Year 11 Eduqas Further Mathematics: Mock Unit Test Walkthrough | 11年级Eduqas进阶数学：单元测试模拟卷解析

📚 Year 11 Eduqas Further Mathematics: Mock Unit Test Walkthrough | 11年级Eduqas进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a typical unit test for the Year 11 Eduqas Level 2 Certificate in Further Mathematics. Each question is carefully broken down to highlight key techniques, common pitfalls, and efficient problem-solving strategies. Use this resource to consolidate your understanding and boost your confidence before the real assessment.

本文详细解析一份典型的 11 年级 Eduqas 进阶数学（Level 2 Certificate）单元测试模拟卷。每道题目都经过逐步拆解，突出核心技巧、常见错误与高效解题策略。考前利用这份材料可巩固知识、提升信心。

1. Polynomials and Factor Theorem | 多项式与因式定理

Question: ) = 2x³ - 5x² + x + 2, prove that (x - 2) is a factor and hence factorise f(x) completely.

题目：已知 f(x) = 2x³ - 5x² + x + 2，证明 (x - 2) 是一个因式，并据此将 f(x) 完全因式分解。

o f(x). f(2) = 2(2)³ - 5(2)² + 2 + 2 = 16 - 20 + 2 + 2 = 0, so (x - 2) is indeed a factor.

首先运用因式定理：将 x = 2 代入 f(x)。f(2) = 2(2)³ - 5(2)² + 2 + 2 = 16 - 20 + 2 + 2 = 0，因此 (x - 2) 确实是一个因式。

ck: (x - 2)(2x² - x - 1) = 2x³ - x² - x - 4x² + 2x + 2 = 2x³ - 5x² + x + 2, which matches.

接着，用多项式长除法或综合除法将 f(x) 除以 (x - 2)。商为 2x² - x - 1。验证：(x - 2)(2x² - x - 1) = 2x³ - x² - x - 4x² + 2x + 2 = 2x³ - 5x² + x + 2，吻合。

= (2x + 1)(x - 1). Therefore the complete factorisation is f(x) = (x - 2)(2x + 1)(x - 1).

现在对二次式因式分解：2x² - x - 1 = (2x + 1)(x - 1)。因此完全因式分解为 f(x) = (x - 2)(2x + 1)(x - 1)。

Key skill: Always test possible integer factors using the factor theorem and perform division accurately. Double-check by expanding your result.

核心技能：务必用因式定理检验整数因式，准确进行除法运算，最后展开结果核对无误。

---

2. Binomial Expansion | 二项式展开

Question: Expand (2 - x)⁵ in ascending powers of x, giving each coefficient in its simplest form.

题目：按 x 的升幂展开 (2 - x)⁵，给出最简形式的各项系数。

Use the binomial theorem: (a + b)ⁿ = Σ (nCk) aⁿ⁻ᵏ bᵏ. Here a = 2, b = -x, n = 5.

应用二项式定理：(a + b)ⁿ = Σ (nCk) aⁿ⁻ᵏ bᵏ。此处 a = 2, b = -x, n = 5。

Term 0: ⁵C₀ (2)⁵ (-x)⁰ = 1 × 32 × 1 = 32

第 0 项：⁵C₀ (2)⁵ (-x)⁰ = 1 × 32 × 1 = 32

Term 1: ⁵C₁ (2)⁴ (-x)¹ = 5 × 16 × (-x) = -80x

第 1 项：⁵C₁ (2)⁴ (-x)¹ = 5 × 16 × (-x) = -80x

Term 2: ⁵C₂ (2)³ (-x)² = 10 × 8 × x² = 80x²

第 2 项：⁵C₂ (2)³ (-x)² = 10 × 8 × x² = 80x²

Term 3: ⁵C₃ (2)² (-x)³ = 10 × 4 × (-x³) = -40x³

第 3 项：⁵C₃ (2)² (-x)³ = 10 × 4 × (-x³) = -40x³

Term 4: ⁵C₄ (2)¹ (-x)⁴ = 5 × 2 × x⁴ = 10x⁴

第 4 项：⁵C₄ (2)¹ (-x)⁴ = 5 × 2 × x⁴ = 10x⁴

Term 5: ⁵C₅ (2)⁰ (-x)⁵ = 1 × 1 × (-x⁵) = -x⁵

第 5 项：⁵C₅ (2)⁰ (-x)⁵ = 1 × 1 × (-x⁵) = -x⁵

The full expansion is 32 - 80x + 80x² - 40x³ + 10x⁴ - x⁵. Pay attention to alternating signs when the second term is negative.

完整展开为 32 - 80x + 80x² - 40x³ + 10x⁴ - x⁵。注意第二项为负时符号交替变化。

---

广告 / Sponsored

3. Solving Rational Inequalities | 有理不等式求解

Question: Solve the inequality (x - 3)/(x + 1) ≤ 2.

题目：解不等式 (x - 3)/(x + 1) ≤ 2。

one side: (x - 3)/(x + 1) - 2 ≤ 0 → (x - 3 - 2(x + 1))/(x + 1) ≤ 0 → (-x - 5)/(x + 1) ≤ 0.

移项使一边为零：(x - 3)/(x + 1) - 2 ≤ 0 → (x - 3 - 2(x + 1))/(x + 1) ≤ 0 → (-x - 5)/(x + 1) ≤ 0。

1) ≥ 0. Critical values: x = -5 (numerator zero) and x = -1 (denominator zero, excluded).

将分子分母同乘 -1（不等号方向改变）：(x + 5)/(x + 1) ≥ 0。临界值：x = -5（分子为零）和 x = -1（分母为零，不可取）。

ffer → fraction negative, not valid. For x > -1, both positive → fraction positive, valid.

在数轴上检验区间：x < -5 时分子分母同负 → 分式为正，满足；-5 < x < -1 时符号相异 → 分式为负，不满足；x > -1 时同正 → 分式为正，满足。

-1, but include x = -5 because the inequality allows equality when the numerator is zero.

解集为 x ≤ -5 或 x > -1。务必排除 x = -1，但包含 x = -5，因为分子为零时不等式可取得等号。

---

4. Matrix Operations and Inverse | 矩阵运算与逆矩阵

Question: [[2,0],[-1,4]]. Calculate AB, find A⁻¹, and hence solve the system 3x + y = 5, 2x - y = 1.

题目：设 A = [[3,1],[2,-1]]，B = [[2,0],[-1,4]]。计算 AB，求 A⁻¹，并由此解方程组 3x + y = 5, 2x - y = 1。

1*4],[2*2 + (-1)*(-1), 2*0 + (-1)*4]] = [[6 - 1, 0 + 4],[4 + 1, 0 - 4]] = [[5,4],[5,-4]].

首先， + 1*4],[2*2 + (-1)*(-1), 2*0 + (-1)*4]] = [[6 - 1, 0 + 4],[4 + 1, 0 - 4]] = [[5,4],[5,-4]]。

(-1) - 1*2 = -3 - 2 = -5. The inverse is (1/det)[[-1,-1],[-2,3]] = [[1/5,1/5],[2/5,-3/5]].

求 A⁻¹，det A = 3*(-1) - 1*2 = -3 - 2 = -5。逆矩阵为 (1/det)[[-1,-1],[-2,3]] = [[1/5,1/5],[2/5,-3/5]]。

; (2/5)*5 + (-3/5)*1 ] = [1 + 1/5; 2 - 3/5] = [6/5; 7/5]. So x = 6/5 = 1.2, y = 7/5 = 1.4.

将方程组写为 A[ x; y ] = [ 5; 1 ]，两边左乘 A⁻¹：[ x; y ] = A⁻¹ [5;1] = [[1/5,1/5],[2/5,-3/5]] [5

Published by TutorHao | Year 11 进阶数学 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）