📚 AQA Mathematics: Worked Examples Explained | AQA 数学：典型例题详解
This revision guide walks you through carefully selected worked examples covering the core topics from the AQA A-Level Mathematics syllabus. Each solution is broken down step by step, with clear reasoning in both English and Chinese, so you can build confidence and master exam technique.
本复习指南通过精心挑选的典型例题，覆盖 AQA A-Level 数学大纲的核心主题。每道题的求解过程被逐步拆解，并用中英双语给出清晰的推理，帮助您建立信心、掌握考试技巧。
1. Solving Quadratic Equations | 解二次方程
Quadratic equations appear frequently on the AQA papers. We start with a standard factorable equation and also show how to use the quadratic formula.
二次方程在 AQA 试卷中频繁出现。我们从一个标准的可因式分解的方程入手，并展示如何使用二次公式求解。
Example: Solve 2x² − 5x − 3 = 0.
例题：解 2x² − 5x − 3 = 0.
By factorising: The product of the coefficient of x² and the constant term is 2 × (−3) = −6. We need two numbers that multiply to −6 and add to −5; they are −6 and +1. Split the middle term: 2x² − 6x + x − 3 = 0, then group: 2x(x − 3) + 1(x − 3) = 0 → (2x + 1)(x − 3) = 0.
因式分解：x² 的系数与常数项的乘积为 2 × (−3) = −6。我们需要找到两个数，它们相乘得 −6、相加得 −5；这两个数是 −6 和 +1。拆分中间项：2x² − 6x + x − 3 = 0，然后分组提取公因式：2x(x − 3) + 1(x − 3) = 0 → (2x + 1)(x − 3) = 0。
Setting each factor to zero gives: 2x + 1 = 0 → x = −½, and x − 3 = 0 → x = 3.
令每个因式为零：2x + 1 = 0 → x = −½，x − 3 = 0 → x = 3。
Now using the quadratic formula: x = [−b ± √(b² − 4ac)] / (2a). Here a = 2, b = −5, c = −3. Substituting: x = [5 ± √(25 − 4×2×(−3))] / (4) = [5 ± √(25 + 24)] / 4 = [5 ± √49] / 4 = [5 ± 7] / 4.
现在使用二次公式：x = [−b ± √(b² − 4ac)] / (2a)。这里 a = 2, b = −5, c = −3。代入得：x = [5 ± √(25 − 4×2×(−3))] / 4 = [5 ± √(25 + 24)] / 4 = [5 ± √49] / 4 = [5 ± 7] / 4。
The two roots are x = (5+7)/4 = 3 and x = (5−7)/4 = −½, confirming the factorised answer.
两个根分别为 x = (5+7)/4 = 3 和 x = (5−7)/4 = −½，验证了因式分解的答案。
Always check by substituting into the original equation: for x = 3, 2(9) − 15 − 3 = 0; for x = −½, 2(¼) + 2.5 − 3 = 0.
一定要代入原方程检验：当 x = 3 时，2(9) − 15 − 3 = 0；当 x = −½ 时，2(¼) + 2.5 − 3 = 0。
2. Solving Exponential and Logarithmic Equations | 解指数与对数方程
Equations involving exponentials and logs are common in Pure Mathematics. Here we use logarithm laws to solve an equation with different bases.
涉及指数和对数的方程在纯数学中很常见。这里我们利用对数运算法则来求解一个底数不同的方程。
Example: Solve 3ˣ⁺¹ = 5²ˣ, giving your answer in exact form.
例题：解 3ˣ⁺¹ = 5²ˣ，给出精确形式的答案。
Take natural logs of both sides: ln(3ˣ⁺¹) = ln(5²ˣ). Using the power rule: (x+1) ln 3 = 2x ln 5.
两边取自然对数：ln(3ˣ⁺¹) = ln(5²ˣ)。运用幂法则：(x+1) ln 3 = 2x ln 5。
Expand and collect x terms: x ln 3 + ln 3 = 2x ln 5 → x ln 3 − 2x ln 5 = − ln 3 → x (ln 3 − 2 ln 5) = − ln 3.
展开并合并含 x 的项：x ln 3 + ln 3 = 2x ln 5 → x ln 3 − 2x ln 5 = − ln 3 → x (ln 3 − 2 ln 5) = − ln 3。
Multiply both sides by −1: x (2 ln 5 − ln 3) = ln 3. Thus x = ln 3 / (2 ln 5 − ln 3).
两边乘以 −1：x (2 ln 5 − ln 3) = ln 3。因此 x = ln 3 / (2 ln 5 − ln 3)。
We can combine the denominator using the quotient law: 2 ln 5 − ln 3 = ln(5²) − ln 3 = ln(25/3). So x = ln 3 / ln(25/3). This is acceptable as an exact answer.
我们可以用商法则合并分母：2 ln 5 − ln 3 = ln(5²) − ln 3 = ln(25/3)。因此 x = ln 3 / ln(25/3)。作为一个精确答案，这样是合格的。
If a calculator value is asked, approximate x ≈ 0.316. But the exact logarithmic form is required here.
如果要求计算器求值，x 约等于 0.316。但这里要求的是精确的对数形式。
3. Differentiation: Tangents and Normals | 微分：切线与法线
AQA questions often ask for the equation of a tangent or normal at a given point. This example demonstrates the standard method.
AQA 的题目经常要求在给定点求切线或法线方程。本例题演示标准方法。
Example: Find the equation of the tangent to the curve y = x³ − 3x² + 2 at the point where x = 1.
例题：求曲线 y = x³ − 3x² + 2 在 x = 1 处的切线方程。
First, find the y-coordinate when x = 1: y = 1³ − 3(1)² + 2 = 1 − 3 + 2 = 0. So the point is (1, 0).
首先，求出 x = 1 时的 y 坐标：y = 1³ − 3(1)² + 2 = 1 − 3 + 2 = 0。因此点为 (1, 0)。
Differentiate to find the gradient function: dy/dx = 3x² − 6x. At x = 1, gradient m = 3(1)² − 6(1) = 3 − 6 = −3.
求导以获得斜率函数：dy/dx = 3x² − 6x。在 x = 1 处，斜率 m = 3(1)² − 6(1) = 3 − 6 = −3。
The tangent equation uses y − y₁ = m(x − x₁): y − 0 = −3(x − 1) → y = −3x + 3.
切线方程使用 y − y₁ = m(x − x₁)：y − 0 = −3(x − 1) → y = −3x + 3。
To find the normal, take the negative reciprocal of the gradient: m_normal = 1/3. The normal equation: y − 0 = (1/3)(x − 1) → y = (1/3)x − 1/3.
若要求法线，取斜率的负倒数：m_normal = 1/3。法线方程：y − 0 = (1/3)(x − 1) → y = (1/3)x − 1/3。
Always check your differentiation by expanding if needed. Here the cubic is already in simple form.
必要时通过展开来检验你的求导。这里的三次式已经是简单形式。
4. Integration: Area Under a Curve | 积分：曲线下的面积
Definite integration is used to find the area between a curve and the x-axis. Care must be taken when the curve crosses the axis.
定积分用于求曲线与 x 轴之间的面积。当曲线穿过坐标轴时需要特别小心。
Example: Find the area bounded by the curve y = x² − 4x + 3 and the x-axis between x = 1 and x = 3.
例题：求由曲线 y = x² − 4x + 3 与 x 轴在 x = 1 和 x = 3 之间围成的面积。
First check if the curve crosses the x-axis within [1,3]. Solve x² − 4x + 3 = 0 → (x−1)(x−3) = 0, so roots are x=1 and x=3. The curve touches the axis at the endpoints and is below the axis for 1 < x < 3. The area is positive but the integral will be negative; we take the absolute value.
首先检查曲线在 [1,3] 内是否穿过 x 轴。解 x² − 4x + 3 = 0 → (x−1)(x−3) = 0，根为 x=1 和 x=3。曲线在区间端点处接触坐标轴，而在 1 < x < 3 位于轴下方。面积应为正，但积分值为负；我们取绝对值。
Set up the definite integral: ∫₁³ (x² − 4x + 3) dx. Integrate: [⅓ x³ − 2x² + 3x] from 1 to 3.
建立定积分：∫₁³ (x² − 4x + 3) dx。积分得：[⅓ x³ − 2x² + 3x] 从 1 到 3。
Evaluate at upper limit: ⅓(27) − 2(9) + 3(3) = 9 − 18 + 9 = 0. Evaluate at lower limit: ⅓(1) − 2(1) + 3(1) = ⅓ − 2 + 3 = 1 ⅓. Subtract: 0 − (4/3) = −4/3. The area is |−4/3| = 4/3 square units.
代入上限：⅓(27) − 2(9) + 3(3) = 9 − 18 + 9 = 0。代入下限：⅓(1) − 2(1) + 3(1) = ⅓ − 2 + 3 = 4/3。相减：0 − (4/3) = −4/3。面积为 |−4/3| = 4/3 平方单位。
If the question asks for the whole area between a curve with multiple intercepts, you would need to split the integral and sum absolute values.
若题目要求曲线有多个截距时的总面积，则需要拆分积分并求绝对值之和。
5. Vectors: Dot Product and Angles | 向量：点积与夹角
Vector geometry is an essential part of the syllabus. The dot product helps find the angle between two vectors and test perpendicularity.
向量几何是教学大纲的重要组成部分。点积可用来求两向量之间的夹角以及检验垂直性。
Example: Given a = 3i − 4j + 2k and b = i + 2j − k, find the angle between a and b, and show that a is perpendicular to c = 2i + j − k.
例题：已知 a = 3i − 4j + 2k，b = i + 2j − k，求 a 与 b 的夹角，并证明 a 垂直于 c = 2i + j − k。
Compute the dot product a·b = (3)(1) + (−4)(2) + (2)(−1) = 3 − 8 − 2 = −7.
计算点积 a·b = (3)(1) + (−4)(2) + (2)(−1) = 3 − 8 − 2 = −7。
Find magnitudes: |a| = √(9 + 16 + 4) = √29; |b| = √(1 + 4 + 1) = √6.
求模长：|a| = √(9 + 16 + 4) = √29；|b| = √(1 + 4 + 1) = √6。
Using cosθ = (a·b)/(|a||b|): cosθ = −7/(√29 √6) = −7/√174. Hence θ = arccos(−7/√174) ≈ arccos(−0.531) ≈ 122.0° (to 1 d.p.).
利用 cosθ = (a·b)/(|a||b|)：cosθ = −7/(√29 √6) = −7/√174。因此 θ = arccos(−7/√174) ≈ arccos(−0.531) ≈ 122.0°（精确到 1 位小数）。
For perpendicular vectors, the dot product must be zero. a·c = (3)(2) + (−4)(1) + (2)(−1) = 6 − 4 − 2 = 0. Hence a ⟂ c.
对于垂直向量，点积必须为零。a·c = (3)(2) + (−4)(1) + (2)(−1) = 6 − 4 − 2 = 0。因此 a ⟂ c。
This method can be extended to find vector equations of lines and planes, but the dot product remains the key tool.
该方法可推广到求直线和平面的向量方程，但点积始终是关键工具。
6. Binomial Distribution | 二项分布
Questions on discrete random variables often use the binomial distribution. You must identify n and p and apply the probability formula correctly.
离散型随机变量的题目常涉及二项分布。您必须正确识别 n 和 p 并应用概率公式。
Example: A fair die is rolled 8 times. Find the probability of obtaining exactly three sixes.
例题：一个均匀的骰子掷 8 次。求恰好得到 3 次六点的概率。
Let X be the number of sixes obtained. X ~ B(8, 1/6). The probability mass function is P(X = r) = C(n, r) p^r (1−p)^(n−r).
设 X 为掷出六点的次数。X ~ B(8, 1/6)。概率质量函数为 P(X = r) = C(n, r) p^r (1−p)^(n−r)。
For r = 3: P(X = 3) = C(8,3) (1/6)^3 (5/6)^5. C(8,3) = 56. (1/6)^3 = 1/216; (5/6)^5 = 3125/7776. Multiply: 56 × (1/216) × (3125/7776) = 56 × 3125 / (216 × 7776). Simplify: 216 = 6³, 7776 = 6⁵, so denominator is 6⁸. Numerator = 56 × 3125 = 175000. Hence P = 175000 / 6⁸ = 175000 / 1679616 ≈ 0.1042 (4 d.p.).
当 r = 3 时：P(X = 3) = C(8,3) (1/6)^3 (5/6)^5。C(8,3) = 56。(1/6)^3 = 1/216；(5/6)^5 = 3125/7776。相乘：56 × (1/216) × (3125/7776) = 56 × 3125 / (216 × 7776)。简化：216 = 6³，7776 = 6⁵，因此分母为 6⁸。分子 = 56 × 3125 = 175000。所以 P = 175000 / 6⁸ = 175000 / 1679616 ≈ 0.1042（4 位小数）。
Cumulative probabilities, such as P(X ≥ 3), would require summing several individual probabilities or using a calculator.
累积概率，如 P(X ≥ 3)，则需要求多个单独概率之和或使用计算器。
7. Kinematics with Calculus | 微积分在运动学中的应用
In mechanics, displacement, velocity and acceleration are linked by differentiation and integration. This example covers a typical particle on a straight line.
在力学中，位移、速度和加速度之间通过微分和积分相互联系。此例涵盖一个典型的质点直线运动。
Example: A particle moves along a straight line such that its displacement s metres from a fixed point O at time t seconds is given by s = t³ − 6t² + 9t (t ≥ 0). Find (a) the times when the particle is at rest, and (b) the acceleration when it passes through O again.
例题：一质点沿直线运动，其从固定点 O 的位移 s 米与时间 t 秒的关系为 s = t³ − 6t² + 9t（t ≥ 0）。求 (a) 质点静止的时刻，(b) 质点再次经过 O 点时的加速度。
Velocity v = ds/dt = 3t² − 12t + 9. At rest, v = 0: 3t² − 12t + 9 = 0 → divide by 3: t² − 4t + 3 = 0 → (t−1)(t−3) = 0. So t = 1 s and t = 3 s.
速度 v = ds/dt = 3t² − 12t + 9。静止时 v = 0：3t² − 12t + 9 = 0 → 除以 3：t² − 4t + 3 = 0 → (t−1)(t−3) = 0。因此 t = 1 秒和 t = 3 秒。
Acceleration a = dv/dt = 6t − 12. To find when the particle passes through O, set s = 0: t³ − 6t² + 9t = 0 → t(t² − 6t + 9) = 0 → t(t−3)² = 0. So t = 0 (initial) or t = 3. At t = 3, acceleration a = 6(3) − 12 = 6 m s⁻².
加速度 a = dv/dt = 6t − 12。要求质点经过 O 的时刻，令 s = 0：t³ − 6t² + 9t = 0 → t(t² − 6t + 9) = 0 → t(t−3)² = 0。因此 t = 0（初始时刻）或 t = 3。当 t = 3 时，加速度 a = 6(3) − 12 = 6 m s⁻²。
Sketching the motion helps visualise changes of direction; at t=1 the velocity changes sign from positive to negative, at t=3 from negative to positive.
画出运动示意图有助于观察方向变化；在 t=1 时速度由正变负，在 t=3 时由负变正。
8. Trigonometric Equations | 三角方程求解
Solving trig equations within a given range requires careful use of identities and quadrant rules. Here we handle a quadratic in sine.
在给定范围内解三角方程，需要巧妙运用恒等式和象限规则。这里我们处理一个关于正弦的二次方程。
Example: Solve 2 sin²θ − sinθ − 1 = 0 for 0° ≤ θ ≤ 360°.
例题：解 2 sin²θ − sinθ − 1 = 0，其中 0° ≤ θ ≤ 360°。
Let y = sinθ. The equation becomes 2y² − y − 1 = 0. Factorise: (2y + 1)(y − 1) = 0 → y = −½ or y = 1.
令 y = sinθ。方程变为 2y² − y − 1 = 0。因式分解：(2y + 1)(y − 1) = 0 → y = −½ 或 y = 1。
Solve sinθ = 1: θ = 90°. (Only one solution in 0° to 360°.)
解 sinθ = 1：θ = 90°。（在 0° 到 360° 内仅此一解。）
Solve sinθ = −½: Since sine is negative, θ lies in the third and fourth quadrants. The related acute angle is sin⁻¹(½) = 30°. So θ = 180° + 30° = 210° and θ = 360° − 30° = 330°.
解 sinθ = −½
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