Case Study Practical Drills for AS Cambridge Further Mathematics | AS剑桥进阶数学：案例分析实战演练

📚 Case Study Practical Drills for AS Cambridge Further Mathematics | AS剑桥进阶数学：案例分析实战演练

Mastering AS Cambridge Further Mathematics demands more than memorising formulas – it requires the ability to apply concepts flexibly across unfamiliar problems. This article presents a series of focused case studies, each designed to mirror the style and challenge of real exam questions. Every drill breaks down a core topic from the syllabus (roots of polynomials, rational functions, summations, matrices, polar coordinates, vectors and proof by induction) into ready‑to‑use reasoning chains. Work through them step by step, and you will build the analytical confidence needed for top marks.

要掌握 AS 剑桥进阶数学，仅靠记住公式远远不够——你需要灵活地将概念运用到陌生的情境中。本文提供一组精心设计的案例实战练习，每个案例都模拟真实考题的风格和难度。这些练习将考纲中的核心主题（多项式根、有理函数、级数求和、矩阵、极坐标、向量和数学归纳法）分解为清晰易懂的推理链条。请一步步完成，你一定能培养出冲击高分的分析能力和自信心。

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1. Roots of Polynomial Equations: Relationships and Transformations | 多项式方程根：关系与变换

The cubic equation 2x³ – 9x² + 12x – 4 = 0 has roots α, β, γ. By using the relationships between roots and coefficients, find the value of α² + β² + γ². Hence determine the cubic equation whose roots are α², β² and γ².

已知三次方程 2x³ – 9x² + 12x – 4 = 0 的根为 α, β, γ。利用根与系数的关系，求 α² + β² + γ² 的值，并由此求出一个以 α², β², γ² 为根的三次方程。

From the given equation we have ∑α = 9/2, ∑αβ = 12/2 = 6 and αβγ = 4/2 = 2.

由给定方程可得 ∑α = 9/2，∑αβ = 6，αβγ = 2。

Now α² + β² + γ² = (∑α)² – 2∑αβ = (9/2)² – 2×6 = 81/4 – 12 = 33/4.

计算 α² + β² + γ² = (∑α)² – 2∑αβ = (9/2)² – 12 = 33/4。

For the new roots α², β², γ² we require their sum, pairwise products and product. Sum = 33/4.

对于新根 α², β², γ²，需要它们的和、两两积之和以及乘积。和为 33/4。

Pairwise sum: ∑α²β² = (∑αβ)² – 2αβγ∑α = 6² – 2×2×(9/2) = 36 – 18 = 18.

两两积之和：∑α²β² = (∑αβ)² – 2αβγ∑α = 36 – 18 = 18。

Product: α²β²γ² = (αβγ)² = 4.

乘积：α²β²γ² = 4。

The required cubic equation is x³ – (33/4)x² + 18x – 4 = 0. Multiplying by 4 gives the integer‑coefficient form 4x³ – 33x² + 72x – 16 = 0.

所求三次方程为 x³ – (33/4)x² + 18x – 4 = 0。两边乘以 4 得到整系数形式 4x³ – 33x² + 72x – 16 = 0。

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2. Rational Functions: Partial Fractions and Inequalities | 有理函数：部分分式与不等式

Consider the function f(x) = (3x² + 5x – 2)/(x² + x – 2). Express f(x) in partial fractions and hence solve the inequality f(x) > 1. Sketch the graph, showing all asymptotes and intercepts.

考虑函数 f(x) = (3x² + 5x – 2)/(x² + x – 2)。将 f(x) 表示成部分分式，并借此解不等式 f(x) > 1。画出图像，标出所有渐近线和截距。

Factorise denominator: x² + x – 2 = (x+2)(x–1). The numerator factorises as (3x–1)(x+2), so f(x) = (3x–1)/(x–1) for x ≠ –2.

分母因式分解：(x+2)(x–1)。分子可分解为 (3x–1)(x+2)，因此当 x ≠ –2 时 f(x) = (3x–1)/(x–1)。

Perform division: (3x–1) ÷ (x–1) gives quotient 3 and remainder 2. Thus f(x) = 3 + 2/(x–1), with a removable discontinuity at x = –2.

做除法：(3x–1) ÷ (x–1)，商为 3，余数为 2。因此 f(x) = 3 + 2/(x–1)，且在 x = –2 处有一个可去间断点。

Solve f(x) > 1: 3 + 2/(x–1) > 1 → 2/(x–1) > –2 → 1/(x–1) > –1. Consider two cases. First, x > 1: 1/(x–1) is positive, so inequality holds for all x > 1. Second, x < 1: multiply by (x–1) negative flips sign → 1 < –1(x–1) → 1 < –x+1 → x < 0. Combine with x < 1 gives x < 0. Excluding x = –2 gives solution set x < –2, –2 < x < 0, x > 1.

解 f(x) > 1：3 + 2/(x–1) > 1 → 2/(x–1) > –2 → 1/(x–1) > –1。分两种情况。当 x > 1 时，1/(x–1) 为正，不等式对所有 x > 1 成立。当 x < 1 时，两边同乘负数 (x–1) 要变号：1 < –1(x–1) → 1 < –x+1 → x < 0。结合 x < 1 得 x < 0。剔除 x = –2，解集为 x < –2, –2 < x < 0, x > 1。

Graph features: vertical asymptote x = 1; horizontal asymptote y = 3; hole at (–2, 7/3); y‑intercept (0,1).

图像特征：垂直渐近线 x = 1；水平渐近线 y = 3；在 (–2, 7/3) 处有一空洞；y 截距 (0,1)。

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3. Summation of Series using Standard Results | 利用标准结果求级数和

Use the standard summations for ∑r, ∑r² and ∑r³ to evaluate S(n) = ∑_{r=1}^{n} r(r+1)(r+2). Express your answer in a fully factorised form.

利用 ∑r, ∑r² 和 ∑r³ 的标准求和公式，求 S(n) = ∑_{r=1}^{n} r(r+1)(r+2)，并将答案写成完全因式分解的形式。

Expand the typical term: r(r+1)(r+2) = r(r²+3r+2) = r³ + 3r² + 2r.

将通项展开：r(r+1)(r+2) = r³ + 3r² + 2r。

Therefore S(n) = ∑r³ + 3∑r² + 2∑r. Standard formulas: ∑r = n(n+1)/2, ∑r² = n(n+1)(2n+1)/6, ∑r³ = [n(n+1)/2]².

因此 S(n) = ∑r³ + 3∑r² + 2∑r。标准公式：∑r = n(n+1)/2，∑r² = n(n+1)(2n+1)/6，∑r³ = [n(n+1)/2]²。

Substituting and taking a common factor of n(n+1)/2:

代入并提取公因子 n(n+1)/2：

S(n) = [n(n+1)/2]² + 3 × n(n+1)(2n+1)/6 + 2 × n(n+1)/2 = n(n+1)/2 [ n(n+1)/2 + (2n+1) + 2 ]

Simplify inside the bracket: n(n+1)/2 + 2n+1+2 = n(n+1)/2 + 2n+3 = (n²+n + 4n+6)/2 = (n²+5n+6)/2 = (n+2)(n+3)/2.

化简括号内：n(n+1)/2 + 2n+1+2 = (n²+n + 4n+6)/2 = (n²+5n+6)/2 = (n+2)(n+3)/2。

Finally S(n) = n(n+1)/2 × (n+2)(n+3)/2 = (1/4)n(n+1)(n+2)(n+3).

最终 S(n) = (1/4)n(n+1)(n+2)(n+3)。

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4. Matrix Algebra: Inverse and Determinant | 矩阵代数：逆与行列式

Let A be the 2×2 matrix

2	1
1	2

. Find A⁻¹ and hence solve the system of equations 2x + y = 5, x + 2y = 4.

设 A 为 2×2 矩阵 (2 1; 1 2)。求 A⁻¹，并由此解方程组 2x + y = 5, x + 2y = 4。

Determinant det(A) = (2)(2) – (1)(1) = 3. Since det(A) ≠ 0, the inverse exists.

行列式 det(A)=4–1=3。非零，逆阵存在。

The adjugate matrix is formed by swapping the diagonal elements and changing the signs of the off‑diagonal elements:

2	–1
–1	2

伴随矩阵：交换主对角线元素，反对角线元素变号，得到 (2 –1; –1 2)。

Thus A⁻¹ = (1/3) × adjugate =

2/3	–1/3
–1/3	2/3

因此 A⁻¹ = (1/3) × 伴随矩阵。

Write the system as A × (x, y)ᵀ = (5, 4)ᵀ. Multiply both sides by A⁻¹: (x, y)ᵀ = A⁻¹ (5, 4)ᵀ.

将方程组写为 A (x, y)ᵀ = (5, 4)ᵀ。两边同乘 A⁻¹ 得 (x, y)ᵀ = A⁻¹ (5, 4)ᵀ。

Compute: x = (2/3)×5 + (–1/3)×4 = 10/3 – 4/3 = 6/3 = 2; y = (–1/3)×5 + (2/3)×4 = –5/3 + 8/3 = 3/3 = 1.

计算得 x=2, y=1。

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5. Matrix Transformations: Invariant Points and Lines | 矩阵变换：不变点与不变直线

For the same matrix A = [[2,1],[1,2]], find all points that are fixed under the transformation. Determine the equations of any lines that are invariant as a whole (each point on the line maps to some point on the same line).

对于同一矩阵 A = [[2,1],[1,2]]，找出变换下所有不变点，并确定整条直线不变的方程（即直线上的每一点都映射到同一直线上某点）。

Invariant points satisfy A(x,y)ᵀ = (x,y)ᵀ. This gives 2x + y = x and x + 2y = y → x + y = 0 and x + y = 0. So the only invariant point is (0,0).

不变点满足 A(x,y)ᵀ = (x,y)ᵀ，即 2x+y=x 且 x+2y=y → x+y=0。唯一不变点为原点。

For invariant lines through the origin we seek vectors such that A v = λ v. The characteristic equation: det(A–λI) = (2–λ)² – 1 = 0 → λ²–4λ+3=0 → λ=1, 3.

求过原点的不变直线，即找特征向量。特征方程：(2–λ)²–1=0 → λ=1,3。

Eigenvectors: λ=1 gives (2–1)x + y = 0 → x+y=0, line y = –x. λ=3 gives (2–3)x + y = 0 → –x+y=0, line y = x. Thus the invariant lines through the origin are y = x and y = –x.

特征向量：λ=1 得 x+y=0，直线 y=–x；λ=3 得 –x+y=0，直线 y=x。因此过原点的不变直线为 y=x 和 y=–x。

A non‑origin invariant line would require each translated point to stay on the line; substituting y = mx + c into the transformation shows that only c=0 works unless the line is vertical. So the only invariant lines are those through the origin found above.

对于不过原点的直线，代入变换方程可证只有 c=0 成立，故不变直线仅为上述过原点的两条。

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6. Polar Coordinates: Sketching and Area | 极坐标：绘图与面积

The polar curve C has equation r = 2 + cos θ for 0 ≤ θ < 2π. Sketch C and find the area of the region enclosed by C in the first quadrant (0 ≤ θ ≤ π/2).

极坐标曲线 C 的方程为 r = 2 + cos θ，0 ≤ θ < 2π。画出 C 的草图，并求第一象限内（0 ≤ θ ≤ π/2）由 C 围成的区域面积。

The curve is a limaçon without an inner loop because 2 > 1. In the first quadrant, r decreases from r(0)=3 to r(π/2)=2. The polar area formula: A = (1/2) ∫ r² dθ.

该曲线为无内环的蜗线。第一象限内 r 从 3 递减至 2。极坐标面积公式：A = ½ ∫ r² dθ。

Compute r² = (2+cos θ)² = 4 + 4cos θ + cos² θ. Using cos² θ = (1+cos 2θ)/2, we have r² = 4 + 4cos θ + ½ + ½ cos 2θ = 9/2 + 4cos θ + (1/2)cos 2θ.

计算 r² = (2+cos θ)² = 4 + 4cos θ + cos² θ。利用降幂公式得 r² = 9/2 + 4cos θ + (1/2)cos 2θ。

Integrate from 0 to π/2: ∫₀^{π/2} r² dθ = [ (9/2)θ + 4sin θ + (1/4)sin 2θ ]₀^{π/2} = (9/2)(π/2) + 4(1) + 0 – 0 = (9π/4) + 4.

从 0 到 π/2 积分：∫ r² dθ = (9π/4) + 4。

Thus the area in the first quadrant = (1/2) × (9π/4 + 4) = (9π/8) +

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