📚 IGCSE CAIE Additional Mathematics: Exam Answer Writing Frameworks and Model Answers | IGCSE CAIE 进阶数学：答题写作框架与范文
In IGCSE CAIE Additional Mathematics (0606), a high-scoring script is not just about correct answers. It is about showing a clear, logical and exam-ready solution. This article gives you reusable writing frameworks for the main question types and a complete model answer so you can see exactly how to present your working.
在 IGCSE CAIE 进阶数学（0606）中，高分答卷不仅取决于答案正确，还取决于你能否写出清晰、有逻辑、符合考试要求的解题过程。本文将给出主要题型的可复用答题写作框架，并提供完整范文，帮助你掌握过程展示的方法。
1. The CAIE Additional Mathematics Paper at a Glance | 试卷结构速览
CAIE IGCSE Additional Mathematics is assessed through two written papers: Paper 1 and Paper 2. Each paper lasts 2 hours and carries 80 marks. Both papers are compulsory and have equal weighting, each contributing 50% to the final grade. Questions are structured, often with several linked parts.
CAIE IGCSE 进阶数学通过两份笔试进行评估：试卷一和试卷二。每份试卷时长 2 小时，满分 80 分。两份试卷均为必考，权重相同，各占总成绩的 50%。题目为结构题，通常包含多个相互关联的小问。
| Component | Paper 1 | Paper 2 |
|---|---|---|
| Duration | 2 hours | 2 hours |
| Total marks | 80 | 80 |
| Weighting | 50% | 50% |
| Question style | Structured and multi-step | Structured and multi-step |
Because many marks are method marks, you must write every key line of working. Even if your final answer is wrong, a correct method can still earn most of the marks.
由于许多分数是方法分，你必须写出每一个关键解题步骤。即使最终答案错误，正确的方法仍可获得大部分分数。
2. The Four-Step Solution Writing Framework | 四步解题写作框架
Use one reliable framework for every question: Read, Plan, Execute, Check. First, identify what is given and what is required. Second, select the relevant formula, identity or strategy. Third, write each line with one operation and state any substitution. Fourth, substitute your answer back and check domain, sign and units.
每个问题都使用一个可靠的框架：审题、规划、执行、检查。首先，明确已知条件和所求内容。其次，选择相关公式、恒等式或策略。第三，逐行写出每一步，每行只做一个运算，并注明代入。第四，将答案代回验证，检查定义域、符号和单位。
| Step | What to write | Why it matters |
|---|---|---|
| 1. Read | Underline given values and target quantity | Prevents misreading |
| 2. Plan | Write the formula or identity first | Secures method marks |
| 3. Execute | One operation per line, clear substitution | Makes the method visible |
| 4. Check | Substitute answer back or test a value | Catches sign and domain errors |
3. Framework for Functions and Quadratic Problems | 函数与二次问题框架
For function questions, always start by stating the domain or the given expression. When finding an inverse, write y = f(x), swap x and y, then solve for y. For composite functions, show the inner substitution clearly before simplifying. For a quadratic inequality, rearrange to make one side zero, factorise or use the quadratic formula, then use a sign diagram or sketch.
函数题要先写出定义域或给定表达式。求反函数时，先写 y = f(x)，交换 x 与 y，再解出 y。复合函数要先明确写出内层代入，再化简。解二次不等式时，先移项使一边为 0，因式分解或用求根公式，然后用符号表或草图判断区间。
The quadratic formula is a key tool. Always write it before substituting:
二次求根公式是关键工具。代入前始终先写出公式：
x = (−b ± √(b² − 4ac)) ÷ (2a)
For example, if x² − 5x + 6 > 0, factorise to (x − 2)(x − 3) > 0. Critical values are x = 2 and x = 3. A sign diagram gives x < 2 or x > 3.
例如，若 x² − 5x + 6 > 0，因式分解为 (x − 2)(x − 3) > 0。临界值为 x = 2 和 x = 3。符号表给出 x < 2 或 x > 3。
4. Framework for Logarithmic and Exponential Equations | 对数与指数方程框架
Start by taking logarithms of both sides when the unknown is in the power. Use the log laws: logₐ(xy) = logₐx + logₐy, logₐ(x ÷ y) = logₐx − logₐy, and logₐ(xⁿ) = n logₐx. Then solve the resulting linear or quadratic equation. Always reject answers that make a logarithm undefined, i.e. when the argument is less than or equal to zero.
当未知数在指数位置时，先对两边取对数。使用对数法则：logₐ(xy) = logₐx + logₐy，logₐ(x ÷ y) = logₐx − logₐy，logₐ(xⁿ) = n logₐx。然后解所得的一次或二次方程。始终舍去使对数无意义的答案，即真数小于或等于 0 的解。
Example: Solve 3^(2x + 1) = 5^(x − 1).
例题：解方程 3^(2x + 1) = 5^(x − 1)。
(2x + 1) ln 3 = (x − 1) ln 5
2x ln 3 + ln 3 = x ln 5 − ln 5
x(2 ln 3 − ln 5) = − ln 5 − ln 3
x = −(ln 5 + ln 3) ÷ (2 ln 3 − ln 5)
Then evaluate the logarithms with a calculator if a decimal answer is required. This line-by-line structure ensures no method marks are lost.
如果题目要求小数答案，再用计算器求出对数值。这种逐行结构可以确保不会丢失方法分。
5. Framework for Trigonometric Equations and Identities | 三角方程与恒等式框架
Transform the equation to a single trigonometric function using identities such as sin²x + cos²x = 1 and tan x = sin x ÷ cos x. Factorise rather than divide by a trigonometric function, because division can lose solutions. Solve in the required interval and list all solutions in ascending order.
利用 sin²x + cos²x = 1 和 tan x = sin x ÷ cos x 等恒等式，将方程化为单一三角函数。应因式分解而不是除以三角函数，因为除法可能丢解。在指定区间内求解，并按升序列出所有解。
Example: Solve 2 sin²x + 3 cos x − 3 = 0 for 0° ≤ x ≤ 360°.
例题：解方程 2 sin²x + 3 cos x − 3 = 0，其中 0° ≤ x ≤ 360°。
2(1 − cos²x) + 3 cos x − 3 = 0
−2 cos²x + 3 cos x − 1 = 0
2 cos²x − 3 cos x + 1 = 0
(2 cos x − 1)(cos x − 1) = 0
So cos x = 1/2 or cos x = 1. In the given interval, the solutions are x = 0°, 60°, 300° and 360°.
因此 cos x = 1/2 或 cos x = 1。在给定区间内，解为 x = 0°、60°、300° 和 360°。
6. Framework for Differentiation and Integration Applications | 微积分应用框架
For stationary points, write dy/dx, set dy/dx = 0, solve for x, then substitute into the original equation to find y. Use the second derivative d²y/dx² or a sign test to determine the nature. For area under a curve, state the integral with limits, integrate term by term, substitute the upper and lower limits, and subtract.
求驻点时，先写 dy/dx，令 dy/dx = 0，解出 x，再代入原方程求 y。用二阶导数 d²y/dx² 或符号法判断极值性质。求曲线下面积时，先写出带上下限的积分，逐项积分，代入上限和下限，再相减。
Example: y = x³ − 6x² + 9x + 1.
例题：y = x³ − 6x² + 9x + 1。
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