📚 IB Math Review Set 8A – Non-Calculator | IB 数学复习题集 8A – 无计算器
This review set is designed to sharpen the essential non-calculator skills required for IB Mathematics, covering algebraic manipulation, functions, trigonometry, sequences, probability, and introductory calculus. Each problem demands analytical reasoning and precise manual computation – exactly the competencies assessed in the paper without a calculator. Work through the exercises systematically, and use the paired English-Chinese explanations to clarify any conceptual difficulties.
这套复习题集旨在磨砺 IB 数学无计算器部分所需的核心技能，涵盖代数运算、函数、三角学、数列、概率和微积分基础。每道题都要求分析性推理和精准的手工计算——正是无计算器试卷所考查的能力。请系统性地完成练习，并借助中英对照的解说扫清概念障碍。
1. Algebraic Simplification & Factorisation | 代数化简与因式分解
Simplify and factorise expressions without a calculator, using common factors, grouping, and special products. This foundational skill is tested heavily in the non-calculator section to ensure fluency with algebraic structures.
不使用计算器化简和因式分解表达式，运用公因式、分组与特殊乘积。这项基础技能在无计算器部分大量考查，以确保对代数结构的熟练掌握。
Example – Factorise completely: 6x³ – 24x. Begin by extracting the greatest common factor 6x, giving 6x(x² – 4). Recognise the difference of two squares: x² – 4 = (x – 2)(x + 2). Thus the fully factorised form is 6x(x – 2)(x + 2).
示例——完全因式分解：6x³ – 24x。先提取最大公因式 6x，得到 6x(x² – 4)。识别平方差：x² – 4 = (x – 2)(x + 2)。因此完全分解形式为 6x(x – 2)(x + 2)。
2. Solving Equations and Inequalities | 解方程与不等式
Linear, quadratic, and radical equations appear frequently; non-calculator problems demand logical steps and checking solutions. Inequalities require sign analysis and careful handling of negative multipliers.
线性、二次和根式方程经常出现；无计算器问题要求逻辑步骤和验根。不等式则需要符号分析并谨慎处理负乘数。
Solve for x: √x + 3 = x – 3. Square both sides: x + 3 = x² – 6x + 9. Rearrange to 0 = x² – 7x + 6, factor as (x – 1)(x – 6) = 0, so x = 1 or x = 6. Verify in the original equation: x = 1 gives √4 = –2 (false), while x = 6 gives √9 = 3 (true). Hence x = 6 is the only solution.
解 x：√x + 3 = x – 3。两边平方：x + 3 = x² – 6x + 9。整理得 0 = x² – 7x + 6，因式分解为 (x – 1)(x – 6) = 0，故 x = 1 或 x = 6。代入原方程检验：x = 1 给出 √4 = –2（假），而 x = 6 给出 √9 = 3（真）。因此唯一解是 x = 6。
3. Functions and Inverse Functions | 函数与反函数
Evaluate composite functions and find inverses algebraically. Remember that the domain of the inverse is the range of the original function; non-calculator tasks often ask for inverse expressions and specific values.
计算复合函数并用代数方法求反函数。记住反函数的定义域是原函数的值域；无计算器题目常要求写出反函数表达式并求特定值。
Given f(x) = 2x + 1 and g(x) = x² – 1 for x ≥ 0, find (g ∘ f)(3) and determine f⁻¹(x). First, f(3) = 7, then g(7) = 48, so (g ∘ f)(3) = 48. For the inverse, set y = 2x + 1, swap x and y: x = 2y + 1, solve for y: y = (x – 1)/2, hence f⁻¹(x) = (x – 1)/2.
若 f(x) = 2x + 1，g(x) = x² – 1（x ≥ 0），求 (g ∘ f)(3) 并确定 f⁻¹(x)。首先 f(3) = 7，然后 g(7) = 48，故 (g ∘ f)(3) = 48。求反函数：设 y = 2x + 1，交换 x 与 y 得 x = 2y + 1，解出 y = (x – 1)/2，因此 f⁻¹(x) = (x – 1)/2。
4. Quadratic Functions & the Discriminant | 二次函数与判别式
Analyse parabolas, vertex form, and the discriminant Δ = b² – solving. Questions often require finding k values that give real, equal, or no real roots.
分析抛物线、顶点式和判别式 Δ = b² – 4ac 以确定根的性质而不必求解。题目常要求找出使方程有实根、等根或无实根的 k 值。
quadratic 2x² + kx + 3 = 0 to have exactly one real root, set the discriminant to zero: k² – 4(2)(3) = 0 → k² – 24 = 0, so k = ±√24 = ±2√6. The unique root then occurs at x = –b/(2a) = –k/4.
要使二次方程 2x² + kx + 3 = 0 恰有一个实根，令判别式为零：k² – 4(2)(3) = 0 → k² – 24 = 0，故 k = ±√24 = ±2√6。此时的唯一根为 x = –b/(2a) = –k/4。
5. Exponents and Logarithms | 指数与对数
te powers and logs without a calculator, applying laws like logₐ(xy) = logₐx + logₐy and alogₐx pect questions that simplify expressions or solve exponential equations by equating bases.
无需计算器处理幂与对数，应用 logₐ(xy) = logₐx + logₐy 和 alogₐx = x 等法则。常见题型为化简表达式或通过化为同底求解指数方程。
Solve 32x – 1 = 27√3. Express both sides as powers of 3: 27√3 = 3³ · 3½ = 33 + ½ = 37/2. Thus 2x – 1 = 7/2, so 2x = 9/2, x = 9/4.
解 32x – 1 = 27√3。将两边表示为 3 的幂：27√3 = 3³ · 3½ = 33 + ½ = 37/2。因此 2x – 1 = 7/2，得 2x = 9/2，x = 9/4。
6. Sequences and Series | 数列与级数
Arithmetic and geometric sequences are core non-calculator topics. You must identify the common difference or ratio, derive the nth term, and sum a finite number of terms using standard formulas – all through symbolic manipulation.
等差与等比数列是无计算器的核心主题。你需要识别公差或公比，推导第 n 项，并使用标准公式手算有限项和——全靠符号运算。
Example – s 22 and the 12th term is 57. Find the first term and common difference. Using uₙ = a + (n – 1)d: a + 4d = 22, a + 11d = 57. Subtract: 7d = 35, so d = 5. Then a = 22 – 4·5 = 2. The sum of the first 20 terms is S₂₀ = 20/2 [2·2 + 19·5] = 10 × (4 + 95) = 990.
示例——某等差数列的第 5 项为 22，第 12 项为 57。求首项与公差。利用 uₙ = a + (n – 1)d：a + 4d = 22，a + 11d = 57。两式相减得 7d = 35，故 d = 5。那么 a = 22 – 4·5 = 2。前 20 项之和 S₂₀ = 20/2 [2·2 + 19·5] = 10 × (4 + 95) = 990。
7. Binomial Theorem | 二项式定理
pecific terms, such as the term independent of x, using the general term Tₖ = nCₖ aⁿ⁻ᵏ bᵏ.
使用二项式系数 nCr 或其公式手算展开 (a + b)ⁿ。重点在于利用通项 Tₖ = nCₖ aⁿ⁻ᵏ bᵏ 找出特定项，如与 x 无关的项。
Find the coefficient of x³ in the expansion of (2 + x)⁵. The general term is 5Cᵣ · 2⁵⁻ʳ · xʳ e need r = 3: 5C₃ = 10, 2² = 4, so the term is 10·4·x³ = 40x³, giving a coefficient of 40.
求 (2 + x)⁵ 展开式中 x³ 的系数。通项为 5Cᵣ · 2⁵⁻ʳ · xʳ。需要 r = 3：5C₃ = 10，2² = 4，该项为 10·4·x³ = 40x³，系数为 40。
8. Trigonometric Identities & Equations | 三角恒等式与方程
ntities such as sin²θ + cos²θ = 1. Solve equations by factoring and using the unit circle.
无计算器的三角学依赖于关键角（0°、30°、45°、60°、90°）的正弦、余弦、正切精确值，以及 sin²θ + cos²θ = 1 等核心恒等式。通过因式分解和单位圆来解方程。
Solve 2sin²θ – sinθ – 1 = 0 for 0° ≤ θ ≤ 360°. Factor as (2sinθ + 1)(sinθ – 1) = 0, so sinθ = –½ or sinθ = 1. From the unit circle, sinθ = –½ at 210° and 330°; sinθ = 1 at 90°. Thus θ = 90°, 210°, 330°.
在 0° ≤ θ ≤ 360° 内解 2sin²θ – sinθ – 1 = 0。因式分解为 (2sinθ + 1)(sinθ – 1) = 0，故 sinθ = –½ 或 sinθ = 1。根据单位圆，sinθ = –½ 位于 210° 和 330°；sinθ = 1 位于 90°。因此 θ = 90°、210°、330°。
9. Probability and Counting Principles | 概率与计数原理
Compute probabilities without a calculator by using tree diagrams, permutations, combinations, and the addition/multiplication rules. Simplify fractions and leave answers in exact form.
利用树状图、排列、组合以及加法/乘法规则手算概率。化简分数并保持精确形式。
en red): P(RB) = (4/10)×(6/9) = 24/90, P(BR) = (6/10)×(4/9) = 24/90. Total = 48/90 = 8/15.
盒中有 4 个红球和 6 个蓝球。无放回地抽取两球，求恰好有一个红球的概率。可能情况为（先红后蓝）或（先蓝后红）：P(RB) = (4/10)×(6/9) = 24/90，P(BR) = (6/10)×(4/9) = 24/90。合计 48/90 = 8/15。
10. Calculus Concepts – Differentiation from First Principles | 微积分概念 – 从第一性原理求导
Non-calculator differentiation emphasises the limit definition: f'(x) = limₕ→₀ [f(x+h) – ent is essential to evaluate the derivative for polynomials and simple rational functions.
无计算器的微分强调极限定义：f'(x) = limₕ→₀ [f(x+h) – f(x)]/h。对多项式和简单有理函数而言，差商的代数化简是求导的关键。
Use first principles to differentiate f(x) = x² + 2x. Set up [ (x+h)² + 2(x+h) – (x² + 2x) ] / h = [ x² + 2xh + h² + 2x + 2h – x² – 2x ] / h = (2xh + h² + 2h)/h = 2x + h + 2. As h → 0, f'(x) = 2x + 2.
用第一性原理求 f(x) = x² + 2x 的导数。构建差商：[ (x+h)² + 2(x+h) – (x² + 2x) ] / h = [ x² + 2xh + h² + 2x + 2h – x² – 2x ] / h = (2xh + h² + 2h)/h = 2x + h + 2。令 h → 0，得 f'(x) = 2x + 2。
11. Integration as Reverse Differentiation | 积分作为微分的逆运算
Basic integration without a calculator involves finding antiderivatives of powers, exponential functions, and simple trigonometric functions. Include the constant of integration, and apply boundary conditions to determine it.
无计算器的基础积分包括求幂函数、指数函数和简单三角函数的反导数。务必加上积分常数，并利用边界条件确定该常数。
Find f(x) if f'(x) = 4x³ – 3x² + 2 and f(1) = 5. Integrate: f(x) = x⁴ – x³ + 2x + C. Substitute x = 1: 1 – 1 + 2 + C = 5, so 2 + C = 5, C = 3. Hence f(x) = x⁴ – x³ + 2x + 3.
已知 f'(x) = 4x³ – 3x² + 2 且 f(1) = 5，求 f(x)。积分得：f(x) = x⁴ – x³ + 2x + C。代入 x = 1：1 – 1 + 2 + C = 5，故 2 + C = 5，C = 3。因此 f(x) = x⁴ – x³ + 2x + 3。
12. Vectors and Geometry | 向量与几何
Non-calculator vector problems involve magnitude, dot product, and angle between vectors. Exact values and simplified surd forms are expected; leave answers such as √13 rather than decimals.
无计算器的向量问题涉及模长、点积和向量间夹角。要求使用精确值与最简根式；答案保留如 √13，而非小数。
Given vectors a = 3i + 4j and b = i – 2j, find the angle between them. Dot product: a·b = 3×1 + 4×(–2) = 3 – 8 = –5. Magnitudes: |a| = √9+16 = 5, |b| = √1+4 = √5. cosθ = –5/(5√5) = –1/√5. Hence θ = arccos(–1/√5), which is an obtuse angle that must be stated exactly.
给定向量 a = 3i + 4j 和 b = i – 2j，求它们的夹角。点积：a·b = 3×1 + 4×(–2) = 3 – 8 = –5。模长：|a| = √9+16 = 5，|b| = √1+4 = √5。cosθ = –5/(5√5) = –1/√5。因此 θ = arccos(–1/√5)，为钝角，须以精确形式表示。
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