IB Mathematics Review Set 1A – Non-Calculator Practice | IB数学复习题集1A – 非计算器练习

📚 IB Mathematics Review Set 1A – Non-Calculator Practice | IB数学复习题集1A – 非计算器练习

This article revisits the essential non-calculator skills tested in IB Mathematics Review Set 1A. We consolidate algebraic manipulation, equation solving, exponential and logarithmic concepts, function evaluation, sequences, binomial expansions, and surd operations. Each section pairs an English explanation with a Chinese equivalent, ensuring clarity for bilingual learners. Work through the examples manually to strengthen the symbolic fluency required for Paper 1 or non-calculator components.

本文回顾了IB数学复习题集1A中必须掌握的非计算器技能。我们整合了代数变形、方程求解、指数与对数概念、函数求值、数列、二项式展开和根式运算。每个部分都将英文讲解与中文对应内容配对，确保双语学习者清晰理解。请手动完成示例，以强化试卷一或其他非计算器部分所需的符号运算能力。

1. Algebraic Expansion and Simplification | 代数展开与化简

Expand products of binomials and collect like terms carefully. Every sign matters when multiplying manually. Watch for the distributive law: a(b + c) = ab + ac. For two binomials, apply the FOIL method (First, Outer, Inner, Last) or the grid approach.

仔细展开二项式的乘积并合并同类项。手动乘法时每个符号都很重要。注意分配律：a(b + c) = ab + ac。对于两个二项式，可以使用FOIL法则（先外内后）或网格法。

Example: Expand and simplify (3x − 2)(x + 5) − 4x(x − 1).
Solution: First expand (3x − 2)(x + 5) = 3x² + 15x − 2x − 10 = 3x² + 13x − 10. Then expand 4x(x − 1) = 4x² − 4x. Subtract to get (3x² + 13x − 10) − (4x² − 4x) = 3x² + 13x − 10 − 4x² + 4x = −x² + 17x − 10.

例题：展开并简化 (3x − 2)(x + 5) − 4x(x − 1)。
解：先展开 (3x − 2)(x + 5) = 3x² + 15x − 2x − 10 = 3x² + 13x − 10。再展开 4x(x − 1) = 4x² − 4x。相减得 (3x² + 13x − 10) − (4x² − 4x) = 3x² + 13x − 10 − 4x² + 4x = −x² + 17x − 10。

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2. Factorisation Techniques | 因式分解技巧

Factorising converts a polynomial into a product of simpler expressions. Always look for a common factor first. For quadratics of the form ax² + bx + c, find two numbers that multiply to ac and add to b. For difference of squares: a² − b² = (a − b)(a + b).

因式分解将多项式转化为较简单表达式的乘积。首先总是寻找公因式。对于形如 ax² + bx + c 的二次式，找出两个数使其乘积为 ac 且和为 b。平方差公式：a² − b² = (a − b)(a + b)。

Example: Factorise fully 6x² − 19x + 10.
Product ac = 60. Seek two numbers with product 60 and sum −19: they are −4 and −15. Split the middle term: 6x² − 4x − 15x + 10. Group: 2x(3x − 2) − 5(3x − 2) = (3x − 2)(2x − 5).

例题：对 6x² − 19x + 10 进行完全因式分解。
乘积 ac = 60。寻找两个数，其乘积为 60 且和为 −19：这两个数是 −4 和 −15。拆分中间项：6x² − 4x − 15x + 10。分组：2x(3x − 2) − 5(3x − 2) = (3x − 2)(2x − 5)。

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3. Solving Linear Equations | 解线性方程

Isolate the variable using inverse operations. Keep the equation balanced by performing the same operation on both sides. Clear fractions by multiplying through by the least common denominator.

通过逆运算分离变量。在方程两边同时执行相同运算以保持平衡。通过乘以最小公分母来消去分数。

Example: Solve (2x + 1)/3 − (x − 2)/4 = 1.
Multiply through by 12: 4(2x + 1) − 3(x − 2) = 12. Expand: 8x + 4 − 3x + 6 = 12. Simplify: 5x + 10 = 12, so 5x = 2 and x = 2/5.

例题：解方程 (2x + 1)/3 − (x − 2)/4 = 1。
两边乘以12：4(2x + 1) − 3(x − 2) = 12。展开：8x + 4 − 3x + 6 = 12。化简：5x + 10 = 12，因此 5x = 2，x = 2/5。

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4. Solving Quadratic Equations | 解二次方程

Three main non-calculator methods: factorising, completing the square, and the quadratic formula. Always set the equation to zero first. The discriminant Δ = b² − 4ac tells you the nature of the roots without solving fully.

三种主要的非计算器方法：因式分解、配方法和二次公式。始终先将方程设为零。判别式 Δ = b² − 4ac 可在不完全求解的情况下告知根的性质。

If ax² + bx + c = 0, then x = [−b ± √(b² − 4ac)] / (2a)

Example: Solve x² − 5x + 3 = 0 using the quadratic formula.
Identify a = 1, b = −5, c = 3. Discriminant: (−5)² − 4·1·3 = 25 − 12 = 13. Thus x = (5 ± √13)/2. Exact values are required without a calculator.

例题：用二次公式解 x² − 5x + 3 = 0。
确定 a = 1, b = −5, c = 3。判别式：(−5)² − 4·1·3 = 25 − 12 = 13。因此 x = (5 ± √13)/2。要求给出精确值，无需计算器。

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5. Laws of Exponents | 指数定律

Review the fundamental rules: aᵐ · aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, aᵐ / aⁿ = aᵐ⁻ⁿ, a⁰ = 1 (a ≠ 0), a⁻ⁿ = 1/aⁿ, and a^(m/n) = ⁿ√(aᵐ). Apply these to simplify expressions with integer and rational exponents.

回顾基本法则：aᵐ · aⁿ = aᵐ⁺ⁿ，(aᵐ)ⁿ = aᵐⁿ，aᵐ / aⁿ = aᵐ⁻ⁿ，a⁰ = 1 (a ≠ 0)，a⁻ⁿ = 1/aⁿ，以及 a^(m/n) = ⁿ√(aᵐ)。将这些法则应用于含有整数和有理数指数的表达式化简。

Example: Simplify (8x⁶)^(2/3) ÷ (4x⁻²).
(8x⁶)^(2/3) = 8^(2/3) · x^(6·2/3). Since 8^(2/3) = (∛8)² = 2² = 4, we get 4x⁴. Then divide by 4x⁻²: (4x⁴)/(4x⁻²) = x^(4 − (−2)) = x⁶.

例题：化简 (8x⁶)^(2/3) ÷ (4x⁻²)。
(8x⁶)^(2/3) = 8^(2/3) · x^(6·2/3)。由于 8^(2/3) = (∛8)² = 2² = 4，得到 4x⁴。再除以 4x⁻²：(4x⁴)/(4x⁻²) = x^(4 − (−2)) = x⁶。

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6. Introduction to Logarithms | 对数入门

Logarithms answer the question: to what power must the base be raised to obtain a given number? If bˣ = y, then x = log_b y. Key properties: log_b (MN) = log_b M + log_b N, log_b (M/N) = log_b M − log_b N, log_b (Mᵏ) = k log_b M. Change of base: log_b a = log_c a / log_c b.

对数回答这样一个问题：底数需要升到多少次幂才能得到给定数值？若 bˣ = y，则 x = log_b y。关键性质：log_b (MN) = log_b M + log_b N，log_b (M/N) = log_b M − log_b N，log_b (Mᵏ) = k log_b M。换底公式：log_b a = log_c a / log_c b。

Example: Given log₂ 3 ≈ 1.585 and log₂ 5 ≈ 2.322, evaluate log₂ 45 without a calculator.
Write 45 = 3² × 5. Then log₂ 45 = log₂ (3²) + log₂ 5 = 2 log₂ 3 + log₂ 5 ≈ 2(1.585) + 2.322 = 3.17 + 2.322 = 5.492. The exact expression is 2 log₂ 3 + log₂ 5.

例题：已知 log₂ 3 ≈ 1.585 且 log₂ 5 ≈ 2.322，不借助计算器求 log₂ 45。
将 45 写为 3² × 5。则 log₂ 45 = log₂ (3²) + log₂ 5 = 2 log₂ 3 + log₂ 5 ≈ 2(1.585) + 2.322 = 3.17 + 2.322 = 5.492。精确表达式为 2 log₂ 3 + log₂ 5。

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7. Function Notation and Evaluation | 函数记号与求值

Functions map inputs to outputs. f(x) denotes the value of function f at x. Composite functions f(g(x)) require working from the inside out. For inverse functions f⁻¹(x), swap x and y and solve for y.

函数将输入映射到输出。f(x) 表示函数 f 在 x 处的值。复合函数 f(g(x)) 需要由内向外计算。对于反函数 f⁻¹(x)，交换 x 和 y 并解出 y。

Example: Let f(x) = 2x − 3 and g(x) = x². Find f(g(4)) and f⁻¹(x).
g(4) = 4² = 16. Then f(g(4)) = f(16) = 2(16) − 3 = 32 − 3 = 29. For the inverse, set y = 2x − 3, swap: x = 2y − 3, solve for y: y = (x + 3)/2. Thus f⁻¹(x) = (x + 3)/2.

例题：设 f(x) = 2x − 3，g(x) = x²。求 f(g(4)) 和 f⁻¹(x)。
g(4) = 4² = 16。那么 f(g(4)) = f(16) = 2(16) − 3 = 32 − 3 = 29。对于反函数，令 y = 2x − 3，交换：x = 2y − 3，解出 y：y = (x + 3)/2。因此 f⁻¹(x) = (x + 3)/2。

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8. Arithmetic and Geometric Sequences | 等差数列与等比数列

An arithmetic sequence has a common difference d: uₙ = u₁ + (n − 1)d. Sum of n terms: Sₙ = n/2 [2u₁ + (n − 1)d] or Sₙ = n/2 (u₁ + uₙ). A geometric sequence has a common ratio r: uₙ = u₁ r^(n−1). Sum of n terms (r ≠ 1): Sₙ = u₁ (1 − rⁿ)/(1 − r). For an infinite geometric series with |r| < 1, S∞ = u₁/(1 − r).

等差数列具有公差 d：uₙ = u₁ + (n − 1)d。前 n 项和：Sₙ = n/2 [2u₁ + (n − 1)d] 或 Sₙ = n/2 (u₁ + uₙ)。等比数列具有公比 r：uₙ = u₁ r^(n−1)。前 n 项和（r ≠ 1）：Sₙ = u₁ (1 − rⁿ)/(1 − r)。对于 |r| < 1 的无穷等比级数，S∞ = u₁/(1 − r)。

Example: Find the sum of the first 8 terms of a geometric sequence with u₁ = 3 and r = −0.5.
S₈ = 3 (1 − (−0.5)⁸) / (1 − (−0.5)). (−0.5)⁸ = 1/256. So S₈ = 3 (1 − 1/256) / 1.5 = 3 × (255/256) ÷ (3/2) = (765/256) × (2/3) = 510/256 = 255/128. Exact fraction is acceptable.

例题：求等比数列前8项和，其中 u₁ = 3，r = −0.5。
S₈ = 3 (1 − (−0.5)⁸) / (1 − (−0.5))。(−0.5)⁸ = 1/256。因此 S₈ = 3 (1 − 1/256) / 1.5 = 3 × (255/256) ÷ (3/2) = (765/256) × (2/3) = 510/256 = 255/128。分数精确值即可。

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9. Binomial Expansion | 二项式展开

For (a + b)ⁿ, the expansion uses binomial coefficients: (n choose r) = nCr = n!/(r!(n−r)!). The general term is nCr a^(n−r) bʳ. In IB non-calculator contexts, expand up to small powers like n = 5 or find a specific term. Use Pascal’s triangle for low n.

对于 (a + b)ⁿ，展开式使用二项式系数：(n 选 r) = nCr = n!/(r!(n−r)!)。通项为 nCr a^(n−r) bʳ。在 IB 非计算器情境中，展开至较小幂次如 n = 5，或求特定项。对于低次幂可使用帕斯卡三角形。

Example: Find the term independent of x in the expansion of (2x² − 1/x)⁶.
General term: T₍ᵣ₊₁₎ = 6Cr (2x²)^(6−r) (−1/x)ʳ = 6Cr 2^(6−r) x^(12−2r) (−1)ʳ x⁻ʳ = 6Cr 2^(6−r) (−1)ʳ x^(12−3r). For the constant term, set exponent zero: 12 − 3r = 0 ⇒ r = 4. The term is 6C4 × 2² × (−1)⁴ = 15 × 4 × 1 = 60.

例题：求 (2x² − 1/x)⁶ 展开式中与 x 无关的项。
通项：T₍ᵣ₊₁₎ = 6Cr (2x²)^(6−r) (−1/x)ʳ = 6Cr 2^(6−r) x^(12−2r) (−1)ʳ x⁻ʳ = 6Cr 2^(6−r) (−1)ʳ x^(12−3r)。常数项需指数为零：12 − 3r = 0 ⇒ r = 4。该项为 6C4 × 2² × (−1)⁴ = 15 × 4 × 1 = 60。

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10. Surds and Rationalisation | 根式与有理化

Simplify radicals by extracting perfect squares, cubes, etc. Rationalise denominators by multiplying numerator and denominator by the conjugate when a surd appears in a binomial denominator. Remember (√a + √b)(√a − √b) = a − b.

化简根式时提取完全平方（或立方等）因子。当分母为含有根式的二项式时，将其乘以共轭式以有理化分母。记住 (√a + √b)(√a − √b) = a − b。

Example: Simplify 3/(√5 − 2) + √20.
First rationalise 3/(√5 − 2): multiply top and bottom by (√5 + 2) → 3(√5 + 2)/(5 − 4) = 3√5 + 6. Next simplify √20 = √(4×5) = 2√5. Sum: (3√5 + 6) + 2√5 = 5√5 + 6.

例题：化简 3/(√5 − 2) + √20。
首先有理化 3/(√5 − 2)：分子分母同乘 (√5 + 2) → 3(√5 + 2)/(5 − 4) = 3√5 + 6。然后化简 √20 = √(4×5) = 2√5。总和：(3√5 + 6) + 2√5 = 5√5 + 6。

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11. Mixed Non‑Calculator Practice Tips | 混合非计算器练习提示

IB non-calculator papers reward precise arithmetic, clear reasoning, and comfort with exact forms. Always express answers using simplified fractions, radicals, or terms of π. Avoid decimals unless the question specifies an approximation. Check factoring by expanding, and verify domains for rational and logarithmic functions.

IB非计算器试卷青睐精确的算术、清晰的推理以及对精确形式的熟练掌握。始终用最简分数、根式或 π 的倍数表示答案。除非题目明确要求，否则避免使用小数。通过展开检验因式分解，并检查有理函数和对数函数的定义域。

• Keep an eye on restrictions: denominators ≠ 0, arguments of logs > 0.
• 注意限制条件：分母 ≠ 0，对数的真数 > 0。
• When solving, show steps to earn method marks even if the final answer is incomplete.
• 求解时展示步骤，即使最终答案未完成也能获得方法分。
• Use estimation to verify that your exact answer is plausible.
• 利用估算验证精确答案是否合理。

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12. Summary and Next Steps | 总结与下一步

Review Set 1A covers fundamental algebra, functions, and sequences that appear repeatedly in IB assessments. Regular practice without a calculator builds speed and confidence. Once you have mastered these topics, attempt timed exercises under exam conditions, and then move to Review Set 1B which introduces higher-order problem solving.

复习题集1A涵盖了IB评估中反复出现的基础代数、函数和数列。定期进行非计算器练习可以提高速度并增强信心。掌握这些主题后，在考试条件下尝试限时练习，然后进入复习题集1B，后者会引入更高阶的问题求解。

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