45° right‑ antly saves time and prevents rounding errors. Extend these values to the unit circle for angles in other quadrants using CAST or symmetry.
非计算器三角问题依赖于 sin、cos、tan 在 0°、30°、45°、60°、90° 及其弧度对应值上的精确记忆。借助特殊三角形（45° 直角三角形，或等边三角形一分为二）可快速回想 sin(π/6)=1/2，cos(π/3)=1/2，tan(π/4)=1。熟记这些值能节省时间，避免舍入误差。利用单位圆和 CAST 法则（或对称性）将这些值推广到其他象限。
Trigonometric identities are essential for simplifying equations. A classic non‑ ≤ x ≤ 2π. Replace sin 2x with 2 sin x cos x, giving 2 sin x cos x = cos x → cos x (2 sin x – derived forms like 1 + tan²θ = sec²θ, can reduce many expressions to a single trig ratio.
三角恒等式是化简方程的关键。一个经典的非计算器问题：求解 sin 2x = cos x，0 ≤ x ≤ 2π。用 2 sin x cos x 替换 sin 2x，得到 2 sin x cos x = cos x → cos x (2 sin x – 1)=0。于是 cos x=0 或 sin x=1/2。精确解为 x=π/2, 3π/2, π/6, 5π/6。答案务必用 π 表示，并检查给定区间。平方恒等式 sin²θ + cos²θ = 1 及其推导形式 1 + tan²θ = sec²θ，能将许多式子化为单一三角函数。
5. Differentiation Techniques | 导数技巧
nctions fluently. Memorise the power rule d/dx [xⁿ] = n xⁿ⁻¹, the exponential rule d/dx [eˣ] = eˣ log rule d/dx [ln x] = 1/x. Trig derivatives such as d/dx[sin x] = cos x and d/dx[cos x] = –sin x should be at your fingertips. For example, differentiate y = 4x³ – 2/x + 5eˣ – 3 ln x. Write 2/x as 2x⁻¹, then y' = 12x² + 2x⁻² + 5eˣ – 3/x.
在没有计算器的情况下，你必须能流畅地对标准函数求导。牢记幂法则 d/dx [xⁿ] = n xⁿ⁻¹，指数法则 d/dx [eˣ] = eˣ，自然对数法则 d/dx [ln x] = 1/x。三角函数的导数如 d/dx[sin x] = cos x 和 d/dx[cos x] = –sin x 必须信手拈来。例如，求 y = 4x³ – 2/x + 5eˣ – 3 ln x 的导数。将 2/x 写作 2x⁻¹，则 y' = 12x² + 2x⁻² + 5eˣ – 3/x。
2x+1)⁴. For y = x²·ln x, product rule: dy/dx = 2x·ln x + x²·(1/x) = 2x ln x + x. For y = eˣ/(x+1), quotient rule: dy/dx = [eˣ(x+1) – eˣ·1]/(x+1)² = x eˣ/(x+1)². Always simplify factorised forms; examiners expect neatly presented derivatives.
链式法则、乘法法则和除法法则是常考内容。对于 y = (2x+1)⁵，链式法则给出 dy/dx = 5(2x+1)⁴·2 = 10(2x+1)⁴。对于 y = x²·ln x，用乘法法则：dy/dx = 2x·ln x + x²·(1/x) = 2x ln x + x。对于 y = eˣ/(x+1)，除法法则：dy/dx = [eˣ(x+1) – eˣ·1]/(x+1)² = x eˣ/(x+1)²。始终保持因式分解形式；阅卷人期望看到整洁的导数结果。
| Function f(x) | Derivative f'(x) |
|---|---|
| k·xⁿ | k·n xⁿ⁻¹ |
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | –sin x |
| tan x | sec² x |
6. Integration Fundamentals | 积分基础
reverse of differentiation. The power rule becomes ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, valid for n ≠ –1. Learn the standard integrals: ∫ eˣ dx = eˣ + C, ∫ 1/x dx = ln|x| + C, ∫ cos x dx = sin x + C, ∫ sin x dx = –cos x + C. A problem like ∫(4x³ – 6x + 5/x) dx yields x⁴ – 3x² + 5 ln|x| + C. Never forget the constant of integration; C is mandatory.
不定积分是微分的逆运算。幂法则变为 ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，n ≠ –1。熟记标准积分：∫ eˣ dx = eˣ + C，∫ 1/x dx = ln|x| + C，∫ cos x dx = sin x + C，∫ sin x dx = –cos x + C。像 ∫(4x³ – 6x + 5/x) dx 这样的题目，结果为 x⁴ – 3x² + 5 ln|x| + C。切勿遗漏积分常数 C，这是必写项。
es. Evaluate ∫₁² (3x²+2) dx: first integrate to x³ + 2x, then substitute limits: [2³+2(2)] – [1³+2(1)] = (8+4) – (1+2) = 9. A non‑ n points algebraically. Always sketch a quick sign diagram to ensure the area is positive.
定积分常用于求曲线下的面积。计算 ∫₁² (3x²+2) dx：先积分得 x³ + 2x，再代入上下限：[2³+2(2)] – [1³+2(1)] = (8+4) – (1+2) = 9。非计算器试卷可能要求计算直线与抛物线之间的面积；用代数方法求出交点后，构造函数差的积分。务必快速画出示意图，确保面积取正值。
7. Sequences, Series and Binomial Expansion | 数列、级数与二项式展开
Arithmetic and geometric sequences appear frequently in non‑calculator sections. For an arithmetic sequence, the nth term is uₙ = u₁ + (n–1)d and sum Sₙ = n/2 [2u₁ + (n–1)d]. A question might ask: find the sum of the first 20 terms of 3, 7, 11, … eep answers exact; no rounding needed. For geometric sequences, uₙ = u₁ rⁿ⁻¹ and Sₙ = u₁(1–rⁿ)/(1–r) for r≠1. Recognise infinite convergent series when |r|<1, where S∞ = u₁/(1–r).
等差数列和等比数列在非计算器部分频繁出现。等差数列的通项为 uₙ = u₁ + (n–1)d，求和公式 Sₙ = n/2 [2u₁ + (n–1)d]。题目可能要求：求数列 3, 7, 11, … 前 20 项的和。这里 u₁=3, d=4，因此 S₂₀ = 20/2 [2(3)+19(4)] = 10(6+76)=820。答案保持精确，无需四舍五入。等比数列的通项为 uₙ = u₁ rⁿ⁻¹，求和为 Sₙ = u₁(1–rⁿ)/(1–r)，r≠1。当 |r|<1 时，识别无穷收敛级数，其和为 S∞ = u₁/(1–r)。
The binomial expansion (a+b)ⁿ = Σ [ⁿCᵣ aⁿ⁻ʳ bʳ] is a key non‑calculator tool. To find the coefficient of x³ in (1+x)⁸, use the general term ⁿCᵣ (1)⁸⁻ʳ xʳ. Set r=3: ⁸C₃ = 56. So the term is 56x³. For more complex expansions like (2–3x)⁵, write the term as ⁵Cᵣ (2)⁵⁻ʳ (–3x)ʳ onal indices for infinite series expansions, ensuring you state the radius of convergence.
二项式展开 (a+b)ⁿ = Σ [ⁿCᵣ aⁿ⁻ʳ bʳ] 是非计算器的重要工具。求 (1+x)⁸ 中 x³ 的系数，使用通项 ⁿCᵣ (1)⁸⁻ʳ xʳ，令 r=3：⁸C₃ = 56，该项为 56x³。对于更复杂的展开如 (2–3x)⁵，写出 ⁵Cᵣ (2)⁵⁻ʳ (–3x)ʳ。练习处理负指数或分数指数的无穷级数展开，并确保给出收敛半径。
8. Vectors Basics | 向量基础
Vector questions on a non‑calculator paper rely on algebraic components. Given a = (2, – nd b = (1, 2, k), find k such that a is perpendicular to b. The dot product a·b = 2(1) + (–1)(2) + 3(k) = 2 – 2 + 3k = 3k. For perpendicularity, a·b=0 ⇒ 3k=0 ⇒ k=0. For parallel vectors, one must be a scalar multiple of the other.
非计算器试卷上的向量题依赖代数分量运算。已知 a = (2, –1, 3) 与 b = (1, 2, k)，求使 a 垂直于 b 的 k 值。点积 a·b = 2(1) + (–1)(2) + 3(k) = 2 – 2 + 3k = 3k。垂直时 a·b=0 ⇒ 3k=0 ⇒ k=0。平行向量要求一个向量是另一个的标量倍数。
The length or magnitude of vector v = (x, y, z) is |v| = √x² + y² + z². The angle between two vectors is found via cos θ = (a·b)/(|a||b| A typical problem gives exact values, such as finding the angle between (1,2,2) and (2,0,–1). Compute a·b = 1·2 + 2·0 + 2·(– immediately. Vector equation of a line r = a + λd is essential for intersection problems.
向量 v = (x, y, z) 的模长 |v| = √x² + y² + z²。两向量夹角由 cos θ = (a·b)/(|a||b|) 求出。典型题目会给出精确数值，例如求 (1,2,2) 与 (2,0,–1) 的夹角。计算点积：1·2 + 2·0 + 2·(–1) = 0，因此直接得夹角为 90°。直线的向量方程 r = a + λd 对于求交点问题至关重要。
9. Complex Numbers (HL) | 复数（高级课程）
uently without a calculator. Be comfortable with the form z = a + bi, its conjugate z* = a – bi, and modulus |z| = √a² + b². Solving quadratics like x² – 4x + 13 = 0 via the quadratic formula yields x = [4 ± √16 – 52]/2 = [4 ± √–36]/2 = 2 ± 3i. Present answers in Cartesian form unless otherwise specified.
对高级课程学生而言，复数常以非计算器形式出现。要熟练处理 z = a + bi 的形式、其共轭 z* = a – bi 以及模 |z| = √a² + b²。求解二次方程如 x² – 4x + 13 = 0，利用求根公式得 x = [4 ± √16 – 52]/2 = [4 ± √–36]/2 = 2 ± 3i。除非另有说明，答案应以代数形式给出。
is done by multiplying numerator and denominator by the conjugate. For example, (1+2i)/(3–i) = (1+2i)(3+i)/(9+1) = (3 + i + 6i – 2)/10 = (1+7i)/10 = 0.1 + 0.7i. Recognising patterns like |z – (a+bi)| = r as a circle centre (a, b) radius r helps solve locus problems without technology.
阿尔冈图推理同样会被考查。乘以 i 会使复数逆时针旋转 90°。复数除法通过分子分母同乘共轭来实现。例如 (1+2i)/(3–i) = (1+2i)(3+i)/(9+1) = (3 + i + 6i – 2)/10 = (1+7i)/10 = 0.1 + 0.7i。识别 |z – (a+bi)| = r 表示圆心为 (a, b)、半径为 r 的圆，有助于在不使用技术的情况下解决轨迹问题。
10. Probability and Combinatorics (Exact Fractions) | 概率与组合（精确分数）
Non‑ probability demands exact fractions or simplified surds. Start with combinatorics: nCr = n!/(r!(n–r)!) and nPr = n!/(n–r)! 4 girls, what is the probability a randomly chosen committee of 3 contains exactly 2 girls? able outcomes = ⁴C₂·⁵C₁ = 6·5 = 30, total outcomes = ⁹C₃ = 84, probability = 30/84 = 5/14.
非计算器概率要求给出精确分数或简化根式。从组合数学入手：nCr = n!/(r!(n–r)!) 与 nPr = n!/(n–r)!。要能通过约分阶乘手动算出诸如 ⁷C₃ = 35 的值。典型题目可能为：从 5 男 4 女中随机选取一个 3 人委员会，求恰好有 2 名女生的概率。有利结果数 = ⁴C₂·⁵C₁ = 6·5 = 30，总结果数 = ⁹C₃ = 84，概率 = 30/84 = 5/14。
For binomial distribution probabilities, the formula P(X=r) = ⁿCᵣ pʳ (1–p)ⁿ⁻ʳ yields exact rational answers when p is a simple fraction. If X~ ith conditional probability; label branches with exact fractions and multiply along paths.
对于二项分布概率，公式 P(X=r) = ⁿCᵣ pʳ (1–p)ⁿ⁻ʳ 当 p 为简单分数时可给出精确有理数答案。若 X~B(4, 1/2)，则 P(X=2) = ⁴C₂ (1/2)² (1/2)² = 6·(1/4)·(1/4) = 6/16 = 3/8。务必彻底化简。树状图有助于处理条件概率；用精确分数标注各分支，并沿路径相乘。
11. Proof Techniques and Logical Reasoning | 证明方法与逻辑推理
IB non‑ assuming the statement holds for n=k, then proving it for n=k+1. A common example: prove ∑r=1n 1)(k/2 + 1) = (k+1)(k+2)/2, which matches the formula. Conclude by principle of induction.
IB 非计算器试卷可能包含简短的证明题，通常涉及数学归纳法或直接推导。数学归纳法需要证明基础情形（如 n=1）成立，假设命题对 n=k 成立，然后证明对 n=k+1 成立。一个常见例子：证明 ∑r=1n r = n(n+1)/2。n=1 时，左=1，右=1(2)/2=1。假设对 k 成立，则对 k+1：和 = k(k+1)/2 + (k+1) = (k+1)(k/2 +
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