A-Level Further Mathematics: Detailed Analysis of Typical Examples | A-Level 进阶数学：典型例题详解

📚 A-Level Further Mathematics: Detailed Analysis of Typical Examples | A-Level 进阶数学：典型例题详解

In A-Level Further Mathematics, mastering problem-solving techniques is essential for success in examinations. This article provides a step-by-step walkthrough of typical examples spanning key topics: complex numbers, matrices, differential equations, hyperbolic functions, polar coordinates, and series. Each example is chosen to illustrate common pitfalls and effective strategies. By studying these solutions carefully, you will deepen your understanding and build confidence in tackling challenging questions.

在A-Level进阶数学中，掌握解题技巧是考试成功的关键。本文循序渐进地剖析了覆盖核心主题的典型例题，包括复数、矩阵、微分方程、双曲函数、极坐标和级数。每个例题都旨在揭示常见的陷阱和高效的策略。通过仔细研读这些解答，你将加深理解，并在应对高难度问题时建立信心。

1. Complex Numbers: Roots of a Complex Equation | 复数：复方程的求根

Problem: Find all four roots of the equation z⁴ = 1 + i√3, giving your answers in the form r e^(iθ) where r > 0 and -π < θ ≤ π.

题目：求方程 z⁴ = 1 + i√3 的全部四个根，答案以 r e^(iθ) 形式给出，其中 r > 0 且 -π < θ ≤ π。

First, express the right-hand side in polar form. Modulus: |1 + i√3| = √(1² + (√3)²) = 2. Argument: arg(1 + i√3) = arctan(√3/1) = π/3. Thus, 1 + i√3 = 2 e^(iπ/3).

首先，将右边表示为极坐标形式。模长：|1 + i√3| = √(1² + (√3)²) = 2。辐角：arg(1 + i√3) = arctan(√3/1) = π/3。因此，1 + i√3 = 2 e^(iπ/3)。

The equation becomes z⁴ = 2 e^(iπ/3). Taking the fourth root gives the general solution: z = 2^(1/4) e^(i(π/3 + 2kπ)/4), k = 0, 1, 2, 3. Here, 2^(1/4) = ⁴√2.

方程变为 z⁴ = 2 e^(iπ/3)。开四次方得到通解：z = 2^(1/4) e^(i(π/3 + 2kπ)/4)，其中 k = 0, 1, 2, 3。这里，2^(1/4) = ⁴√2。

For k = 0: θ = (π/3)/4 = π/12. z₀ = ⁴√2 e^(iπ/12).

当 k = 0：θ = (π/3)/4 = π/12，z₀ = ⁴√2 e^(iπ/12)。

For k = 1: θ = (π/3 + 2π)/4 = (7π/3)/4 = 7π/12. z₁ = ⁴√2 e^(i7π/12).

当 k = 1：θ = (π/3 + 2π)/4 = (7π/3)/4 = 7π/12，z₁ = ⁴√2 e^(i7π/12)。

For k = 2: θ = (π/3 + 4π)/4 = (13π/3)/4 = 13π/12. This exceeds π, so we subtract 2π to bring it into the principal range: 13π/12 – 2π = -11π/12. z₂ = ⁴√2 e^(-i11π/12).

当 k = 2：θ = (π/3 + 4π)/4 = (13π/3)/4 = 13π/12，超出主值范围，减去 2π 得到 -11π/12，因此 z₂ = ⁴√2 e^(-i11π/12)。

For k = 3: θ = (π/3 + 6π)/4 = (19π/3)/4 = 19π/12, or -5π/12. z₃ = ⁴√2 e^(-i5π/12).

当 k = 3：θ = (π/3 + 6π)/4 = (19π/3)/4 = 19π/12，或 -5π/12，z₃ = ⁴√2 e^(-i5π/12)。

Notice the symmetrical distribution of arguments around the circle – a hallmark of n-th roots of a complex number.

请注意辐角在圆周上呈对称分布——这是复数 n 次方根的典型特征。

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2. Matrices: Inverse and System of Linear Equations | 矩阵：逆矩阵与线性方程组

Problem: For the matrix A = [ [2, 1, 1], [1, 3, 2], [1, 0, 1] ], find A⁻¹ and hence solve the system:
2x + y + z = 5
x + 3y + 2z = 10
x + z = 3

题目：对于矩阵 A = [ [2, 1, 1], [1, 3, 2], [1, 0, 1] ]，求 A⁻¹，并由此解方程组：
2x + y + z = 5
x + 3y + 2z = 10
x + z = 3

We first compute the determinant of A. Expanding along row 3: det(A) = 1*(1*2 – 1*3) – 0 + 1*(2*3 – 1*1) = 1*(2 – 3) + 1*(6 – 1) = -1 + 5 = 4. Since det(A) ≠ 0, the inverse exists.

先计算 A 的行列式。沿第三行展开：det(A) = 1*(1*2 – 1*3) – 0 + 1*(2*3 – 1*1) = 1*(2 – 3) + 1*(6 – 1) = -1 + 5 = 4。因为 det(A) ≠ 0，逆矩阵存在。

Next, find the matrix of cofactors. C₁₁ = +(3*1 – 2*0) = 3; C₁₂ = -(1*1 – 2*1) = 1; C₁₃ = +(1*0 – 3*1) = -3; C₂₁ = -(1*1 – 1*0) = -1; C₂₂ = +(2*1 – 1*1) = 1; C₂₃ = -(2*0 – 1*1) = 1; C₃₁ = +(1*2 – 1*3) = -1; C₃₂ = -(2*2 – 1*1) = -3; C₃₃ = +(2*3 – 1*1) = 5.

接下来求余子式矩阵。C₁₁ = +(3*1 – 2*0) = 3；C₁₂ = -(1*1 – 2*1) = 1；C₁₃ = +(1*0 – 3*1) = -3；C₂₁ = -(1*1 – 1*0) = -1；C₂₂ = +(2*1 – 1*1) = 1；C₂₃ = -(2*0 – 1*1) = 1；C₃₁ = +(1*2 – 1*3) = -1；C₃₂ = -(2*2 – 1*1) = -3；C₃₃ = +(2*3 – 1*1) = 5。

The cofactor matrix is [[3, 1, -3], [-1, 1, 1], [-1, -3, 5]]. Transpose it to get the adjugate: adj(A) = [[3, -1, -1], [1, 1, -3], [-3, 1, 5]]. Then A⁻¹ = (1/det(A)) adj(A) = 1/4 * [[3, -1, -1], [1, 1, -3], [-3, 1, 5]].

余子式矩阵为 [[3, 1, -3], [-1, 1, 1], [-1, -3, 5]]。转置得到伴随矩阵：adj(A) = [[3, -1, -1], [1, 1, -3], [-3, 1, 5]]。因此 A⁻¹ = (1/4) [[3, -1, -1], [1, 1, -3], [-3, 1, 5]]。

To solve the system, write it as A * [x, y, z]ᵀ = [5, 10, 3]ᵀ. Multiply both sides by A⁻¹: [x, y, z]ᵀ = A⁻¹ * [5, 10, 3]ᵀ. Compute:
x = 1/4 (3*5 + (-1)*10 + (-1)*3) = 1/4 (15 – 10 – 3) = 2/4 = 0.5;
y = 1/4 (1*5 + 1*10 + (-3)*3) = 1/4 (5 + 10 – 9) = 6/4 = 1.5;
z = 1/4 (-3*5 + 1*10 + 5*3) = 1/4 (-15 + 10 + 15) = 10/4 = 2.5.

要解方程组，将其写作 A * [x, y, z]ᵀ = [5, 10, 3]ᵀ。两边同乘 A⁻¹：计算得 x = 0.5，y = 1.5，z = 2.5。

Always verify by substitution: 2(0.5)+1.5+2.5=1+1.5+2.5=5, 0.5+3(1.5)+2(2.5)=0.5+4.5+5=10, 0.5+2.5=3. The solution is consistent.

务必代入原方程验证：计算均成立，解是一致的。

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3. Differential Equations: Second-Order Linear ODE with Constant Coefficients | 微分方程：常系数二阶线性常微分方程

Problem: Solve the differential equation y” – 3y’ + 2y = e^x, with initial conditions y(0) = 1, y'(0) = 0.

题目：求解微分方程 y” – 3y’ + 2y = e^x，并满足初始条件 y(0) = 1，y'(0) = 0。

First, find the complementary function (CF) by solving the homogeneous equation y” – 3y’ + 2y = 0. The auxiliary equation is m² – 3m + 2 = 0, giving m = 1, m = 2. Therefore, CF: y_c = A e^x + B e^(2x).

首先求补函数 (CF)，解齐次方程 y” – 3y’ + 2y = 0。辅助方程为 m² – 3m + 2 = 0，解得 m = 1，m = 2。因此，CF：y_c = A e^x + B e^(2x)。

For the particular integral (PI), the right-hand side is e^x. However, e^x is already part of the complementary function, so we must multiply by x. Try y_p = C x e^x. Then y_p’ = C e^x (1 + x), y_p” = C e^x (2 + x). Substitute into the ODE: C e^x (2 + x) – 3C e^x (1 + x) + 2C x e^x = e^x. Simplify: C e^x [(2 + x) – 3(1 + x) + 2x] = C e^x (2 + x – 3 – 3x + 2x) = C e^x (-1) = -C e^x. Set this equal to e^x ⇒ -C = 1 ⇒ C = -1. Hence, y_p = -x e^x.

对于特解 (PI)，右边为 e^x。但由于 e^x 已是补函数的一部分，需乘以 x。试设 y_p = C x e^x。求导后代回原方程，简化得到 -C e^x = e^x，因此 C = -1，y_p = -x e^x。

The general solution is y = y_c + y_p = A e^x + B e^(2x) – x e^x.

通解为 y = A e^x + B e^(2x) – x e^x。

Apply initial conditions. y(0) = A + B = 1. y’ = A e^x + 2B e^(2x) – e^x(1 + x). Then y'(0) = A + 2B – 1 = 0 ⇒ A + 2B = 1. Solve simultaneously: from A + B = 1, A = 1 – B. Substitute: (1 – B) + 2B = 1 ⇒ 1 + B = 1 ⇒ B = 0, then A = 1.

应用初始条件：y(0) = A + B = 1；y'(0) = A + 2B – 1 = 0。解得 B = 0，A = 1。

Thus, the particular solution satisfying the initial conditions is y = e^x – x e^x = (1 – x)e^x.

因此满足初始条件的特解为 y = e^x – x e^x = (1 – x)e^x。

The key was recognising the overlap between the forcing term and the complementary function, necessitating the multiplication by x.

关键点在于识别出强迫项与补函数的重复，从而需要乘以 x。

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4. Hyperbolic Functions: Solving Equations and Identities | 双曲函数：解方程与恒等式

Problem: Given that sinh x = 3/4, find the exact values of cosh x and tanh x without using a calculator. Hence solve the equation 2 cosh x + 3 sinh x = 5.

题目：已知 sinh x = 3/4，不用计算器求 cosh x 和 tanh x 的精确值。并由此解方程 2 cosh x + 3 sinh x = 5。

We use the identity cosh² x – sinh² x = 1. Since cosh x ≥ 1 always, cosh x = √(1 + sinh² x) = √(1 + (3/4)²) = √(1 + 9/16) = √(25/16) = 5/4. Then tanh x = sinh x / cosh x = (3/4) / (5/4) = 3/5.

利用恒等式 cosh² x – sinh² x = 1。因 cosh x ≥ 1 始终成立，cosh x = √(1 + (3/4)²) = 5/4。进而 tanh x = (3/4) / (5/4) = 3/5。

Now consider the equation 2 cosh x + 3 sinh x = 5. Substitute the exponential definitions: cosh x = (e^x + e⁻ˣ)/2, sinh x = (e^x – e⁻ˣ)/2. Then 2*(e^x + e⁻ˣ)/2 + 3*(e^x – e⁻ˣ)/2 = 5 → (2e^x + 2e⁻ˣ + 3e^x – 3e⁻ˣ)/2 = 5 → (5e^x – e⁻ˣ)/2 = 5 → multiply by 2: 5e^x – e⁻ˣ = 10. Multiply by e^x: 5e^(2x) – 1 = 10 e^x → 5e^(2x) – 10 e^x – 1 = 0.

现在考虑方程 2 cosh x + 3 sinh x = 5。代入指数定义式后化简得 5e^(2x) – 10 e^x – 1 = 0。

Let u = e^x (>0). Then 5u² – 10u – 1 = 0. Use quadratic formula: u = [10 ± √(100 + 20)] / 10 = [10 ± √120] / 10 = [10 ± 2√30] / 10 = 1 ± √30 / 5. Since u > 0, we take the positive root: u = 1 + √30/5. Thus e^x = 1 + √30/5, so x = ln(1 + √30/5).

令 u = e^x (>0)，得 5u² – 10u – 1 = 0。用求根公式得 u = 1 ± √30/5，取正根 u = 1 + √30/5，从而 x = ln(1 + √30/5)。

Alternatively, the initial given condition sinh x = 3/4 could be used to find x directly, but the above method illustrates the power of exponential substitution for equations mixing hyperbolic functions.

另一种方法是利用已知的 sinh x = 3/4 直接求解 x，但上述方法展示了指数代换在处理混合双曲函数方程时的威力。

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5. Polar Coordinates: Area Enclosed by a Cardioid | 极坐标：心形线围成的面积

Problem: The curve C has polar equation r = a(1 + cos θ), for 0 ≤ θ < 2π, where a > 0 is a constant. Find the total area enclosed by C.

题目：曲线 C 的极坐标方程为 r = a(1 + cos θ)，0 ≤ θ < 2π，其中 a > 0 为常数。求 C 所围成的总面积。

The area in polar coordinates is given by A = ½ ∫ r² dθ. Due to symmetry about the initial line (θ = 0), we can integrate from 0 to π and double the result. Thus, A = 2 * ½ ∫₀^π [a(1 + cos θ)]² dθ = ∫₀^π a² (1 + 2cos θ + cos² θ) dθ.

极坐标中的面积公式为 A = ½ ∫ r² dθ。由于曲线关于初始线 (θ = 0) 对称，可对 0 到 π 积分并将结果加倍。因此 A = ∫₀^π a² (1 + 2cos θ + cos² θ) dθ。

Use the identity cos² θ = ½(1 + cos 2θ). Then A = a² ∫₀^π [1 + 2cos θ + ½ + ½ cos 2θ] dθ = a² ∫₀^π [3/2 + 2cos θ + ½ cos 2θ] dθ.

利用恒等式 cos² θ = ½(1 + cos 2θ)，得 A = a² ∫₀^π [3/2 + 2cos θ + ½ cos 2θ] dθ。

Integrate term by term: ∫ 3/2 dθ = 3θ/2; ∫ 2cos θ dθ = 2 sin θ; ∫ ½ cos 2θ dθ = (½)( sin 2θ / 2) = (¼) sin 2θ. Evaluate from 0 to π:
[3θ/2 + 2 sin θ + ¼ sin 2θ]₀^π = (3π/2 + 0 + 0) – (0) = 3π/2.

逐项积分并代入上下限，结果等于 3π/2。

Therefore, A = a² * (3π/2) = (3π a²)/2. This is the total area enclosed by the cardioid.

因此，A = a² * (3π/2) = (3π a²)/2，这就是心形线围成的总面积。

It’s instructive to note that for a simple circle r = a, the area is πa²; the cardioid encloses 1.5 times that area for the same a.

值得留意的是，对于简单的圆 r = a，面积为 πa²；而对于相同的 a，心形线的面积是其 1.5 倍。

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6. Series: Method of Differences | 级数：差分法

Problem: Use the method of differences to find an expression for ∑ᵣ₌₁ⁿ 1/(r(r+1)). Hence evaluate the sum to infinity.

题目：用差分法求 ∑ᵣ₌₁ⁿ 1/(r(r+1)) 的表达式，并由此求无穷级数的和。

Start by expressing the general term in partial fractions: 1/(r(r+1)) ≡ A/r + B/(r+1). Multiply by r(r+1): 1 ≡ A(r+1) + Br. Set r = 0 → 1 = A; set r = -1 → 1 = -B ⇒ B = -1. So 1/(r(r+1)) = 1/r – 1/(r+1).

先将通项用部分分式表示：1/(r(r+1)) = 1/r – 1/(r+1)。

Now write the sum: Sₙ = ∑ᵣ₌₁ⁿ (1/r – 1/(r+1)). Expand terms:
r=1: 1/1 – 1/2
r=2: 1/2 – 1/3
r=3: 1/3 – 1/4
…
r=n-1: 1/(n-1) – 1/n
r=n: 1/n – 1/(n+1).

写出求和式 Sₙ = ∑ᵣ₌₁ⁿ (1/r – 1/(r+1))，并逐项展开。

Notice the diagonal cancellation: all intermediate fractions cancel, leaving only the first part of the first term and the second part of the last term. Thus Sₙ = 1 – 1/(n+1) = n/(n+1).

观察斜对角线上的相消：所有中间项均被抵消，仅余第一项的第一部分和最后一项的第二部分。因此 Sₙ = 1 – 1/(n+1) = n/(n+1)。

As n → ∞, 1/(n+1) → 0, so the sum to infinity is S∞ = 1.

当 n → ∞ 时，1/(n+1) → 0，故无穷和为 S∞ = 1。

The method of differences is a powerful tool for summing rational series where the general term can be decomposed into differences of successive terms of a sequence.

差分法是求和有理型级数的有力工具，前提是通项可拆分为某序列连续项的差。

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