Exam-style Practice: Paper 2 | 模拟考试练习：试卷二

📚 Exam-style Practice: Paper 2 | 模拟考试练习：试卷二

Paper 2 of the Edexcel A Level Mathematics qualification tests pure mathematics in a structured, problem-solving format. This revision set focuses on the most frequently examined skills and gives you exam-style prompts to work through under timed conditions.

Edexcel A Level 数学资格考试卷二以结构化的问题解决形式考查纯数学。本复习专题聚焦最常考的技能，并提供模拟考试风格的题目，供你在计时条件下练习。

1. Algebraic Methods and Proof | 代数方法与证明

Algebraic manipulation is often the hidden engine behind many marks. You need to be quick with factor theorem, long division, and partial fractions.

代数运算是许多分数背后的隐藏引擎。你需要快速掌握因式定理、长除法和部分分式。

Exam-style question: Given f(x) = 2x³ – 5x² + x + 2, show that (x – 2) is a factor and hence express f(x) in fully factorised form.

模拟考题：已知 f(x) = 2x³ – 5x² + x + 2，证明 (x – 2) 是一个因式，并据此将 f(x) 写成完全分解形式。

Solution outline: Evaluate f(2) = 16 – 20 + 2 + 2 = 0, so the factor theorem applies. Division gives 2x² – x – 1, which factorises to (2x + 1)(x – 1). Hence f(x) = (x – 2)(2x + 1)(x – 1).

解题框架：计算 f(2) = 16 – 20 + 2 + 2 = 0，所以可应用因式定理。除法后得 2x² – x – 1，继续分解为 (2x + 1)(x – 1)。因此 f(x) = (x – 2)(2x + 1)(x – 1)。

Common pitfall: Do not attempt long division before checking the remainder; if f(2) ≠ 0, the factor theorem cannot be used.

常见错误：在验证余数之前不要贸然进行长除法；若 f(2) ≠ 0，则不能使用因式定理。

2. Functions and Graphs | 函数与图像

Functions questions test domain, range, inverse functions, composite functions, and transformations. Sketching is essential for justifying your answers.

函数题考查定义域、值域、反函数、复合函数和图像变换。画图对于证明答案非常重要。

Exam-style question: The function g is defined by g(x) = ln(2x – 1), x > 1/2. Find g⁻¹(x) and state its domain.

模拟考题：函数 g 定义为 g(x) = ln(2x – 1)，x > 1/2。求 g⁻¹(x) 并写出其定义域。

Solution outline: Let y = ln(2x – 1). Then eʸ = 2x – 1, so x = (eʸ + 1)/2. Therefore g⁻¹(x) = (eˣ + 1)/2, with domain x ∈ ℝ.

解题框架：设 y = ln(2x – 1)。则 eʸ = 2x – 1，所以 x = (eʸ + 1)/2。因此 g⁻¹(x) = (eˣ + 1)/2，定义域为 x ∈ ℝ。

Common pitfall: Remember that the range of g becomes the domain of g⁻¹. Here g has range ℝ, so g⁻¹ is defined for all real x.

常见错误：记住 g 的值域会变成 g⁻¹ 的定义域。本题中 g 的值域为 ℝ，因此 g⁻¹ 对所有实数 x 都有定义。

3. Sequences and Series | 数列与级数

Arithmetic and geometric sequences appear regularly, often in context. Be ready to derive the sum formulae and to use sigma notation fluently.

等差和等比数列经常出现，常结合实际背景。要能推导求和公式，并熟练使用 Σ 记号。

Exam-style question: An arithmetic sequence has first term 7 and common difference 4. Find the smallest n such that the sum of the first n terms exceeds 1000.

模拟考题：一个等差数列首项为 7，公差为 4。求最小的 n，使得前 n 项和大于 1000。

Sₙ = n/2 × (2a + (n – 1)d)

Solution outline: Substitute a = 7 and d = 4 to get Sₙ = n/2 × (14 + 4(n – 1)) = 2n² + 5n. Solve 2n² + 5n > 1000. Testing n = 21 and n = 22 gives S₂₁ = 987 and S₂₂ = 1078, so the smallest n is 22.

解题框架：代入 a = 7 和 d = 4，得 Sₙ = n/2 × (14 + 4(n – 1)) =

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