📚 Mixed Edexcel A-Level Maths Practice from PDF Joiner (4)-202 | 来自PDF合并(4)-202的Edexcel A-Level数学混合练习
This article draws on the rich collection of exercises compiled in the file PDF Joiner (4)-202, a resource widely used by Edexcel A-Level students for targeted revision. The mixed set of questions covers key Pure Mathematics topics from quadratics and exponentials to calculus, sequences, and vectors. Each worked example is explained step by step, providing a clear model for exam-style solutions and reinforcing the methods required to achieve top marks.
本文精选自《PDF Joiner (4)-202》汇编习题集，这是Edexcel A-Level 学生广泛使用的专题复习资料。题目涵盖纯数学的核心模块，包括二次函数、指数与对数、微积分、级数以及向量。每道例题均给出详细的分步解析，示范如何规范书写解答，帮助同学们扎实掌握考点方法，向A*冲刺。
1. Factoring Quadratic Expressions | 因式分解二次表达式
Factorising is one of the most fundamental skills in A-Level algebra. The PDF Joiner (4)-202 includes numerous quadratic expressions where the leading coefficient is greater than 1. A typical example requires splitting the middle term or using the AC method.
因式分解是A-Level代数中最基本的能力之一。《PDF Joiner (4)-202》中收录了大量二次项系数大于1的表达式，需要同学们熟练运用“十字相乘法”或“AC 方法”。
Example: Factorise 2x² − 7x − 15.
Step 1: Multiply a and c: 2 × (−15) = −30. Find two numbers that multiply to −30 and add to −7. The pair is −10 and +3.
例题：因式分解 2x² − 7x − 15。
第一步：将 a 与 c 相乘：2 × (−15) = −30。寻找两个数，其乘积为 −30，且和为 −7。这对数是 −10 和 3。
Step 2: Rewrite the middle term: 2x² − 10x + 3x − 15. Now factor by grouping: 2x(x − 5) + 3(x − 5) = (2x + 3)(x − 5).
第二步：将中间项拆开：2x² − 10x + 3x − 15。分组提取公因式：2x(x − 5) + 3(x − 5) = (2x + 3)(x − 5)。
Check: Expand to confirm: (2x + 3)(x − 5) = 2x² − 10x + 3x − 15 = 2x² − 7x − 15. The factorisation is correct.
验证：将因式展开：(2x + 3)(x − 5) = 2x² − 10x + 3x − 15 = 2x² − 7x − 15，结果正确。
2. Completing the Square and Vertex Form | 完成平方与顶点式
Completing the square is essential for finding the vertex of a quadratic graph and for solving equations where factoring is not straightforward. Many questions in the PDF compilation test this technique alongside graph interpretations.
完成平方是寻找二次函数图像顶点以及解不易因式分解的方程的核心方法。《PDF汇编》中的不少题目都考查了这一技巧，并常与图像解读结合。
| General Form 一般式 | Completed Square Form 顶点式 |
|---|---|
| f(x) = ax² + bx + c | f(x) = a(x − h)² + k |
Example: Write f(x) = x² − 6x + 10 in the form (x − p)² + q, and state the coordinates of the vertex.
Step 1: Take the coefficient of x, divide by 2, and square it: (−6/2)² = 9. Add and subtract 9: x² − 6x + 9 − 9 + 10 = (x − 3)² + 1.
例题：将 f(x) = x² − 6x + 10 写成 (x − p)² + q 的形式，并写出顶点坐标。
第一步：取 x 的系数 −6，除以 2 得 −3，平方得 9。在式子中加上并减去 9：x² − 6x + 9 − 9 + 10 = (x − 3)² + 1。
Step 2: Hence p = 3, q = 1. The vertex is at (3, 1). The graph is a U-shaped parabola shifted 3 units right and 1 unit up from the standard y = x².
第二步：因此 p = 3, q = 1。顶点坐标为 (3, 1)。图像为开口向上的抛物线，相对于标准 y = x² 向右平移 3 个单位，向上平移 1 个单位。
3. Solving Exponential Equations | 解指数方程
Exponential equations appear frequently, often requiring identical bases or the use of logarithms. The PDF Joiner (4)-202 provides several graded examples to ensure fluency with the index laws.
指数方程在考试中很常见，通常需要统一底数或者借助对数求解。《PDF Joiner (4)-202》通过分层练习帮助同学们熟练掌握指数运算法则。
Example: Solve 3²ˣ⁺¹ = 27.
Step 1: Express 27 as a power of 3: 27 = 3³. The equation becomes 3²ˣ⁺¹ = 3³.
例题：解方程 3²ˣ⁺¹ = 27。
第一步：将 27 写成 3 的幂：27 = 3³。原方程化为 3²ˣ⁺¹ = 3³。
Step 2: Equate the exponents, since the bases are equal: 2x + 1 = 3 ⇒ 2x = 2 ⇒ x = 1. Always check: 3²⁽¹⁾⁺¹ = 3³ = 27.
第二步：底数相同，比较指数：2x + 1 = 3 ⇒ 2x = 2 ⇒ x = 1。验算：3²⁽¹⁾⁺¹ = 3³ = 27，成立。
For more complex exponentials like 2ˣ = 7, take logarithms: x log 2 = log 7 ⇒ x = log 7 / log 2.
对于更复杂的指数方程如 2ˣ = 7，可两边取对数：x log 2 = log 7 ⇒ x = log 7 / log 2。
4. Logarithmic Equations and Properties | 对数方程与性质
Logarithms often intimidate students, but the rules are straightforward once mastered. The PDF collection includes equations requiring combination, change of base, and careful domain checks.
对数常令学生望而生畏，但掌握其运算法则后便十分清晰。《PDF汇编》中的题目涉及对数合并、换底公式以及定义域检验。
Example: Solve log₂(x) + log₂(x − 2) = 3.
Step 1: Use the product rule: log₂[x(x − 2)] = 3.
例题：解方程 log₂(x) + log₂(x − 2) = 3。
第一步：运用积的对数法则：log₂[x(x − 2)] = 3。
Step 2: Rewrite in exponential form: x(x − 2) = 2³ = 8 ⇒ x² − 2x − 8 = 0 ⇒ (x − 4)(x + 2) = 0. Potential solutions: x = 4, x = −2.
第二步：转化为指数形式：x(x − 2) = 2³ = 8 ⇒ x² − 2x − 8 = 0 ⇒ (x − 4)(x + 2) = 0。可能解为 x = 4 和 x = −2。
Step 3: Check the domain: arguments of logarithms must be positive. For x = −2, log₂(−2) is undefined. Hence the only valid solution is x = 4.
第三步：检验定义域：对数的真数必须为正数。x = −2 时，log₂(−2) 无意义，因此唯一解为 x = 4。
5. Trigonometric Equations in Given Intervals | 给定区间上的三角方程
Solving trig equations requires knowledge of exact values and the CAST diagram. The PDF Joiner (4)-202 supplies many practice equations over 0° to 360° (or in radians), helping students build speed in finding all solutions.
解三角方程需要熟记特殊角精确值并灵活运用 CAST 图。《PDF Joiner (4)-202》提供了大量在 0° 到 360°（或弧度制）内的练习，帮助学生快速找出所有解。
Example: Solve 2 sin θ = √3 for 0° ≤ θ ≤ 360°.
Step 1: Sin θ = √3/2. The reference angle is 60°, where sin 60° = √3/2.
例题：在 0° ≤ θ ≤ 360° 内解 2 sin θ = √3。
第一步：sin θ = √3/2。锐角为 60°，sin 60° = √3/2。
Step 2: Sine is positive in the first and second quadrants. Therefore θ = 60° and θ = 180° − 60° = 120°.
第二步：正弦在第一和第二象限为正。因此 θ = 60° 和 θ = 180° − 60° = 120°。
Always write the solution set as {60°, 120°} and ensure the values are within the specified interval.
解集应写成 {60°, 120°}，并确认所有角都在给定区间内。
6. Differentiation: Tang
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