📚 Mock Unit Test Walkthrough for CAIE Year 11 Further Mathematics | CAIE 11年级进阶数学单元测试模拟卷解析
This mock unit test is crafted to help Year 11 students consolidate the core topics of the CAIE Further Mathematics syllabus (AS level), including polynomials, series, matrices, induction, polar coordinates, and vectors. Each question is followed by a detailed bilingual solution that highlights key techniques and common pitfalls.
本模拟单元测试专为11年级学生设计，旨在巩固CAIE进阶数学（AS阶段）的核心主题，涵盖多项式、级数、矩阵、归纳法、极坐标与向量。每道题后附有详细的中英双语解析，突出关键技巧和常见陷阱。
1. Question 1: Roots and Coefficients of Polynomial Equations | 问题1：多项式方程的根与系数
as roots α and β. (a) Without solving the equation, find the values of α² + β² and (α - β)². (b) Form a quadratic equation with integer coefficients whose roots are α/β and β/α. (c) Evaluate α⁴ + β⁴.
已知二次方程 2x² - 8x + 3 = 0 的根为 α 和 β。(a) 不解方程，求 α² + β² 和 (α - β)² 的值。(b) 构造一个以 α/β 和 β/α 为根且系数为整数的二次方程。(c) 计算 α⁴ + β⁴。
dentify the sum and product of the roots from 2x² - 8x + 3 = 0: α + β = 8/2 = 4, αβ = 3/2.
首先，由方程 2x² - 8x + 3 = 0 得到根的和与积：α + β = 8/2 = 4，αβ = 3/2。
² - 2 × (3/2) = 16 - 3 = 13. For (α - β)² = (α + β)² - 4αβ = 16 - 4 × (3/2) = 16 - 6 = 10.
第(a)部分：α² + β² = (α + β)² - 2αβ = 4² - 2 × (3/2) = 16 - 3 = 13。(α - β)² = (α + β)² - 4αβ = 16 - 4 × (3/2) = 16 - 6 = 10。
x + P = 0 → x² - (26/3)x + 1 = 0. Multiplying through by 3 gives the integer-coefficient form: 3x² - 26x + 3 = 0.
第(b)部分：新根的和 S = α/β + β/α = (α² + β²)/(αβ) = 13 ÷ (3/2) = 26/3。积 P = (α/β) × (β/α) = 1。因此所求方程为 x² - S x + P = 0 → x² - (26/3)x + 1 = 0。两边同乘3，得整系数方程：3x² - 26x + 3 = 0。
ng the known values: 13² - 2 × (3/2)² = 169 - 2 × (9/4) = 169 - 9/2 = (338 - 9)/2 = 329/2.
第(c)部分：α⁴ + β⁴ = (α² + β²)² - 2(αβ)²。代入已知值：13² - 2 × (3/2)² = 169 - 2 × (9/4) = 169 - 9/2 = (338 - 9)/2 = 329/2。
2. Question 2: Summation of Series | 问题2：级数求和
(a) Use the standard results for ∑r, ∑r² and ∑r³ to show that ∑r=1n r(r+1)(r+2) = (1/4)n(n+1)(n+2)(n+3). (b) Hence evaluate ∑r=1030 r(r+1)(r+2).
(a) 利用 ∑r、∑r² 和 ∑r³ 的标准结果，证明 ∑r=1n r(r+1)(r+2) = (1/4)n(n+1)(n+2)(n+3)。(b) 由此计算 ∑r=1030 r(r+1)(r+2)。
Expand r(r+1)(r+2): r(r² + 3r + 2) = r³ + 3r² + 2r. Therefore the sum is ∑r³ + 3∑r² + 2∑r.
展开 r(r+1)(r+2)：r(r² + 3r + 2) = r³ + 3r² + 2r。因此求和式等于 ∑r³ + 3∑r² + 2∑r。
Standard results: ∑r=1n r = n(n+1)/2, ∑r=1n r² = n(n+1)(2n+1)/6, ∑r=1n r³ = [n(n+1)/2]².
标准结果：∑r=1n r = n(n+1)/2，∑r=1n r² = n(n+1)(2n+1)/6，∑r=1n r³ = [n(n+1)/2]²。
2 × n(n+1)/2 = n²(n+1)²/4 + n(n+1)(2n+1)/2 + n(n+1). Take out common factor n(n+1): n(n+1)
Published by TutorHao | Year 11 进阶数学 Revision Series | aleveler.com
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