📚 IGCSE CAIE Additional Mathematics: Past Paper Deep-Dive | IGCSE CAIE 进阶数学：历年真题深度解析
Past papers are the most reliable map to the real IGCSE CAIE Additional Mathematics examination. This deep-dive article uses years of Paper 1 and Paper 2 questions to reveal recurring question types, examiner expectations and high-leverage revision targets.
历年真题是了解 IGCSE CAIE 进阶数学考试最可靠的地图。本文通过多年 Paper 1 与 Paper 2 真题，揭示反复出现的题型、评分标准与高效率复习重点。
Instead of reading notes again, the efficient candidate reverse-engineers the paper: which topics appear in every session, which command words decide method marks, and which short questions quietly separate Grade A from A*. The twelve sections below follow that logic from syllabus structure to final exam technique.
高效率的考生不会重复读笔记，而是"逆向拆解"试卷：哪些主题每场必考、哪些指令词决定方法分、哪些"小题"在悄悄区分 A 与 A*。下面十二个小节就按这个逻辑，从考纲结构一直讲到最后的应试技巧。
1. Syllabus Blueprint and Topic Weighting | 考纲蓝图与主题权重
The 0606 syllabus is examined through two equally weighted papers, each 2 hours long and worth 80 marks. Paper 1 and Paper 2 have the same style: no choice of questions, a calculator allowed throughout, and a mixture of short and extended items. Functions, trigonometry and calculus dominate, but a structured topic analysis shows that "small-topic" questions such as circular measure and vectors are often the ones that separate Grade A from A*.
0606 大纲由两份权重相同的试卷组成，每卷 2 小时、80 分。Paper 1 与 Paper 2 风格一致：没有选题、全程允许使用计算器、题型长短结合。函数、三角与微积分占比最高，但对主题的结构化分析显示，弧度制、向量等"小题"往往是区分 A 与 A* 的关键。
| Topic | Typical weight | Past-paper trend |
|---|---|---|
| Functions | 10–15% | Inverse and composite functions appear almost every session. |
| Trigonometry | 15–20% | Identity-based equation solving is highly predictable. |
| Calculus | 20–25% | Stationary points, kinematics and area are recurring. |
| Circular measure / vectors | 8–12% | Short but discriminating questions. |
| Algebra: quadratics, polynomials, logs | 15–20% | Embedded inside almost every longer question. |
| Series, permutations and combinations | 5–10% | One or two self-contained items per paper. |
These percentages are indicative, not official, but they match the balance seen in recent papers.
这些百分比为指示性数据，并非官方比例，但与近年试卷的分布高度一致。
A practical use of this table is planning: if your weakest row is calculus at 20–25%, that single row is worth more than the whole of series and permutations combined. Revision should be weighted in exactly the same proportions as the paper.
这张表的实用价值在于规划复习：如果你最弱的是占 20–25% 的微积分，那么这一行的价值超过"数列 + 排列组合"的总和。复习的时间分配，应当与试卷的权重完全对齐。
2. Functions, Domain and Inverse Functions | 函数、定义域与反函数
A function f(x) only has an inverse if it is one-to-one. Questions frequently ask for f⁻¹(x), then give a restricted domain, expecting you to state the range from the original domain and swap x and y systematically.
函数 f(x) 只有在一一对应时才有反函数。真题常要求先求 f⁻¹(x)，再给出限制定义域，要求你根据原定义域写出值域，并系统地交换 x 与 y。
, rearrange to make x the subject, then interchange x and y. Every mark scheme rewards a clear statement of the domain of f⁻¹, which equals the range of f. If the question says "find f⁻¹(x) and state its domain", omitting the domain usually costs one mark.
标准做法是：写 y = f(x)，把 x 变为因变量进行变形，再交换 x 与 y。所有评分标准都会给"写出 f⁻¹ 的定义域"这一步分，而 f⁻¹ 的定义域就是 f 的值域。若题目要求"求 f⁻¹(x) 并写出其定义域"，漏写定义域通常直接丢 1 分。
n a full domain it has no inverse; after restricting to x ≥ 2, the inverse is f⁻¹(x) = 2 + √x − 1.
最常见错误是忘记对二次函数如 f(x) = (x − 2)² + 1 限制定义域。在完整定义域上它没有反函数；限制为 x ≥ 2 后，其反函数为 f⁻¹(x) = 2 + √x − 1。
f(x) = (x − 2)² + 1, x ≥ 2 ⟹ f⁻¹(x) = 2 + √x − 1, x ≥ 1
e ± sign of the square root is decided by the restriction, so always quote the domain before choosing the sign.
注意两个相互关联的结论：f 的值域是 f(x) ≥ 1，而它正是 f⁻¹ 的定义域。出题人刻意让根号前的正负号由限制条件决定，因此务必先写定义域、再决定取正号还是负号。
3. Composite Functions and Graph Transformations | 复合函数与图像变换
Composite functions fg(x) and gf(x) are generally different, and the order matters. The standard check question is "state the value of x for which fg(x) is not defined", which forces you to compare the range of g with the domain of f.
复合函数 fg(x) 与 gf(x) 通常不相等，顺序会影响结果。常见追问是"写出使 fg(x) 无定义的 x 值"，这迫使你比较 g 的值域与 f 的定义域。
For fg to exist at a point, the output of g must lie inside the domain of f. Solving g(x) ∉ domain(f) gives the excluded values. Writing out the two ranges and domains side by side before substituting saves far more time than trial and error.
要让 fg 在某点有意义，g 的输出必须落在 f 的定义域内。解 g(x) ∉ domain(f) 即可得到被排除的取值。先把两个函数的值域与定义域并列写出，再代入计算，比反复试错省时得多。
= a·f(b(x + c)) + d, apply the translations inside the bracket first, then the stretches.
图像变换几乎每场必考，只要顺序正确就是"送分题"。对于 y = a·f(b(x + c)) + d，先处理括号内的平移，再处理伸缩。
- y = f(x) + d — translation of d units in the y-direction.
- y = f(x + c) — translation of c units in the negative x-direction.
- y = a·f(x) — stretch parallel to the y-axis with scale factor a.
- y = f(bx) — stretch parallel to the x-axis with scale factor 1 ÷ b.
- y = f(x) + d —— 沿 y 方向平移 d 个单位。
- y = f(x + c) —— 沿 x 轴负方向平移 c 个单位。
- y = a·f(x) —— 平行于 y 轴拉伸，比例为 a。
- y = f(bx) —— 平行于 x 轴拉伸，比例为 1 ÷ b。
4. Quadratic Theory, Discriminant and Inequalities | 二次理论、判别式与不等式
Completing the square converts ax² + bx + c into a(x + p)² + q, which simultaneously gives the vertex, the minimum value and the range. Past papers combine this with the discriminant, so both techniques must be fluent.
配方法把 ax² + bx + c 化为 a(x + p)² + q，一次性给出顶点、最小值与值域。真题常把它与判别式结合考查，因此两种技术都要熟练。
ax² + bx + c = a(x + b ÷ 2a)² + c − b² ÷ 4a
ated root when Δ = 0, and no real roots when Δ < 0. Typical questions use it in reverse — "find the values of k for which the line is a tangent to the curve" means set the equation equal and require Δ = 0.
判别式 Δ = b² − 4ac 用于判断根的个数：Δ > 0 有两个不同实根，Δ = 0 有重根，Δ < 0 无实根。常见考法是反向使用——"求使直线与曲线相切的 k 值"，即令两式相等并要求 Δ = 0。
Quadratic inequalities require a sketch or sign diagram rather than a memorised rule. Solve the equation first, then decide whether the solution set is "between" the roots (for < 0 with a > 0) or "outside" them (for > 0 with a > 0).
二次不等式不能靠背结论，而要靠草图或符号图。先解方程，再判断解集是"两根之间"（a > 0 且不等式为 < 0）还是"两根之外"（a > 0 且不等式为 > 0）。
Also useful: the sum of the roots is −b ÷ a and the product is c ÷ a. These identities appear in questions that give one root and ask for the other, or give the sum and product and ask for the equation.
同样实用的还有：两根之和为 −b ÷ a，两根之积为 c ÷ a。这类恒等式常用于"已知一根求另一根"或"已知和与积求方程"的题型。
5. Indices, Surds and Logarithms | 指数、根式与对数
Index laws and surd manipulation appear as an embedded step in larger questions, so errors here contaminate the whole item. Learn to rationalise a denominator and to simplify nested surds such as √50 = 5√2 without hesitation.
指数法则与根式化简通常作为大题中的一步出现，这里出错会污染整道题。要练到能毫不犹豫地分母有理化，并化简 √50 = 5√2 这类嵌套根式。
- aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1 ÷ aⁿ.
- √a × √b = √ab, and a ÷ √b is rationalised as a√b ÷ b.
- aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ，a⁰ = 1，a⁻ⁿ = 1 ÷ aⁿ。
- √a × √b = √ab，而 a ÷ √b 有理化为 a√b ÷ b。
Logarithms reverse indices. The key laws are heavily examined, and equations that cannot be solved by matching bases must be handled with logs on both sides.
对数是指数的逆运算，其核心法则高频出现；而无法通过"同底"解决的方程，必须在两边同时取对数。
logₐ x = n ⇌ aⁿ = x
- logₐ xy = logₐ x + logₐ y and logₐ (x ÷ y) = logₐ x − logₐ y.
- logₐ xⁿ = n·logₐ x, logₐ a = 1, logₐ 1 = 0.
- Change of base: logₐ x = log_b x ÷ log_b a.
- logₐ xy = logₐ x + logₐ y，logₐ (x ÷ y) = logₐ x − logₐ y。
- logₐ xⁿ = n·logₐ x，logₐ a = 1，logₐ 1 = 0。
- 换底公式：logₐ x = log_b x ÷ log_b a。
The classic equation is of the form 2x + 1 = 3x o expect exponential growth and decay models, where the answer must be rounded in context.
经典方程形如 2x + 1 = 3x，两边取对数得 (x + 1)log 2 = x log 3，再合并含 x 的项。此外还要准备指数增长与衰减模型，答案需结合情境取近似值。
6. Polynomials, Remainder and Factor Theorem | 多项式、余数定理与因式定理
) is a factor. These two statements answer a large share of polynomial questions directly.
余数定理指出：f(x) 除以 (x − a) 的余数为 f(a)。因式定理是其特例，即 f(a) = 0 时 (x − a) 为因式。这两句话直接解决了大量多项式题目。
olution of f(x) = 0. Marks are awarded step by step, so always show the substitution f(a) = 0 explicitly rather than jumping to the answer.
典型多问题目先给出未知常数，要求用已知因式求出它，再要求完全因式分解，最后解 f(x) = 0。分数是分步给的，因此必须显式写出 f(a) = 0 的代换过程，不能直接跳到结论。
ivisor is (ax + b), the factor theorem still works, but remember to substitute x = −b ÷ a.
多项式长除法值得练习，因为它同时给出后续因式分解所需的二次因式。当除式为 (ax + b) 时，因式定理依然适用，但要注意代入 x = −b ÷ a。
Some sessions include partial fractions with linear factors in the denominator. The method of comparing coefficients or substituting convenient values of x is quick, and final answers must be written in the requested form.
部分场次还会考查分母为一次因式的部分分式。用比较系数或代入便于计算的 x 值都很快，最终答案必须写成题目要求的形式。
7. Trigonometric Identities and Equation Solving | 三角恒等式与解三角方程
the reciprocal ratios sec θ, cosec θ and cot θ, and solve equations over a stated interval — usually 0 ≤ θ ≤ 2π or 0° ≤ θ ≤ 360°.
0606 的三角远超直角三角形范围。你需要处理弧度、倒数关系 sec θ、cosec θ、cot θ，并在给定区间（通常是 0 ≤ θ ≤ 2π 或 0° ≤ θ ≤ 360°）内解方程。
sin²θ + cos²θ = 1, tan θ = sin θ ÷ cos θ, sec²θ = 1 + tan²θ
Most exam equations are solved by converting everything to a single ratio. For example, replace cos²θ by 1 − sin²θ to obtain a quadratic in sin θ, factorise, and then find all angles in the interval from the sine graph rather than from memory.
考试中的多数方程都通过"化归为同一函数"求解。例如把 cos²θ 换成 1 − sin²θ，得到关于 sin θ 的二次方程，因式分解后，依据正弦函数图像而非死记来找出区间内所有角。
s also examined. An expression a·sin θ + b·cos θ can be rewritten as R·sin(θ + α) with R = √a² + b², which immediately gives the maximum and minimum values and allows equations to be solved in one step.
合角形式同样在考查范围内。表达式 a·sin θ + b·cos θ 可改写为 R·sin(θ + α)，其中 R = √a² + b²，这能立刻给出最大值与最小值，并使方程一步可解。
- Exact values to memorise: sin 30° = ½, cos 60° = ½, tan 45° = 1.
- Always give answers to the accuracy stated, and check whether the interval is in degrees or radians.
- If the interval is 0 ≤ θ ≤ 2π, an answer in degrees loses accuracy marks.
- 必背精确值：sin 30° = ½，cos 60° = ½，tan 45° = 1。
- 答案精度按题目要求给出，并确认区间单位是角度还是弧度。
- 若区间写的是 0 ≤ θ ≤ 2π，用角度作答会丢掉精度分。
8. Circular Measure and Radian Geometry | 弧度制与圆弧几何
Circular measure is a short topic but a discriminating one: the formulas are simple, yet the geometry questions hide a triangle inside a sector. Learn the three core results and the segment area that follows from them.
弧度制内容不多，却是"分水岭"题型：公式简单，但几何题往往在扇形里藏一个三角形。要掌握三个核心结论以及由它们推出的弓形面积。
s = rθ, A = ½ r²θ, A(segment) = ½ r²(θ − sin θ)
of a sector and its radius, requiring you to find θ from s = rθ and then compute the area — whether the question wants the arc length only or the whole perimeter including two radii.
此处 θ 必须用弧度。常见题型给出扇形的周长与半径，要求先用 s = rθ 求 θ 再算面积，或反向求解。要仔细辨别题目问的是弧长，还是包含两条半径的整个周长。
Harder variants combine a sector with a triangle: the shaded region equals sector minus triangle, so you need ½ r²θ − ½ r² sin θ. Showing the intermediate values of θ, the triangle area and the sector area secures the method marks even if the final arithmetic slips.
较难的变式把扇形与三角形结合：阴影面积等于扇形减三角形，即 ½ r²θ − ½ r² sin θ。把 θ、三角形面积与扇形面积的中间值写清楚，即使最后计算失误也能保住方法分。
9. Differentiation: Stationary Points and Kinematics | 微分：驻点与运动学
Differentiation is the largest single block in the paper, and questions are usually
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