📚 GCSE WJEC Further Mathematics: Case Study Practical Drills | GCSE WJEC 进阶数学：案例分析实战演练
In the WJEC GCSE Further Mathematics examination, applying your knowledge to solve complex, multi-step problems is crucial. These case study drills are designed to build your problem-solving skills by walking through real exam-style scenarios, connecting different topics such as algebra, trigonometry, calculus, matrices and vectors. Each case study breaks down a problem, explains the reasoning step by step, and highlights key techniques. By working through these examples, you will gain confidence and fluency in tackling challenging questions.
在 WJEC GCSE 进阶数学考试中，将知识应用于解决复杂的多步骤问题至关重要。这些案例分析实战演练旨在通过真实的考试风格情景，将代数、三角、微积分、矩阵和向量等不同主题连接起来，培养你的解题能力。每个案例分解问题，逐步解释推理，并突出关键技巧。通过这些例子，你将获得信心和流利度，应对具有挑战性的问题。
1. Algebra Foundations: Quadratic Equations and Graphs | 代数基础：二次方程与函数图像
Case: Solve 2x² − 5x − 3 = 0 and sketch the graph, marking the vertex and intercepts.
案例：解方程 2x² − 5x − 3 = 0，绘制图像并标出顶点和截距。
− 3 = 0. Group: (2x² − 6x) + (x − 3) = 2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3). Set each factor to zero: x = −½ or x = 3.
步骤1：通过拆分中项进行因式分解。寻找乘积为 2×(−3)=−6 且和为 −5 的两个数。这两个数是 −6 和 1。重写为 2x² − 6x + x − 3 = 0。分组并提取公因子：(2x² − 6x) + (x − 3) = 2x(x − 3) + 1(x − 3) = (2x+1)(x−3)。令各因子为零，得 x = −½ 或 x = 3。
Step 2: The y-intercept occurs when x = 0: y = 2(0)² − 5(0) − 3 = −3.
步骤2：y 轴截距出现在 x = 0 时：y = 2×0² − 5×0 − 3 = −3。
− 3 = 3.125 − 6.25 − 3 = −6.125. Vertex: (1.25, −6.125). The axis of symmetry is x = 1.25.
步骤3：顶点的 x 坐标由 x = −b/(2a) = 5/4 = 1.25 给出。代入计算 y 值：y = 2×(1.25)² − 5×1.25 − 3 = 2×1.5625 − 6.25 − 3 = 3.125 − 6.25 − 3 = −6.125。顶点坐标为 (1.25, −6.125)，对称轴为 x = 1.25。
Step 4: Plot the intercepts (−0.5,0), (3,0), (0,−3) and the vertex, then draw a smooth upward-opening parabola.
步骤4：描出截距点 (−0.5,0), (3,0), (0,−3) 以及顶点，然后绘制一条平滑的开口向上的抛物线。
2. Function Transformations and Inequalities | 函数变换与不等式
ansformed to g(x) = 2(x+3)² − 4. Describe the transformations in order and solve g(x) ≥ 0.
案例：f(x) = x² 变换为 g(x) = 2(x+3)² − 4。依序描述变换，并解不等式 g(x) ≥ 0。
Step 1: Starting from f(x), replace x by (x+3) for a horizontal translation of −3 (left shift). Then multiply by 2 for a vertical stretch with scale factor 2. Finally subtract 4 for a vertical translation down by 4. The sequence: translation by vector (−3, 0), stretch parallel to y-axis factor 2, translation by (0, −4).
步骤1：从 f(x) 出发，将 x 替换为 (x+3) 实现水平平移 −3（左移3个单位）。然后乘以 2 实现垂直拉伸，比例因子为 2。最后减去 4 实现垂直方向下移 4。变换顺序为：平移向量 (−3, 0)，平行于 y 轴、比例因子 2 的拉伸，平移 (0, −4)。
Step 2: Solve 2(x+3)² − 4 ≥ 0 → 2(x+3)² ≥ 4 → (x+3)² ≥ 2. Take square root: |x+3| ≥ √2. Thus x+3 ≤ −√2 or x+3 ≥ √2, giving x ≤ −3 − √2 or x ≥ −3 + √2.
步骤2：解不等式 2(x+3)² − 4 ≥ 0 → 2(x+3)² ≥ 4 → (x+3)² ≥ 2。开平方得 |x+3| ≥ √2。因此 x+3 ≤ −√2 或 x+3 ≥ √2，解得 x ≤ −3 − √2 或 x ≥ −3 + √2。
3. Trigonometric Applications: Solving Triangles | 三角学应用：解三角形
= 7 cm, b = 10 cm. Find the two possible values for angle B and the corresponding side c.
案例：在三角形 ABC 中，A = 30°，a = 7 cm，b = 10 cm。求角 B 的两个可能值以及对应边 c。
− 45.6° = 134.4°. Both are valid because the sum of angles would still be less than 180°.
步骤1：使用正弦定理：n B / b = sin A / a → sin B = (b sin A)/a = (10 × sin 30°)/7 = (10 × 0.5)/7 = 5/7 ≈ 0.7143。B = arcsin(0.7143) ≈ 45.6° 或 B = 180° − 45.6° = 134.4°。两者均有效，因为三角形内角和仍小于 180°。
₂ = 134.4°, C₂ = 180° − 30° − 134.4° = 15.6°, c₂ = (7 × sin 15.6°)/0.5 ≈ (7 × 0.2689)/0.5 ≈ 3.76 cm.
步骤2：对于 B₁ = 45.6°，C₁ = 180° − 30° − 45.6° = 104.4°。应用正弦定理：c₁ / sin C₁ = a / sin A → c₁ = (7 × sin 104.4°)/ sin 30° ≈ (7 × 0.9689)/0.5 ≈ 13.56 cm。对于 B₂ = 134.4°，C₂ = 15.6°，c₂ = (7 × sin 15.6°)/0.5 ≈ (7 × 0.2689)/0.5 ≈ 3.76 cm。
4. Sequences and Series: Arithmetic and Geometric | 数列与级数：等差与等比
Case: (a) In an arithmetic sequence the first term is 5 and the 10th term is 32. Find the common difference and the sum of the first 20 terms. (b) In a geometric sequence the second term is 6 and the fifth term is 162. Find the first term and common ratio.
案例：(a) 一个等差数列首项为 5，第 10 项为 32。求公差及前 20 项之和。(b) 一个等比数列第二项为 6，第五项为 162。求首项和公比。
2: Sum of first 20 terms S₂₀ = n/2 [2a + (n−1)d] = 20/2 [2×5 + 19×3] = 10 [10 + 57] = 670.
(a) 步骤1：a = 5，a + 9d = 32 → 9d = 27 → d = 3。步骤2：前 20 项之和 S₂₀ = n/2 [2a + (n−1)d] = 20/2 [2×5 + 19×3] = 10 [10 + 57] = 670。
= 162. Divide the equations: (ar⁴)/(ar) = 162/6 → r³ = 27 → r = 3. Then a × 3 = 6 → a = 2.
(b) 步骤1：设首项为 a，公比为 r。第二项 ar = 6，第五项 ar⁴ = 162。两式相除：(ar⁴)/(ar) = 162/6 → r³ = 27 → r = 3。代入得 a×3 = 6 → a = 2。
5. Introduction to Differentiation: Tangents and Normals | 微分入门：切线与法线
Find the equation of the tangent to the curve y = x³ − 3x² + 2x at the point where x = 2.
案例：求曲线 y = x³ − 3x² + 2x 在 x = 2 处的切线方程。
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