IGCSE Edexcel Further Pure Maths: Unit Test Mock Paper Analysis | IGCSE Edexcel 进阶数学：单元测试模拟卷解析

📚 IGCSE Edexcel Further Pure Maths: Unit Test Mock Paper Analysis | IGCSE Edexcel 进阶数学：单元测试模拟卷解析

Unit tests are an essential checkpoint in mastering the IGCSE Edexcel Further Pure Mathematics syllabus. This mock paper analysis walks you through a carefully designed test covering key topics: complex numbers, polynomials, inequalities, series, matrices, polar coordinates, hyperbolic functions, differentiation, integration, and differential equations. Each worked solution not only shows the correct answer but also highlights the reasoning and typical examiner expectations. By studying these detailed explanations, you will strengthen your problem‑solving skills and avoid common pitfalls.

单元测试是掌握 IGCSE Edexcel 进阶纯数课程的关键环节。这份模拟卷解析逐题带你复习精心设计的测试，覆盖复数、多项式、不等式、级数、矩阵、极坐标、双曲函数、微分、积分和微分方程等核心主题。每道题目的详细解答不仅给出正确答案，还突出解题思路与典型考试要求。通过研读这些解析，你将提升解题能力并避开常见错误。

1. Complex Numbers: Solving Quadratic Equations | 复数：解二次方程

Question 1: Solve the equation z² – 4z + 13 = 0, giving your answers in the form a + bi.

问题1：解方程 z² – 4z + 13 = 0，答案以 a + bi 的形式给出。

The quadratic is z² – 4z + 13 = 0. We find the discriminant: Δ = (–4)² – 4(1)(13) = 16 – 52 = –36.

二次方程为 z² – 4z + 13 = 0。计算判别式：Δ = (–4)² – 4(1)(13) = 16 – 52 = –36。

Since Δ < 0, the roots are complex. √Δ = √–36 = 6i. Applying the quadratic formula,

因为 Δ < 0，根为复数。√Δ = √–36 = 6i。应用求根公式，

z = 4 ± 6i2 = 2 ± 3i

z = (4 ± 6i)/2 = 2 ± 3i

Thus the solutions are z = 2 + 3i and z = 2 – 3i. Always remember to simplify the expression fully and present both conjugate pairs.

因此解为 z = 2 + 3i 和 z = 2 – 3i。务必彻底化简表达式并写出一对共轭复数。

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2. Polynomials: Factor and Remainder Theorem | 多项式：因式与余式定理

Question 2: Given f(x) = 2x³ – 7x² + 3x + 8, show that (x – 2) is a factor and hence factorise f(x) completely.

问题2：已知 f(x) = 2x³ – 7x² + 3x + 8，证明 (x – 2) 是一个因式，并由此将 f(x) 完全分解因式。

By the Factor Theorem, if (x – 2) is a factor then f(2) = 0. Evaluate f(2) = 2(8) – 7(4) + 3(2) + 8 = 16 – 28 + 6 + 8 = 2. Wait, 16 – 28 = –12, +6 = –6, +8 = 2, not zero. Let me check the polynomial: perhaps it should be f(x) = 2x³ – 7x² f(2)=16-28+6+8=2, indeed not zero. So we need to make it a factor. Maybe change to f(x)=2x³ – 3x² – 3x + 2, with factor (x-2). Let's redesign to be correct and typical. I'll use a standard polynomial: f(x) = 2x³ – 3x² – 3x + 2, check f(2)=16-12-6+2=0. Good. So I'll write that. I'll adjust question accordingly.

Based on the correction: f(x) = 2x³ – 3x² – 3x + 2. f(2) = 2(8) – 3(4) – 3(2) + 2 = 16 – 12 – 6 + 2 = 0, confirming (x – 2) is a factor. Then perform polynomial division by (x – 2) to get 2x² + x – 1, and factorise further as (2x – 1)(x + 1). So complete factorisation: (x – 2)(2x – 1)(x + 1). I'll present with English and Chinese pairs.

To avoid confusion, I'll write the intended question and solution. The actual answer should be consistent. I'll fabricate the question: f(x) = 2x³ – 3x² – 3x + 2. Show (x – 2) is a factor, then factorise completely. This is a valid problem.

Now output the content.

Question 2: Let f(x) = 2x³ – 3x² – 3x + 2. Use the Factor Theorem to verify that (x – 2) is a factor of f(x). Hence factorise f(x) completely.

问题2：设 f(x) = 2x³ – 3x² – 3x + 2。利用因式定理验证 (x – 2) 是 f(x) 的因式，并由此将 f(x) 完全分解因式。

First, substitute x = 2: f(2) = 2(8) – 3(4) – 3(2) + 2 = 16 – 12 – 6 + 2 = 0. Hence (x – 2) is a factor.

首先代入 x = 2：f(2) = 2(8) – 3(4) – 3(2) + 2 = 16 – 12 – 6 + 2 = 0。因此 (x – 2) 是一个因式。

Perform polynomial division: (2x³ – 3x² – 3x + 2) ÷ (x – 2) yields 2x² + x – 1.

进行多项式除法：(2x³ – 3x² – 3x + 2) ÷ (x – 2) 得到商 2x² + x – 1。

Now factorise the quadratic: 2x² + x – 1 = (2x – 1)(x + 1). Therefore, f(x) = (x – 2)(2x – 1)(x + 1).

现在分解二次式：2x² + x – 1 = (2x – 1)(x + 1)。因此 f(x) = (x – 2)(2x – 1)(x + 1)。

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3. Inequalities: Rational Functions | 不等式：有理函数

Question 3: Find the set of values of x for which x+3x-2 ≥ 0, x ≠ 2.

问题3：求满足 (x+3)/(x-2) ≥ 0 且 x ≠ 2 的 x 的取值范围。

Critical points where the expression can change sign are at x = –3 (numerator zero) and x = 2 (denominator zero). Test intervals: x < –3, –3 < x < 2, x > 2.

表达式的符号可能发生变化的临界点为 x = –3（分子为零）和 x = 2（分母为零）。检验区间：x < –3, –3 < x < 2, x > 2。

For x = –4, -4+3-4-2 = -1-6 = 1/6 > 0. For x = 0, 3-2 = –1.5 < 0. For x = 3, 61 = 6 > 0. The inequality includes equality where the numerator is zero, so x = –3 is included. x = 2 is always excluded.

取 x = –4，(-4+3)/(-4-2) = (-1)/(-6) = 1/6 > 0。取 x = 0，(3)/(-2) = –1.5 < 0。取 x = 3，(6)/(1) = 6 > 0。不等式包含等号，因此分子为零的点包括在内，即 x = –3，而 x = 2 总是排除。

Solution: x ≤ –3 or x > 2

解：x ≤ –3 或 x > 2

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4. Summation of Series: Using Standard Results | 级数求和：使用标准结果

Question 4: Evaluate \sum_r=1¹⁵ (2r² + 3r - 1).

问题4：计算 ∑_r=1¹⁵ (2r² + 3r – 1)。

Use the standard summation formulae: Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6, Σ1 = n. For n = 15:

使用标准求和公式：Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6, Σ1 = n。令 n = 15：

Σ r² = (15×16×31)/6 = 1240, Σ r = (15×16)/2 = 120, Σ 1 = 15.

∑ r² = (15×16×31)/6 = 1240，∑ r = (15×16)/2 = 120，∑ 1 = 15。

Thus the required sum = 2(1240) + 3(120) – 15 = 2480 + 360 – 15 = 2825.

所求总和 = 2(1240) + 3(120) – 15 = 2480 + 360 – 15 = 2825。

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5. Matrices: Determinant and Inverse | 矩阵：行列式与逆矩阵

Question 5: Given A = (2, 5; 1, 3), find det(A) and A^-1. Hence solve the system: 2x + 5y = 7, x + 3y = 4.

问题5：已知 A = (2 5; 1 3) （表示 2×2 矩阵），求 det(A) 和 A^-1，并由此解方程组：2x + 5y = 7，x + 3y = 4。

det(A) = (2)(3) – (5)(1) = 6 – 5 = 1. Since the determinant is non‑zero, the inverse exists.

det(A) = 2×3 – 5×1 = 6 – 5 = 1。行列式不为零，逆矩阵存在。

The inverse is A^-1 = 11 (3, -5; -1, 2) = (3, -5; -1, 2).

逆矩阵为 A^-1 = (1/1) (3 -5; -1 2) = (3 -5; -1 2)。

The system can be written as A (x y)^T = (7 4)^T. Multiplying by A^-1: (x y)^T = (3 -5; -1 2)(7 4) = (3×7 + (-5)×4, -1×7 + 2×4) = (21-20, -7+8) = (1, 1).

方程组可写成 A (x y)^T = (7 4)^T。左乘 A^-1 得：(x y)^T = (3 -5; -1 2)(7 4) = (21-20, -7+8) = (1, 1)。

Thus x = 1, y = 1.

因此 x = 1, y = 1。

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6. Polar Coordinates: Converting Equations | 极坐标：方程转换

Question 6: Convert the polar equation r = 6cosθ to Cartesian form, and identify the shape.

问题6：将极坐标方程 r = 6cosθ 转换为直角坐标方程，并识别其形状。

Recall: x = r cosθ, y = r sinθ, and r² = x² + y². Multiply both sides by r: r² = 6r cosθ → x² + y² = 6x.

注意：x = r cosθ，y = r sinθ，且 r² = x² + y²。等式两端同乘 r：r² = 6r cosθ → x² + y² = 6x。

Complete the square for x: x² – 6x + y² = 0 → (x – 3)² – 9 + y² = 0 → (x – 3)² + y² = 9.

对 x 配方：x² – 6x + y² = 0 → (x – 3)² – 9 + y² = 0 → (x – 3)² + y² = 9。

This is a circle with centre (3, 0) and radius 3.

这是一个圆心在 (3, 0)、半径为 3 的圆。

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7. Hyperbolic Functions: Solving Equations | 双曲函数：解方程

Question 7: Solve the equation 2cosh x – 5sinh x = 7, giving your answers in logarithmic form.

问题7：解方程 2cosh x – 5sinh x = 7，答案以对数形式表示。

Use definitions: cosh x = (e^x + e^-x)/2, sinh x = (e^x – e^-x)/2. Substitute: 2*(e^x+e^-x)/2 – 5*(e^x–e^-x)/2 = 7 → (e^x+e^-x) – (5/2)(e^x–e^-x) = 7.

利用定义：cosh x = (e^x + e^-x)/2, sinh x = (e^x – e^-x)/2。代入：2×(e^x+e^-x)/2 – 5×(e^x–e^-x)/2 = 7 → (e^x+e^-x) – (5/2)(e^x–e^-x) = 7。

Multiply by 2: 2e^x + 2e^-x – 5e^x + 5e^-x = 14 → –3e^x + 7e^-x = 14. Multiply by e^x: –3e^2x + 7 = 14e^x. Rearrange: 3e^2x + 14e^x – 7 = 0.

两边乘 2：2e^x + 2e^-x – 5e^x + 5e^-x = 14 → –3e^x + 7e^-x = 14。两边同乘 e^x：–3e^2x + 7 = 14e^x。整理得：3e^2x + 14e^x – 7 = 0。

Let y = e^x > 0. Then 3y² + 14y – 7 = 0. Discriminant = 196 + 84 = 280 = 4×70, so y = [–14 ± √280]/(6) = [–14 ± 2√70]/6 = [–7 ± √70]/3. Since y > 0, take y = (√70 – 7)/3.

令 y = e^x > 0，得 3y² + 14y – 7 = 0。判别式 = 196 + 84 = 280 = 4×70，故 y = [–14 ± √280]/6 = [–14 ± 2√70]/6 = [–7 ± √70]/3。因 y > 0，取 y = (√70 – 7)/3。

Thus x = ln((√70 – 7)/3).

所以 x = ln((√70 – 7)/3)。

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8. Differentiation: Implicit and Parametric | 微分：隐函数与参数方程

Question 8: A curve is defined implicitly by x²y + xy² = 6. Find dy/dx at the point (1, 2).

问题8：一条曲线由隐函数 x²y + xy² = 6 定义。求在点 (1, 2) 处的 dy/dx。

Differentiate both sides with respect to x: d/dx (x²y) + d/dx (xy²) = 0. Using the product rule, 2xy + x² dy/dx + (1·y² + x·2y dy/dx) = 0.

两边对 x 求导：d/dx (x²y) + d/dx (xy²) = 0。运用乘法法则，2xy + x² dy/dx + (y² + 2xy dy/dx) = 0。

Collect dy/dx terms: (x² + 2xy) dy/dx = –2xy – y². At (1,2): x=1, y=2, so x²+2xy = 1+4=5, –2xy – y² = –4 – 4 = –8. Thus dy/dx = –8/5.

合并含 dy/dx 的项：(x² + 2xy) dy/dx = –2xy – y²。在点 (1,2) 处：x=1, y=2，x²+2xy = 1+4=5，–2xy – y² = –4 – 4 = –8。因此 dy/dx = –8/5。

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9. Integration: Using Substitution | 积分：使用代换法

Question 9: Use the substitution u = x² + 1 to evaluate ∫₀¹ 2x√x²+1 dx.

问题9：利用代换 u = x² + 1 计算 ∫₀¹ 2x√x²+1 dx。

Let u = x² + 1, then du/dx = 2x, so du = 2x dx. When x = 0, u = 1; when x = 1, u = 2.

令 u = x² + 1，则 du/dx = 2x，即 du = 2x dx。当 x = 0 时 u = 1；当 x = 1 时 u = 2。

The integral becomes ∫_u=1² √u du = ∫₁² u^1/2 du = [ (2/3) u^3/2 ]₁² = (2/3)(2^3/2 – 1^3/2) = (2/3)(2√2 – 1).

积分变为 ∫_u=1² √u du = ∫₁² u^1/2 du = [ (2/3) u^3/2 ]₁² = (2/3)(2^3/2 – 1 ^{Published by TutorHao | IGCSE 进阶数学 Revision Series | aleveler.com}

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