📚 Summary and Mixed Examination Practice | IB数学总结与混合考试练习
This comprehensive revision guide brings together all major topics from the IB Diploma Mathematics courses – both Analysis and Approaches (AA) and Applications and Interpretation (AI) – with a focus on summarising key concepts and providing mixed examination-style practice. Whether you are aiming for Standard Level or Higher Level mastery, working through summaries and integrated examples will reinforce your understanding and prepare you for the variety of problem types you will face in the final examinations. Each section revisits core theory and then challenges you with a carefully chosen practice problem, blending different techniques just as real exam questions often do.
这份综合复习指南汇集了IB文凭数学课程（分析与方法 AA 以及应用与解析 AI）的所有主要专题，重点总结核心概念并提供混合考试风格的练习。无论你的目标是标准级别还是高级别，通过总结和综合例题的训练，都将加深理解，为你应对期末考试中各种题型做好准备。每一节都回顾核心理论，然后提出一个精选练习，融合不同技巧，正如真实考题中常见的那样。
1. Core Algebra and Equations | 核心代数与方程
atics. Master the laws of exponents: aᵐ × aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁻ⁿ = 1/aⁿ, and a¹/ⁿ = ⁿ√a. Logarithms are the inverses: if aˣ +bx+c=0 are solved by factoring, completing the square, or the quadratic formula x = [-b ± √b²-4ac ells the nature of roots. For inequalities, always consider critical values and test intervals. Systems of linear equations may be solved by substitution, elimination, or matrix methods.
代数是一切进阶数学的基础。掌握指数定律：aᵐ × aⁿ = aᵐ⁺ⁿ，(aᵐ)ⁿ = aᵐⁿ，a⁻ⁿ = 1/aⁿ，以及 a¹/ⁿ = ⁿ√a。对数为逆运算：若 aˣ = y，则 x = logₐ y，关键的换底公式为 logₐ b = log b / log a。二次方程 ax²+bx+c=0 可通过因式分解、配方法或求根公式 x = [-b ± √b²-4ac]/(2a) 求解。判别式 Δ = b²-4ac 决定根的性质。解不等式时，必须考虑临界值并检验区间。线性方程组可通过代入法、消元法或矩阵方法求解。
Mixed Practice Example: Solve for x in 3²ˣ⁺¹ = 27ˣ⁻¹ and state the nature of the roots of x² - 4x + 5 = 0.
混合练习示例：解方程 3²ˣ⁺¹ = 27ˣ⁻¹，并判断 x² - 4x + 5 = 0 的根的性质。
Rewrite 27 as 3³: 3²ˣ⁺¹ = (3³)ˣ⁻¹ = 3³⁽ˣ This illustrates how real-life problems often merge exponential equations with polynomial analysis.
将27写成 3³：3²ˣ⁺¹ = (3³)ˣ⁻¹ = 3³⁽ˣ⁻¹⁾。指数相等：2x+1 = 3(x-1) → 2x+1 = 3x-3 → x = 4。对于二次方程，Δ = (-4)² - 4(1)(5) = 16 - 20 = -4 < 0，因此有两个共轭复根。这展示了实际题目如何综合指数方程与多项式分析。
2. Functions and Their Transformations | 函数及其变换
ering: y = a f(b(x - h)) + k applies a vertical stretch/compression (a), horizontal stretch/compression (1/b), shift right h, and shift up k. Always perform stretches and reflections before translations. Rational functions of the form (ax+b)/(cx+d) have vertical asymptote x = -d/c and horizontal asymptote y = a/c.
函数 f 将定义域中的每个 x 映射到值域中唯一的 y。核心概念包括定义域、值域、复合函数 f(g(x)) 和反函数 f⁻¹(x)，其图像关于直线 y = x 对称。求反函数时，交换 x 和 y 再解出 y。图像变换遵循严格顺序：y = a f(b(x - h)) + k 施加垂直拉伸/压缩(a)、水平拉伸/压缩(1/b)、右移 h、上移 k。务必先进行伸缩和反射，再进行平移。形如 (ax+b)/(cx+d) 的有理函数具有垂直渐近线 x = -d/c 和水平渐近线 y = a/c。
Mixed Practice Example: Given f(x) = √x-2 with domain x ≥ 2, find the inverse function and state its domain and range.
混合练习示例：已知 f(x) = √x-2，定义域 x ≥ 2，求其反函数，并给出其定义域和值域。
Let y = √x-2. Swap x and y: x = √y-2 how the square root and square are inverses only under restricted domain/range conditions.
令 y = √x-2。交换 x 和 y：x = √y-2。平方得 x² = y-2 → y = x²+2，因此反函数为 f⁻¹(x) = x²+2。原定义域 x≥2 对应值域 y≥0，故反函数定义域为 x≥0，值域 y≥2。注意平方根与平方仅在定义域和值域受限时才互为反函数。
3. Exponentials and Logarithms | 指数与对数
The natural base e ≈ 2.718 appears throughout calculus and modelling. f(x) = eˣ has derivative eˣ uations, isolate the exponential or logarithm and beware of extraneous solutions from taking logs of negative numbers.
自然底数 e ≈ 2.718 贯穿微积分与建模。f(x) = eˣ 的导数为自身，其反函数 ln x 的导数为 1/x。对数法则与指数法则呼应：ln(ab) = ln a + ln b，ln(a/b) = ln a - ln b，ln aᵇ = b ln a。指数模型 P = P₀ eᵏᵗ 描述增长(k>0)或衰减(k<0)。半衰期或倍增时间满足 eᵏᵗ = 2。解方程时，隔离指数或对数，并注意从负数取对数可能产生增根。
Mixed Practice Example: Solve ln(2x+1) - ln(x-3) = 1 for x, checking validity.
混合练习示例：解方程 ln(2x+1) - ln(x-3) = 1，并检验根的合理性。
/(2.718-2)=10.154/0.718≈14.14 >3, it is valid. This exercise combines log laws with rational solving.
合并对数：ln[(2x+1)/(x-3)] = 1 → (2x+1)/(x-3) = e¹ = e。交叉相乘：2x+1 = e(x-3) → 2x+1 = ex - 3e → 移项：2x - ex = -3e - 1 → x(2-e) = -3e-1 → x = (-3e-1)/(2-e)。分子分母同乘 -1：x = (3e+1)/(e-2)。检验定义域：2x+1>0 且 x-3>0 → x>3。由于 (3e+1)/(e-2) ≈ (9.154+1)/(2.718-2)=10.154/0.718≈14.14 >3，该解有效。此题综合了对数法则与有理方程求解。
4. Sequences, Series, and the Binomial Theorem | 数列、级数与二项式定理
sum Sₙ = u₁(1-rⁿ)/(1-r) for r≠1. An infinite geometric series converges to S = u₁/(1-r) if |r|<1. The binomial expansion for (1+x)ⁿ, where n is rational, is 1 + nx + n(n-1)x²/2! + ... valid for |x| (a+b)ⁿ is C(n,k) aⁿ⁻ᵏ bᵏ (n integer). Apply these to find coefficients or specific terms.
等差数列具有公差 d：uₙ = u₁ + (n-1)d，和 Sₙ = n/2 (u₁+uₙ)。等比数列具有公比 r：uₙ = u₁ rⁿ⁻¹，当 r≠1 时和 Sₙ = u₁(1-rⁿ)/(1-r)。若 |r|<1，无穷等比级数收敛于 S = u₁/(1-r)。对于指数为有理数 n 的二项展开式，(1+x)ⁿ = 1 + nx + n(n-1)x²/2! + ...，该展开在 |x|<1 时有效。(a+b)ⁿ 的通项为 C(n,k) aⁿ⁻ᵏ bᵏ（n 为整数）。运用这些求系数或特定项。
Mixed Practice Example: Find the coefficient of x³ in the expansion of (2 - 3x)⁵. Also, find the sum of the infinite series 12 + 6 + 3 + …
混合练习示例：求 (2 - 3x)⁵ 展开式中 x³ 的系数，并求无穷级数 12 + 6 + 3 + … 的和。
his demonstrates the power of simple formula application when you recognise the structure.
含 x³ 的项：C(5,3) (2)² (-3x)³ = 10 × 4 × (-27x³) = -1080 x³，系数为 -1080。等比级数：u₁=12，r=6/12=0.5，和 S = 12/(1-0.5) = 24。这展示了识别结构后套用公式的强大之处。
5. Trigonometry and Circular Functions | 三角学与圆函数
0°. Exact values for sin, cos, tan of multiples of π/6, π/4, π/3 must be memorised. Basic identities:
| Identity | Formula |
|---|---|
| Pythagorean | sin²θ + cos²θ = 1 |
| Tangent | tanθ = sinθ/cosθ |
| Double angle | sin2θ = 2 sinθ cosθ, cos2θ = cos²θ - sin²θ |
IB 要求使用弧度制：π rad = 180°。必须熟记 π/6、π/4、π/3 倍数角的正弦、余弦、正切精确值。基本恒等式：
| 恒等式 | 公式 |
|---|---|
| 毕达哥拉斯 | sin²θ + cos²θ = 1 |
| 正切 | tanθ = sinθ/cosθ |
| 倍角 | sin2θ = 2 sinθ cosθ，cos2θ = cos²θ - sin²θ |
Mixed Practice Example: Solve 2 sin²x - cos x = 1 for 0 ≤ x ≤ 2π, giving answers in radians.
混合练习示例：在 0 ≤ x ≤ 2π 上解方程 2 sin²x - cos x = 1，结果以弧度表示。
(cos x + 1) = 0. cos x = 1/2 → x = π/3, 5π/3; cos x = -1 → x = π. Solutions: π/3, π, 5π/3.
代换 sin²x = 1 - cos²x：2(1 - cos²x) - cos x = 1 → 2 - 2 cos²x - cos x = 1 → -2 cos²x - cos x + 1 = 0 → 两边乘 -1：2 cos²x + cos x - 1 = 0。解二次方程：(2 cos x - 1)(cos x + 1) = 0。cos x = 1/2 → x = π/3, 5π/3；cos x = -1 → x = π。解为 π/3, π, 5π/3。
6. Calculus: Differentiation | 微积分：微分
Derivative as rate of change and slope of tangent: f'(x) = lim (h→0) [f(x+h)-f(x)]/h. Core
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