Mock Exam Walkthrough: Year 13 CIE Further Pure Mathematics | 模拟卷解析：Year 13 CIE 进阶纯数学

📚 Mock Exam Walkthrough: Year 13 CIE Further Pure Mathematics | 模拟卷解析：Year 13 CIE 进阶纯数学

Mock exams are a crucial part of preparation for CIE Further Pure Mathematics. This walkthrough analyses selected questions from a typical unit test, covering high-yield topics such as roots of polynomials, rational functions, matrices, complex numbers, polar coordinates, series, induction, Maclaurin series, hyperbolic functions, and integration. By dissecting each solution, we aim to deepen conceptual understanding and sharpen problem-solving techniques for the final examination.

模拟考试是备考 CIE 进阶纯数学的重要环节。本解析精选了一份典型单元测试中的题目，涵盖多项式根、有理函数、矩阵、复数、极坐标、级数、归纳法、麦克劳林级数、双曲函数和积分等高频考点。通过逐题拆解解题思路，我们旨在深化概念理解，提升解题技巧，为最终考试做好充分准备。

1. Roots of a Cubic Equation | 三次方程根的问题

Consider the cubic equation x³ – 4x² + x + 6 = 0 with roots α, β, γ. Using Vieta’s formulas, the sum of the roots α+β+γ equals -(-4)/1 = 4, the sum of pairwise products αβ+βγ+γα equals 1/1 = 1, and the product αβγ equals -6/1 = -6.

考虑三次方程 x³ – 4x² + x + 6 = 0，根为 α, β, γ。由韦达定理，根的和 α+β+γ = 4，两两乘积之和 αβ+βγ+γα = 1，根的积 αβγ = -6。

To find α²+β²+γ², we square the sum of roots: (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα). Substituting known values: 4² = α²+β²+γ² + 2×1 ⇒ 16 = α²+β²+γ² + 2. Therefore α²+β²+γ² = 14.

为求 α²+β²+γ²，我们将根的和平方： (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα)。代入已知值得 16 = α²+β²+γ² + 2，故 α²+β²+γ² = 14。

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2. Rational Functions and Oblique Asymptotes | 有理函数与斜渐近线

We are given f(x) = (x²+2x-3)/(x-1). Factorising the numerator yields (x+3)(x-1). For x ≠ 1, the function simplifies to f(x) = x+3. This simplification reveals that the graph is essentially the straight line y = x+3, but with a hole (removable discontinuity) at x = 1.

给定 f(x) = (x²+2x-3)/(x-1)。分子因式分解得 (x+3)(x-1)。当 x ≠ 1 时，函数可化简为 f(x) = x+3。这表明图像本质上是直线 y = x+3，但在 x = 1 处存在一个可去间断点（空圈）。

Since the simplified form is linear, there are no vertical asymptotes; the line itself can be considered an oblique asymptote. At x = 1, f(1) would be 4, but the original function is undefined, so we place an open circle at (1, 4). The sketch is simply the line y = x+3 with this hole indicated.

由于化简后为线性函数，不存在垂直渐近线；该直线本身可视为斜渐近线。在 x = 1 处，函数值应为 4，但原函数未定义，因此我们在 (1, 4) 画一个空圈。草图即直线 y = x+3 并标出间断点。

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3. Inverse of a 3×3 Matrix | 三阶矩阵求逆

We need the inverse of matrix A = [[2,1,1],[1,3,2],[1,0,0]]. First, compute the determinant using expansion along the third row: det(A) = 1×(1×0 – 1×2) – 0 + 0 = -2. Since the determinant is non-zero, the inverse exists.

我们需要求矩阵 A = [[2,1,1],[1,3,2],[1,0,0]] 的逆矩阵。首先，沿第三行展开计算行列式：det(A) = 1×(1×0 – 1×2) = -2。行列式不为零，因此逆矩阵存在。

Next, obtain the cofactor matrix. For each element aᵢⱼ, compute Cᵢⱼ = (-1)ⁱ⁺ʲMᵢⱼ. The cofactor matrix is: [[ (3×0-2×0), -(1×0-2×1), (1×0-3×1) ]; [ -(1×0-1×0), (2×0-1×1), -(2×0-1×1) ]; [ (1×2-1×3), -(2×2-1×1), (2×3-1×1) ]] which simplifies to [[0, 2, -3]; [0, -1, 1]; [-1, -3, 5]]. Transposing yields the adjugate matrix: [[0, 0, -1]; [2, -1, -3]; [-3, 1, 5]].

接下来求余子式矩阵。各元素代数余子式矩阵为 [[0, 2, -3]; [0, -1, 1]; [-1, -3, 5]]。将其转置得到伴随矩阵：[[0, 0, -1]; [2, -1, -3]; [-3, 1, 5]]。

The inverse is (1/det) × adjugate = -1/2 × [[0, 0, -1]; [2, -1, -3]; [-3, 1, 5]] = [[0, 0, 1/2]; [-1, 1/2, 3/2]; [3/2, -1/2, -5/2]]. This can be verified by multiplying A A⁻¹ = I.

逆矩阵为 A⁻¹ = -1/2 × 伴随矩阵 = [[0, 0, 1/2]; [-1, 1/2, 3/2]; [3/2, -1/2, -5/2]]。可通过 A A⁻¹ = I 验证。

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4. Complex Numbers: Locus and Argument | 复数：轨迹与辐角

We are asked to find the locus satisfying |z – 2| = |z + 2i| and then determine the minimum argument of z on this locus with Re(z) ≥ 0. Let z = x + iy.

我们要求满足 |z – 2| = |z + 2i| 的轨迹，并在 Re(z) ≥ 0 的条件下求轨迹上辐角的最小值。设 z = x + iy。

The condition becomes |(x-2) + iy| = |x + i(y+2)|. Squaring both sides: (x-2)² + y² = x² + (y+2)². Expanding gives x² – 4x + 4 + y² = x² + y² + 4y + 4. Cancelling x², y² and 4 simplifies to -4x = 4y, hence y = -x. The locus is the straight line y = -x.

条件化为 |(x-2) + iy| = |x + i(y+2)|。两边平方得 (x-2)² + y² = x² + (y+2)²。展开并消去相同项得 -4x = 4y，故 y = -x。轨迹为直线 y = -x。

Points on the line can be expressed as z = t – it for real t. If t > 0, Re(z) = t ≥ 0, and arg(z) = -π/4 (fourth quadrant). If t = 0, the point is the origin, argument undefined. If t < 0, Re(z) < 0, which is excluded. Thus on the allowed half-line, the argument is constant at -π/4. The minimum argument is -π/4.

线上点可表示为 z = t – it (t 为实数)。当 t > 0 时，Re(z) = t ≥ 0，辐角为 -π/4；t = 0 时辐角无定义；t < 0 时实部为负，不合条件。因此允许的半直线上辐角恒为 -π/4，最小辐角即 -π/4。

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5. Polar Coordinates: Area of a Loop | 极坐标：环面积

A curve has polar equation r = a sin 2θ, with a > 0. One complete loop is traced for θ from 0 to π/2. The area A enclosed by a polar curve is given by A = ½ ∫ r² dθ over one period of the loop.

曲线极坐标方程为 r = a sin 2θ (a > 0)。一个完整的环对应 θ 从 0 到 π/2。极坐标下环的面积为 A = ½ ∫ r² dθ。

Substituting r: A = ½ ∫₀^{π/2} a² sin² 2θ dθ. Using the identity sin² 2θ = (1 – cos 4θ)/2, we obtain A = (a²/2) × (1/2) ∫₀^{π/2} (1 – cos 4θ) dθ = (a²/4) [θ – (1/4) sin 4θ] from 0 to π/2.

代入 r：A = ½ ∫₀^{π/2} a² sin² 2θ dθ。利用 sin² 2θ = (1 – cos 4θ)/2，得 A = (a²/4) ∫₀^{π/2} (1 – cos 4θ) dθ = (a²/4) [θ – (1/4) sin 4θ]₀^{π/2}。

Evaluating the bounds: at θ = π/2, sin 4θ = sin 2π = 0; at 0, both terms 0. Thus A = (a²/4) × (π/2) = π a² / 8. A concise closed form.

代入上下限：θ = π/2 时 sin 4θ = sin 2π = 0；θ = 0 时各项为零。故 A = (a²/4) × (π/2) = π a² / 8。结果简洁。

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6. Summation of Series by Method of Differences | 级数求和的差分法

We need to evaluate Σ_{r=1}^{n} 1/(r(r+1)). Express the term in partial fractions: 1/(r(r+1)) ≡ 1/r – 1/(r+1).

求 Σ_{r=1}^{n} 1/(r(r+1))。将通项裂项：1/(r(r+1))

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