📚 Core SL Content: Review Exercise | IB数学核心SL复习练习
This comprehensive review exercise consolidates the essential topics from the IB Mathematics: Analysis and Approaches SL core syllabus. Work through each section carefully, checking your understanding with the provided examples and solutions. The aim is to build fluency in algebraic manipulation, function behaviour, trigonometric identities, calculus techniques, and statistical reasoning, all of which are fundamental to success in the final examination.
这份综合性复习练习整合了IB数学分析与和方法SL核心课程的重要主题。请认真完成每一部分，并通过所给例题和解答来检查自己的理解程度。目标是提高代数运算、函数性质分析、三角恒等式运用、微积分技巧以及统计推理的熟练程度，这些都是最终考试成功的基础。
1. Algebraic Expressions and Equations | 代数表达式与方程
Simplify expressions involving integer exponents, expand binomials, factorise quadratics, and solve linear and quadratic equations. Always check for restrictions when dealing with rational expressions.
化简含有整数指数的表达式，展开二项式，分解二次三项式，并求解一次和二次方程。在处理有理表达式时务必检查定义域限制。
Example 1: Solve the equation 2x² − 5x − 3 = 0.
例题1：解方程 2x² − 5x − 3 = 0。
Factorising gives (2x + 1)(x − 3) = 0, so 2x + 1 = 0 or x − 3 = 0. Hence x = −½ or x = 3.
因式分解得 (2x + 1)(x − 3) = 0，因此 2x + 1 = 0 或 x − 3 = 0。故 x = −½ 或 x = 3。
Example 2: Simplify (x² − 4) / (x² − x − 6).
例题2：化简 (x² − 4) / (x² − x − 6)。
Factor numerator and denominator: (x − 2)(x + 2) over (x − 3)(x + 2). Cancel the common factor (x + 2), provided x ≠ −2. The simplified form is (x − 2) / (x − 3), x ≠ 3, x ≠ −2.
分子分母因式分解：(x − 2)(x + 2) 除以 (x − 3)(x + 2)。约去公因式 (x + 2)，前提是 x ≠ −2。化简结果为 (x − 2) / (x − 3)，且 x ≠ 3，x ≠ −2。
2. Functions and Graphs | 函数与图像
Understand domain, range, composite and inverse functions. Sketch graphs of linear, quadratic, cubic, reciprocal, exponential and trigonometric functions, identifying key points including intercepts, vertices and asymptotes.
理解定义域、值域、复合函数和反函数。画出一元一次函数、二次函数、三次函数、反比例函数、指数函数以及三角函数的草图，并标出截距、顶点和渐近线等关键点。
Example: Given f(x) = 3x + 2 and g(x) = x² − 1, find (f ∘ g)(x) and state its range for x ∈ ℝ.
例题：已知 f(x) = 3x + 2 且 g(x) = x² − 1，求 (f ∘ g)(x) 并指出当 x ∈ ℝ 时的值域。
(f ∘ he smallest value of 3x² is 0, so the minimum output is −1. The range is therefore y ≥ −1.
(f ∘ g)(x) = f(g(x)) = 3(x² − 1) + 2 = 3x² − 1。因为 x² ≥ 0，所以 3x² 的最小值为0，因此输出的最小值为 −1。所以值域为 y ≥ −1。
3. Sequences and Series | 数列与级数
n/2 (2u₁ + (n−1)d). For geometric sequences: uₙ = u₁ rⁿ⁻¹ and Sₙ = u₁ (1 − rⁿ) / (1 − r), |r| ≠ 1. Apply infinite geometric series sum S = u₁ / (1 − r) for |r| < 1.
运用等差数列公式：uₙ = u₁ + (n−1)d 以及 Sₙ = n/2 (2u₁ + (n−1)d)。等比数列公式：uₙ = u₁ rⁿ⁻¹ 以及 Sₙ = u₁ (1 − rⁿ) / (1 − r)，|r| ≠ 1。当 |r| < 1 时，可应用无穷等比级数求和公式 S = u₁ / (1 − r)。
Example: The 5th term of an arithmetic sequence is 14 and the 12th term is 35. Find the common difference and the first term.
例题：一等差数列的第5项为14，第12项为35。求公差和首项。
= 14 and u₁ + 11d = 35. Subtract: 7d = 21 → d = 3. Substitute back: u₁ + 12 = 14 → u₁ = 2.
列出方程：u₁ + 4d = 14，u₁ + 11d = 35。两式相减得 7d = 21 → d = 3。代回得 u₁ + 12 = 14 → u₁ = 2。
4. Trigonometry | 三角学
g identities: sin²θ + cos²θ = 1, tanθ = sinθ / cosθ. Use the sine and cosine rules for non‑right‑angled triangles.
使用弧度制进行运算，弧长公式 s = rθ，扇形面积公式 A = ½ r²θ。利用恒等式 sin²θ + cos²θ = 1 以及 tanθ = sinθ / cosθ 来解三角方程。在非直角三角形中灵活运用正弦定理和余弦定理。
Example: Solve sin 2x = cos x for 0 ≤ x ≤ π.
例题：在 0 ≤ x ≤ π 范围内解方程 sin 2x = cos x。
½. Solutions: x = π/2 from cos x = 0; from sin x = ½, x = π/6 or 5π/6. All within domain.
利用倍角公式：2 sin x cos x = cos x → cos x (2 sin x − 1) = 0。因此 cos x = 0 或 sin x = ½。由 cos x = 0 得 x = π/2；由 sin x = ½ 得 x = π/6 或 5π/6，所有解均在定义域内。
5. Exponents and Logarithms | 指数与对数
xponents: aᵐ × aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ. Understand logarithms as inverse of exponentials: aˣ = b ⇔ g(AB), log A − log B = log(A/B), log Aⁿ = n log A. Change of base: logₐ b = log b / log a.
运用指数律化简表达式：aᵐ × aⁿ = aᵐ⁺ⁿ，(aᵐ)ⁿ = aᵐⁿ。理解对数作为指数运算的逆运算：aˣ = b ⇔ x = logₐ b。应用对数律：log A + log B = log(AB)，log A − log B = log(A/B)，log Aⁿ = n log A。换底公式：logₐ b = log b / log a。
Example: Solve log₂(x + 1) + log₂(x − 1) = 3.
例题：解方程 log₂(x + 1) + log₂(x − 1) = 3。
= ±3. Check domain: x + 1 > 0 and x − 1 > 0 → x > 1. So x = 3 is the only valid solution.
合并对数：log₂[(x + 1)(x − 1)] = 3 → log₂(x² − 1) = 3 → x² − 1 = 2³ = 8 → x² = 9 → x = ±3。检查定义域：x + 1 > 0 且 x − 1 > 0 → x > 1，因此 x = 3 是唯一有效解。
6. Calculus: Differentiation | 微分
= 0). Determine the nature of stationary points using the sign test or second derivative.
对多项式、指数函数、三角函数以及乘积/商使用链式法则求导。运用微分求切线斜率、变化率，并通过令 dy/dx = 0 求驻点。采用一阶导数符号检验法或二阶导数法判断驻点性质。
Example: nates of the stationary point on the curve y = x³ − 6x² + 9x + 2 and determine its nature.
例题：求曲线 y = x³ − 6x² + 9x + 2 上驻点的坐标并判断其性质。
ve: d²y/dx² = 6x − 12. At x=1, negative → maximum (1,6); at x=3, positive → minimum (3,2).
dy/dx = 3x² − 12x + 9。令其为零：3(x² − 4x + 3) = 0 → (x − 1)(x − 3) = 0 → x = 1 或 x = 3。求 y 坐标：x=1 时 y = 1 − 6 + 9 + 2 = 6；x=3 时 y = 27 − 54 + 27 + 2 = 2。二阶导数：d²y/dx² = 6x − 12。当 x=1 时为负 → 极大值点 (1,6)；当 x=3 时为正 → 极小值点 (3,2)。
7. Calculus: Integration | 积分
Integrate polynomials, 1/x, eˣ, sin x and cos x. Find the constant of integration using boundary conditions. Use definite integrals to compute areas under curves and areas between two curves.
对多项式函数、1/x、eˣ、sin x 和 cos x 进行积分。利用边界条件确定积分常数。运用定积分计算曲线下方面积以及两曲线之间的面积。
Example: Find the area enclosed by y = x² and y = x + 2.
例题：求由 y = x² 与 y = x + 2 所围成区域的面积。
x=−1: ½ − 2 + 1/3 = −7/6. Subtract: 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 square units.
求交点：x² = x + 2 → x² − x − 2 = 0 → (x−2)(x+1)=0，所以 x = −1 或 2。上方函数为 y = x + 2。面积 = ∫₋₁² [(x+2) − x²] dx = [½x² + 2x − (1/3)x³] 从 −1 到 2。计算：x=2 时 2 + 4 − 8/3 = 10/3；x=−1 时 ½ − 2 + 1/3 = −7/6。相减：10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 平方单位。
8. Statistics and Probability | 统计与概率
iation. Understand mutually exclusive and independent events. Apply probability rules: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A|B) = P(A ∩ B)/P(B). Use normal distribution to model continuous data.
用平均数、中位数、众数、方差和标准差概括数据。理解互斥事件与独立事件。应用概率公式：P(A ∪ B) = P(A) + P(B) − P(A ∩ B) 以及 P(A|B) = P(A ∩ B)/P(B)。使用正态分布对连续数据进行建模。
Example: A fair six‑sided die is rolled twice. Find the probability that the sum of the scores is 8 given that the first roll is an even number.
例题：抛掷一枚均匀的六面骰子两次。已知第一次抛掷结果为偶数，求两次点数之和为8的概率。
comes. Favorable sums of 8: (2,6), (4,4), (6,2) → 3 outcomes. So probability = 3/18 = 1/6.
已知第一次为偶数的样本空间：(2,1) 至 (2,6)，(4,1) 至 (4,6)，(6,1) 至 (6,6) → 共18个等可能结果。和为8的有利结果：(2,6), (4,4), (6,2) → 3个。因此概率 = 3/18 = 1/6。
9. Probability Distributions | 概率分布
Use the binomial distribution X ~ Apply the normal distribution with standardisation z = (x − μ)/σ and use the normal table.
使用二项分布 X ~ B(n, p)，其中 P(X = k) = C(n, k) pᵏ (1−p)ⁿ⁻ᵏ。求期望值 E(X) = np 以及方差 Var(X) = np(1−p)。应用正态分布并通过标准化 z = (x − μ)/σ 查正态分布表。
Example: In a factory, 10% of products are defective. A sample of 20 is taken. Find the probability exactly 3 are defective.
例题：某工厂产品中次品率为10%。随机抽取20件样本，求恰好有3件次品的概率。
X ~ ⁷ ≈ 1140 × 0.001 × 0.1668 ≈ 0.190 (exact to 3 d.p.) This is a typical calculator question.
X ~ B(20, 0.1)。P(X = 3) = C(20,3) × (0.1)³ × (0.9)¹⁷。C(20,3) = 1140。因此 P = 1140 × 0.001 × 0.9¹⁷ ≈ 1140 × 0.001 × 0.1668 ≈ 0.190（精确至三位小数）。此题通常使用计算器完成。
10. Vectors in Two Dimensions | 平面向量
Represent vectors as column vectors or in component form i, j. Calculate magnitude |v| = √x² + y². Add and subtract vectors, multiply by a scalar. Use the dot product v·w = |v||w| cos θ to find the angle between vectors, and check for perpendicularity (v·w = 0).
用列向量或分量形式 i, j 表示向量。计算模长 |v| = √x² + y²。进行向量的加法、减法以及数乘运算。使用点积 v·w = |v||w| cos θ 求两向量间夹角，并判断垂直（即 v·w = 0）。
Example: Given a = 3i − 2j and b = −i + 4j, find the angle between a and b.
例题：已知 a = 3i − 2j，b = −i + 4j，求 a 与 b 之间的夹角。
Dot product: a·b = 3(−1) + (−2)(4) = −3 −8 = −11. Magnitudes: |a| = √(3²+(−2)²) = √13, |b| = √((−1)²+4²) = √17. cos θ = (−11) / (√13 × √17) ≈ −11/14.866 ≈ −0.740. θ ≈ cos⁻¹(−0.740) ≈ 138° (to nearest degree).
点积：a·b = 3(−1) + (−2)(4) = −3 −8 = −11。模长：|a| = √(3²+(−2)²) = √13，|b| = √((−1)²+4²) = √17。因此 cos θ = (−11) / (√13 × √17) ≈ −11/14.866 ≈ −0.740。θ ≈ cos⁻¹(−0.740) ≈ 138°（精确到度）。
11. Mathematical Models and Review Strategy | 数学模型与复习策略
Combine algebraic, graphical and numerical approaches to model real‑world contexts: linear regression, exponential growth/decay and simple trigonometric models. Always interpret your answer in context and comment on the reasonableness of results.
综合运用代数、图像和数值方法建立实际情境的数学模型：线性回归、指数增长/衰减以及简单的三角函数模型。始终在具体情境中解释答案，并对结果的合理性进行评价。
When reviewing, practise past paper questions under timed conditions, identify common errors (e.g. forgetting to check domain restrictions or misapplying log laws), and create concise summary cards for each topic. Consistent problem‑solving and reflection are key to mastering the IB SL core content.
复习时，请在计时条件下练习往年真题，找出常见错误（例如忘记检查定义域限制或错误运用对数律），并为每个主题制作简洁的总结卡片。持续的解题练习与反思是掌握IB SL核心内容的关键所在。
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