Exam-style practice: Paper 2 | 考试风格练习：试卷二

📚 Exam-style practice: Paper 2 | 考试风格练习：试卷二

This resource provides 10 exam-style questions tailored to the Edexcel A-Level Mathematics Paper 2 (Pure Mathematics). Each question is followed by a fully worked solution, allowing you to test your understanding of key topics such as algebra, trigonometry, exponentials, calculus, numerical methods, vectors and parametric equations. Use the bilingual step-by-step breakdown to strengthen both your problem-solving skills and mathematical fluency.

本资源提供10道与爱德思A-Level数学试卷二（纯数学）风格一致的练习题。每道题均配有详细解答，帮助你检验对代数、三角、指数、微积分、数值方法、向量及参数方程等核心主题的理解。利用中英双语的逐步解析，提升解题能力与数学表达的准确性。

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1. Question 1 – Factor theorem and proof | 问题1 – 因式定理与证明

factorise f(x) completely. (c) Prove that the equation f(x) = 0 has exactly one real root.

设 f(x) = x³ + x² + x + 1。(a) 证明 (x + 1) 是 f(x) 的因式。(b) 由此将 f(x) 完全分解因式。(c) 证明方程 f(x) = 0 恰好有一个实根。

−1) = (−1)³ + (−1)² + (−1) + 1 = −1 + 1 − 1 + 1 = 0. Since f(−1) = 0, (x + 1) is a factor.

(a) 使用因式定理：计算 f(−1) = (−1)³ + (−1)² + (−1) + 1 = −1 + 1 − 1 + 1 = 0。因为 f(−1) = 0，所以 (x + 1) 是因式。

) using algebraic long division. The quotient is x² + 1. Therefore f(x) = (x + 1)(x² + 1).

(b) 用代数长除法将 f(x) 除以 (x + 1)，商为 x² + 1。因此 f(x) = (x + 1)(x² + 1)。

1, which has no real solutions (discriminant < 0). Hence the equation has exactly one real root.

(c) 方程 (x + 1)(x² + 1) = 0 推出 x + 1 = 0 或 x² + 1 = 0。前者得到实根 x = −1；后者 x² = −1 无实数解（判别式 < 0）。因此方程恰好有一个实根。

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2. Question 2 – Binomial expansion | 问题2 – 二项式展开

Find the first four terms, in ascending powers of x, of the binomial expansion of √1 + 3x. State the set of values of x for which the expansion is valid.

求 √1 + 3x 按 x 的升幂排列的前四项。说明该展开有效的 x 的取值范围。

Rewrite as (1 + 3x)^½. Using the binomial expansion: (1 + u)ⁿ = 1 + n u + [n(n−1)/2!] u² + [n(n−1)(n−2)/3!] u³ + … with u = 3x, n = ½.

将原式写为 (1 + 3x)^½。利用二项展开式：(1 + u)ⁿ = 1 + n u + [n(n−1)/2!] u² + [n(n−1)(n−2)/3!] u³ + …，其中 u = 3x, n = ½。

3/2)]/6 × (3x)³ = (3/8)/6 × 27x³ = (3/8)×(1/6)×27x³ = (3/48)×27x³ = (81/48)x³ = (27/16)x³.

第1项：1。第2项：½ × 3x = (3/2)x。第3项：[½ × (−½)]/2 × (3x)² = (−¼)/2 × 9x² = −(9/8)x²。第4项：[½ × (−½) × (−3/2)]/6 × (3x)³ = (3/8)/6 × 27x³ = (3/48)×27x³ = (81/48)x³ = (27/16)x³。

The expansion is valid when |3x| < 1, i.e. |x| < ⅓, so −⅓ < x < ⅓.

当 |3x| < 1 时展开有效，即 |x| < ⅓，所以 −⅓ < x < ⅓。

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3. Question 3 – Trigonometric equation | 问题3 – 三角方程

Solve the equation 2 sin²θ − cos θ − 1 = 0 for 0° ≤ θ ≤ 360°.

解方程 2 sin²θ − cos θ − 1 = 0，其中 0° ≤ θ ≤ 360°。

Use the identity sin²θ ≡ cos²θ − cos θ − 1 = 0 → −2 cos²θ − cos θ + 1 = 0. Multiply by −1: 2 cos²θ + cos θ − 1 = 0.

利用恒等式 sin²θ ≡ 1 − cos²θ。代入得 2(1 − cos²θ) − cos θ − 1 = 0 → 2 − 2 cos²θ − cos θ − 1 = 0 → −2 cos²θ − cos θ + 1 = 0。两边乘以−1：2 cos²θ + cos θ − 1 = 0。

actorise the quadratic in cos θ: (2 cos θ − 1)(cos θ + 1) = 0. So cos θ = ½ or cos θ = −1.

对关于 cos θ 的二次式进行因式分解：(2 cos θ − 1)(cos θ + 1) = 0。因此 cos θ = ½ 或 cos θ = −1。

60°, θ = 60°, 300°. For cos θ = −1, θ = 180°. Hence the solution set is {60°, 180°, 300°}.

对于 cos θ = ½ 且 0° ≤ θ ≤ 360°，θ = 60°, 300°。对于 cos θ = −1，θ = 180°。因此解集为 {60°, 180°, 300°}。

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4. Question 4 – Exponential and logarithmic equation | 问题4 – 指数与对数方程

Solve the equation 3 × 2ˣ = 5ˣ⁺¹, giving your answer in the form x = ln a / ln b, where a and b are constants.

解方程 3 × 2ˣ = 5ˣ⁺¹，并将答案写成 x = ln a / ln b 的形式，其中 a 和 b 为常数。

Take natural logarithms of both sides: ln(3 × 2ˣ) = ln(5ˣ⁺¹). Using laws of logs: ln 3 + x ln 2 = (x + 1) ln 5.

两边取自然对数：ln(3 × 2ˣ) = ln(5ˣ⁺¹)。利用对数法则：ln 3 + x ln 2 = (x + 1) ln 5。

right side: ln 3 + x ln 2 = x ln 5 + ln 5. Collect x terms: x ln 2 − x ln 5 = ln 5 − ln 3.

展开右边：ln 3 + x ln 2 = x ln 5 + ln 5。将含 x 的项移到一边：x ln 2 − x ln 5 = ln 5 − ln 3。

ain the standard form: x = ln(3/5) / (ln 5 − ln 2), or equivalently x = ln(3/5) / ln(5/2).

提取 x：x (ln 2 − ln 5) = ln(5/3)。故 x = ln(5/3) / (ln 2 − ln 5)。将分子分母同乘 −1 得标准形式：x = ln(3/5) / (ln 5 − ln 2)，即 x = ln(3/5) / ln(5/2)。

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5. Question 5 – Differentiation and stationary points | 问题5 – 微分与驻点

+ 12x + 7. Find the coordinates of the stationary points of C and determine their nature.

曲线 C 的方程为 y = 2x³ − 9x² + 12x + 7。求 C 的驻点坐标，并判断其性质。

3x + 2) = 6(x − 1)(x − 2). Stationary points occur when dy/dx = 0, giving x = 1 or x = 2.

求导：dy/dx = 6x² − 18x + 12。因式分解：6(x² − 3x + 2) = 6(x − 1)(x − 2)。令 dy/dx = 0 得

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