📚 Exam-style Practice: Paper 2 Pure Mathematics (Edexcel A-Level) | 考试风格练习：试卷2 纯数学（爱德思 A-Level）
This article provides a set of exam-style practice questions for Edexcel A-Level Mathematics Paper 2 (Pure Mathematics). Working through these examples will help you revise core techniques in algebra, functions, coordinate geometry, sequences and series, trigonometry, exponentials and logarithms, differentiation, integration, vectors and proof. Each topic includes a typical question followed by a step-by-step solution, together with the exam technique points that examiners reward most often. Every section is written so that you can cover the solution, attempt the question yourself, then compare your working line by line.
本文提供了一套针对爱德思 A-Level 数学试卷2（纯数学）的考试风格练习题。通过这些例题，你可以巩固代数、函数、坐标几何、数列与级数、三角学、指数与对数、微分、积分、向量和证明等核心技巧。每个主题都包含一道典型题目及其分步解答，并附上考官最看重的考试技巧要点。每一节的写法都方便你先遮住解答、自己动手尝试，再逐行对照检查。
1. Algebraic Methods | 代数方法
w that (x − 2) is a factor of f(x) = x³ − 4x² + x + 6 and hence factorise f(x) completely.
题目：证明 (x − 2) 是 f(x) = x³ − 4x² + x + 6 的因式，并由此将 f(x) 完全分解因式。
em confirms that (x − 2) is a factor of f(x). The key point is that you must substitute, not divide, in order to earn the first mark.
第一步：计算 f(2)。代入 x = 2 得 f(2) = 2³ − 4 × 2² + 2 + 6 = 8 − 16 + 2 + 6 = 0。由因式定理，f(2) = 0 说明 (x − 2) 是 f(x) 的因式。关键点是：要拿第一分必须做代入检验，而不是直接做除法。
f(2) = 0 ⟹ (x − 2) is a factor of f(x)
panding backwards: (x − 2)(x² − 2x − 3) = x³ − 2x² − 3x − 2x² + 4x + 6 = x³ − 4x² + x + 6.
第二步：用 f(x) 除以 (x − 2)。通过多项式长除法或系数比较，得到商式 x² − 2x − 3。可以反向展开快速检验：(x − 2)(x² − 2x − 3) = x³ − 2x² − 3x − 2x² + 4x + 6 = x³ − 4x² + x + 6，与题目一致。
x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3)
bers with product −3 and sum −2 are −3 and +1. Therefore the complete factorisation is shown below.
第三步：分解二次式。x² − 2x − 3 = (x − 3)(x + 1)，因为两数之积为 −3、之和为 −2 的正是 −3 与 +1。因此完全分解结果如下。
f(x) = (x − 2)(x − 3)(x + 1)
The three roots of f(x) = 0 are x = 2, x = 3 and x = −1. Note the command word "hence" u have just verified, and it also tells you the division step is expected in your working.
f(x) = 0 的三个根为 x = 2、x = 3 和 x = −1。注意指令词 "hence"（由此）：它要求你必须使用刚刚验证过的因式 (x − 2)，同时暗示解答中应出现除法这一步。
2. Functions and Graphs | 函数与图像
− 4x + 7 for x ≥ 2. Find the range of f and the inverse function f⁻¹, stating its domain.
题目：函数 f 定义为 f(x) = x² − 4x + 7，x ≥ 2。求 f 的值域以及反函数 f⁻¹，并写出其定义域。
cause the domain is restricted to x ≥ 2, the vertex is included, so the range is f(x) ≥ 3.
第一步：配方。x² − 4x + 7 = (x − 2)² + 3。因为对一切实数 x 都有 (x − 2)² ≥ 0，f(x) 的最小值在 x = 2 处取得，等于 3。由于定义域限制为 x ≥ 2，顶点被包含在内，所以值域为 f(x) ≥ 3。
f(x) = (x − 2)² + 3, range: f(x) ≥ 3
Step 2: Find the inverse. Write y = (x − 2)² + 3, so (x − 2)² = y − 3 and x − 2 = ±√y − 3. Since the domain of f has x ≥ 2, we take the positive root: x = 2 + √y − 3. Swapping x and y gives f⁻¹(x) = 2 + √x − 3, with domain x ≥ 3.
第二步：求反函数。令 y = (x − 2)² + 3，则 (x − 2)² = y − 3，x − 2 = ±√y − 3。因为 f 的定义域是 x ≥ 2，所以取正根：x = 2 + √y − 3。交换 x 与 y 得 f⁻¹(x) = 2 + √x − 3，其定义域为 x ≥ 3。
f⁻¹(x) = 2 + √x − 3, domain x ≥ 3
mations also appear frequently: y = f(x) + a shifts up by a; y = f(x + a) shifts left by a; y = af(x) stretches vertically by factor a; y = f(ax) stretches horizontally by factor 1/a.
第三步：牢记图像联系。y = f⁻¹(x) 的图像是 y = f(x) 关于直线 y = x 的反射，且 f⁻¹ 的定义域等于 f 的值域。图像变换也是高频考点：y = f(x) + a 向上平移 a；y = f(x + a) 向左平移 a；y = af(x) 纵向拉伸 a 倍；y = f(ax) 横向拉伸 1/a 倍。
3. Coordinate Geometry | 坐标几何
r bisector of AB, and find the centre and radius of the circle x² + y² − 6x + 8y − 11 = 0.
题目：已知点 A(1, 2) 与 B(5, 10)。求线段 AB 的垂直平分线方程，并求圆 x² + y² − 6x + 8y − 11 = 0 的圆心与半径。
2)/(5 − 1) = 8/4 = 2. The perpendicular gradient is the negative reciprocal, so m = −1/2.
第一步：中点与斜率。中点 M 为 ((1 + 5)/2, (2 + 10)/2) = (3, 6)。AB 的斜率为 (10 − 2)/(5 − 1) = 8/4 = 2。垂直线的斜率取其负倒数，所以 m = −1/2。
y − 6 = −½(x − 3), i.e. x + 2y − 15 = 0
y + 4)² − 16 = 11, hence (x − 3)² + (y + 4)² = 36. The centre is (3, −4) and the radius is √36 = 6.
第二步：对圆的方程配方。将 x 项与 y 项分别归类：(x² − 6x) + (y² + 8y) = 11。配方得 (x − 3)² − 9 + (y + 4)² − 16 = 11，即 (x − 3)² + (y + 4)² = 36。所以圆心为 (3, −4)，半径为 √36 = 6。
(x − 3)² + (y + 4)² = 36, centre (3, −4), radius 6
atic with discriminant Δ = b² − 4ac equal to zero. If Δ > 0 the line is a chord (two intersections); if Δ < 0 the line does not meet the circle. Examiners expect you to state this condition explicitly rather than merely computing.
第三步：与切线和交点问题联系。把直线方程代入圆方程后得到二次方程，若判别式 Δ = b² − 4ac 等于零，则直线与圆相切；Δ > 0 为割线（两个交点）；Δ < 0 则不相交。答题时要明确写出这一判据，而不是只做计算。
4. Sequences and Series | 数列与级数
Question: An arithmetic sequence has first term 7 and common difference 4. Find the 20th term and the sum of the first 20 terms. Hence state the sum to infinity of the geometric series with first term 7 and common ratio 0.5.
题目：等差数列首项为 7，公差为 4。求第 20 项与前 20 项之和。并由此写出首项为 7、公比为 0.5 的等比级数的无穷和。
m formula uₙ = a + (n − 1)d. With a = 7, d = 4 and n = 20, u₂₀ = 7 + 19 × 4 = 7 + 76 = 83.
第一步：使用通项公式 uₙ = a + (n − 1)d。代入 a = 7、d = 4、n = 20，得 u₂₀ = 7 + 19 × 4 = 7 + 76 = 83。
a Sₙ = n/2[2a + (n − 1)d]. S₂₀ = 20/2 × [2 × 7 + 19 × 4] = 10 × (14 + 76) = 10 × 90 = 900.
第二步：使用求和公式 Sₙ = n/2[2a + (n − 1)d]。S₂₀ = 20/2 × [2 × 7 + 19 × 4] = 10 × (14 + 76) = 10 × 90 = 900。
u₂₀ = 83, S₂₀ = 900
p 3: Geometric series. For a geometric series, uₙ = arⁿ⁻¹ and Sₙ = a(1 − rⁿ)/(1 − r). When |r| es converges and S∞ = a/(1 − r). Here a = 7 and r = 0.5, so S∞ = 7/(1 − 0.5) = 7/0.5 = 14.
第三步：等比级数。等比数列 uₙ = arⁿ⁻¹，前 n 项和 Sₙ = a(1 − rⁿ)/(1 − r)。当 |r| < 1 时级数收敛，S∞ = a/(1 − r)。本题 a = 7、r = 0.5，故 S∞ = 7/(1 − 0.5) = 7/0.5 = 14。
efficient of x² is 720. Always quote the binomial coefficient explicitly to secure the method mark.
第四步：二项式展开。试卷2 中常见如 (2 + 3x)⁵ 的展开。x² 项为 C(5, 2) × 2³ × (3x)² = 10 × 8 × 9x² = 720x²，故 x² 的系数为 720。务必明确写出二项式系数，以拿到方法分。
5. Trigonometry | 三角学
θ + 30°) = 0.5 for 0° ≤ θ < 360°, and state the cosine rule and the arc length formula.
题目：在 0° ≤ θ < 360° 范围内解方程 sin(2θ + 30°) = 0.5，并写出余弦定理与弧长公式。
nd u = 150° + 360°k. Because θ is in [0°, 360°), u lies in [30°, 750°), so the valid values are u = 30°, 150°, 390°, 510°.
第一步：先求括号内的角。令 u = 2θ + 30°。由 sin u = 0.5 得主值 u = 30°。通解为 u = 30° + 360°k 和 u = 150° + 360°k。因 θ 属于 [0°, 360°)，故 u 属于 [30°, 750°)，符合条件的值为 u = 30°、150°、390°、510°。
nswers lie in the required interval, and writing them in ascending order is good practice.
第二步：换回 θ。由 2θ = u − 30° 得 2θ = 0°、120°、360°、480°，所以 θ = 0°、60°、180°、240°。四个解都在所需区间内，按从小到大排列是好习惯。
θ = 0°, 60°, 180°, 240°
hagorean identity sin²θ + cos²θ = 1 and the relation tan θ = sin θ/cos θ are used constantly in proof and in solving equations.
第三步：回顾常用结论。余弦定理为 a² = b² + c² − 2bc cos A。在弧度制下，弧长 s = rθ，扇形面积 = ½r²θ。恒等式 sin²θ + cos²θ = 1 以及 tan θ = sin θ/cos θ 在证明与解方程中反复使用。
asured in radians, the interval and the values would both change, and many marks are lost simply by mixing degrees with radians. Read the question's instruction once more before you convert anything.
第四步：警惕失分陷阱。若题目中 θ 以弧度给出，区间与解的数值都会改变，很多学生仅仅因为角度制与弧度制混用而失分。转换之前请再读一遍题目说明。
6. Exponentials and Logarithms | 指数与对数
Question: Solve 32x − 1 = 5x + 2, giving your answer to three decimal places. Also explain how y = aekt can be linearised.
题目：解方程 32x − 1 = 5x + 2，答案保留三位小数；并说明如何将 y = aekt 线性化。
. This is the standard first step whenever the unknown appears in two different exponents.
第一步：两边取自然对数。由于两边均为正，得 (2x − 1) ln 3 = (x + 2) ln 5。当未知数出现在两个不同底数的指数上，这是标准的第一步。
3 − ln 3 = x ln 5 + 2 ln 5, so x(2 ln 3 − ln 5) = 2 ln 5 + ln 3, giving the result below.
第二步：展开并合并同类项。2x ln 3 − ln 3 = x ln 5 +
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