📚 Case Study Practice in Year 10 WJEC Further Mathematics | Year 10 WJEC 进阶数学：案例分析实战演练
In this article, we bring together key topics from the WJEC Level 2 Additional Mathematics specification through a series of worked case studies. Each section presents a complete problem solution with step‑by‑step reasoning in both English and Chinese, reinforcing essential skills in algebra, calculus, matrices, logarithms, trigonometry and more.
本文通过一系列完整的案例分析，将 WJEC Level 2 进阶数学课程中的核心主题串联起来。每个小节都展示一道典型题目的求解过程，并用英中双语逐步讲解推理思路，旨在巩固代数、微积分、矩阵、对数、三角学等重要技能。
1. Solving Quadratic Inequalities | 解二次不等式
Problem: Solve the inequality x² – 5x + 6 > 0.
题目：解不等式 x² – 5x + 6 > 0。
First factorise the quadratic expression: (x – 2)(x – 3) > 0. The critical values are x = 2 and x = 3. These divide the number line into three intervals: x < 2, 2 < x < 3, and x > 3.
首先分解因式：(x – 2)(x – 3) > 0。关键值为 x = 2 和 x = 3。这两个值将数轴分成三个区间：x < 2、2 < x < 3 和 x > 3。
Test a value from each interval. For x = 1, the product ( –1)( –2) is positive; for x = 2.5, (0.5)( –0.5) is negative; for x = 4, (2)(1) is positive. The inequality requires the product to be greater than zero, so we take the intervals where the expression is positive.
在每个区间取一个测试值。当 x = 1 时，乘积 ( –1)( –2) 为正；x = 2.5 时，(0.5)( –0.5) 为负；x = 4 时，(2)(1) 为正。题目要求乘积大于零，因此取表达式为正的区间。
The solution is x < 2 or x > 3. We write this in set notation as {x : x < 2} ∪ {x : x > 3}.
解集为 x < 2 或 x > 3，用集合表示为 {x : x < 2} ∪ {x : x > 3}。
2. Differentiation from First Principles | 从第一原理求导数
Problem: Use the limit definition to find the derivative of f(x) = x².
题目：利用极限定义求函数 f(x) = x² 的导数。
Recall the definition: f'(x) = lim(h→0) [f(x + h) – f(x)] / h. Substitute f(x) = x²: f(x + h) = (x + h)² = x² + 2xh + h².
回顾定义：f'(x) = lim(h→0) [f(x + h) – f(x)] / h。代入 f(x) = x²：f(x + h) = (x + h)² = x² + 2xh + h²。
Then f(x + h) – f(x) = (x² + 2xh + h²) – x² = 2xh + h². Divide by h: (2xh + h²)/h = 2x + h. Taking the limit as h → 0, we obtain f'(x) = 2x.
于是 f(x + h) – f(x) = (x² + 2xh + h²) – x² = 2xh + h²。除以 h 得 (2xh + h²)/h = 2x + h。令 h → 0 取极限，即得 f'(x) = 2x。
This process confirms the power rule and deepens understanding of instantaneous rate of change.
这一过程验证了幂函数求导法则，也加深了对瞬时变化率概念的理解。
3. Matrix Multiplication | 矩阵乘法
Problem: Given A = [1 2; 3 4] and B = [5 6; 7 8], calculate the product AB.
题目：已知 A = [1 2; 3 4]，B = [5 6; 7 8]，计算乘积 AB。
Matrix multiplication is defined by row‑by‑column dot products. The element in row i, column j of AB is the sum of the products of corresponding entries from row i of A and column j of B.
矩阵乘法定义为行乘列的点积。AB 中第 i 行第 j 列的元素，是 A 的第 i 行与 B 的第 j 列对应元素乘积之和。
Row 1 × Column 1: (1×5) + (2×7) = 5 + 14 = 19. Row 1 × Column 2: (1×6) + (2×8) = 6 + 16 = 22. Row 2 × Column 1: (3×5) + (4×7) = 15 + 28 = 43. Row 2 × Column 2: (3×6) + (4×8) = 18 + 32 = 50.
第一行乘第一列：(1×5) + (2×7) = 5 + 14 = 19。第一行乘第二列：(1×6) + (2×8) = 6 + 16 = 22。第二行乘第一列：(3×5) + (4×7) = 15 + 28 = 43。第二行乘第二列：(3×6) + (4×8) = 18 + 32 = 50。
Therefore, AB = [19 22; 43 50]. Always check that the number of columns in A equals the number of rows in B; here both are 2 × 2, so multiplication is valid.
因此，AB = [19 22; 43 50]。务必检查 A 的列数与 B 的行数相等；此处均为 2×2 矩阵，故可相乘。
4. Solving Logarithmic Equations | 解对数方程
Problem: Solve log₂(x) + log₂(x – 2) = 3.
题目：解方程 log₂(x) + log₂(x – 2) = 3。
Apply the product rule for logarithms: log₂(x) + log₂(x – 2) = log₂[x(x – 2)]. The equation becomes log₂[x(x – 2)] = 3.
利用对数的乘法法则：log₂(x) + log₂(x – 2) = log₂[x(x – 2)]。方程化为 log₂[x(x – 2)] = 3。
Rewrite in exponential form: x(x – 2) = 2³ = 8. Expand: x² – 2x – 8 = 0. This quadratic factorises as (x – 4)(x + 2) = 0, giving potential solutions x = 4 and x = –2.
改写成指数形式：x(x – 2) = 2³ = 8。展开得 x² – 2x – 8 = 0。因式分解为 (x – 4)(x + 2) = 0，得到可能的解 x = 4 和 x = –2。
Check the domain restrictions: the arguments of the logarithms must be positive. For x = –2, log₂(–2) is undefined. Therefore, only x = 4 is valid.
检查定义域限制：对数的真数必须大于零。对于 x = –2，log₂(–2) 无定义。因此唯有 x = 4 是有效解。
5. Binomial Expansion | 二项式展开
Problem: Expand (1 + 2x)⁴ in ascending powers of x, simplifying each coefficient.
题目：按 x 的升幂展开 (1 + 2x)⁴，并化简各项系数。
Use the binomial theorem: (a + b)ⁿ = Σ [nCk aⁿ⁻ᵏ bᵏ] from k=0 to n. Here a = 1, b = 2x, and n = 4.
使用二项式定理：(a + b)ⁿ = Σ [nCk aⁿ⁻ᵏ bᵏ]，k 从 0 到 n。此处 a = 1，b = 2x，n = 4。
Calculate the terms: k=0: 4C0 (1)⁴(2x)⁰ = 1. k=1: 4C1 (1)³(2x)¹ = 4 × 2x = 8x. k=2: 4C2 (1)²(2x)² = 6 × 4x² = 24x². k=3: 4C3 (1)¹(2x)³ = 4 × 8x³ = 32x³. k=4: 4C4 (1)⁰(2x)⁴ = 1 × 16x⁴ = 16x⁴.
计算各项：k=0: 4C0 (1)⁴(2x)⁰ = 1。k=1: 4C1 (1)³(2x)¹ = 4 × 2x = 8x。k=2: 4C2 (1)²(2x)² = 6 × 4x² = 24x²。k=3: 4C3 (1)¹(2x)³ = 4 × 8x³ = 32x³。k=4: 4C4 (1)⁰(2x)⁴ = 1 × 16x⁴ = 16x⁴。
The full expansion is 1 + 8x + 24x² + 32x³ + 16x⁴. This is valid for all real x, but if |2x| < 1 the series would converge rapidly in further work.
完整的展开式为 1 + 8x + 24x² + 32x³ + 16x⁴。此展开对所有实数 x 均成立，但在后续学习中若 |2x| < 1，级数会快速收敛。
6. Trigonometric Identities and Equations | 三角恒等式与方程
Problem: Given that sin²θ + cos²θ = 1, solve 2sin²θ – cosθ = 1 for 0° ≤ θ ≤ 360°.
题目：已知 sin²θ + cos²θ = 1，在区间 0° ≤ θ ≤ 360° 内解方程 2sin²θ – cosθ = 1。
Replace sin²θ with 1 – cos²θ. The equation becomes 2(1 – cos²θ) – cosθ = 1 → 2 – 2cos²θ – cosθ = 1. Rearranging gives –2cos²θ – cosθ + 1 = 0, or multiply by –1: 2cos²θ + cosθ – 1 = 0.
将 sin²θ 替换为 1 – cos²θ。方程变为 2(1 – cos²θ) – cosθ = 1 → 2 – 2cos²θ – cosθ = 1。移项得 –2cos²θ – cosθ + 1 = 0，或两边乘 –1：2cos²θ + cosθ – 1 = 0。
Treat this as a quadratic in cosθ. Factorise: (2cosθ – 1)(cosθ + 1) = 0. So cosθ = ½ or cosθ = –1.
将此看作关于 cosθ 的二次方程。因式分解：(2cosθ – 1)(cosθ + 1) = 0。因此 cosθ = ½ 或 cosθ = –1。
For cosθ = ½ in the given interval, θ = 60° and 300°. For cosθ = –1, θ = 180°. The solution set is {60°, 180°, 300°}.
在给定区间内，cosθ = ½ 的解为 θ = 60° 和 300°；cosθ = –1 的解为 θ = 180°。解集为 {60°, 180°, 300°}。
7. Polynomial Division | 多项式除法
Problem: Divide the polynomial f(x) = 2x³ – 5x² + 3x – 7 by (x – 3), and state the quotient and remainder.
题目：用 (x – 3) 除以多项式 f(x) = 2x³ – 5x² + 3x – 7，写出商式和余数。
We use long division. Write the dividend in descending powers: 2x³ – 5x² + 3x – 7. Divide the leading term: (2x³) ÷ (x) = 2x². Multiply (x – 3) by 2x² and subtract: (2x³ – 5x²) – (2x³ – 6x²) = x². Bring down +3x.
我们采用长除法。按降幂排列被除式：2x³ – 5x² + 3x – 7。首项相除：(2x³) ÷ (x) = 2x²。用 2x² 乘 (x – 3) 并相减：(2x³ – 5x²) – (2x³ – 6x²) = x²。移下 +3x。
Now divide x² by x, giving +x. Multiply (x – 3) by x → x² – 3x. Subtract: (x² + 3x) – (x² – 3x) = 6x. Bring down –7. Divide 6x by x → +6. Multiply (x – 3) by 6 → 6x – 18. Subtract: (6x – 7) – (6x – 18) = 11.
接着用 x² ÷ x 得 +x。以 x 乘 (x – 3) 得 x² – 3x。相减：(x² + 3x) – (x² – 3x) = 6x。移下 –7。6x ÷ x 得 +6。以 6 乘 (x – 3) 得 6x – 18。相减：(6x – 7) – (6x – 18) = 11。
The quotient is 2x² + x + 6 and the remainder is 11. Thus f(x) = (x – 3)(2x² + x + 6) + 11.
商式为 2x² + x + 6，余数为 11。因此 f(x) = (x – 3)(2x² + x + 6) + 11。
8. Finding the Equation of a Tangent | 求切线方程
Problem: Find the equation of the tangent to the curve y = x³ – 3x + 2 at the point where x = 1.
题目：求曲线 y = x³ – 3x + 2 在 x = 1 处的切线方程。
Differentiate to obtain the gradient function: dy/dx = 3x² – 3. At x = 1, the gradient m = 3(1)² – 3 = 0. This indicates a horizontal tangent.
求导得梯度函数：dy/dx = 3x² – 3。在 x = 1 处，梯度 m = 3(1)² – 3 = 0。这表明切线是水平的。
Find the y‑coordinate: y = (1)³ – 3(1) + 2 = 0. So the point of contact is (1, 0). With m = 0, the tangent is a horizontal line passing through y = 0, so its equation is y = 0.
求 y 坐标：y = (1)³ – 3(1) + 2 = 0。因此切点为 (1, 0)。由于 m = 0，切线是一条通过 y = 0 的水平线，故方程为 y = 0。
Even when the gradient is zero, the method remains identical: differentiate, substitute, then apply y – y₁ = m(x – x₁).
即便梯度为零，方法也完全相同：求导、代入，然后利用点斜式 y – y₁ = m(x – x₁)。
9. Arithmetic Sequences and Series | 等差数列与级数
Problem: The 5th term of an arithmetic sequence is 12 and the 12th term is 5. Find the first term and the common difference.
题目：一等差数列的第 5 项为 12，第 12 项为 5。求首项和公差。
Recall the nth term formula: uₙ = a + (n – 1)d, where a is the first term and d is the common difference. Write two equations: u₅ = a + 4d = 12 and u₁₂ = a + 11d = 5.
回顾通项公式：uₙ = a + (n – 1)d，其中 a 为首项，d 为公差。列出两个方程：u₅ = a + 4d = 12，u₁₂ = a + 11d = 5。
Subtract the first equation from the second: (a + 11d) – (a + 4d) = 5 – 12 → 7d = –7, so d = –1. Substitute back into a + 4(–1) = 12 → a – 4 = 12 → a = 16.
用第二式减去第一式：(a + 11d) – (a + 4d) = 5 – 12 → 7d = –7，因此 d = –1。代回 a + 4(–1) = 12 → a – 4 = 12 → a = 16。
The first term is 16 and the common difference is –1. This means the sequence decreases by 1 each step: 16, 15, 14, 13, 12, …
首项为 16，公差为 –1。这意味着数列每次减 1：16, 15, 14, 13, 12, ……
10. Applying Integration to Areas | 积分在面积中的应用
Problem: Find the area bounded by the curve y = 4x – x² and the x‑axis.
题目：求曲线 y = 4x – x² 与 x 轴所围区域的面积。
First determine where the curve meets the x‑axis. Solve 4x – x² = 0 → x(4 – x) = 0, so x = 0 and x = 4. These are the limits of integration.
首先确定曲线与 x 轴的交点。解 4x – x² = 0 → x(4 – x) = 0，得 x = 0 和 x = 4。这两个值就是积分限。
Set up the definite integral: Area = ∫₀⁴ (4x – x²) dx. Integrate term by term: ∫4x dx = 2x², ∫x² dx = (1/3)x³, so the antiderivative is 2x² – (1/3)x³.
建立定积分：面积 = ∫₀⁴ (4x – x²) dx。逐项积分：∫4x dx = 2x²，∫x² dx = (1/3)x³，因此原函数为 2x² – (1/3)x³。
Evaluate from 0 to 4: [2(4)² – (1/3)(4)³] – [0] = [2×16 – (1/3)×64] = 32 – 64/3 = (96/3 – 64/3) = 32/3. The exact area is 32/3 square units.
代入上下限： [2(4)² – (1/3)(4)³] – [0] = [32 – 64/3] = 96/3 – 64/3 = 32/3。精确面积为 32/3 平方单位。
Always sketch the graph to confirm the region lies entirely above the x‑axis; if part lies below, the integral must be split.
建议先绘制草图，确认区域完全在 x 轴上方；若有部分在下方，则需拆分积分区间。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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