IGCSE CCEA Maths: Worked Examples Explained | IGCSE CCEA 数学：典型例题详解

📚 IGCSE CCEA Maths: Worked Examples Explained | IGCSE CCEA 数学：典型例题详解

This article provides a selection of carefully worked examples covering core topics in the CCEA IGCSE Mathematics specification. Each question is broken down step-by-step to reinforce understanding and prepare you for the exam.

本文精选了CCEA IGCSE数学考试中的核心典型例题，并详细分解解题步骤，帮助你加深理解、做好考前准备。

1. Solving Linear Equations | 解一元一次方程

Example: Solve 3x – 7 = 2x + 5.

例题：解方程 3x – 7 = 2x + 5。

Step 1: Subtract 2x from both sides to eliminate the x-term on the right: 3x – 2x – 7 = 5 → x – 7 = 5.

步骤1：两边减去 2x，消去右边的 x 项：3x – 2x – 7 = 5 → x – 7 = 5。

Step 2: Add 7 to both sides to isolate x: x = 12.

步骤2：两边加上 7，解出 x：x = 12。

Step 3: Check by substituting x = 12 back into the original equation. LHS = 3 × 12 – 7 = 36 – 7 = 29, RHS = 2 × 12 + 5 = 24 + 5 = 29. Both sides are equal, so the solution is correct.

步骤3：将 x = 12 代入原方程检验。左边 = 3 × 12 – 7 = 36 – 7 = 29，右边 = 2 × 12 + 5 = 24 + 5 = 29。两边相等，解正确。

2. Factorising Quadratic Expressions | 二次式的因式分解

Example: Factorise x² + 5x + 6.

例题：因式分解 x² + 5x + 6。

We need two numbers that multiply to the constant term (+6) and add to the coefficient of x (+5). The pairs of factors of 6 are 1 and 6, or 2 and 3. Only 2 and 3 sum to 5.

我们需要找到两个数，它们的乘积等于常数项 (+6)，和等于 x 的系数 (+5)。6 的因数对有 1 与 6，或 2 与 3。只有 2 和 3 的和为 5。

Thus, x² + 5x + 6 = (x + 2)(x + 3). Check by expanding: (x+2)(x+3) = x² + 3x + 2x + 6 = x² + 5x + 6.

因此，x² + 5x + 6 = (x + 2)(x + 3)。展开验证：(x+2)(x+3) = x² + 3x + 2x + 6 = x² + 5x + 6。

3. Solving Quadratic Equations by Factorisation | 用因式分解法解二次方程

Example: Solve x² – 5x + 6 = 0.

例题：解方程 x² – 5x + 6 = 0。

First, factorise the quadratic expression. Look for two numbers that multiply to +6 and add to -5. They are -2 and -3.

首先，对二次式进行因式分解。寻找乘积为 +6、和为 -5 的两个数。这两个数是 -2 和 -3。

So x² – 5x + 6 = (x – 2)(x – 3). Now set each factor equal to zero: x – 2 = 0 → x = 2; x – 3 = 0 → x = 3.

因此 x² – 5x + 6 = (x – 2)(x – 3)。然后令每个因式等于零：x – 2 = 0 → x = 2；x – 3 = 0 → x = 3。

The solutions are x = 2 or x = 3. Always quickly substitute back to verify: 2² – 5×2 + 6 = 4 – 10 + 6 = 0; 3² – 5×3 + 6 = 9 – 15 + 6 = 0.

解为 x = 2 或 x = 3。快速代回验证：2² – 5×2 + 6 = 4 – 10 + 6 = 0；3² – 5×3 + 6 = 9 – 15 + 6 = 0。

4. Solving Linear Inequalities | 解一元一次不等式

Example: Solve 2x – 3 < 7.

例题：解不等式 2x – 3 < 7。

Step 1: Add 3 to both sides: 2x < 10.

步骤1：两边加上 3，得 2x < 10。

Step 2: Divide both sides by 2: x < 5. The inequality sign does not change because we divided by a positive number.

步骤2：两边除以 2，得 x < 5。因为除以正数，不等号方向不变。

We can represent the solution on a number line as an open circle at 5 with an arrow pointing left. In set notation: {x : x < 5}.

我们可以在数轴上用空心圆点表示 5，并向左画箭头。用集合符号表示为 {x : x < 5}。

5. Equation of a Straight Line | 直线方程

Example: A line passes through the points (1, 3) and (4, 9). Find its equation in the form y = mx + c.

例题：一条直线经过点 (1, 3) 和 (4, 9)。求其方程为 y = mx + c 的形式。

First, calculate the gradient m: m = (9 – 3) / (4 – 1) = 6 / 3 = 2.

首先计算斜率 m：m = (9 – 3) / (4 – 1) = 6 / 3 = 2。

Now use the point-slope form with one of the points, say (1, 3): y – 3 = 2(x – 1).

现在使用点斜式，代入其中一个点，比如 (1, 3)：y – 3 = 2(x – 1)。

Expand and rearrange: y – 3 = 2x – 2 → y = 2x + 1. The y-intercept c is 1. Check with the other point (4,9): 2×4 + 1 = 9. The equation is y = 2x + 1.

展开并整理：y – 3 = 2x – 2 → y = 2x + 1。y 轴截距 c = 1。用另一点 (4,9) 验证：2×4 + 1 = 9。方程为 y = 2x + 1。

6. Basic Trigonometry in Right-Angled Triangles | 直角三角形的基本三角比

Example: In a right-angled triangle, the side opposite angle θ is 3 cm, the adjacent side is 4 cm. Find sin θ, cos θ and tan θ.

例题：在一个直角三角形中，角 θ 的对边长为 3 cm，邻边长为 4 cm。求 sin θ、cos θ 和 tan θ。

First, find the hypotenuse using Pythagoras’ theorem: hypotenuse = √(3² + 4²) = √(9 + 16) = √25 = 5 cm.

首先，用勾股定理求斜边：斜边 = √(3² + 4²) = √(9 + 16) = √25 = 5 cm。

Now apply the definitions: sin θ = opposite / hypotenuse = 3/5. cos θ = adjacent / hypotenuse = 4/5. tan θ = opposite / adjacent = 3/4.

现在应用定义：sin θ = 对边 / 斜边 = 3/5。cos θ = 邻边 / 斜边 = 4/5。tan θ = 对边 / 邻边 = 3/4。

These ratios are often left as fractions or given as decimals. For CCEA, both exact and decimal answers may be accepted, but exact values are preferred unless stated otherwise.

通常这些比值保留为分数或写成小数。CCEA 考试中一般两种形式均接受，但除非特别说明，优先使用精确值。

7. Probability of Independent Events | 独立事件的概率

Example: A fair coin is tossed twice. Find the probability of getting two heads.

例题：抛掷一枚均匀硬币两次。求两次都是正面的概率。

The probability of a head on a single toss is 1/2. Since the tosses are independent, we multiply the probabilities: P(head and head) = 1/2 × 1/2 = 1/4.

单次抛掷出现正面的概率为 1/2。因为两次抛掷相互独立，我们将概率相乘：P(正面且正面) = 1/2 × 1/2 = 1/4。

This can also be represented in a sample space diagram: {HH, HT, TH, TT}. There are 4 equally likely outcomes, and only HH gives two heads, so probability = 1/4.

这也可以用样本空间图表示：{HH, HT, TH, TT}。总共有 4 个等可能结果，只有 HH 满足条件，因此概率为 1/4。

8. Mean and Median | 平均数和中位数

Example: Find the mean and median of the data set: 3, 7, 5, 12, 2.

例题：求数据集 3, 7, 5, 12, 2 的平均值和中位数。

Mean: Add all values and divide by the number of data points. Sum = 3 + 7 + 5 + 12 + 2 = 29. Number of data points = 5. Mean = 29 / 5 = 5.8.

平均值：将所有数值相加后除以数据个数。总和 = 3 + 7 + 5 + 12 + 2 = 29。数据个数 = 5。平均值 = 29 / 5 = 5.8。

Median: Arrange the numbers in ascending order: 2, 3, 5, 7, 12. The median is the middle value, which is 5.

中位数：将数字按升序排列：2, 3, 5, 7, 12。中位数是中间值，即 5。

For an odd number of values, the median is straightforward. For an even number, you average the two middle numbers.

对于奇数个数值，中位数很简单；对于偶数个，则需要取中间两个数的平均数。

9. Area and Circumference of a Circle | 圆的面积和周长

Example: A circle has a radius of 7 cm. Find its circumference and area. Leave your answer in terms of π, then approximate using π ≈ 3.14.

例题：一个圆的半径为 7 cm。求其周长和面积。先用 π 表示结果，然后用 π ≈ 3.14 求出近似值。

Circumference formula: C = 2πr = 2 × π × 7 = 14π cm. Approximating: 14 × 3.14 = 43.96 cm.

周长公式：C = 2πr = 2 × π × 7 = 14π cm。近似计算：14 × 3.14 = 43.96 cm。

Area formula: A = πr² = π × 7² = 49π cm². Approximating: 49 × 3.14 = 153.86 cm².

面积公式：A = πr² = π × 7² = 49π cm²。近似计算：49 × 3.14 = 153.86 cm²。

Remember to use the correct units and that radius is half the diameter. CCEA often asks for exact answers in terms of π unless a decimal is specified.

注意使用正确的单位，并且半径为直径的一半。CCEA 常要求保留 π 的精确值，除非题目明确要求给出小数。

10. Functions and Inverse Functions | 函数与反函数

Example: f(x) = 2x + 3. Find the inverse function f⁻¹(x).

例题：已知 f(x) = 2x + 3，求反函数 f⁻¹(x)。

Step 1: Write f(x) as y: y = 2x + 3.

步骤1：将 f(x) 写成 y：y = 2x + 3。

Step 2: Swap x and y to reflect the inverse relation: x = 2y + 3.

步骤2：交换 x 和 y 以体现反函数关系：x = 2y + 3。

Step 3: Solve for y: x – 3 = 2y → y = (x – 3) / 2.

步骤3：解出 y：x – 3 = 2y → y = (x – 3) / 2。

Therefore, f⁻¹(x) = (x – 3) / 2. You can verify by checking that f(f⁻¹(x)) = x: f((x-3)/2) = 2((x-3)/2) + 3 = x – 3 + 3 = x.

因此，f⁻¹(x) = (x – 3) / 2。你可以通过验证 f(f⁻¹(x)) = x 来确认：f((x-3)/2) = 2((x-3)/2) + 3 = x – 3 + 3 = x。

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IGCSE CCEA Maths: Worked Examples Explained | IGCSE CCEA 数学：典型例题详解