Year 11 CIE Further Mathematics: Case Study Practical Drills | Year 11 CIE 进阶数学：案例分析实战演练

📚 Year 11 CIE Further Mathematics: Case Study Practical Drills | Year 11 CIE 进阶数学：案例分析实战演练

Welcome to this intensive case study drill session designed for Year 11 students preparing for CIE Further Mathematics (Additional Mathematics 0606). These practical exercises will sharpen your problem-solving skills across key topics: quadratics, logarithms, trigonometry, calculus, vectors, polynomials, binomial expansion, kinematics, and functions. Each case is presented with step-by-step reasoning, linking concepts to typical exam questions.

欢迎参加专为 Year 11 学生设计的 CIE 进阶数学（附加数学 0606）强化案例分析演练。这些实战练习将提升你在核心主题上的解题能力：二次函数、对数、三角学、微积分、向量、多项式、二项式展开、运动学以及函数。每个案例均配有逐步推理，紧密联系典型考试题型。

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1. Quadratic Equations with Parameters | 含参数的二次方程

Problem: Find the range of values of k for which the quadratic equation x² + kx + k + 2 = 0 has real roots.

问题：求使得二次方程 x² + kx + k + 2 = 0 有实数根的 k 的取值范围。

Step 1: A quadratic equation ax² + bx + c = 0 has real roots if and only if its discriminant b² – 4ac ≥ 0.

步骤1：二次方程 ax² + bx + c = 0 有实数根当且仅当其判别式 b² – 4ac 大于或等于零。

Here a = 1, b = k, c = k + 2. The discriminant is Δ = k² – 4(1)(k+2) = k² – 4k – 8.

此处 a = 1, b = k, c = k + 2。判别式为 Δ = k² – 4(k+2) = k² – 4k – 8。

Set the condition Δ ≥ 0: k² – 4k – 8 ≥ 0.

令 Δ ≥ 0：k² – 4k – 8 ≥ 0。

To solve the quadratic inequality, first find the roots of k² – 4k – 8 = 0 using the quadratic formula: k = [4 ± √(16 + 32)] / 2 = [4 ± √48] / 2 = [4 ± 4√3] / 2 = 2 ± 2√3.

求解二次不等式，先求方程 k² – 4k – 8 = 0 的根，使用求根公式：k = [4 ± √(16+32)] / 2 = [4 ± √48] / 2 = [4 ± 4√3] / 2 = 2 ± 2√3。

Since the coefficient of k² is positive, the parabola opens upward. The inequality k² – 4k – 8 ≥ 0 holds for k ≤ 2 – 2√3 or k ≥ 2 + 2√3.

由于 k² 系数为正，抛物线开口向上。不等式 k² – 4k – 8 ≥ 0 的解为 k ≤ 2 – 2√3 或 k ≥ 2 + 2√3。

Answer: k ∈ (-∞, 2 – 2√3] ∪ [2 + 2√3, ∞)

答案：k ∈ (-∞, 2 – 2√3] ∪ [2 + 2√3, ∞)

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2. Exponential Growth Modeling | 指数增长建模

Case: A bacterial colony starts with 500 cells and triples every 4 hours. The population after t hours is modelled by P = 500 × 3^(t/4). After how many hours will the colony reach 20,000 cells?

案例：一个细菌群体初始有500个细胞，每4小时数量增至三倍。t 小时后的数量模型为 P = 500 × 3^(t/4)。问经过多少小时后群体达到20,000个细胞？

Solution: Set P = 20,000 and solve for t.

解法：令 P = 20,000 并求解 t。

20,000 = 500 × 3^(t/4) ⇒ 3^(t/4) = 40.

20,000 = 500 × 3^(t/4) ⇒ 3^(t/4) = 40。

Take common logarithms: log(3^(t/4)) = log 40. Apply the power rule: (t/4) log 3 = log 40.

取常用对数：log(3^(t/4)) = log 40。应用幂法则：(t/4) log 3 = log 40。

Rearrange: t/4 = log 40 / log 3, so t = 4 × (log 40 / log 3).

整理得：t/4 = log 40 / log 3，因此 t = 4 × (log 40 / log 3)。

Using a calculator: log 40 ≈ 1.60206, log 3 ≈ 0.47712. Then t ≈ 4 × (1.60206 / 0.47712) ≈ 4 × 3.357 ≈ 13.43 hours.

使用计算器：log 40 ≈ 1.60206, log 3 ≈ 0.47712。则 t ≈ 4 × (1.60206 / 0.47712) ≈ 4 × 3.357 ≈ 13.43 小时。

The colony reaches 20,000 cells after approximately 13.4 hours.

大约13.4小时后，细菌群体达到20,000个细胞。

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3. Trigonometric Equations | 三角方程求解

Problem: Solve the equation 2 sin²x – sin x – 1 = 0 for 0° ≤ x ≤ 360°.

问题：在 0° ≤ x ≤ 360° 范围内解方程 2 sin²x – sin x – 1 = 0。

Step 1: Treat as a quadratic in sin x. Factorise: (2 sin x + 1)(sin x – 1) = 0.

步骤1：将其视为 sin x 的二次方程。因式分解：(2 sin x + 1)(sin x – 1) = 0。

Step 2: Set each factor equal to zero: 2 sin x + 1 = 0 or sin x – 1 = 0.

步骤2：令每个因式等于零：2 sin x + 1 = 0 或 sin x – 1 = 0。

From 2 sin x + 1 = 0: sin x = -1/2. The reference angle is 30°. In the given interval, sin is negative in the third and fourth quadrants, so x = 180° + 30° = 210° and x = 360° – 30° = 330°.

由 2 sin x + 1 = 0 得 sin x = -1/2。参考角为30°。给定区间内正弦在第三和第四象限为负，因此 x = 180° + 30° = 210° 以及 x = 360° – 30° = 330°。

From sin x – 1 = 0: sin x = 1, giving x = 90°.

由 sin x – 1 = 0 得 sin x = 1，即 x = 90°。

Solution set: x = 90°, 210°, 330°.

解集：x = 90°, 210°, 330°。

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4. Tangents and Normals via Differentiation | 利用微分求切线与法线

Case: For the curve y = x³ – 3x + 2, find the equation of the tangent and the normal at the point where x = 2.

案例：对于曲线 y = x³ – 3x + 2，求在 x = 2 处的切线方程和法线方程。

First, find the y-coordinate: y(2) = 2³ – 3(2) + 2 = 8 – 6 + 2 = 4. The point is (2, 4).

首先求 y 坐标：y(2) = 8 – 6 + 2 = 4。点为 (2, 4)。

Differentiate to find the gradient function: dy/dx = 3x² – 3.

微分得到斜率函数：dy/dx = 3x² – 3。

At x = 2, the gradient of the tangent is m_t = 3(2)² – 3 = 12 – 3 = 9.

在 x = 2 处，切线的斜率 m_t = 3×4 – 3 = 9。

Equation of tangent: y – 4 = 9(x – 2) → y = 9x – 14.

切线方程：y – 4 = 9(x – 2) → y = 9x – 14。

The gradient of the normal is the negative reciprocal: m_n = -1/9.

法线的斜率为负倒数：m_n = -1/9。

Equation of normal: y – 4 = -1/9 (x – 2) → 9y – 36 = -x + 2 → x + 9y = 38.

法线方程：y – 4 = -1/9 (x – 2) → x + 9y = 38。

Tangent: y = 9x – 14 | Normal: x + 9y = 38

切线：y = 9x – 14 | 法线：x + 9y = 38

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5. Definite Integration and Area | 定积分与面积

Problem: Find the area enclosed between the curve y = x² + 1, the x-axis, and the lines x = 1 and x = 3.

问题：求曲线 y = x² + 1、x 轴以及直线 x = 1 和 x = 3 所围成的面积。

The area is given by the definite integral A = ∫₁³ (x² + 1) dx.

面积由定积分 A = ∫₁³ (x² + 1) dx 给出。

Integrate: ∫ (x² + 1) dx = x³/3 + x.

积分：∫ (x² + 1) dx = x³/3 + x。

Evaluate from 1 to 3: F(3) = 3³/3 + 3 = 9 + 3 = 12; F(1) = 1³/3 + 1 = 1/3 + 1 = 4/3.

求值：F(3) = 27/3 + 3 = 9 + 3 = 12；F(1) = 1/3 + 1 = 4/3。

Area = 12 – 4/3 = 32/3 square units.

面积 = 12 – 4/3 = 32/3 平方单位。

Area = 32/3 units²

面积 = 32/3 单位²

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6. Vector Geometry: Proving Collinearity | 向量几何：证明共线

Given points A(1, 2), B(3, 5), and C(5, 8). Show that these three points are collinear and find the ratio AB : BC.

已知点 A(1, 2)、B(3, 5) 和 C(5, 8)。证明这三点共线，并求比值 AB : BC。

Form vectors: AB = (3-1, 5-2) = (2, 3). AC = (5-1, 8-2) = (4, 6).

构造向量：AB = (2, 3)。AC = (4, 6)。

Observe that AC = 2 × AB, which means AC is a scalar multiple of AB. Therefore, A, B, and C lie on the same straight line.

注意到 AC = 2 × AB，即 AC 是 AB 的标量倍数。因此 A、B、C 三点共线。

Since B lies between A and C, AB = (2, 3) and BC = C – B = (5-3, 8-5) = (2, 3) = AB. Hence AB : BC = 1 : 1.

由于 B 在 A 和 C 之间，BC = (2,

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