A-Level Edexcel Further Mathematics: Worked Examples Explained | A-Level Edexcel 进阶数学：典型例题详解

📚 A-Level Edexcel Further Mathematics: Worked Examples Explained | A-Level Edexcel 进阶数学：典型例题详解

This article presents a selection of typical exam-style questions from the Edexcel A-Level Further Mathematics syllabus, complete with fully annotated solutions. Each core topic is illustrated by a worked example that highlights key techniques and common pitfalls. The bilingual explanations are designed to strengthen both your conceptual understanding and your ability to communicate mathematical reasoning clearly.

本文精选了 Edexcel A-Level 进阶数学大纲中的典型考题，并给出了详细的解题分析。每个核心主题都通过一道例题来展示关键技巧和常见易错点。中英双语讲解旨在帮助你加深概念理解，同时提升清晰表达数学推理的能力。

1. Complex Numbers: Solving Equations and Polar Form | 复数：解方程与极坐标形式

Example: 3 + 2i)z + (1 + 3i) = 0. Hence express the solutions in polar form and find their product.

例题： 求解二次方程 z² − (3 + 2i)z + (1 + 3i) = 0。然后将解表示为极坐标形式，并求这两个解的乘积。

9 + 12i + 4i² = 5 + 12i. Then 4ac = 4(1 + 3i) = 4 + 12i. So Δ = (5 + 12i) − (4 + 12i) = 1.

我们使用复数系数的求根公式。其中 a = 1，b = −(3 + 2i)，c = 1 + 3i。判别式 Δ = b² − 4ac。计算 b² = (3+2i)² = 9 + 12i + 4i² = 5 + 12i。而 4ac = 4(1 + 3i) = 4 + 12i。因此 Δ = (5 + 12i) − (4 + 12i) = 1。

s are z = [ (3+2i) ± 1 ] / 2. This gives z₁ = (4+2i)/2 = 2 + i, and z₂ = (2+2i)/2 = 1 + i.

因为 Δ = 1，其平方根为 ±1。于是解为 z = [ (3+2i) ± 1 ] / 2。由此可得 z₁ = (4+2i)/2 = 2 + i，z₂ = (2+2i)/2 = 1 + i。

te in polar form r(cos θ + i sin θ), we compute modulus and argument. For z₁ = 2 + i: r₁ = √2²+1² = √5; θ₁ = arctan(1/2) ≈ 0.4636 rad. So z₁ = √5 (cos 0.4636 + i sin 0.4636). For z₂ = 1 + i: r₂ = √2; θ₂ = π/4. Hence z₂ = √2 (cos π/4 + i sin π/4).

要写成极坐标形式 r(cos θ + i sin θ)，需要计算模长和辐角。对于 z₁ = 2 + i：r₁ = √2²+1² = √5；θ₁ = arctan(1/2) ≈ 0.4636 弧度。故 z₁ = √5 (cos 0.4636 + i sin 0.4636)。对于 z₂ = 1 + i：r₂ = √2；θ₂ = π/4。于是 z₂ = √2 (cos π/4 + i sin π/4)。

3i, which matches the constant term in the equation. In polar form the product has modulus √5 × √2 = √10 and argument 0.4636 + π/4 ≈ 1.249 rad, which is indeed the argument of 1+3i.

乘积 z₁z₂ = (2+i)(1+i) = 2 + 2i + i + i² = 1 + 3i，正好等于方程中的常数项。在极坐标形式下，乘积的模长为 √5 × √2 = √10，辐角为 0.4636 + π/4 ≈ 1.249 弧度，这与 1+3i 的辐角一致。

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2. Matrices: Inverse and Linear Systems | 矩阵：逆与线性方程组

Example: 1], [0, 2, -1]], find A⁻¹. Hence solve the system 2x − y = 4, x + 3y + z = 2, 2y − z = −3.

例题： 对于矩阵 A = [[2, -1, 0], [1, 3, 1], [0, 2, -1]]，求 A⁻¹。然后解方程组 2x − y = 4, x + 3y + z = 2, 2y − z = −3。

e find the inverse by augmenting with the identity and using row operations. Start with [A|I]. R₁: 2 -1 0 | 1 0 0; R₂: 1 3 1 | 0 1 0; R₃: 0 2 -1 | 0 0 1. Swap R₁↔R₂ to get a leading 1: R₁ = [1 3 1 | 0 1 0], R₂ = [2 -1 0 | 1 0 0], R₃ unchanged. Then R₂ − 2R₁ → R₂: [0 -7 -2 | 1 -2 0]. Multiply R₂ by −1/7: [0 1 2/7 | −1/7 2/7 0]. Clear above and below: R₁ − 3R₂ → R₁: [1 0 1/7 | 3/7 1/7 0]; R₃ − 2R₂ → R₃: [0 0 −11/7 | 2/7 −4/7 1]. Multiply R₃ by −7/11: [0 0 1 | −7/11]. Finally back-substitute to get the left block as identity. The right block is A⁻¹.

我们通过将矩阵与单位矩阵并列、进行行变换来求逆。初始矩阵 [A|I] 为：R₁: 2 -1 0 | 1 0 0; R₂: 1 3 1 | 0 1 0; R₃: 0 2 -1 | 0 0 1。交换 R₁↔R₂ 使首元为 1：R₁ = [1 3 1 | 0 1 0], R₂ = [2 -1 0 | 1 0 0], R₃ 不变。然后 R₂ − 2R₁ → R₂: [0 -7 -2 | 1 -2 0]。将 R₂ 乘以 −1/7: [0 1 2/7 | −1/7 2/7 0]。清除上下的元素：R₁ − 3R₂ → R₁: [1 0 1/7 | 3/7 1/7 0]; R₃ − 2R₂ → R₃: [0 0 −11/7 | 2/7 −4/7 1]。将 R₃ 乘以 −7/11: [0 0 1 | −2/11 4/11 −7/11]。最后回代使左侧成为单位矩阵，右侧即为 A⁻¹。

The resulting inverse is A⁻¹ = (1/11) × [[5, 1, 1], [−1, 2, 2], [−2, 4, −7]]? Let's verify carefully. After full elimination, the accurate inverse is: A⁻¹ = ⎡ 3/11 1/11 1/11 ⎤; ⎢ −1/11 2/11 2/11 ⎥; ⎣ −2/11 4/11 −7/11 ⎦. But multiplying by 11 we can keep fractions. The system is A [x y z]ᵀ = [4 2 −3]ᵀ. So [x y z]ᵀ = A⁻¹ [4 2 −3]ᵀ.

最终得到的逆矩阵为 A⁻¹ = (1/11) × [[5, 1, 1], [−1, 2, 2], [−2, 4, −7]]？我们仔细验证。完成所有消元后，准确的逆矩阵是：A⁻¹ = ⎡ 3/11 1/11 1/11 ⎤; ⎢ −1/11 2/11 2/11 ⎥; ⎣ −2/11 4/11 −7/11 ⎦。为方便可以使用分数。方程组为 A [x y z]ᵀ = [4 2 −3]ᵀ。因此 [x y z]ᵀ = A⁻¹ [4 2 −3]ᵀ。

1)/11 = 21/11. Thus the solution is x = 1, y = −6/11, z = 21/11. Check in the original equations to confirm.

计算可得：x = (3×4 + 1×2 + 1×(−3))/11 = (12+2−3)/11 = 11/11 = 1。y = (−1×4 + 2×2 + 2×(−3))/11 = (−4+4−6)/11 = −6/11。z = (−2×4 + 4×2 + (−7)×(−3))/11 = (−8+8+21)/11 = 21/11。因此解为 x = 1, y = −6/11, z = 21/11。可代回原方程进行验证。

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3. Vectors: Dot Product, Cross Product and Planes | 向量：点积、叉积与平面

Example: Given points A(1, 2, 3), B(4, 0, 5) and C(2, 3, 1), find the vector equation of the plane passing through A, B and C. Then find the shortest distance from the point P(5, 1, −2) to this plane.

例题： 已知点 A(1, 2, 3)、B(4, 0, 5) 和 C(2, 3, 1)，求过这三点的平面向量方程。然后求点 P(5, 1, −2) 到该平面的最短距离。

) − (2)(1), (2)(1) − (3)(−2), (3)(1) − (−2)(1) ) = (4−2, 2−(−6), 3−(−2)) = (2, 8, 5). So n = 2i + 8j + 5k.

首先求出平面内的两个方向向量：AB = B − A = (3, −2, 2)，AC = C − A = (1, 1, −2)。法向量 n = AB × AC。计算叉积： −2)(−2) − (2)(1), (2)(1) − (3)(−2), (3)(1) − (−2)(1) ) = (4−2, 2−(−6), 3−(−2)) = (2, 8, 5)。因此 n = 2i + 8j + 5k。

ying to 2x + 8y + 5z = 2 + 16 + 15 = 33. Thus the Cartesian equation is 2x + 8y + 5z = 33.

平面方程为 n·(r − A) = 0，其中 r = (x, y, z)。即 2(x−1) + 8(y−2) + 5(z−3) = 0，化简得 2x + 8y + 5z = 2 + 16 + 15 = 33。于是笛卡尔方程为 2x + 8y + 5z = 33。

Shortest distance from P(5,1,−2) to the plane is |2(5)+8(1)+5(−2)−33| / √2²+8²+5² = |10+8−10−33| / √4+64+25 = |−25| / √93 = 25/√93. The distance is exactly 25/√93 units, which can be rationalised if required.

点 P(5,1,−2) 到该平面的最短距离为 |2(5)+8(1)+5(−2)−33| / √2²+8²+5² = |10+8−10−33| / √4+64+25 = |−25| / √93 = 25/√93。距离精确为 25/√93 单位长度，必要时可将其有理化。

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4. Hyperbolic Functions: Identities and Differentiation | 双曲函数：恒等式与微分

Example: at : Prove that cosh²x − sinh²x = 1. Differentiate f(x) = sinh(2x) cosh x and find the exact value of f'(ln 2).

例题： 证明 cosh²x − sinh²x = 1。对 f(x) = sinh(2x) cosh x 求导，并求 f'(ln 2) 的精确值。

Recall definitions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ − e⁻ˣ)/2. Then cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4, sinh²x = (e²ˣ − 2 + e⁻²ˣ)/4. Subtracting gives (e²ˣ+2+e⁻²ˣ − e²ˣ+2−e⁻²ˣ)/4 = 4/4 = 1. The identity holds for all real x.

回顾定义：cosh x = (eˣ + e⁻ˣ)/2，sinh x = (eˣ − e⁻ˣ)/2。那么 cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4，sinh²x = (e²ˣ − 2 + e⁻²ˣ)/4。相减得到 (e²ˣ+2+e⁻²ˣ − e²ˣ+2−e⁻²ˣ)/4 = 4/4 = 1。该恒等式对所有实数 x 成立。

For f(x) = sinh(2x) cosh x, use the product rule: f' nh²x, but we can keep it as is. Alternatively express in exponentials to evaluate at ln 2.

对于 f(x) = sinh(2x) cosh x，使用乘法法则：f'(x) = 2cosh(2x)·cosh x + sinh(2x)·sinh x。我们知道 cosh(2x) = cosh²x + sinh²x，但可以保留原式。也可以用指数形式来表示，以便在 x = ln 2 处求值。

At x = ln 2: eˣ = 2, e⁻ˣ 5/4. Also sinh(2 ln 2) = sinh(ln 4) = (4 − 1/4)/2 = 15/8; cosh(2 ln 2) = (4+1/4)/2 = 17/8.

在 x = ln 2 处：eˣ = 2，e⁻ˣ = 1/2。于是 sinh(ln 2) = (2 − 1/2)/2 = 3/4；cosh(ln 2) = (2+1/2)/2 = 5/4。同时 sinh(2 ln 2) = sinh(ln 4) = (4 − 1/4)/2 = 15/8；cosh(2 ln 2) = (4+1/4)/2 = 17/8。

Substitute: 2cosh(2x) cosh x = 2×(17/8)×(5/4) = (34/8)×(5/4)? sinh(2x) sinh x = (15/8)×(3/4) = 45/32. Sum: 85/16 + 45/32 = 170/32 + 45/32 = 215/32. So f'(ln 2) = 215/32.

代入计算：2cosh(2x) cosh x = 2×(17/8)×(5/4) = (34/8)×(5/4)？注意，2 × 17/8 = 34/8 = 17/4。然后 17/4 × 5/4 = 85/16。第二项：sinh(2x) sinh x = (15/8)×(3/4) = 45/32。求和：85/16 + 45/32 = 170/32 + 45/32 = 215/32。因此 f'(ln 2) = 215/32。

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5. Polar Coordinates: Area and Arc Length | 极坐标：面积与弧长

Example: = 1 + cos θ for 0 ≤ θ ≤ 2π. Find the total area enclosed by C and the length of the curve.

例题： 曲线 C 的极坐标方程为 r = 1 + cos θ，定义域为 0 ≤ θ ≤ 2π。求 C 所围成的总面积以及曲线的长度。

θ. Use cos²θ = (1 + cos 2θ)/2. So integrand becomes 1 + 2cos θ + ½ + ½ cos 2θ = 3/2 + 2cos θ + ½ cos 2θ.

这是一条心脏线，关于极轴对称。所围面积为 A = ½ ∫₀²π r² dθ = ½ ∫₀²π (1 + cos θ)² dθ。展开：(1 + cos θ)² = 1 + 2cos θ + cos²θ。利用 cos²θ = (1 + cos 2θ)/2，被积函数变为 1 + 2cos θ + ½ + ½ cos 2θ = 3/2 + 2cos θ + ½ cos 2θ。

0. So ∫ r² dθ = 3π. Thus A = ½ × 3π = 3π/2. The total area enclosed is 3π/2 square units.

逐项积分：∫₀²π (3/2) dθ = 3π；∫₀²π 2cos θ dθ = 0；∫₀²π ½ cos 2θ dθ = 0。因此 ∫ r² dθ = 3π。于是 A = ½ × 3π = 3π/2。所围总面积为 3π/2 平方单位。

s θ) = 4 cos²(θ/2). Since 0 ≤ θ ≤ 2π, cos(θ/2) ≥ 0 for θ in [0,π] and negative in [π,2π]. Actually we integrate over the whole domain, √(4 cos²(θ/2)) = 2|cos(θ/2)| os(θ/2) dθ = 4 ∫₀π cos(θ/2) dθ = 4 [2 sin(θ/2)]₀π = 8. So the total arc length is 8 units.

极坐标曲线的弧长公式为 L = ∫ √(r² + (dr/dθ)²) dθ。此处 dr/dθ = −sin θ。因此 = (1+cos θ)² + sin²θ = 1 + 2cos θ + cos²θ + sin²θ = 2 + 2cos θ = 2(1+cos θ) = 4 cos²(θ/2)。由于 0 ≤ θ ≤ 2π，在 [0,π] 上 cos(θ/2) ≥ 0，而在 [π,2π] 上为负。实际上在整个区间上积分时，√(4 cos²(θ/2)) = 2|cos(θ/2)|。由对称性，L = 2∫₀π 2 cos(θ/2) dθ = 4 ∫₀π cos(θ/2) dθ = 4 [2 sin(θ/2)]₀π = 8。因此总弧长为 8 单位。

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6. First-Order Differential Equations: Integrating Factor | 一阶微分方程：积分因子

Example: Solve the differential equation dy/dx + 2y = e⁻ˣ, given the initial condition y(0) = 1.

例题： 求解微分方程 dy/dx + 2y = e⁻ˣ，并满足初始条件 y(0) = 1。

This is a linear first-order ODE. The integrating factor is μ(x) = e^{∫2 dx} = e^2x. Multiply both sides of the equation by e^2x: e^2x dy/dx + 2e^2x y = e^x. The left side is the derivative of y e^2x. So d/dx (y e^2x) = e^x.

这是一阶线性常微分方程。积分因子为 μ(x) = e^{∫2 dx} = e^2x。将方程两边同乘 e^2x：e^2x dy/dx + 2e^2x y = e^x。左边正是 y e^2x 的导数。因此 d/dx (y e^2x) = e^x。

Integrate both sides with respect to x: y e^2x = ∫ e^x dx = e^x + C, where C is an arbitrary constant. Hence y = e^−2x (e^x + C) = e^−x + C e^−2x.

两边关于 x 积分：y e^2x = ∫ e^x dx = e^x + C，其中 C 为任意常数。于是 y = e^−2x (e^x + C) = e^−x + C e^−2x。

Apply the initial condition y(0) = 1: 1 = e⁰ + C e⁰ = 1 + C ⇒ C = 0. Thus the particular solution is y = e^−x.

代入初始条件 y(0) = 1：1 = e⁰ + C e⁰ = 1 + C ⇒ C = 0。因此特解为 y = e^−x。

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7. Maclaurin Series: Expansion of Functions | 麦克劳林级数：函数展开

Example: Find the Maclaurin series for ln(1 + sin x) up to the term in x³.

例题： 求 ln(1 + sin x) 的麦克劳林级数，展开至含 x³ 的项。

We need derivatives evaluated at x = 0. Let f(x) = ln(1 + sin x). f(0) = ln(1+0) = 0. f'(x) = cos x / (1 + sin x). f'(0) = 1 / 1 = 1. Second derivative: f'' ) / (1+sin x)² = −1 / (1+sin x). Wait, check: sin²x+cos²x=1, so numerator becomes −sin x −

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