IB Math Exam Practice and Key Concept Analysis | IB数学真题演练与考点解析

📚 IB Math Exam Practice and Key Concept Analysis | IB数学真题演练与考点解析

IB Mathematics exams test not only computational skills but also a deep understanding of concepts and the ability to apply them in unfamiliar contexts. This article walks through ten key topics, each illustrated with a typical past-paper-style question and a step‑by‑step solution. By working through these examples, you will consolidate the essential techniques required for both Standard Level and Higher Level papers.

IB数学考试不仅考查计算能力，更检验对概念的深刻理解以及在陌生情境中应用它们的能力。本文梳理了十个核心考点，每个考点都配有一道典型的真题风格例题，并给出逐步解析。通过这些例子的演练，你将巩固标准级别和高级别试卷中必备的关键技巧。

1. Algebra and Equation Solving | 代数与方程求解

Algebraic manipulation stays at the heart of IB Math. You are expected to solve linear, quadratic, and simultaneous equations fluently, often as a first step in larger problems. A solid grasp of factorising, completing the square, and the quadratic formula is essential.

代数运算始终是IB数学的核心。你需要熟练求解线性、二次和联立方程，这往往是解答大题的第一步。扎实掌握因式分解、配方法以及二次公式至关重要。

Example (IB past paper): Solve 2x² – 5x – 3 = 0.

例题（IB真题）：解方程 2x² – 5x – 3 = 0。

Factorising the quadratic gives (2x + 1)(x – 3) = 0. Setting each factor to zero yields 2x + 1 = 0 ⇒ x = –1/2, and x – 3 = 0 ⇒ x = 3.

对二次式进行因式分解得 (2x + 1)(x – 3) = 0。令每个因式为零：2x + 1 = 0 解得 x = –1/2，x – 3 = 0 解得 x = 3。

2. Functions and Graph Transformations | 函数与图像变换

Understanding the language of functions—domain, range, composite, and inverse—is fundamental. Graph transformations such as translations, stretches, and reflections let you quickly sketch complicated functions from a parent function.

理解函数的语言——定义域、值域、复合函数和反函数——是基础。通过平移、伸缩和反射等图像变换，你能从基本函数快速勾画出复杂函数的图像。

Example: The graph of y = f(x) is translated 3 units right and 2 units up, then reflected in the x‑axis. Write the equation of the transformed graph in the form y = –f(x – a) + b, and state a and b.

例题：已知 y = f(x) 的图像向右平移 3 个单位，向上平移 2 个单位，再关于 x 轴对称。写出变换后图像形如 y = –f(x – a) + b 的方程，并给出 a 和 b 的值。

Translation right by 3 replaces x with (x – 3): y = f(x – 3). Translation up by 2 adds 2: y = f(x – 3) + 2. Reflection in the x‑axis multiplies the whole function by –1: y = –[f(x – 3) + 2] = –f(x – 3) – 2. So a = 3, b = –2.

向右平移 3 个单位：将 x 替换为 (x – 3)，得到 y = f(x – 3)。向上平移 2 个单位：加上 2，y = f(x – 3) + 2。关于 x 轴反射：整个函数乘以 –1，y = –[f(x – 3) + 2] = –f(x – 3) – 2。因此 a = 3，b = –2。

3. Exponents and Logarithms | 指数与对数

Exponential and logarithmic functions model growth and decay processes. You must be confident switching between exponential and log forms, applying log laws, and solving equations where the unknown appears in the exponent.

指数和对数函数用于模拟增长和衰减过程。你必须熟练在指数形式与对数形式之间转换，会应用对数运算法则，并会求解未知数出现在指数上的方程。

Example: Solve 3^2x+1 = 5, giving your answer in the form x = (ln p)/(ln q).

例题：解方程 3^2x+1 = 5，答案写成 x = (ln p)/(ln q) 的形式。

Take natural logarithms of both sides: ln(3^2x+1) = ln 5. Using the power rule, (2x+1) ln 3 = ln 5. Expand: 2x ln 3 + ln 3 = ln 5. Rearrange: 2x ln 3 = ln 5 – ln 3 = ln(5/3). Divide by 2 ln 3: x = ln(5/3) / (2 ln 3) = ln(5/3) / ln(3²) = ln(5/3) / ln 9. So p = 5/3, q = 9, or directly x = (ln(5/3)) / (2 ln 3) is acceptable.

两边取自然对数：ln(3^2x+1) = ln 5。利用幂法则，(2x+1) ln 3 = ln 5。展开：2x ln 3 + ln 3 = ln 5。移项：2x ln 3 = ln 5 – ln 3 = ln(5/3)。两边同时除以 2 ln 3：x = ln(5/3) / (2 ln 3) = ln(5/3) / ln(3²) = ln(5/3) / ln 9。因此 p = 5/3，q = 9，或直接写 x = (ln(5/3)) / (2 ln 3) 均可。

4. Trigonometry | 三角学

Trigonometry covers radian measure, the unit circle, trigonometric identities, and solving equations within given intervals. The sine and cosine rules, along with area formulas, are tested frequently in triangle problems.

三角学涵盖弧度制、单位圆、三角恒等式以及在给定区间内解三角方程。正弦定理、余弦定理以及三角形面积公式经常在三角形问题中考到。

Example: In triangle ABC, AB = 7 cm, AC = 9 cm, and angle A = 40°. Find BC and the area of the triangle.

例题：在三角形 ABC 中，AB = 7 cm，AC = 9 cm，∠A = 40°。求 BC 及三角形的面积。

Use the cosine rule: BC² = AB² + AC² – 2·AB·AC·cos A = 7² + 9² – 2·7·9·cos 40° = 49 + 81 – 126 cos 40° ≈ 130 – 96.529 = 33.471. So BC = √33.471 ≈ 5.79 cm. The area uses Area = (1/2)·AB·AC·sin A = 0.5 × 7 × 9 × sin 40° ≈ 31.5 × 0.6428 = 20.25 cm².

用余弦定理：BC² = AB² + AC² – 2·AB·AC·cos A = 7² + 9² – 2·7·9·cos 40° = 49 + 81 – 126 cos 40° ≈ 130 – 96.529 = 33.471。所以 BC = √33.471 ≈ 5.79 cm。面积用公式：面积 = (1/2)·AB·AC·sin A = 0.5 × 7 × 9 × sin 40° ≈ 31.5 × 0.6428 = 20.25 cm²。

5. Calculus – Differentiation | 微积分——导数

Differentiation techniques—power rule, product rule, quotient rule, and chain rule—are tested directly and through applications like tangents, normals, and optimisation. You must also know how to find stationary points and determine their nature.

求导技巧——幂法则、乘法法则、除法法则和链式法则——会直接考查，也会结合切线、法线和最优化等应用。你还必须会求驻点并判断其性质。

Example: Find the coordinates of the stationary point on the curve y = x³ – 3x + 2 and determine whether it is a maximum or minimum.

例题：求曲线 y = x³ – 3x + 2 上驻点的坐标，并判断它是极大值还是极小值。

Differentiate: dy/dx = 3x² – 3. Set dy/dx = 0: 3x² – 3 = 0 ⇒ x² = 1, x = ±1. When x = 1, y = 1 – 3 + 2 = 0; when x = –1, y = –1 + 3 + 2 = 4. Second derivative: d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0 ⇒ minimum at (1, 0). At x = –1, d²y/dx² = –6 < 0 ⇒ maximum at (–1, 4).

求导：dy/dx = 3x² – 3。令 dy/dx = 0：3x² – 3 = 0 ⇒ x² = 1，x = ±1。当 x = 1，y = 1 – 3 + 2 = 0；当 x = –1，y = –1 + 3 + 2 = 4。二阶导数：d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0 ⇒ 极小值点 (1, 0)。在 x = –1 处，d²y/dx² = –6 < 0 ⇒ 极大值点 (–1, 4)。

6. Calculus – Integration | 微积分——积分

Integration is used for finding areas under curves, volumes of revolution, and solving differential equations. You must be comfortable with indefinite and definite integrals, integration by substitution, and recognising the link between a function and its derivative.

积分用于求曲线下方面积、旋转体体积以及解微分方程。你必须熟练掌握不定积分和定积分、换元积分法，并能识别函数与其导数之间的联系。

Example: Find the area enclosed by y = x² – 4x + 5, the x‑axis, and the lines x = 1 and x = 3.

例题：求 y = x² – 4x + 5 的图像与 x 轴以及直线 x = 1 和 x = 3 所围成的区域的面积。

Area = ∫₁³ (x² – 4x + 5) dx. Integrate term by term: ∫x² dx = x³/3, ∫–4x dx = –2x², ∫5 dx = 5x. So the antiderivative is (1/3)x³ – 2x² + 5x. Evaluate from 1 to 3: F(3) = (27/3) – 18 + 15 = 9 – 18 + 15 = 6; F(1) = (1/3) – 2 + 5 = (1/3) + 3 = 10/3. Area = 6 – 10/3 = 8/3 square units.

面积 = ∫₁³ (x² – 4x + 5) dx。逐项积分：∫x² dx = x³/3，∫–4x dx = –2x²，∫5 dx = 5x。原函数为 (1/3)x³ – 2x² + 5x。从 1 到 3 计算定积分：F(3) = (27/3) – 18 + 15 = 9 – 18 + 15 = 6；F(1) = (1/3) – 2 + 5 = (1/3) + 3 = 10/3。面积为 6 – 10/3 = 8/3 平方单位。

7. Vectors | 向量

Vector questions involve operations like addition, scalar multiplication, dot product, and finding the angle between vectors. Vector equations of lines and planes are HL core content, while SL focuses on 2D and basic 3D vector geometry.

向量题涉及向量的加法、数乘、点积以及求向量间的夹角。直线的向量方程和平面方程是 HL 的核心内容，SL 则侧重二维向量和基础的三维向量几何。

Example: Given vectors a = 2i – j + 3k and b = –i + 4j + 2k, find the angle between a and b.

例题：已知向量 a = 2i – j + 3k，b = –i + 4j + 2k，求 a 与 b 的夹角。

Dot product: a·b = (2)(–1) + (–1)(4) + (3)(2) = –2 – 4 + 6 = 0. Magnitudes: |a| = √(2² + (–1)² + 3²) = √(4+1+9) = √14; |b| = √((–1)² + 4² + 2²) = √(1+16+4) = √21. cosθ = (a·b) / (|a||b|) = 0 / (√14·√21) = 0. Thus θ = 90° or π/2 radians.

点积：a·b = (2)(–1) + (–1)(4) + (3)(2) = –2 – 4 + 6 = 0。模长：|a| = √(2² + (–1)² + 3²) = √(4+1+9) = √14；|b| = √((–1)² + 4² + 2²) = √(1+16+4) = √21。cosθ = (a·b) / (|a||b|) = 0 / (√14·√21) = 0。因此 θ = 90° 或 π/2 弧度。

8. Probability and Statistics | 概率与统计

IB expects you to calculate probabilities using tree diagrams, Venn diagrams, and conditional probability formulas. In statistics, you work with measures of central tendency, dispersion, linear regression, and the normal distribution.

IB 要求你使用树状图、韦恩图和条件概率公式计算概率。在统计部分，你将处理集中趋势的度量、离散程度、线性回归以及正态分布。

Example: Bag X contains 3 red and 2 blue marbles. Bag Y contains 4 red and 1 blue marble. A bag is chosen at random, and then a marble is drawn from it. Given that the marble is red, what is the probability it came from Bag X?

例题：袋 X 中有 3 个红球和 2 个蓝球，袋 Y 中有 4 个红球和 1 个蓝球。随机选一个袋子，再从该袋中随机摸出一个球。已知摸出的是红球，求它来自袋 X 的概率。

Let events: X = choosing Bag X, Y = choosing Bag Y, R = drawing red. P(X) = P(Y) = 1/2. P(R|X) = 3/5, P(R|Y) = 4/5. Total probability of red: P(R) = P(R|X)P(X) + P(R|Y)P(Y) = (3/5)(1/2) + (4/5)(1/2) = (3/10) + (4/10) = 7/10. By Bayes’ theorem, P(X|R) = [P(R|X)P(X)] / P(R) = (3/10) / (7/10) = 3/7.

9. Complex Numbers | 复数

Complex numbers appear mainly in the HL course. You need to perform arithmetic in Cartesian form, convert between forms, use De Moivre’s theorem, and find roots of complex polynomials.

复数主要出现在 HL 课程中。你需要以笛卡尔形式进行运算，在三种形式之间转换，使用棣莫弗定理，以及求复系数多项式的根。

Example: Express the complex number z = (1 + i)⁴ in the form a + bi, where a, b ∈ ℝ.

例题：将复数 z = (1 + i)⁴ 写成 a + bi 的形式，其中 a, b ∈ ℝ。

First, write 1 + i in modulus‑argument form: |1+i| = √(1²+1²) = √2, argument θ = arctan(1/1) = π/4. So 1+i = √2 (cos π/4 + i sin π/4). Using De Moivre: (1+i)⁴ = (√2)⁴ [cos(4×π/4) + i sin(4×π/4)] = 4 (cos π + i sin π) = 4(–1 + 0i) = –4. Thus a = –4, b = 0.

首先将 1 + i 写成模—辐角形式：|1+i| = √(1²+1²) = √2，辐角 θ = arctan(1/1) = π/4。所以 1+i = √2 (cos π/4 + i sin π/4)。利用棣莫弗定理：(1+i)⁴ = (√2)⁴ [cos(4×π/4) + i sin(4×π/4)] = 4 (cos π + i sin π) = 4(–1 + 0i) = –4。因此 a = –4，b = 0。

10. Sequences and Series | 数列与级数

Arithmetic and geometric sequences are typical in IB. You must be able to identify the type, find the nth term, sum a given number of terms, and handle infinite geometric series when the common ratio satisfies |r| < 1.

等差数列和等比数列是 IB 的典型内容。你必须能识别数列类型，求第 n 项，求前 n 项和，并在公比满足 |r| < 1 时处理无穷等比级数。

Example: The third term of a geometric sequence is 12, and the sixth term is 96. Find the first term and the common ratio. Then find the sum to infinity if it exists.

例题：一个等比数列的第三项是 12，第六项是 96。求首项和公比。若无穷级数存在，请求其和。

For a geometric sequence, u_n = u₁ r^n-1. Given u₃ = u₁ r² = 12 and u₆ = u₁ r⁵ = 96. Divide the equations: (u₁ r⁵)/(u₁ r²) = 96/12 ⇒ r³ = 8 ⇒ r = 2. Substitute back: u₁ × 2² = 12 ⇒ u₁ × 4 = 12 ⇒ u₁ = 3. Since |r| = 2 ≥ 1, the sum to infinity does not exist (it diverges).

等比数列中 u_n = u₁ r^n-1。已知 u₃ = u₁ r² = 12，u₆ = u₁ r⁵ = 96。两式相除：(u₁ r⁵)/(u₁ r²) = 96/12 ⇒ r³ = 8 ⇒ r = 2。代回得：u₁ × 2² = 12 ⇒ u₁ × 4 = 12 ⇒ u₁ = 3。由于 |r| = 2 ≥ 1，无穷和不存在（发散）。

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IB Math Exam Practice and Key Concept Analysis | IB数学真题演练与考点解析