📚 Exam-Style Practice: A Level | 考试风格练习：A Level
Welcome to this comprehensive exam-style practice for Edexcel A Level Mathematics. This article provides a selection of typical exam questions covering pure mathematics, mechanics, and statistics. Each question is followed by a clear, step-by-step solution, allowing you to test your understanding, identify common pitfalls, and improve your exam technique. Work through each question before checking the solution, and use the bilingual explanations to deepen your grasp of the key concepts.
欢迎来到这份全面的 Edexcel A Level 数学考试风格练习。本文精选了涵盖纯数学、力学和统计的典型考题。每道题后附有清晰、逐步的解答，帮助你检验理解、识别常见错误并提升应试技巧。请先独立完成每道题，再核对解答，并利用中英对照的解释加深对核心概念的掌握。
1. Algebraic Division and Partial Fractions | 代数除法与部分分式
Question: Express f(x) = (3x³ – 5x² + 2x – 7) ÷ (x – 2) in the form Ax² + Bx + C + D/(x – 2), where A, B, C and D are constants.
问题：将 f(x) = (3x³ – 5x² + 2x – 7) ÷ (x – 2) 表示成 Ax² + Bx + C + D/(x – 2) 的形式，其中 A、B、C、D 为常数。
Solution: We perform polynomial long division. Step 1: Divide the leading term 3x³ by x to obtain 3x². This is the first term of the quotient.
解答：进行多项式长除法。步骤1：将首项 3x³ 除以 x 得 3x²，即为商的第一项。
Step 2: Multiply (x – 2) by 3x² to get 3x³ – 6x². Subtract this from the original cubic: (3x³ – 5x²) – (3x³ – 6x²) = x². Bring down the next term +2x, giving x² + 2x.
步骤2：将 (x – 2) 乘以 3x² 得 3x³ – 6x²。从原三次式中减去：(3x³ – 5x²) – (3x³ – 6x²) = x²。降下下一项 +2x，得 x² + 2x。
Step 3: Divide x² by x to obtain x. This becomes the next part of the quotient.
步骤3：将 x² 除以 x 得 x，作为商的下一部分。
Step 4: Multiply (x – 2) by x to get x² – 2x. Subtract from (x² + 2x) to obtain 4x. Bring down the constant –7 to form 4x – 7.
步骤4：将 (x – 2) 乘以 x 得 x² – 2x。从 (x² + 2x) 中减去得 4x。降下常数 –7，形成 4x – 7。
Step 5: Divide 4x by x to get 4. This is the constant term of the quotient.
步骤5：将 4x 除以 x 得 4，即商的常数项。
Step 6: Multiply (x – 2) by 4 to get 4x – 8. Subtract from (4x – 7) to find the remainder 1. Therefore, f(x) = 3x² + x + 4 + 1/(x – 2).
步骤6：将 (x – 2) 乘以 4 得 4x – 8。从 (4x – 7) 中减去得余数 1。因此，f(x) = 3x² + x + 4 + 1/(x – 2)。
Answer: A = 3, B = 1, C = 4, D = 1.
答案：A = 3，B = 1，C = 4，D = 1。
2. Differentiation with Chain Rule and Product Rule | 链式法则与乘积法则求导
Question: Differentiate y = e²ˣ sin(3x) with respect to x, giving your answer in its simplest factorised form.
问题：对 y = e²ˣ sin(3x) 关于 x 求导，并以最简因式形式给出答案。
Solution: Recognise that this is a product of two functions. Let u = e²ˣ and v = sin(3x). Then y' = u' v + u v' (product rule).
解答：识别出这是两个函数的乘积。设 u = e²ˣ、v = sin(3x)。则 y' = u' v + u v'（乘积法则）。
Find u': using the chain rule, derivative of e²ˣ is 2e²ˣ. Find v': derivative of sin(3x) is 3cos(3x).
求 u'：使用链式法则，e²ˣ 的导数为 2e²ˣ。求 v'：sin(3x) 的导数为 3cos(3x)。
Substitute into the product rule: dy/dx = (2e²ˣ)sin(3x) + e²ˣ(3cos(3x)) = e²ˣ [2 sin(3x) + 3 cos(3x)].
代入乘积法则：dy/dx = (2e²ˣ)sin(3x) + e²ˣ(3cos(3x)) = e²ˣ [2 sin(3x) + 3 cos(3x)]。
This is the fully factorised form and cannot be simplified further.
这是完全因式分解后的形式，无法进一步化简。
3. Integration and Area Under a Curve | 积分与曲线下的面积
Question: The curve C has equation y = x³ – 4x. Find the total area of the region enclosed between the curve and the x-axis from x = –2 to x = 2.
问题：曲线 C 的方程为 y = x³ – 4x。求在 x = –2 到 x = 2 之间，曲线与 x 轴所围成的区域的总面积。
Solution: First, find where the curve crosses the x-axis by solving x³ – 4x = 0 => x(x² – 4) = 0 => x = –2, 0, 2. The region is split into two parts: [–2, 0] and [0, 2].
解答：首先求曲线与 x 轴的交点：解 x³ – 4x = 0 ⇒ x(x² – 4) = 0 ⇒ x = –2, 0, 2。区域分为两部分：[–2, 0] 和 [0, 2]。
Because the curve may be above or below the axis, we compute the absolute area. Integrate separately and take absolute values.
由于曲线可能在轴上方或下方，我们需计算绝对面积。分别积分并取绝对值。
Integrate f(x) = x³ – 4x: ∫ (x³ – 4x) dx = (1/4)x⁴ – 2x². Compute the definite integral from –2 to 0: [ (1/4)(0)⁴ – 2(0)² ] – [ (1/4)(–2)⁴ – 2(–2)² ] = 0 – (4 – 8) = 4. So on [–2, 0] the integral is 4 (positive, area = 4).
对 f(x) = x³ – 4x 积分：∫ (x³ – 4x) dx = (1/4)x⁴ – 2x²。计算从 –2 到 0 的定积分：[ (1/4)(0)⁴ – 2(0)² ] – [ (1/4)(–2)⁴ – 2(–2)² ] = 0 – (4 – 8) = 4。因此在 [–2, 0] 上积分为 4（正值，面积 = 4）。
Now compute from 0 to 2: [ (1/4)(2)⁴ – 2(2)² ] – 0 = (4 – 8) – 0 =
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