📚 Exam-style Practice: Paper 1: Pure Mathematics | 考试风格练习：卷1：纯数学
Welcome to this focused revision resource for Edexcel A-Level Pure Mathematics Paper 1. The questions selected here mirror the style, difficulty, and coverage of the real exam, helping you build fluency in algebraic techniques, calculus, trigonometry, vectors and more. Each problem is followed by a fully worked solution with bilingual commentary to strengthen your understanding and boost your exam confidence.
欢迎使用本套针对Edexcel A-Level纯数学卷1的考试风格练习。本文精选的题目在风格、难度和知识覆盖上都高度仿照真实考试，帮助你提升代数技巧、微积分、三角学、向量等核心能力。每道题都配有完整的详细解答与中英双语解析，旨在强化理解并提升你的应试信心。
1. Partial Fractions and Binomial Expansion | 分式分解与二项式展开
Question: (a) Express (4x+3)/((x-2)(x+1)) in partial fractions. (b) Hence expand (4x+3)/((x-2)(x+1)) in ascending powers of x up to and including the term in x², giving each coefficient in simplest form and stating the range of values of x for which the expansion is valid.
题目：(a) 将 (4x+3)/((x-2)(x+1)) 分解为部分分式。(b) 据此将 (4x+3)/((x-2)(x+1)) 按 x 的升幂展开，写到含 x² 项为止，系数需化为最简并写出展开有效的 x 取值范围。
Solution (a): Write (4x+3)/((x-2)(x+1)) = A/(x-2) + B/(x+1). Multiply by (x-2)(x+1) to obtain 4x+3 = A(x+1) + B(x-2). Substitute x=2 ⇒ 11 = 3A ⇒ A = 11/3. Substitute x=-1 ⇒ -1 = -3B ⇒ B = 1/3. Therefore the partial fractions are 11/(3(x-2)) + 1/(3(x+1)).
解答(a)：设 (4x+3)/((x-2)(x+1)) = A/(x-2) + B/(x+1)。两边乘以 (x-2)(x+1) 得 4x+3 = A(x+1) + B(x-2)。代入 x=2 得 11=3A ⇒ A=11/3；代入 x=-1 得 -1=-3B ⇒ B=1/3。故部分分式为 11/(3(x-2)) + 1/(3(x+1))。
(b) Rewrite each term with a factorised constant: 11/(3(x-2)) = 11/3 · (x-2)⁻¹ = 11/3 · (-2)⁻¹ (1 – x/2)⁻¹ = -11/6 (1 – x/2)⁻¹. And 1/(3(x+1)) = 1/3 · (x+1)⁻¹ = 1/3 · (1 + x)⁻¹. Expand using the binomial series (1 + u)⁻¹ = 1 – u + u² – … valid for |u| < 1. For -11/6 (1 - x/2)⁻¹, use u = -x/2 so |x/2| < 1 ⇒ |x| < 2; expansion: -11/6 [1 + (x/2) + (x/2)² + ...] = -11/6 [1 + x/2 + x²/4 + ...]. For 1/3 (1 + x)⁻¹, valid for |x| < 1, expansion: 1/3 [1 - x + x² - ...]. Adding the series up to x² gives (-11/6 - 11x/12 - 11x²/24) + (1/3 - x/3 + x²/3) = (-11/6 + 1/3) + (-11/12 - 1/3)x + (-11/24 + 1/3)x². Simplify: -11/6 + 2/6 = -9/6 = -3/2; -11/12 - 4/12 = -15/12 = -5/4; -11/24 + 8/24 = -3/24 = -1/8. The overall range is the intersection of |x| < 2 and |x| < 1, so |x| < 1. Expansion: -3/2 - (5/4)x - (1/8)x² + ... valid for |x| < 1.
解答(b)：将每一项改写为可展开形式：11/(3(x-2)) = 11/3 · (-2)⁻¹ (1 – x/2)⁻¹ = -11/6 (1 – x/2)⁻¹；1/(3(x+1)) = 1/3 (1 + x)⁻¹。利用二项式展开 (1 + u)⁻¹ = 1 – u + u² – … (|u|<1)。对 -11/6 (1 - x/2)⁻¹，取 u = -x/2，要求 |x/2| < 1 ⇒ |x| < 2，得 -11/6 [1 + x/2 + x²/4 + ...]。对 1/3 (1 + x)⁻¹，|x| < 1，得 1/3 [1 - x + x² - ...]。相加到 x² 项：(-11/6 + 1/3) + (-11/12 - 1/3)x + (-11/24 + 1/3)x² = -3/2 - (5/4)x - (1/8)x²。公共范围为 |x| < 1。最终展开：-3/2 - (5/4)x - (1/8)x² + ... 有效域 |x| < 1。
2. Functions and Inverse Functions | 函数与反函数
Question: Given f(x) = ln(3x – 1), x > 1/3, and g(x) = e^(2x) – 4, x ∈ ℝ. (a) Find f⁻¹(x) and state its domain. (b) Solve the equation f(g(x)) = 0, giving your answer in exact form.
题目：已知 f(x) = ln(3x – 1)，x > 1/3；g(x) = e^(2x) – 4，x ∈ ℝ。(a) 求 f⁻¹(x) 并写出其定义域；(b) 解方程 f(g(x)) = 0，答案保留精确形式。
(a) Let y = ln(3x – 1). Swap x and y: x = ln(3y – 1) ⇒ e^x = 3y – 1 ⇒ y = (e^x + 1)/3. Therefore f⁻¹(x) = (e^x + 1)/3. The domain of f⁻¹ is the range of f. Since x > 1/3, 3x – 1 > 0, and ln(3x – 1) can take all real values, so range of f is ℝ. Thus domain of f⁻¹ is x ∈ ℝ.
(a) 设 y = ln(3x – 1)，交换 x 与 y 得 x = ln(3y – 1) ⇒ e^x = 3y – 1 ⇒ y = (e^x + 1)/3。因此 f⁻¹(x) = (e^x + 1)/3。f 的值域为全体实数（因为 3x-1 > 0 可趋近 0⁺ 和 +∞），故 f⁻¹ 的定义域为 x ∈ ℝ。
(b) f(g(x)) = ln(3(e^(2x) – 4) – 1) = ln(3e^(2x) – 13). Set equal to 0: ln(3e^(2x) – 13) = 0 ⇒ 3e^(2x) – 13 = 1 ⇒ 3e^(2x) = 14 ⇒ e^(2x) = 14/3 ⇒ 2x = ln(14/3) ⇒ x = 1/2 ln(14/3). This is the exact solution.
(b) 计算复合：f(g(x)) = ln(3(e^(2x)-4)-1) = ln(3e^(2x)-13)。令其等于0：ln(3e^(2x)-13)=0 ⇒ 3e^(2x)-13=1 ⇒ e^(2x)=14/3 ⇒ x = ½ ln(14/3)。
3. Trigonometric Identities and Equations | 三角恒等式与方程
Question: Given that sinθ = 3/5 and θ is an obtuse angle, find the exact values of cosθ and tanθ. Hence solve the equation 5sin²x – 3cos²x = 0 for 0 ≤ x ≤ 2π.
题目：已知 sinθ = 3/5 且 θ 为钝角，求 cosθ 和 tanθ 的精确值。据此解方程 5sin²x – 3cos²x = 0，其中 0 ≤ x ≤ 2π。
For obtuse θ (90° < θ < 180°), cosθ is negative. Using sin²θ + cos²θ = 1: cos²θ = 1 - (3/5)² = 1 - 9/25 = 16/25 ⇒ cosθ = -4/5. Then tanθ = sinθ/cosθ = (3/5)/(-4/5) = -3/4.
由 θ 为钝角，cosθ 为负。利用恒等式 sin²θ + cos²θ = 1，得 cos²θ = 1 – 9/25 = 16/25 ⇒ cosθ = -4/5。于是 tanθ = sinθ/cosθ = -3/4。
For the equation 5sin²x – 3cos²x = 0, rewrite as 5sin²x = 3cos²x ⇒ tan²x = 3/5 ⇒ tanx = ±√(3/5). Let tanα = √(3/5). Principal value α = arctan√(0.6) ≈ 0.659 rad. All solutions in [0,2π]: x = α, π – α, π + α, 2π – α. Exact forms can be expressed as arctan(±√(3/5)) but decimal approximations may be required. The question expects exact radian answers with arctan if calculator not allowed, but typical answer: x = arctan(√(3/5)), π – arctan(√(3/5)), π + arctan(√(3/5)), 2π – arctan(√(3/5)).
对方程 5sin²x – 3cos²x = 0，移项得 tan²x = 3/5 ⇒ tanx = ±√(3/5)。设主值 α = arctan√(3/5) ≈ 0.659 rad，在 [0,2π] 内共有四个解：x = α, π-α, π+α, 2π-α。结果可用反三角函数精确表示。
4. Exponential and Logarithmic Equations | 指数与对数方程
Question: Solve the equation 2^(2x+1) – 5·2^x + 2 = 0, giving your answers in exact logarithmic form.
题目：解方程 2^(2x+1) – 5·2^x + 2 = 0，答案用精确的对数形式表示。
Rewrite 2^(2x+1) as 2·(2^x)². Let y = 2^x, then equation becomes 2y² – 5y + 2 = 0. Solve quadratic: (2y – 1)(y – 2) = 0 ⇒ y = 1/2 or y = 2. Return to 2^x: 2^x = 1/2 = 2⁻¹ ⇒ x = -1. And 2^x = 2 ⇒ x = 1. Solutions: x = -1, x = 1.
将 2^(2x+1) 改写为 2·(2^x)²。设 y = 2^x，方程化为 2y² – 5y + 2 = 0，分解得 (2y-1)(y-2)=0 ⇒ y=1/2 或 y=2。代回：2^x = 2⁻¹ ⇒ x=-1；2^x = 2 ⇒ x=1。解为 x=-1, x=1。
Verification: For x=-1, LHS = 2^(-1) – 5·2⁻¹ + 2 = 1/2 – 5/2 + 2 = 0. For x=1, LHS = 2^3 – 10 + 2 = 8 – 8 = 0.
验证：x=-1 时，左边=½ – 5/2 + 2 = 0；x=1 时，左边=8 – 10 + 2 = 0，均成立。
5. Differentiation and Stationary
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