A-Level OCR Mathematics: Worked Examples Explained | A-Level OCR 数学：典型例题详解

📚 A-Level OCR Mathematics: Worked Examples Explained | A-Level OCR 数学：典型例题详解

This article provides a collection of fully worked examples spanning the OCR A-Level Mathematics specification (H240), covering Pure, Statistics, and Mechanics. Each example is carefully chosen to illustrate key techniques required for exam success.

本文提供了一组涵盖OCR A-Level数学大纲（H240）的典型例题详解，包括纯数、统计和力学。每个例题都经过精心挑选，旨在展示考试成功所需的关键技巧。

1. Quadratic Discriminant and Range of k | 二次判别式与 k 的取值范围

The equation x² + (k – 3)x + 4 = 0 has no real roots. Find the range of possible values for the constant k. [OCR style]

方程 x² + (k – 3)x + 4 = 0 没有实根，求常数 k 的取值范围。

For no real roots, the discriminant must be less than zero: b² – 4ac < 0. Here a = 1, b = k – 3, c = 4.

方程没有实根，判别式必须小于零：b² – 4ac < 0。此时 a = 1, b = k – 3, c = 4。

(k – 3)² – 4×1×4 < 0 → (k – 3)² – 16 < 0

Expand and simplify: k² – 6k + 9 – 16 < 0 → k² – 6k – 7 < 0. Factorise: (k – 7)(k + 1) < 0.

展开并化简：k² – 6k + 9 – 16 < 0 → k² – 6k – 7 < 0。因式分解：(k – 7)(k + 1) < 0。

Sketching the quadratic with roots –1 and 7 shows the inequality is satisfied between the roots: –1 < k < 7.

画出二次函数示意图，根为 –1 和 7，不等式在两根之间成立：–1 < k < 7。

2. Differentiation and Tangent Equation | 微分与切线方程

Find the equation of the tangent to the curve y = x³ – 4x² + 5 at the point where x = 2. [Typical OCR Pure]

求曲线 y = x³ – 4x² + 5 在 x = 2 处的切线方程。

First differentiate: dy/dx = 3x² – 8x. At x = 2, gradient m = 3(2)² – 8(2) = 12 – 16 = –4.

首先求导：dy/dx = 3x² – 8x。在 x = 2 处，斜率 m = 3(2)² – 8(2) = 12 – 16 = –4。

When x = 2, y = (2)³ – 4(2)² + 5 = 8 – 16 + 5 = –3. So the point is (2, –3).

当 x = 2 时，y = 2³ – 4×2² + 5 = 8 – 16 + 5 = –3。因此切点为 (2, –3)。

Using y – y₁ = m(x – x₁): y – (–3) = –4(x – 2) → y + 3 = –4x + 8 → y = –4x + 5.

利用点斜式 y – y₁ = m(x – x₁)：y – (–3) = –4(x – 2) → y + 3 = –4x + 8 → y = –4x + 5。

3. Trigonometric Equations | 三角方程求解

Solve the equation 2 sin² θ + 3 cos θ = 0 for 0° ≤ θ ≤ 360°. [OCR Pure]

解方程 2 sin² θ + 3 cos θ = 0，其中 0° ≤ θ ≤ 360°。

Use identity sin² θ = 1 – cos² θ: 2(1 – cos² θ) + 3 cos θ = 0 → 2 – 2 cos² θ + 3 cos θ = 0 → –2 cos² θ + 3 cos θ + 2 = 0. Multiply by –1: 2 cos² θ – 3 cos θ – 2 = 0.

利用恒等式 sin² θ = 1 – cos² θ：2(1 – cos² θ) + 3 cos θ = 0 → 2 – 2 cos² θ + 3 cos θ = 0 → –2 cos² θ + 3 cos θ + 2 = 0。两边乘 –1：2 cos² θ – 3 cos θ – 2 = 0。

Let u = cos θ. Then 2u² – 3u – 2 = 0. Factorise: (2u + 1)(u – 2) = 0 → u = –½ or u = 2.

令 u = cos θ，则 2u² – 3u – 2 = 0。因式分解：(2u + 1)(u – 2) = 0 → u = –½ 或 u = 2。

cos θ = 2 has no solution since –1 ≤ cos θ ≤ 1. So cos θ = –½. For 0° ≤ θ ≤ 360°, θ = 120°, 240°.

cos θ = 2 无解，因为 –1 ≤ cos θ ≤ 1。故 cos θ = –½。在 0° 到 360° 范围内，θ = 120°, 240°。

4. Logarithmic Equations | 对数方程

Solve log₂ (x + 1) + log₂ (x – 2) = 3. [OCR exam style]

解方程 log₂ (x + 1) + log₂ (x – 2) =

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