📚 Review Set 14A - Non-Calculator | 复习题集 14A - 非计算器
This article provides a comprehensive walkthrough of IB Mathematics Review Set 14A, designed as a non-calculator practice set covering essential algebraic, trigonometric, and calculus skills. The problems in this set are carefully chosen to sharpen your mental computation, reinforce foundational concepts, and prepare you for Paper 1 sections where no calculator is allowed. Each section below presents a typical question type, followed by step-by-step strategies in both English and Chinese. Mastering these exercises will build the confidence to handle symbolic manipulation, equation solving, and function analysis without relying on technology.
本文全面解析 IB 数学复习题集 14A，该题集为非计算器练习设计，涵盖代数、三角和微积分等核心技能。题目经过精选，旨在锻炼心算能力、巩固基础概念，并为不允许使用计算器的试卷做好准备。以下每个小节都给出一个典型题型，并用中英文提供逐步解题策略。掌握这些练习将帮助你自信地处理符号运算、方程求解和函数分析，而无需依赖计算器。
1. Simplifying Algebraic Expressions | 代数表达式化简
Simplify the expression (3x² - 2x + 1) - (x² - 4x - 3) + 2(x - 1).
化简表达式 (3x² - 2x + 1) - (x² - 4x - 3) + 2(x - 1)。
) = 2x²; (-2x + 4x + 2x) = 4x; (1 + 3 - 2) = 2. The simplified expression is 2x² + 4x + 2.
首先注意去括号时符号的变化：3x² - 2x + 1 - x² + 4x + 3 + 2x - 2。合并同类项可得：(3x² - x²) = 2x²; (-2x + 4x + 2x) = 4x; (1 + 3 - 2) = 2。化简结果为 2x² + 4x + 2。
2. Solving Equations and Inequalities | 解方程与不等式
2(x + 3) - 5 = 3(x - 1) + 4. Also find the solution set of the inequality x² - 5x + 6 ≤ 0.
解方程：2(x + 3) - 5 = 3(x - 1) + 4。并求不等式 x² - 5x + 6 ≤ 0 的解集。
chart, the product is negative or zero between the roots. Hence the solution is 2 ≤ x ≤ 3.
先展开两边：2x + 6 - 5 = 3x - 3 + 4，化简得 2x + 1 = 3x + 1。两边减2x得 1 = x + 1，故 x = 0。对于不等式，将二次式分解为 (x - 2)(x - 3) ≤ 0。临界点为 2 和 3。使用区间测试或符号表可知，两根之间乘积为负或零，因此解集为 2 ≤ x ≤ 3。
3. Functions and Composite Functions | 函数与复合函数
Given f(x) = 2x + 3 and g(x) = x² - 1, find (f ∘ g)(x) and (g ∘ the domain of the composite function f(g(x)) when the domain of g is restricted to x ≥ 0.
已知 f(x) = 2x + 3，g(x) = x² - 1，求 (f ∘ g)(x) 与 (g ∘ f)(x)。若将 g 的定义域限制为 x ≥ 0，试确定 f(g(x)) 的定义域。
By definition, (f ∘ g)(x) = f(g(x)) = 2(x² - 1) + 3 = 2x² - 2 + 3 = 2x² + 1. Similarly, (g ∘ 2x² + 1 is defined for all real x. Since g(x) is defined for x ≥ 0, the composite function's domain remains [0, ∞).
根据定义，(f ∘ g)(x) = f(g(x)) = 2(x² - 1) + 3 = 2x² + 1。(g ∘ f)(x) = g(f(x)) = (2x + 3)² - 1 = 4x² + 12x + 8。对于 f(g(x))，若 x ≥ 0，则 2x² + 1 对所有实数均有定义，且g(x)在x≥0时有定义，故复合函数的定义域为 [0, ∞)。
4. Exponents and Logarithms | 指数与对数
Simplify (8²ˣ × 4ˣ⁺¹) / 2³ˣ your answer as a single power of 2. Then solve the equation log₂(x - 1) + log₂(x + 3) = 3.
化简 (8²ˣ × 4ˣ⁺¹) / 2³ˣ⁻²，并将结果写成 2 的幂。再解方程 log₂(x - 1) + log₂(x + 3) = 3。
Express all bases as powers of 2: 8 = 2³, so 8²ˣ = (2³)²ˣ = 2⁶ˣ; 4ˣ⁺¹ = (2²)ˣ⁺¹ = 2²ˣ⁺². Multiply them: 2⁶ˣ · 2²ˣ⁺² = 2⁸ˣ⁺². Dividing by 2³ˣ⁻² gives 2⁽⁸ˣ⁺²⁾ ⁻ ⁽³ˣ⁻²⁾ = 2⁵ˣ = 8. Expand: x² + 2x - 3 = 8 → x² + 2x - 11 = 0. Discriminant = 4 + 44 = 48, so x = [-2 ± √48]/2 = [-2 ± 4√3]/2 = -1 ± 2√3. Check domain: x - 1 > 0 ⇒ x > 1, and x + 3 > 0 always true for x > 1. Thus only x = -1 + 2√3 is valid (approximately 2.464).
将所有底数化为 2 的幂：8 = 2³，故 8²ˣ = (2³)²ˣ = 2⁶ˣ；4ˣ⁺¹ = (2²)ˣ⁺¹ = 2²ˣ⁺²。相乘得 2⁶ˣ · 2²ˣ⁺² = 2⁸ˣ⁺²。除以 2³ˣ⁻² 得 2⁽⁸ˣ⁺²⁾ ⁻ ⁽³ˣ⁻²⁾ = 2⁵ˣ⁺⁴。对数方程合并：log₂[(x - 1)(x + 3)] = 3，则 (x - 1)(x + 3) = 8。展开：x² + 2x - 3 = 8 → x² + 2x - 11 = 0。判别式 = 4 + 44 = 48，解为 x = [-2 ± √48]/2 = -1 ± 2√3。检查定义域：x > 1，故仅 x = -1 + 2√3 满足要求。
5. Trigonometric Identities and Equations | 三角恒等式与方程
in θ + cos θ)² = 1 + sin 2θ. Hence solve sin 2θ = 0.5 for 0 ≤ θ ≤ 2π without a calculator.
证明 (sin θ + cos θ)² = 1 + sin 2θ。然后不解计算器求出在 0 ≤ θ ≤ 2π 内方程 sin 2θ = 0.5 的解。
e know sin(π/6) = 0.5. The general solution for 2θ is 2θ = π/6 + 2kπ or 2θ = 5π/6 + 2kπ, k ∈ ℤ hin [0, 2π], take k = 0, 1: θ = π/12, 13π/12, 5π/12, 17π/12. These are the four solutions.
左边展开：sin²θ + 2 sin θ cos θ + cos²θ。利用恒等式 sin²θ + cos²θ = 1 与 2 sin θ cos θ = sin 2θ，即得 1 + sin 2θ。对于 sin 2θ = 0.5，已知 sin(π/6) = 0.5。2θ 的通解为 2θ = π/6 + 2kπ 或 2θ = 5π/6 + 2kπ，k∈ℤ。除以2得 θ = π/12 + kπ 或 θ = 5π/12 + kπ。在 [0, 2π] 内取 k = 0, 1 可得 θ = π/12, 13π/12, 5π/12, 17π/12。
6. Differentiation and Tangent Lines | 导数与切线
7 using first principles. Then determine the equation of the tangent line to the curve y = √x at x = 4.
用第一原理求 f(x) = x³ - 4x² + 2x - 7 的导数。再求曲线 y = √x 在 x = 4 处的切线方程。
First principles: f'(x) = limh→0 ²h + 3xh² + h³ - 8xh - 4h² + 2h]. Divide by h: 3x² + 3xh + h² - 8x - 4h + 2. Take h → 0: f'(x) = 3x² - 8x + 2. For y = √x = x1/2, y' = (1/2)x-1/2 = 1/(2√x m = 1/(2·2) = 1/4, and point (4, 2). Tangent line: y - 2 = (1/4)(x - 4) → y = (1/4)x + 1.
第一原理：f'(x) = limh→0 [f(x+h) - f(x)]/h。展开 f(x+h) 并化简，差除以h后令 h→0 可得 f'(x) = 3x² - 8x + 2。对于 y = √x，其导数为 y' = 1/(2√x)。在 x = 4 处，斜率 m = 1/(2·2) = 1/4，点坐标为 (4, 2)。切线方程为 y - 2 = (1/4)(x - 4)，即 y = (1/4)x + 1。
7. Basic Integration | 基本积分
. Also find the area enclosed between the curve y = x³ and the x-axis from x = 0 to x = 2.
计算定积分 ∫₁³ (2x² - 3x + 1) dx。并求曲线 y = x³ 在 x = 0 到 x = 2 之间与 x 轴围成的面积。
bove x-axis (positive). Area = ∫₀² x³ dx = [x⁴/4]₀² = 16/4 = 4. So area is 4 square units.
逐步积分：∫ (2x² - 3x + 1) dx = (2/3)x³ - (3/2)x² + x + C。代入上下限，上限 x=3 得 7.5，下限 x=1 得 1/6，两者相减为 22/3。对于 y = x³ 在 [0,2] 上的面积，函数非负，面积 = ∫₀² x³ dx = [x⁴/4]₀² = 4 平方单位。
8. Vectors in the Plane | 平面向量
ind the magnitude of a + b, and the angle between a and b (to the nearest degree without a calculator using exact values).
已知向量 a = 3i - 2j，b = -i + 4j，求 a + b 的模长，以及 a 与 b 之间的夹角（不用计算器，给出精确值的近似度数）。
a + b = (3 - 1)i + (-2 + 4)j = 2i + 2j. Its magnitude is √2² + 2² = √8 = 2√2. For the angle, use dot product: a·b = (3)(-1) + (-2)(4) = -3 - 8 = -11. |a| = √9 + 4 = √13; |b| = √1 + 16 = √17. cos θ = a·b / (|a||b|) = -11 / (√13 √17 angle with cos⁻¹(0.74): cos 42° ≈ 0.74, so θ ≈ 180° - 42° = 138° (to the nearest degree).
a + b = (3 - 1)i + (-2 + 4)j = 2i + 2j，模长为 √2² + 2² = 2√2。利用点积求角：a·b = -11，|a| = √13，|b| = √17，cos θ = -11 / (√13 √17) = -11 / √221。√221 约等于 14.866，故 cos θ ≈ -0.74。参考角满足 cos 42° ≈ 0.74，因此 θ ≈ 180° - 42° = 138°。
9. Complex Numbers | 复数
-2 + i. Compute z + w, z · w, and the modulus of z. Also express the complex number (1 + i√3) in polar form.
设 z = 3 - 4i，w = -2 + i。计算 z + w、z · w 以及 z 的模。并将复数 (1 + i√3) 表示为极形式。
= (3)(-2) + (3)(i) + (-4i)(-2) + (-4i)(i) = -6 + 3i + 8i - 4i² = -6 + 11i + 4 = -2 + 11i. |z| = √(3² + (-4)²) = √9 + 16 = 5. For 1 + i√3, modulus r = √(1² + (√3)²) = √4 = 2. Argument θ: tan θ = √3 / 1 = √3 3 (since real and imaginary positive). Polar form: 2(cos(π/3) + i sin(π/3)) or 2 cis(π/3).
加减法直接合并：z + w = 1 - 3i。乘法：z · w = (3)(-2) + 3i + 8i - 4i² = -2 + 11i。|z| = √9+16=5。对于 1 + i√3，模 r = √1 + 3 = 2，辐角 θ 满足 tan θ = √3，且实虚部皆正，故 θ = π/3。极形式为 2(cos(π/3) + i sin(π/3))。
10. Sequences and Series | 序列与级数
An arithmetic sequence has first term 7 and common difference -3. Find the 20th term and the sum of the first 20 terms. For the geometric sequence 3, 6, 12, ..., find the smallest n such that the sum of the first n terms exceeds 1000.
某等差数列首项为 7，公差为 -3，求第 20 项及前 20 项之和。对于等比数列 3, 6, 12, ...，求最小的 n 使得前 n 项之和超过 1000。
Arithmetic: n-th term un = 7 + (n-1)(-3). For n=20, u20 = 7 - 57 = -50. Sum S20 rst + last) = 10(7 + (-50)) = -430. Geometric: first term a = 3, common ratio r = 2. Sum Sn = a(rn - 1)/(r - 1) = 3(2n - 1)/(2 - 1) = 3(2n - 1). We need 3(2n - 1) > 1000 ⇒ 2n - 1 > 1000/3 ≈ 333.33 ⇒ 2n > 334.33. Test powers: 28 = 256, 29 = 512. So 29 = 512 > 334.33, thus n = 9. Check S9 = 3(512 - 1) = 1533 > 1000. Hence smallest n is 9.
等差数列通项 un = 7 + (n-1)(-3)，第20项为 -50。前20项和 S20 = 10(7 - 50) = -430。等比数列 a = 3，r = 2，前 n 项和 Sn = 3(2n - 1)。解不等式 3(2n - 1) > 1000 得 2n > 334.33，28=256，29=512，故 n=9 时和为 1533 大于1000，最小 n 为9。
11. Polynomial Division and Remainder Theorem | 多项式除法与余数定理
Divide the polynomial 2x³ - 5x² + 3x - 7 by (x - 2). Find the quotient and remainder. Then use the Remainder Theorem to verify the remainder when divided by (x - 2). Factorise completely 2x³ - 3x² - 3x + 2 given that (x - 2) is a factor.
用 (x - 2) 去除多项式 2x³ - 5x² + 3x - 7，求商式和余数。然后用余数定理验证除以 (x - 2) 的余数。已知 (x - 2) 是 2x³ - 3x² - 3x + 2 的一个因式，对其进行因式分解。
16 - 20 + 6 - 7 = -5, matches. For the second polynomial, since (x-2) is a factor, P(2)=0? Let' n factor 2x² + x - 1 = (2x - 1)(x + 1). So complete factorisation: (x - 2)(2x - 1)(x + 1).
用综合除法以根2运算：系数2, -5, 3, -7，带入后商式为 2x² - x + 1，余数 -5。余数定理验证：P(2) = 16 - 20 + 6 - 7 = -5，一致。对第二个多项式，用 (x-2) 试根得 P(2)=0，故 (x-2) 为因式。除法得商式 2x² + x - 1，再分解二次式得 (2x - 1)(x + 1)。整体因式分解为 (x - 2)(2x - 1)(x + 1)。
12. Permutations and Combinations | 排列与组合
A committee of 4 members is to be selected from 10 people. How many different committees are possible? If the committee must include a specific person and one of two other specific people, how many selections are possible? Also find the number of ways to arrange the letters of the word 'REVIEW' such that the vowels are together.
从 10 人中选出一个 4 人委员会，共有多少种不同的选法？如果委员会必须包含某特定人以及另外两个特定人之一，有多少种选法？并求将 'REVIEW' 一词的字母重新排列且元音字母相邻的方法数。
{B,C} is 2, and choose 2 from the other 7 is C(7,2)=21. Total = 2 × 21 = 42. For the word 'REVIEW' owels: E, I, E (three vowels). Treat the vowel block as a single unit. Within the block, E's repeated, number of arrangements: 3!/2! = 3. Now we have block + R, V, W (4 items), all distinct: 4! = 24. Total = 3 × 24 = 72 ways.
10人选4人：C(10,4) = 210。受限委员会：必须包含A，且从B、C中恰好选一人，其余2人从剩余7人中选。选B或C有2种，选其余2人有C(7,2)=21种，共42种。单词 'REVIEW' 有6个字母，元音 E, I, E 三字母相邻。将元音视为一个整体，内部排列数为 3!/2! = 3（E重复）。整体包括这一块和 R, V, W 共4个不同对象，排列数 4! = 24。总方法数 3 × 24 = 72。
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