IGCSE OCR Maths: Worked Example Explained | IGCSE OCR 数学：典型例题详解

📚 IGCSE OCR Maths: Worked Example Explained | IGCSE OCR 数学：典型例题详解

This article presents a selection of typical IGCSE OCR Mathematics questions with detailed solutions. Each worked example is chosen from core topics such as algebra, geometry, trigonometry, statistics, and probability, reflecting the style and difficulty of the real exam. By working through these solutions, you will strengthen your problem-solving skills and build confidence for the final papers.

本文精选了IGCSE OCR数学中具有代表性的典型例题，并提供详细解答。每个例题都来自核心主题，如代数、几何、三角函数、统计和概率，真实反映了考试的题型与难度。通过这些解题过程，你将巩固解题技巧，增强应对大考的信心。

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

Solve the quadratic equation: 2x² – 5x – 3 = 0.

解二次方程：2x² – 5x – 3 = 0。

Step 1: Write the equation in standard form. It is already given as 2x² – 5x – 3 = 0.

步骤1：将方程写成标准形式。原方程已给出为 2x² – 5x – 3 = 0。

Step 2: Factorise the quadratic. We need two numbers that multiply to 2 × (–3) = –6 and add to –5. The numbers are –6 and +1. Split the middle term: 2x² – 6x + x – 3.

步骤2：因式分解二次式。需找两个数，乘积为 2 × (–3) = –6，和为 –5。这两个数是 –6 和 +1。拆分中间项：2x² – 6x + x – 3。

Step 3: Factor by grouping: (2x² – 6x) + (x – 3) = 2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3).

步骤3：分组分解：(2x² – 6x) + (x – 3) = 2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3)。

Step 4: Set each factor to zero: 2x + 1 = 0 → x = –½, and x – 3 = 0 → x = 3.

步骤4：令每个因式等于零：2x + 1 = 0 → x = –½，x – 3 = 0 → x = 3。

The solutions are x = –½ and x = 3.

解为 x = –½ 和 x = 3。

2. Function Transformations | 函数图像变换

The point P(2, 5) lies on the graph of y = f(x). Find the coordinates of the image of P on the graph of y = f(x – 4) + 3.

点 P(2, 5) 在 y = f(x) 的图像上。求点 P 在 y = f(x – 4) + 3 图像上的对应点的坐标。

The transformation from y = f(x) to y = f(x – 4) + 3 is a translation by vector (4, 3). This means every point (x, y) is mapped to (x + 4, y + 3).

从 y = f(x) 到 y = f(x – 4) + 3 的变换是沿向量 (4, 3) 的平移。这意味着每个点 (x, y) 映射到 (x + 4, y + 3)。

So the image of P(2, 5) is (2 + 4, 5 + 3) = (6, 8).

因此，P(2, 5) 的像点为 (2 + 4, 5 + 3) = (6, 8)。

3. Similar Triangles and Proportion | 相似三角形与比例

Triangle ABC is similar to triangle PQR. AB = 6 cm, BC = 8 cm, CA = 10 cm. PQ = 9 cm. Find the lengths of QR and RP.

三角形 ABC 相似于三角形 PQR。AB = 6 cm，BC = 8 cm，CA = 10 cm，PQ = 9 cm。求 QR 和 RP 的长度。

Since the triangles are similar, corresponding sides are in the same ratio. AB corresponds to PQ, so scale factor = PQ ÷ AB = 9 ÷ 6 = 1.5.

由于三角形相似，对应边成比例。AB 对应 PQ，因此比例因子 = PQ ÷ AB = 9 ÷ 6 = 1.5。

Then QR = BC × 1.5 = 8 × 1.5 = 12 cm, and RP = CA × 1.5 = 10 × 1.5 = 15 cm.

因此 QR = BC × 1.5 = 8 × 1.5 = 12 cm，RP = CA × 1.5 = 10 × 1.5 = 15 cm。

4. Cosine Rule in Trigonometry | 余弦定理

In triangle ABC, side a = 7 cm, side b = 9 cm, and angle C = 40°. Find side c.

在三角形 ABC 中，边 a = 7 cm，边 b = 9 cm，角 C = 40°。求边 c。

Using the cosine rule: c² = a² + b² – 2ab cos C.

使用余弦定理：c² = a² + b² – 2ab cos C。

Substitute the values: c² = 7² + 9² – 2 × 7 × 9 × cos 40°.

代入数值：c² = 7² + 9² – 2 × 7 × 9 × cos 40°。

Calculate: 49 + 81 – 126 cos 40°. cos 40° ≈ 0.7660, so 126 × 0.7660 ≈ 96.516, thus c² ≈ 130 – 96.516 = 33.484.

计算：49 + 81 – 126 cos 40°。cos 40° ≈ 0.7660，因此 126 × 0.7660 ≈ 96.516，所以 c² ≈ 130 – 96.516 = 33.484。

c = √33.484 ≈ 5.79 cm (to 3 significant figures).

c = √33.484 ≈ 5.79 cm（保留三位有效数字）。

5. Histograms and Frequency Density | 直方图与频数密度

In a histogram, the class interval 20 ≤ t < 30 has a frequency density of 2.4. Calculate the frequency for this interval.

在直方图中，组距 20 ≤ t < 30 的频数密度为 2.4。计算该组的频数。

Frequency = frequency density × class width. The class width = 30 – 20 = 10.

频数 = 频数密度 × 组距。组距 = 30 – 20 = 10。

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IGCSE OCR Maths: Worked Example Explained | IGCSE OCR 数学：典型例题详解