Year 12 Cambridge Further Mathematics: Mock Test Paper Analysis | 剑桥Year12进阶数学：单元测试模拟卷解析

📚 Year 12 Cambridge Further Mathematics: Mock Test Paper Analysis | 剑桥Year12进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test designed for Year 12 Cambridge Further Mathematics. The paper covers core topics such as polynomial roots, summation of series, matrices and induction, rational functions, polar coordinates, complex numbers, hyperbolic functions, vectors, and proof by induction. Each question is analysed step by step, highlighting key concepts and common errors, making it an ideal revision resource for students preparing for their AS Level Further Pure Mathematics examinations.

本文详细解析了一套为剑桥Year12进阶数学设计的单元测试模拟卷。试卷涵盖多项式根、级数求和、矩阵与归纳法、有理函数、极坐标、复数、双曲函数、向量以及数学归纳法等核心专题。每道题目均逐步分析，强调关键概念与常见错误，为准备AS阶段进阶纯数学考试的学生提供理想的复习参考。

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1. Question 1: Polynomial Roots | 问题1：多项式根

Question: 3x − 1 = 0 has roots α, β, γ. Without solving the equation, find (i) Σα² and (ii) Σ(1/α).

题目：三次方程 2x³ − 5x² + 3x − 1 = 0 的根为 α, β, γ。不解方程，求 (i) Σα² 和 (ii) Σ(1/α)。

ymmetric sums are Σα = −b/a, Σαβ = c/a, and αβγ = −d/a. Here a = 2, b = −5, c = 3, d = −1.

对于三次方程 ax³ + bx² + cx + d = 0，对称和为 Σα = −b/a，Σαβ = c/a，αβγ = −d/a。本题中 a = 2, b = −5, c = 3, d = −1。

Σα = 5/2, Σαβ = 3/2, αβγ = 1/2

To find Σα², use the identity Σα² = (Σα)² − 2Σαβ.

求 Σα²，使用恒等式 Σα² = (Σα)² − 2Σαβ。

Σα² = (5/2)² − 2(3/2) = 25/4 − 3 = 13/4

For Σ(1/α), combine fractions: Σ(1/α) = (αβ + βγ + γα) / (αβγ) = Σαβ / (αβγ).

对于 Σ(1/α)，通分得：Σ(1/α) = (αβ + βγ + γα) / (αβγ) = Σαβ / (αβγ)。

Σ(1/α) = (3/2) / (1/2) = 3

Common mistake: forgetting to change the sign for the product term. Always check the standard form carefully.

常见错误：忘记乘积项的符号变化。务必仔细核对标准形式。

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2. Question 2: Summation of Series | 问题2：级数求和

Question: ess Σ (from r=1 to n) r(r+1) in terms of n. (b) Hence evaluate Σ (from r=10 to 20) r(r+1).

题目：(a) 用 n 表示 Σ (r=1 到 n) r(r+1)。(b) 由此计算 Σ (r=10 到 20) r(r+1) 的值。

. Then split the summation using standard formulae Σr² = n(n+1)(2n+1)/6 and Σr = n(n+1)/2.

首先展开通项：r(r+1) = r² + r。然后利用标准公式拆分求和：Σr² = n(n+1)(2n+1)/6，Σr = n(n+1)/2。

Σ r(r+1) = Σr² + Σr = [n(n+1)(2n+1)/6] + [n(n+1)/2]

n(n+1)/6 to simplify. The sum becomes n(n+1)(2n+1 + 3)/6 = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3.

提取公因式 n(n+1)/6 化简。总和变为 n(n+1)(2n+1 + 3)/6 = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3。

Σ (r=1 to n) r(r+1) = n(n+1)(n+2)/3

is the difference of the sum up to 20 and up to 9. Compute S(20) − S(9) using the formula.

对于 (b) 部分，从 r=10 到 20 的和等于前 20 项和减去前 9 项和。代入公式计算 S(20) − S(9)。

S(20) = 20×21×22/3 = 3080, S(9) = 9×10×11/3 = 330

Required sum = 3080 − 330 = 2750

Always check the lower limit when using the subtraction method. This technique is very efficient for partial sums.

使用差值法时务必检查下限。这种技巧对于求部分和非常高效。

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3. Question 3: Matrices and Induction | 问题3：矩阵与归纳

Question: Let M be the 2×2 matrix [cosθ −sinθ; sinθ essing entries as single trigonometric functions. (b) Prove by induction that Mⁿ = [cos nθ −sin nθ; sin nθ cos nθ] for all positive integers n.

题目：设 M 为 2×2 矩阵 [cosθ −sinθ; sinθ cosθ]。(a) 求 M² 和 M³，并将元素表示为单个三角函数。(b) 用数学归纳法证明对所有正整数 n，有 Mⁿ = [cos nθ −sin nθ; sin nθ cos nθ]。

cos²θ − sin²θ = cos 2θ. Similarly, other entries simplify using compound angle identities.

求 M²，将 M 自乘。左上角元素为 cosθ·cosθ + (−sinθ)·sinθ = cos²θ − sin²θ = cos 2θ。其余元素可类似用复合角恒等式化简。

M² = [cos2θ −sin2θ; sin2θ cos2θ]

by M to get M³, applying cos2θ·cosθ − sin2θ·sinθ = cos3θ. The result confirms the pattern.

用 M² 乘 M 得 M³，应用 cos2θ·cosθ − sin2θ·sinθ = cos3θ。结果验证了规律。

M³ = [cos3θ −sin3θ; sin3θ cos3θ]

For the induction proof: Base case n=1 holds by definition. Assume true for n=k. Then M^k+1 = M^k osθ − sin kθ·sinθ, ...] = [cos(k+1)θ, ...]. The inductive step uses the addition formulas.

归纳证明：当 n=1 时由定义成立。假设 n=k 时成立，则 M^k+1 = M^k M = [cos kθ·cosθ − sin kθ·sinθ, ...] = [cos(k+1)θ, ...]。归纳步骤运用了加法公式。

This matrix represents a rotation by angle θ about the origin. The result shows that applying it n times gives a rotation by nθ.

该矩阵表示绕原点旋转 θ 角。结果表明，应用 n 次即为旋转 nθ 角。

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4. Question 4: Rational Functions and Asymptotes | 问题4：有理函数与渐近线

Question: Find equations of all asymptotes. (ii) Find intersections with the axes. (iii) Sketch the curve.

题目：已知 f(x) = (2x² − 3x − 2) / (x² − 1)。(i) 求所有渐近线方程。(ii) 求与坐标轴的交点。(iii) 绘制曲线草图。

occur where the denominator is zero and the numerator is non-zero, i.e. x = 1 and x = −1.

对分子分母因式分解：f(x) = (2x+1)(x−2) / [(x−1)(x+1)]。垂直渐近线出现在分母为零且分子不为零处，即 x = 1 和 x = −1。

e is y = 2. Check for oblique asymptotes: since degrees are equal, only horizontal exists.

当 x → ±∞ 时，f(x) → 2x² / x² = 2，因此水平渐近线为 y = 2。因为分子分母次数相同，斜渐近线不存在。

Axial intercepts: set y=0 ⇒ numerator=0 ⇒ x = −½, 2. y-intercept: set x=0 ⇒ f(0) = (−2)/(−1) = 2. So intersections: (−½,0), (2,0), (0,2).

坐标轴交点：令 y=0 ⇒ 分子=0 ⇒ x = −½, 2。y轴截距：令 x=0 ⇒ f(0) = (−2)/(−1) = 2。因此交点为 (−½,0), (2,0), (0,2)。

gn of f(x) in intervals between vertical asymptotes. The curve approaches y=2 at extremes.

绘制草图时，用虚线标出渐近线，标记交点，并考虑各区间 f(x) 的正负号。曲线在无穷远处逼近 y=2。

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5. Question 5: Polar Coordinates | 问题5：极坐标

Question: (1 + cos θ), 0 ≤ θ < 2π. (a) Sketch the curve. (b) Show that the area enclosed by C is 6π.

题目：曲线 C 的极坐标方程为 r = 2(1 + cos θ)，0 ≤ θ < 2π。(a) 绘制曲线草图。(b) 证明 C 所围面积为 6π。

θ = 0). The curve passes through (4,0), (2,π/2), (0,π). It has a cusp at the pole when θ =

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