📚 Review Set 2C: Mixed IB Math Practice | IB数学综合复习题集2C
Review Set 2C is crafted to provide a rigorous mixed-topic workout, covering the full spectrum of concepts assessed in the IB Mathematics: Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. It unites algebra, functions, trigonometry, calculus, vectors, probability and statistics into cohesive, exam‑style problems. By working through this set, students sharpen their ability to move fluently between topics—exactly the skill demanded by Paper 1 and Paper 2.
IB数学复习题集2C旨在提供高强度、跨主题的综合训练，覆盖分析与方法（AA）以及应用与解释（AI）课程中的全部核心概念。它将代数、函数、三角、微积分、向量、概率与统计融为一体，设计成考试风格的习题。通过完成这套练习，学生可以提升在不同主题间灵活切换的能力，这正是IB数学卷一和卷二所要求的关键技能。
1. Overview and Structure of the Set | 题集概览与结构
Review Set 2C contains approximately 20–25 multi‑part questions, each bridging two or more syllabus topics. Questions are ordered from straightforward recall to complex, unfamiliar scenarios, mirroring the progression of an IB examination. A built‑in time allocation guide encourages students to practise under timed conditions, with marks indicated for each sub‑question.
复习题集2C包含约20–25道多部分习题，每道题都融合了两个或以上的课程主题。题目由浅入深，从直接回忆型逐步过渡到复杂的、未见过的情境，完全仿照IB考试的难度递进特点。题集内嵌了时间分配指南，并标注了每个子问题的分值，鼓励学生进行限时模拟训练。
The topics covered typically include polynomial algebra, exponential and logarithmic equations, trigonometric identities and equations, differentiation (including product, quotient and chain rules), integration (both definite and indefinite, with substitution), vector geometry in two and three dimensions, probability distributions (binomial and normal), and kinematics. This broad coverage ensures that no major area is left unreviewed.
题集覆盖的主题通常包括多项式代数、指数与对数方程、三角恒等式与方程、微分（含乘积法则、商法则和链式法则）、积分（定积分与不定积分，含换元法）、二维与三维向量几何、概率分布（二项分布与正态分布）以及运动学。如此广泛的覆盖确保不遗漏任何重要知识领域。
Key tip: Attempt Review Set 2C after you have revised each individual topic, then use your errors to guide focused review.
关键提示：在完成各章节单独复习后再尝试复习题集2C，然后利用错题来指导针对性复习。
2. Algebraic Manipulation and Polynomials | 代数运算与多项式
mials, factorisation and use of the Remainder and Factor Theorems. A typical problem asks: ' − 3x² − 11x + 6 and that (x − 3) is a factor, express f(x) as a product of linear factors.' 3x − 2, which further factorises into (2x − 1)(x + 2). Thus f(x) = (x − 3)(2x − 1)(x + 2).
复习题集2C中有若干题目要求进行多项式除法、因式分解，并应用余式定理和因式定理。一道常见题目是：“已知f(x) = 2x³ − 3x² − 11x + 6且 (x − 3) 是一个因式，请将f(x) 表示为一次因式的乘积。” 通过综合除法或长除法可得到商式2x² + 3x − 2，再因式分解为 (2x − 1)(x + 2)。因此 f(x) = (x − 3)(2x − 1)(x + 2)。
icient of x³ is 270. Using nCr, the x³ term is ¹⁰C₃·(1)²·(px)³ = 10p³x³, giving 10p³ = 270 ⇒ ems test algebraic precision and understanding of the general term formula: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ.
二项式定理也时有出现，通常是给定某一项的系数，要求展开 (2 + kx)ⁿ 形式。学生必须将系数对应相等来求出未知常数。例如，在 (1 + px)⁵ 的展开式中，x³ 的系数为270。利用 nCr，x³ 项为 ¹⁰C₃·1²·(px)³ = 10p³x³，因此 10p³ = 270 ⇒ p = 3。这类题目检验代数精确性以及对通项公式 Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ 的理解。
3. Exponential and Logarithmic Functions | 指数与对数函数
Equations such as e²ˣ + 5eˣ − 6 = 0 are disguised quadratics. Substituting u = eˣ rms the equation into u² + 5u − 6 = 0, which factorises to (u + 6)(u − 1) = 0. Since u = eˣ > 0, the only valid solution is eˣ = 1, so x = 0. Ignoring the domain restriction eˣ > 0 is a common pitfall that Review Set 2C highlights.
像 e²ˣ + 5eˣ − 6 = 0 这样的方程是隐藏的二次型。令 u = eˣ，方程化为 u² + 5u − 6 = 0，因式分解得 (u + 6)(u − 1) = 0。由于 u = eˣ > 0，唯一有效解为 eˣ = 1，故 x = 0。忽略 eˣ > 0 的定义域限制是常见错误，复习题集2C特别加以强调。
− 2) = 2³ = 8. Expanding and rearranging yields x² − x − 10 = 0, with solutions x = ½(1 ± √41). We must check the domain: x > 2, so only x = ½(1 + √41) is valid. This reinforces the habit of checking arguments of logs are positive.
对数问题常涉及对数的合并与展开。例如，求解 log₂ (x + 1) + log₂ (x − 2) = 3。利用乘积法则得到 log₂[(x + 1)(x − 2)] = 3，因此 (x + 1)(x − 2) = 2³ = 8。展开整理得 x² − x − 10 = 0，解为 x = ½(1 ± √41)。必须检验定义域：x > 2，因此只有 x = ½(1 + √41) 是有效解。这强化了检验对数真数恒为正的习惯。
Key formula: logₐ M + logₐ N = logₐ (MN) and aˣ = eˣˡⁿᵃ
关键公式：logₐ M + logₐ N = logₐ (MN) 且 aˣ = eˣˡⁿᵃ
4. Trigonometric Functions and Equations | 三角函数与方程
ce cos x = 0 or sin x = ½. Within the interval, cos x = 0 at x = π/2, 3π/2; sin x = ½ at x = π/6, 5π/6. All four solutions must be stated in radians unless degrees are specified.
复习题集2C中的三角问题要求学生熟练掌握恒等式、弧度制，以及在有限定义域内解方程。典型例题：在 0 ≤ x ≤ 2π 内解 sin 2x = cos x。利用倍角公式 sin 2x = 2 sin x cos x，方程化为 2 sin x cos x = cos x，即 cos x (2 sin x − 1) = 0。因此 cos x = 0 或 sin x = ½。在所给区间内，cos x = 0 的解为 x = π/2, 3π/2；sin x = ½ 的解为 x = π/6, 5π/6。除非指定角度制，所有解都必须以弧度表示。
d to find amplitude, period, phase shift and vertical shift from a graph or data table. This links trigonometric parameters to real-world contexts like tide heights or Ferris wheel motion.
正弦函数建模也时有出现：给定函数 h(t) = A sin(B(t − C)) + D，要求学生根据图形或数据表求出振幅、周期、相移和垂直位移。这将三角参数与现实情境（如潮汐高度或摩天轮运动）联系起来。
5. Calculus: Differentiation Techniques | 微积分：微分技巧
confidently find derivatives of composite functions such as f(x) = ln(3x² + 1) or g(x) = esin x in rule is tested repeatedly: if y = (5x − 2)⁴, then dy/dx = 4(5x − 2)³ × 5 = 20(5x − 2)³.
复习题集2C中的微分题从基本幂函数法则延伸到链式法则、乘积法则和商法则。学生必须熟练求出复合函数的导数，例如 f(x) = ln(3x² + 1) 或 g(x) = esin x。链式法则是反复考查的要点：若 y = (5x − 2)⁴，则 dy/dx = 4(5x − 2)³ × 5 = 20(5x − 2)³。
s require evaluating the derivative and then using point-gradient form y − y₁ = m(x − x₁).
题集中还包含求切线和法线方程的问题。对于曲线 y = x³ − 3x + 2，在 x = 1 处的梯度为 3(1)² − 3 = 0，表明该点是驻点。在点 (1,0) 处的切线方程为 y = 0，即一条水平线。这类题型需要先求出导数值，再代入点斜式 y − y₁ = m(x − x₁)。
| f(x) | f'(x) |
| xⁿ | nxⁿ⁻¹ |
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
Table of standard derivatives commonly needed in Set 2C | 题集2C中常用的标准导数表
6. Calculus: Integration and Area | 微积分：积分与面积
s are found by solving x² = 2 − x → x² + x − 2 = 0 → (x + 2)(x − 1) = 0, giving x = −2, x = 1. The area is ∫₋₂¹ [(2 − x) − x²] dx.
不定积分题从基本幂法则开始：∫ (3x² + 2x) dx = x³ + x² + C。学生一定不能忘记积分常数。随后，定积分被用来计算曲线下方面积。一道典型问题：求 y = x² 和 y = 2 − x 所围成的面积。首先求交点：解 x² = 2 − x → x² + x − 2 = 0 → (x + 2)(x − 1) = 0，得 x = −2 和 x = 1。面积即为 ∫₋₂¹ [(2 − x) − x²] dx。
Integration by substitution is tested with integrals such as ∫ x√x² + 1 dx. Let u = x² + 1, so du = 2x dx and x dx = ½ du. The integral becomes ½ ∫ √u du = ½ × (2/3) u3/2 = (1/3)(x² + 1)3/2 + C. Review Set 2C often embeds substitution within longer multi-step questions, requiring students to choose an appropriate u and adjust the limits for definite integrals.
换元积分法以 ∫ x√x² + 1 dx 这类题目进行考查。设 u = x² + 1，则 du = 2x dx，从而 x dx = ½ du。积分变为 ½ ∫ √u du = ½ × (2/3) u3/2 = (1/3)(x² + 1)3/2 + C。复习题集2C常将换元法嵌套在较长的多步问题中，要求学生选择合适的 u，并针对定积分调整积分限。
7. Vectors in Two and Three Dimensions | 二维与三维向量
and b = i + 4j + 2k, the dot product a·b = 3(1) + (−2)(4) + 1(2) = −3. The magnitudes are |a| = √9 + 4 + 1 = √14 and |b| = √1 + 16 + 4 = √21. Hence cos θ = (−3)/(√14·√21) = −3/√294 = −3/(7√6) after simplification, and the acute angle is π − arccos(3/(7√6)). This type of calculation tests both algebraic manipulation and conceptual understanding of direction.
向量题复习了模长、标量积和夹角计算。给定 a = 3i − 2j + k 和 b = i + 4j + 2k，点积 a·b = 3(1) + (−2)(4) + 1(2) = −3。模长 |a| = √9+4+1 = √14，|b| = √1+16+4 = √21。于是 cos θ = (−3)/(√14·√21) = −3/√294 = −3/(7√6)，简化后锐角为 π − arccos(3/(7√6))。这类计算既检验代数运算，也检验对方向概念的理解。
tudents may be asked to show that two lines are skew, or to find the intersection point of two lines by equating components. This reinforces the idea that a consistent value of the parameter t must satisfy all three component equations simultaneously.
空间直线问题采用向量形式 r = a + td。学生可能需要证明两条直线异面，或通过分量等式求出两直线的交点。这强化了一个概念：参数 t 的一致取值必须同时满足三个分量方程。
8. Probability and Statistics | 概率与统计
Probability distributions form a core part of Review Set 2C. For a binomial variable X ~ the complement rule. The set also tests the use of the GDC’s binomCdf or binomPdf functions, but analytical understanding remains essential for exam success.
概率分布是复习题集2C的核心部分。对于二项变量 X ~ B(8, 0.4)，学生需要计算 P(X = 3) = ⁸C₃ (0.4)³ (0.6)⁵，并利用补集规则计算 P(X ≥ 2)。题集也考查如何运用图形计算器的 binomCdf 或 binomPdf 功能，但分析理解力对于取得好成绩仍然必不可少。
0)/20 = −0.5 and z₂ = 0.75, then P(−0.5 < Z < 0.75) is found using standard normal tables or GDC. Inverse normal calculations (finding a value given a probability) also appear regularly.
正态分布问题通常涉及标准化：Z = (X − μ)/σ。题目可能陈述苹果质量服从平均值为150克、标准差为20克的正态分布，求苹果质量介于140克与165克之间的概率。计算 z 值得 z₁ = (140−150)/20 = −0.5，z₂ = 0.75，然后利用标准正态表或计算器求出 P(−0.5 < Z < 0.75)。反向正态计算（已知概率求取值）也频繁出现。
9. Kinematics and Applications | 运动学与应用题
t 4 seconds requires splitting the integral at the turning points t = 1 and t = 3 to account for changes in direction.
复习题集2C将微积分与运动学结合：给定位移 s(t) = t³ − 6t² + 9t（米，t 以秒计），学生求出速度 v(t) = ds/dt = 3t² − 12t + 9 和加速度 a(t) = 6t − 12。当 v(t) = 0 时质点静止，故 3t² − 12t + 9 = 0 → t² − 4t + 3 = 0 → t = 1, 3。在前4秒内所经过的总路程需要在转折点 t = 1 和 t = 3 处分段积分，以考虑方向变化。
Optimisation
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