Year 10 CAIE Further Maths: Unit Test Mock Paper Walkthrough | 十年级 CAIE 进阶数学：单元测试模拟卷解析

📚 Year 10 CAIE Further Maths: Unit Test Mock Paper Walkthrough | 十年级 CAIE 进阶数学：单元测试模拟卷解析

Welcome to our comprehensive walkthrough of a Year 10 CAIE Further Maths unit test mock paper. This resource is designed to help students identify common pitfalls, reinforce key concepts, and build confidence ahead of formal assessments. We will cover topics ranging from quadratic discriminants and simultaneous equations to calculus and trigonometry, reflecting the typical scope of the Cambridge IGCSE Additional Mathematics (0606) syllabus. Each worked example is accompanied by bilingual explanations to clarify reasoning and methodology.

欢迎阅读十年级 CAIE 进阶数学单元测试模拟卷的详细解析。本资源旨在帮助学生发现常见错误、巩固核心概念，并在正式评估前建立信心。我们将涵盖从二次判别式、联立方程到微积分和三角学的多个主题，反映剑桥 IGCSE 附加数学 (0606) 教学大纲的典型范围。每个范例都配有双语解释，以阐明推理过程和方法。

1. Quadratic Equations and the Discriminant | 二次方程与判别式

Question: Find the range of values of k for which the quadratic equation 2x² + 3x + (5 – k) = 0 has no real roots.

题目：求 k 的取值范围，使二次方程 2x² + 3x + (5 – k) = 0 没有实数根。

Step 1: Recall that for a quadratic ax² + bx + c = 0, the discriminant is given by Δ = b² – 4ac. Here a = 2, b = 3, c = 5 – k.

步骤1：对于二次方程 ax² + bx + c = 0，判别式 Δ = b² – 4ac。此处 a = 2, b = 3, c = 5 – k。

Compute the discriminant: Δ = 3² – 4(2)(5 – k) = 9 – 40 + 8k = 8k – 31.

计算判别式：Δ = 3² – 4(2)(5 – k) = 9 – 40 + 8k = 8k – 31。

For no real roots, the condition is Δ < 0. Thus 8k – 31 < 0.

没有实根的条件是 Δ < 0。因此 8k – 31 < 0。

Solve the inequality: 8k < 31 ⇒ k < 31/8.

解不等式：8k < 31 ⇒ k < 31/8。

Conclusion: The equation has no real roots when k < 31/8 (i.e. k < 3.875).

结论：当 k < 31/8（即 k < 3.875）时，方程没有实数根。

Extension: If Δ > 0, there are two distinct real roots; if Δ = 0, there is one repeated real root. Understanding the discriminant allows you to quickly determine the nature of the roots without solving the equation fully.

拓展：若 Δ > 0，则有两个不等实根；若 Δ = 0，则有一个重根。理解判别式能让你无需完整求解方程即可快速判断根的性质。

2. Simultaneous Equations and Quadratic Inequalities | 联立方程与二次不等式

Question: Solve the simultaneous equations: y = x² – 2x – 3 and y = 4 – x. Hence, determine the solution set for the inequality x² – 2x – 3 > 4 – x.

题目：解联立方程组：y = x² – 2x – 3 和 y = 4 – x。由此求不等式 x² – 2x – 3 > 4 – x 的解集。

Set the two expressions for y equal: x² – 2x – 3 = 4 – x.

令两个 y 表达式相等：x² – 2x – 3 = 4 – x。

Rearrange to standard form: x² – x – 7 = 0.

化为标准形式：x² – x – 7 = 0。

Apply the quadratic formula: x = [1 ± √(1 + 28)] / 2 = (1 ± √29) / 2.

使用求根公式：x = [1 ± √(1 + 28)] / 2 = (1 ± √29) / 2。

For the inequality x² – 2x – 3 > 4 – x, first bring all terms to one side: x² – x – 7 > 0.

对于不等式 x² – 2x – 3 > 4 – x，先将所有项移到一边：x² – x – 7 > 0。

The corresponding quadratic has roots α = (1 – √29)/2 and β = (1 + √29)/2. Since the coefficient of x² is positive, the parabola opens upward, so the inequality holds for x < α or x > β.

对应的二次方程根为 α = (1 – √29)/2 和 β = (1 + √29)/2。由于 x² 系数为正，抛物线开口向上，因此不等式在 x < α 或 x > β 时成立。

Hence the solution set is x < (1 – √29)/2 or x > (1 + √29)/2.

因此解集为 x < (1 – √29)/2 或 x > (1 + √29)/2。

It is often helpful to sketch the line and parabola to see intersection points and the region where the quadratic lies above the line.

绘图常有助于观察直线与抛物线的交点以及二次函数高于直线的区域。

3. Polynomials and the Remainder Theorem | 多项式与余数定理

Question: The polynomial f(x) = x³ + ax² + bx – 10 has a factor (x + 2) and leaves a remainder of 12 when divided by (x – 1). Find the values of a and b.

题目：多项式 f(x) = x³ + ax² + bx – 10 含有因式 (x + 2)，且除以 (x – 1) 时余数为 12。求 a 与 b 的值。

Using the Factor Theorem, since (x + 2) is a factor, f(-2) = 0. Compute f(-2): (-2)³ + a(-2)² + b(-2) – 10 = -8 + 4a – 2b – 10 = 4a – 2b – 18 = 0. Simplify: 2a – b = 9. (Equation 1)

运用因式定理，因 (x + 2) 是因式，故 f(-2) = 0。计算 f(-2)：(-2)³ + a(-2)² + b(-2) – 10 = -8 + 4a – 2b – 10 = 4a – 2b – 18 = 0。简化得 2a – b = 9。（式1）

By the Remainder Theorem, the remainder when divided by (x – 1) is f(1) = 12. Substitute x=1: (1)³ + a(1)² + b(1) – 10 = 1 + a + b – 10 = a + b – 9 = 12, giving a + b = 21. (Equation 2)

根据余数定理，除以 (x – 1) 的余数为 f(1) = 12。代入 x=1：1 + a + b – 10 = a + b – 9 = 12，得 a + b = 21。（式2）

Solve simultaneously: add Equation 1 and Equation 2: (2a – b) + (a + b) = 9 + 21 ⇒ 3a = 30, so a = 10. Substitute back: 10 + b = 21 ⇒ b = 11.

联立求解：(2a – b) + (a + b) = 9 + 21 ⇒ 3a = 30，得 a = 10。代回：10 + b = 21 ⇒ b = 11。

Thus a = 10, b = 11. The polynomial is f(x) = x³ + 10x² + 11x – 10.

因此 a = 10，b = 11。该多项式为 f(x) = x³ + 10x² + 11x – 10。

4. Binomial Expansion | 二项式展开

Question: Find the coefficient of x&sup4; in the expansion of (2 – 3x)&sup6;.

题目：求 (2 – 3x)&sup6; 展开式中 x&sup4; 的系数。

The general term in the binomial expansion of (a + b)¹ is given by ¹C𝑟 a¹&supminus;¹𝑟 b¹𝑟. Here a = 2, b = -3x, n = 6. The term containing x¹𝑟 is T𝑟₊₁ = &sup6;C𝑟 2&sup6;&supminus;𝑟 (-3x)¹𝑟 = &sup6;C𝑟 2&sup6;&supminus;𝑟 (-3)¹𝑟 x¹𝑟.

二项式 (a + b)¹ 展开的通项为 ¹C𝑟 a¹&supminus;¹𝑟 b¹𝑟。这里 a = 2, b = -3x, n = 6。包含 x¹𝑟 的项为 T𝑟₊₁ = &sup6;C𝑟 2&sup6;&supminus;𝑟 (-3x)¹𝑟 = &sup6;C𝑟 2&sup6;&supminus;𝑟 (-3)¹𝑟 x¹𝑟。

We need the power of x to be 4, so r = 4. Substitute r = 4: coefficient = &sup6;C₄ × 2&sup6;&supminus;&sup4; × (-3)&sup4; = 15 × 2² × 81.

我们需要 x 的指数为 4，因此 r = 4。代入 r = 4：系数 = &sup6;C₄ × 2&sup6;&supminus;&sup4; × (-3)&sup4; = 15 × 2² × 81。

Calculate: 15 × 4 = 60; 60 × 81 = 4860. So the coefficient of x&sup4; is 4860.

计算：15 × 4 = 60；60 × 81 = 4860。因此 x&sup4; 的系数为 4860。

Always careful with signs: (-3)&sup4; is positive, but an odd power would introduce a negative sign.

注意符号：(-3)&sup4; 为正，但奇次幂会引入负号。

5. Exponential and Logarithmic Equations | 指数与对数方程

Question: Solve the equation 2²&#x

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Year 10 CAIE Further Maths: Unit Test Mock Paper Walkthrough | 十年级 CAIE 进阶数学：单元测试模拟卷解析