Review Set 18A – Non-Calculator Practice for IB Mathematics | 复习题集18A – IB数学非计算器部分练习

📚 Review Set 18A – Non-Calculator Practice for IB Mathematics | 复习题集18A – IB数学非计算器部分练习

This article provides a detailed walkthrough of a typical non-calculator review set (18A) for IB Mathematics. Non-calculator problems test your ability to manipulate expressions, recall exact values, and apply logical reasoning without technological aid. Working through these examples will strengthen your algebraic fluency and conceptual understanding, essential for success in Paper 1 for both Analysis & Approaches (AA) and Applications & Interpretation (AI).

本文详细讲解了IB数学中典型的非计算器复习题集18A。非计算器题型旨在检验你的表达式变形能力、精确值记忆以及逻辑推理，不使用任何计算工具。通过以下范例的训练，你将增强代数流畅度和概念理解，这对分析与方法(AA)和应用与解释(AI)的试卷一成功至关重要。

1. Polynomial Factorisation | 多项式因式分解

Problem: Factorise fully 3x³ − 12x. Solution: First factor out the greatest common factor, 3x: 3x(x² − 4). Then recognise x² − 4 as a difference of squares, which factors to (x − 2)(x + 2). Hence the fully factorised form is 3x(x − 2)(x + 2).

题目：完全因式分解 3x³ − 12x。解答：先提取最大公因式 3x：3x(x² − 4)。然后将 x² − 4 视为平方差，分解为 (x − 2)(x + 2)。因此完全分解式为 3x(x − 2)(x + 2)。

2. Solving Equations | 方程求解

Problem: Solve for x: √(2x+3) − 1 = x. Solution: Isolate the radical: √(2x+3) = x + 1. Square both sides: 2x + 3 = (x + 1)² = x² + 2x + 1. Simplify to obtain x² − 2 = 0, so x = ±√2. Check for extraneous roots. For x = √2, the original equation holds. For x = −√2, LHS = √(3 − 2√2) − 1. Since √(3 − 2√2) simplifies to √((√2 − 1)²) = √2 − 1, LHS becomes (√2 − 1) − 1 = √2 − 2, which does not equal RHS −√2. Therefore the only solution is x = √2.

题目：解方程 √(2x+3) − 1 = x。解答：移项得 √(2x+3) = x + 1。两边平方：2x+3 = x²+2x+1，整理得 x²=2，x=±√2。检验增根：x = √2 时成立；x = −√2 时，左边 √(3 − 2√2) − 1 可化为 (√2−1)−1=√2−2，不等于右边 −√2。故解为 x = √2。

3. Function Transformation | 函数变换

Problem: Given f(x) = 1/x, write the equation of the graph after a vertical stretch by factor 3, a translation of 2 units right, and 1 unit up. Solution: Start with y = 1/x. Vertical stretch by 3 gives y = 3/x. Translate 2 units right: replace x with (x−2), yielding y = 3/(x−2). Finally, translate 1 unit up: y = 3/(x−2) + 1. The transformed function is g(x) = 3/(x−2) + 1.

题目：已知 f(x)=1/x，写出图像经过垂直拉伸3倍、向右平移2单位、向上平移1单位后的方程。解答：垂直拉伸3倍得 y=3/x；右移2单位替换 x 为 (x−2) 得 y=3/(x−2)；上移1单位得 y=3/(x−2)+1。变换后函数为 g(x)=3/(x−2)+1。

4. Arithmetic Sequences | 等差数列

Problem: The third term of an arithmetic sequence is 7 and the eighth term is 22. Find the first term u₁ and the common difference d. Solution: Using uₙ = u₁ + (n−1)d: u₃ = u₁ + 2d = 7, u₈ = u₁ + 7d = 22. Subtracting the first equation from the second gives 5d = 15, so d = 3. Substituting back yields u₁ = 7 − 2(3) = 1. Hence u₁ = 1, d = 3.

题目：等差数列第3项为7，第8项为22。求首项 u₁ 和公差 d。解答：通项 uₙ = u₁ + (n−1)d。得方程组 u₁+2d=7, u₁+7d=22，两式相减得 5d=15，d=3；回代得 u₁=1。

5. Trigonometric Exact Values | 三角函数精确值

Problem: Without a calculator, evaluate sin 150° + cos 240° − tan 315°. Solution: sin 150° = sin(180°−30°) = sin 30° = ½. cos 240° = cos(180°+60°) = −cos 60° = −½. tan 315° = tan(360°−45°) = −tan 45° = −1. Thus the expression becomes ½ + (−½) − (−1) = 0 + 1 = 1.

题目：不用计算器，求 sin 150° + cos 240° − tan 315°。解答：sin150°=sin30°=½, cos240°=−cos60°=−½, tan315°=−tan45°=−1。原式 = ½ − ½ + 1 = 1。

6. Exponential and Logarithmic Equations | 指数与对数方程

Problem: Solve 2ˣ⁺¹ = 8²ˣ⁻³. Solution: Express 8 as a power of 2: 8 = 2³. Then the equation becomes 2ˣ⁺¹ = (2³)²ˣ⁻³ = 2³⁽²ˣ⁻³⁾ = 2⁶ˣ⁻⁹. Since the bases are equal, equate exponents: x + 1 = 6x − 9 → 1 + 9 = 6x − x → 10 = 5x → x = 2.

题目：解 2ˣ⁺¹ = 8²ˣ⁻³。解答：将 8 写为 2³：2ˣ⁺¹ = 2³⁽²ˣ⁻³⁾ = 2⁶ˣ⁻⁹。指数相等得 x+1=6x−9，解得 x=2。

7. Binomial Expansion | 二项式展开

Problem: Find the coefficient of x³ in the expansion of (2x − 1)⁵. Solution: The general term is ⁵Cᵣ (2x)⁵⁻ʳ (−1)ʳ. We require the power of x to be 3, so 5 − r = 3 ⇒ r = 2. The corresponding term is ⁵C₂ (2x)³ (−1)² = 10 × 8x³ × 1 = 80x³. Therefore the coefficient is 80.

题目：求 (2x − 1)⁵ 展开式中 x³ 的系数。解答：通项为 ⁵Cᵣ (2x)⁵⁻ʳ (−1)ʳ。令 x 的指数为 3，即 5−r=3 ⇒ r=2。该项为 ⁵C₂ (2x)³ (−1)² = 10×8x³ = 80x³，系数为80。

8. Basic Differentiation | 基础求导

Problem: Find dy/dx if y = (3x² − 2√x + 5)/x, for x > 0. Solution: Simplify first by dividing each term: y = 3x − 2x¹⁄²⁻¹ + 5x⁻¹ = 3x − 2x⁻¹⁄² + 5x⁻¹. Differentiate term by term: d/dx(3x) = 3; d/dx(−2x⁻¹⁄²) = −2 × (−½) x⁻³⁄² = x⁻³⁄²; d/dx(5x⁻

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