Edexcel AS Maths Exam-Style Practice Questions | 爱德思AS数学考试风格练习题

📚 Edexcel AS Maths Exam-Style Practice Questions | 爱德思AS数学考试风格练习题

This collection of exam-style questions targets the core Pure Mathematics topics in the Edexcel AS Mathematics specification. Each question is followed by a step-by-step solution to help you build confidence and refine your problem-solving skills. Work through the questions independently before checking the solutions.

本系列考试风格练习题针对爱德思AS数学大纲中的核心纯数主题。每道题后附有逐步解答，帮助你建立信心并提高解题能力。先独立完成题目，再查看解答。

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1. Algebraic Manipulation and Polynomials | 代数运算与多项式

Given that f(x) = 2x³ - 3x² - 3x + 2 and that (x - 2) is a factor, fully factorise f(x).

已知 f(x) = 2x³ - 3x² - 3x + 2，且 (x - 2) 是一个因子，彻底因式分解 f(x)。

tor. f(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0, so (x - 2) is indeed a factor.

步骤1：验证因子。f(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0，因此 (x - 2) 确实是因子。

Step 2: Divide f(x) by (x - 2) using polynomial long division (or synthetic division). The result is a quadratic quotient: 2x² + x - 1.

步骤2：用多项式长除法（或综合除法）将 f(x) 除以 (x - 2)。得到商式为二次式：2x² + x - 1。

Step 3: Factorise the quadratic. 2x² + x - 1 = (2x - 1)(x + 1).

步骤3：分解二次式。2x² + x - 1 = (2x - 1)(x + 1)。

Final answer: f(x) = (x - 2)(2x - 1)(x + 1).

最终答案：f(x) = (x - 2)(2x - 1)(x + 1)。

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2. Quadratic Inequalities and the Discriminant | 二次不等式与判别式

values of k for which the quadratic equation x² + kx + 9 = 0 has two distinct real roots.

求使二次方程 x² + kx + 9 = 0 有两个不同实根的 k 的取值范围。

nt b² - 4ac must be positive: b² - 4ac > 0. Here a = 1, b = k, c = 9, so k² - 4(1)(9) > 0.

要有两个不同实根，判别式 b² - 4ac 必须大于0：b² - 4ac > 0。此处 a = 1, b = k, c = 9，所以 k² - 4(1)(9) > 0。

Simplify: k² - 36 > 0 => k² > 36.

化简：k² - 36 > 0 => k² > 36。

Taking square roots: |k| > 6, so k < -6 or k > 6.

开方：|k| > 6，故 k < -6 或 k > 6。

Solution set: {k : k < -6} ∪ {k : k > 6}.

解集：{k : k < -6} ∪ {k : k > 6}。

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3. Coordinate Geometry – Equation of a Circle | 坐标几何——圆的方程

A circle has its centre at the point (3, -2) and passes through the point (6, 2). Find the equation of the circle.

一个圆的圆心在点 (3, -2) 且经过点 (6, 2)。求该圆的方程。

r is the distance between (3, -2) and (6, 2): r = √[(6 - 3)² + (2 - (-2))²] = √[3² + 4²] = √9 + 16 = √25 = 5.

半径 r 为点 (3, -2) 与 (6, 2) 之间的距离：r = √[(6 - 3)² + (2 - (-2))²] = √[3² + 4²] = √9 + 16 = √25 = 5。

Hence the equation is (x - 3)² + (y + 2)² = 5² = 25.

因此方程为 (x - 3)² + (y + 2)² = 5² = 25。

it becomes x² - 6x + 9 + y² + 4y + 4 = 25, which simplifies to x² + y² - 6x + 4y - 12 = 0.

若展开，得到 x² - 6x + 9 + y² + 4y + 4 = 25，化简为 x² + y² - 6x + 4y - 12 = 0。

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4. Trigonometric Equations | 三角方程

Solve the equation 2 sin θ = √3 for 0° ≤ θ < 360°.

解方程 2 sin θ = √3，其中 0° ≤ θ < 360°。

Divide both sides by 2: sin θ = √3 / 2.

两边同除以2：sin θ = √3 / 2。

The reference acute angle for sin⁻¹(√3/2) is 60°.

对应 sin⁻¹(√3/2) 的锐角为 60°。

Sine is positive in the first and second quadrants, so θ = 60° and θ = 180° - 60° = 120°.

正弦在第一和第二象限为正，故 θ = 60° 和 θ = 180° - 60° = 120°。

Solutions: θ = 60°, 120°.

解：θ = 60°, 120°。

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5. Exponential and Logarithmic Equations | 指数与对数方程

Solve 3²ˣ = 5, giving your answer to 3 significant figures.

解 3²ˣ = 5，答案保留三位有效数字。

Take natural logarithms: ln(3²ˣ) = ln 5 => 2x ln 3 = ln 5.

取自然对数：ln(3²ˣ) = ln 5 => 2x ln 3 = ln 5。

Rearrange: x = ln 5 / (2 ln 3).

整理：x = ln 5 / (2 ln 3)。

lculator: ln 5 ≈ 1.60944, ln 3 ≈ 1.09861, so x ≈ 1.60944 / (2 × 1.09861) ≈ 0.732 (3 s.f.).

计算器计算：ln 5 ≈ 1.60944，ln 3 ≈ 1.09861，故 x ≈ 1.60944 / (2 × 1.09861) ≈ 0.732（三位有效数字）。

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6. Differentiation – Finding a Tangent | 微分——求切线

Find the equation of the tangent to the curve y = x³ - 4x² + 7 at the point where x = 2.

求曲线 y = x³ - 4x² + 7 在 x = 2 处的切线方程。

First differentiate: dy/dx = 3x² - 8x.

首先求导：dy/dx = 3x² - 8x。

At x = 2, the gradient m = 3(2)² - 8(2) = 12 - 16 = -4.

在 x = 2 处，梯度 m = 3(2)² - 8(2) = 12 - 16 = -4。

Find the y-coordinate: y = (2)³ - 4(2)² + 7 = 8 - 16 + 7 = -1, so the point is (2, -1).

求 y 坐标：y = (2)³ - 4(2)² + 7 = 8 - 16 + 7 = -1，故点为 (2, -1)。

t equation using point-slope form: y - (-1) = -4(x - 2) => y + 1 = -4x + 8 => y = -4x + 7.

用点斜式写切线方程：y - (-1) = -4(x - 2) => y + 1 = -4x + 8 => y = -4x + 7。

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7. Integration – Indefinite Integral and Area | 积分——不定积分与面积

2x + 3) dx. Hence find the area under the curve y = 6x² - 2x + 3 between x = 1 and x = 2.

求 ∫ (6x² - 2x + 3) dx。由此计算曲线 y = 6x² - 2x + 3 在 x = 1 到 x = 2 之间的面积。

Integrate term by term: ∫ 6x² dx = 2x³, ∫ -2x dx = -x², ∫ 3 dx = 3x, plus the constant C.

逐项积分：∫ 6x² dx = 2x³，∫ -2x dx = -x²，∫ 3 dx = 3x，加上常数 C。

So the indefinite integral is 2x³ - x² + 3x + C.

因此不定积分为 2x³ - x² + 3x + C。

For the area, use the definite integral F(x) = 2x³ - x² + 3x. Evaluate F(2) and F(1).

求面积，用定积分 F(x) = 2x³ - x² + 3x。计算 F(2) 和 F(1)。

F(2) = 2(8) - 4 + 6 = 18; F(1) = 2 - 1 + 3 = 4. Area = F(2) - F(1) = 14 square units.

F(2) = 2(8) - 4 + 6 = 18；F(1) = 2 - 1 + 3 = 4。面积 = F(2) - F(1) = 14 平方单位。

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8. Arithmetic Sequences | 等差数列

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