Review Set 5A - NON-CALCULATOR | 复习题集5A – 非计算器部分

📚 Review Set 5A – NON-CALCULATOR | 复习题集5A – 非计算器部分

This article revisits the core skills tested in a typical IB Mathematics non‑calculator review set, covering algebra, functions, trigonometry, exponentials, logarithms, and basic calculus. You will find worked examples, key facts, and strategies to handle common question types without a calculator.

本文回顾IB数学非计算器复习题集中测试的核心技能，涵盖代数、函数、三角学、指数、对数以及基础微积分。您将看到解题示例、关键事实以及在无计算器情况下处理常见题型的策略。

---

1. Algebraic Expansion and Factorisation | 代数展开与因式分解

Non‑calculator papers place a strong emphasis on accurate expansion and factorisation, often involving polynomials of degree 2 or 3. You must be able to multiply brackets systematically and recognise standard forms such as the difference of two squares or perfect squares.

非计算器试卷非常强调准确的展开与因式分解，常涉及二次或三次多项式。你必须能够系统地相乘括号，并识别平方差或完全平方等标准形式。

Example: Expand and simplify (x + 2)(x² – ultiply each term in the first bracket by every term in the second bracket: x·x² = x³, x·(–3x) = –3x², x·1 = x; 2·x² = 2x², 2·(–3x) = –6x, 2·1 = 2. Combining like terms gives x³ – x² – 5x + 2.

示例：展开并化简 (x + 2)(x² – 3x + 1)。将第一个括号中的每一项乘以第二个括号中的每一项：x·x² = x³，x·(–3x) = –3x²，x·1 = x；2·x² = 2x²，2·(–3x) = –6x，2·1 = 2。合并同类项得到 x³ – x² – 5x + 2。

orisation, grouping is often required when a cubic polynomial has four terms. Factorise x³ – 3x² – 4x + 12 by grouping: (x³ – 3x²) + (–4x + 12) = x²(x – 3) – 4(x – 3) = (x – 3)(x² – 4). Recognise x² – 4 as a difference of squares, giving the fully factorised form (x – 3)(x – 2)(x + 2).

因式分解时，当三次多项式有四项时常需分组。分解 x³ – 3x² – 4x + 12：分组为 (x³ – 3x²) + (–4x + 12) = x²(x – 3) – 4(x – 3) = (x – 3)(x² – 4)。将 x² – 4 视为平方差，得到完全分解式 (x – 3)(x – 2)(x + 2)。

---

2. Rationalising Denominators | 有理化分母

Leaving surds in the denominator is generally not accepted in simplified answers. Rationalise the denominator by multiplying the numerator and denominator by the conjugate when a binomial surd is present, or by the single surd for a monomial denominator.

化简后的答案通常不允许分母中保留根式。当分母为二项根式时，将分子分母同乘以共轭式进行有理化；若为单项根式则乘以该根式自身。

For instance, simplify 1/(√5 – √3). Multiply top and bottom by (√5 + √3): the denominator becomes (√5)² – (√3)² = 5 – 3 = 2, giving (√5 + √3)/2. Similarly, to simplify 3/√2, multiply by √2/√2 to obtain (3√2)/2.

例如，化简 1/(√5 – √3)。将分子分母同乘以 (√5 + √3)：分母变为 (√5)² – (√3)² = 5 – 3 = 2，得到 (√5 + √3)/2。类似地，化简 3/√2，乘以 √2/√2 可得 (3√2)/2。

Rationalising is also necessary when dealing with expressions like 4/(1 + √3). The conjugate is (1 – √3); the numerator becomes 4(1 – √3) and the denominator 1 – 3 = –2, so the expression simplifies to –2(1 – √3) = 2√3 – 2.

在处理如 4/(1 + √3) 的表达式时也需要有理化。共轭式为 (1 – √3)；分子变为 4(1 – √3)，分母为 1 – 3 = –2，因此表达式化简为 –2(1 – √3) = 2√3 – 2。

---

3. Laws of Exponents and Logarithms | 指数与对数运算法则

Mastery of exponent rules allows simplification without a calculator. Recall that a^m × aⁿ = a^m+n, (a^m)ⁿ = a^mn, and a^–n = 1/aⁿ. These are essential when simplifying expressions such as (2x³)² × (x⁻¹)³.

掌握指数法则可以在无计算器的情况下进行化简。回忆 a^m × aⁿ = a^m+n，(a^m)ⁿ = a^mn，以及 a^–n = 1/aⁿ。这些在化简如 (2x³)² × (x⁻¹)³ 的表达式时至关重要。

Simplify: ys apply the power to both the coefficient and the variable part, then combine like bases.

化简：(2x³)² × (x⁻¹)³ = 4x⁶ × x⁻³ = 4x³。务必将幂同时应用于系数和变量部分，然后合并同底数的指数。

Logarithm laws mirror exponent rules. The key relationships are log_a(xy) = log_ax + log_ay, log_a(x/y) = log_ax – log_ay, and log_a(xⁿ) = n log_ax. Without a calculator, exact values are obtained by rewriting arguments as powers of the base, e.g. log₂8 = 3 because 2³ = 8.

对数法则与指数法则相对应。关键关系式为 log_a(xy) = log_ax + log_ay，log_a(x/y) = log_ax – log_ay，以及 log_a(xⁿ) = n log_ax。无计算器时，通过将真数写为底数的幂来得到精确值，例如 log₂8 = 3，因为 2³ = 8。

log_aa = 1, log_a1 = 0

---

4. Solving Exponential and Logarithmic Equations | 求解指数与对数方程

Non‑calculator exponential equations often rely on writing both sides with the same base. Solve 2^x+1 = 8^x. Rewrite 8 as 2³, giving 2^x+1 = (2³)^x = 2^3x. Equating exponents yields x + 1 = 3x, so 2x = 1 and x = ½.

非计算器指数方程通常依赖将两边写成同底数。解方程 2^x+1 = 8^x。将 8 重写为 2³，得到 2^x+1 = (2³)^x = 2^3x。令指数相等得 x + 1 = 3x，因此 2x = 1，x = ½。

Logarithmic equations require checking domain restrictions. Solve log₂(x + 1) + log₂(x – 1) = 3. Combine logs: log₂[(x+1)(x–1)] = 3 → (x² – 1) = 2³ = 8 → x² = 9 → x = ±3. Reject x = –3 because x – 1 would be negative, so x = 3.

对数方程需要检查定义域限制。解方程 log₂(x + 1) + log₂(x – 1) = 3。合并对数：log₂[(x+1)(x–1)] = 3 → (x² – 1) = 2³ = 8 → x² = 9 → x = ±3。舍去 x = –3，因为 x – 1 为负数，所以 x = 3。

Equations involving the natural base e are solved similarly: e^2x = 5 would leave the answer as x = ½ ln 5 or express it in terms of natural logarithms exactly.

涉及自然底数 e 的方程类似求解：e^2x = 5 给出 2x = ln 5，但在无计算器情况下，答案保留为 x = ½ ln 5 或用自然对数精确表示。

---

广告 / Sponsored

5. Function Transformations and Compositions | 函数变换与复合

Understanding how graphs shift, stretch, and reflect is crucial. Given a parent function f(x), the transformed function a·f(b(x – h)) + k involves a vertical stretch by factor |a|, a horizontal stretch by factor 1/|b|, a horizontal translation by h, and a vertical translation by k. Reflections occur if a or b is negative.

理解函数图像如何平移、伸缩和反射至关重要。给定母函数 f(x)，变换后的函数 a·f(b(x – h)) + k 涉及垂直方向伸缩 |a| 倍、水平方向伸缩 1/|b| 倍、水平平移 h 个单位以及垂直平移 k 个单位。若 a 或 b 为负则发生反射。

Example: Describe the sequence of transformations that maps f(x) = √x to g(x) = 2√x – 3 + 1. First, translate 3 units right: √x – 3; then stretch vertically by factor 2: 2√x – 3; finally translate 1 unit up: 2√x – 3 + 1.

示例：描述将 f(x) = √x 变为 g(x) = 2√x – 3 + 1 的变换顺序。首先，向右平移 3 个单位：√x – 3；然后垂直拉伸 2 倍：2√x – 3；最后向上平移 1 个单位：2√x – 3 + 1。

Composition of functions (f ∘ g)(x) = f(g(x)) must be evaluated by substitution. If f(x) = x² + 1 and g(x) = 2x – 3, then (f ∘ g)(x) = (2x – 3)² + 1 = 4x² – + 10. Domain considerations are important when square roots or denominators are involved.

复合函数 (f ∘ g)(x) = f(g(x)) 必须通过代入计算。若 f(x) = x² + 1 且 g(x) = 2x – 3，则 (f ∘ g)(x) = (2x – 3)² + 1 = 4x² – 12x + 10。当涉及平方根或分母时，需考虑定义域。

---

6. Exact Trigonometric Values and Identities | 三角函数的精确值与恒等式

Without a calculator, you must know the exact sine, cosine, and tangent values for 0°, 30°, 45°, 60°, and 90° (or 0, π/6, π/4, π/3, π/2 radians). These are derived from special right‑angled triangles and the unit circle.

无计算器时，你必须牢记 0°、30°、45°、60° 和 90°（或 0、π/6、π/4、π/3、π/2 弧度）的正弦、余弦和正切的精确值。这些值源自特殊直角三角形和单位圆。

θ (deg)	θ (rad)	sin θ	cos θ	tan θ
0°	0	0	1	0
30°	π/6	½	½√3	√3/3
45°	π/4	½√2	½√2	1
60°	π/3	½√3	½	√3
90°	π/2	1	0	undefined

frequently used to simplify expressions or prove relationships. For instance, simplify (1 – sin²θ)/cos θ gives cos²θ/cos θ = cos θ.

勾股恒等式如 sin²θ + cos²θ = 1 和 1 + tan²θ = sec²θ 常用于化简表达式或证明关系。例如，化简 (1 – sin²θ)/cos θ 得 cos²θ/cos θ = cos θ。

---

7. Solving Trigonometric Equations | 求解三角方程

Non‑ ey rely on recognising exact values and understanding the periodic nature of trig functions. Solve 2 sin x = 1 for 0 ≤ x ≤ 2π. the exact values, x = π/6 and x = 5π/6 are the principal solutions in the given interval.

非计算器的三角方程通常要求在特定区间内求解，例如 0 ≤ x ≤ 2π。解题依赖于识别精确值并理解三角函数的周期性。在 0 ≤ x ≤ 2π 内解 2 sin x = 1。首先，sin x = ½。根据精确值，在给定区间内的主解为 x = π/6 和 x = 5π/6。

values within 0 ≤ x ≤ 2π: π/6, 7π/6, 5π/6, 11π/6. Always list all solutions in the domain.

对于包含复合角的方程，如 cos(2x) = ½，先求一般角：2x = π/3 + 2kπ 或 2x = 5π/3 + 2kπ。然后除以 2 得到 x = π/6 + kπ 和 x = 5π/6 + kπ。最后筛选 0 ≤ x ≤ 2π 内的所有值：π/6, 7π/6, 5π/6, 11π/6。务必列出定义域内的所有解。

---

8. Differentiation from First Principles | 从第一原理求导

The derivative from first principles is defined as the limit of the difference quotient. This foundational concept often appears in non‑calculator proofs or to differentiate simple polynomials.

f '(x) = lim_h→0 [f(x + h) – f(x)] / h

从第一原理求导定义为差商的极限。这一基本概念常出现在非计算器的证明题中，或用于对简单多项式求导。

Example: ples to differentiate f(x) = x². Set up f(x + h) = (x + h)² = x² + 2xh + h². Then f(x + h) – ) = 2xh + h². Divide by h to get (2xh + h²)/h = 2x + h. Taking the limit as h → 0 yields f '(x) = 2x.

示例：用第一原理求 f(x) = x² 的导数。设定 f(x + h) = (x + h)² = x² + 2xh + h²。则 f(x + h) – f(x) = 2xh + h²。除以 h 得 (2xh + h²)/h = 2x + h。取 h → 0 时的极限得到 f '(x) = 2x。

For f(x) = 1/x, the algebra involves finding a common denominator; the limit evaluates to – eing comfortable expanding (x + h)³ or rationalising numerators with radicals is expected.

这一过程加深了对变化率的理解。对于 f(x) = 1/x，代数运算涉及通分；极限求值得出 –1/x²。需能熟练展开 (x + h)³ 或有理化带根式的分子。

---

9. Basic Derivative Rules and Tangents | 基本求导法则与切线

Once first principles are understood, the power rule, sum rule, and constant multiple rule make differentiation efficient: d/dx (xⁿ) = n x^n–1. For non‑calculator exams

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）