Case Study Practice for GCSE Eduqas Further Mathematics | GCSE Eduqas进阶数学案例分析实战演练

📚 Case Study Practice for GCSE Eduqas Further Mathematics | GCSE Eduqas进阶数学案例分析实战演练

This article presents a series of real‑world case studies designed to sharpen your problem‑solving skills for the GCSE Eduqas Further Mathematics examination. Each scenario integrates multiple topics — from calculus and optimisation to trigonometry, matrices, and financial mathematics — demanding the analytical approach expected in the higher tier. Work through each example carefully, comparing your reasoning with the step‑by‑step solutions provided.

本文通过一系列真实情境的案例分析，帮助你提升GCSE Eduqas进阶数学考试所需的解题能力。每个场景综合了微积分、最优化、三角学、矩阵以及金融数学等多个主题，要求运用高阶思维进行分析。请仔细研究每一个例题，对照给出的分步解答检验自己的推理过程。

1. Quadratic Optimisation: Enclosing a Riverbank Field | 二次优化：河岸围栏问题

A farmer has 120 m of fencing and wishes to fence off a rectangular field along a straight river, using the river as one side (so no fencing is needed on that side). Find the dimensions that give the maximum area for the field.

一位农民有 120 米长的篱笆，他想沿着笔直的河岸围出一个矩形田地，利用河流作为其中一条边（河岸一侧无需篱笆）。求能使田地面积最大的长宽尺寸。

o widths and one length, we have length = 120 − 2x. The area A = x(120 − 2x) = 120x − 2x².

设垂直于河岸的宽度为 x m。因为篱笆总长覆盖两条宽和一条长，所以长 = 120 − 2x。面积 A = x(120 − 2x) = 120x − 2x²。

Rewrite in standard quadratic form: A = −2x² + 120x. The graph of this quadratic is a ∩‑ e vertex is at x = −b/(2a). Here a = −2, b = 120, so x = −120 / (2 × −2) = 120 / 4 = 30 m.

整理成标准二次式：A = −2x² + 120x。该二次函数图像为开口向下的抛物线，因此最大值出现在顶点。对于 ax² + bx + c，顶点横坐标为 x = −b/(2a)。这里 a = −2, b = 120，故 x = −120 / (2 × −2) = 120 / 4 = 30 m。

tep is recognising the constraint and expressing one variable in terms of the other before optimising.

于是长 = 120 − 2(30) = 60 m。最大面积 = 30 × 60 = 1800 m²。关键在于识别约束条件，在优化前将一个变量用另一个变量表示。

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2. Financial Mathematics: Compound Interest with Regular Saving | 金融数学：定期储蓄复利计算

Ella invests £1 500 at the beginning of each year in an account paying 3.5% interest compounded annually. Calculate the total value of her investment after 4 complete years, just after the fourth deposit has been made.

艾拉每年年初投资 1500 英镑到一个年利率 3.5%、复利计算的账户中。计算在完成 4 年后，即刚存入第 4 笔款项后，她的总投资价值。

s, and the fourth earns no interest in the same calendar year. Use the compound formula A = P(1 + r)ⁿ.

每笔存款获得利息的年数不同。第一笔赚取 4 年利息，第二笔 3 年，第三笔 2 年，第四笔在同年末不产生利息。使用复利公式 A = P(1 + r)ⁿ。

Deposit	Years of interest (n)	Value at end
1st (£1 500)	4	1500 × 1.035⁴
2nd (£1 500)	3	1500 × 1.035³
3rd (£1 500)	2	1500 × 1.035²
4th (£1 500)	0	1500 × 1.035⁰ = 1500

Compute each term: 1.035⁴ ≈ 1.1475, so 1500×1.1475 = £1 721.25; 1.035³ ≈ 1.1087, gives £1 663.05; 1.035² = 1.071225, gives £1 606.84; fourth remains £1 500. Sum ≈ £1 721.25 + £1 663.05 + £1 606.84 + £1 500 = £6 491.14.

计算各项：1.035⁴ ≈ 1.1475，得 £1 721.25；1.035³ ≈ 1.1087，得 £1 663.05；1.035² = 1.071225，得 £1 606.84；第四笔 £1 500。总和 ≈ £6 491.14。

Geometric series knowledge can also be used: the sum is 1500(1.035⁴ + 1.035³ + 1.035² + 1), but the table method makes the different growth periods clear.

也可以运用等比数列求和：1500(1.035⁴ + 1.035³ + 1.035² + 1)，但表格法能更清晰地展示不同增长期。

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3. Kinematics: Stopping Distance from a Velocity Function | 运动学：由速度函数求刹车距离

econds. The car comes to rest when v = 0. Determine the distance travelled during braking.

一辆汽车刹车后的速度由 v(t) = 20 − 4t m/s 给出，其中 t 为时间（秒）。当 v = 0 时汽车静止。求刹车过程中行驶的距离。

First find the time to stop: set 20 − 4t = 0 ⇒ t = 5 s. The distance is the area under the velocity‑time graph, which is a straight line. Using calculus, distance = ∫ v dt from t=0 to t=5.

首先求静止时刻：令 20 − 4t = 0 ⇒ t = 5 s。距离即为速度‑时间图像下的面积，这是一条直线。用微积分计算，距离 = ∫ v dt 从 t=0 到 t=5。

s = ∫₀⁵ (20 − 4t) dt = [20t − 2t²]₀⁵ = (20×5 − 2×5²) − (0 − 0) = 100 − 50 = 50 m

0 m. Both methods confirm the result and reinforce the link between calculus and geometry.

另一种方法，利用三角形面积：底为 5 s，高为 20 m/s，面积 = ½ × 5 × 20 = 50 m。两种方法相互印证，强化了微积分与几何之间的联系。

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4. Matrices: Transforming a Shape in the Plane | 矩阵：平面图形的变换

A triangle has vertices A(2, 1), B(5, 1) and C(3, 4). Apply a rotation of 90° anticlockwise about the origin. Use matrix multiplication to find the new coordinates.

一个三角形的顶点坐标为 A(2, 1), B(5, 1), C(3, 4)。绕原点逆时针旋转 90°。使用矩阵乘法求出新坐标。

The rotation matrix for 90° anticlockwise is R = [0 −1; 1 0]. Each point is represented as a column vector and pre‑multiplied by R.

逆时针旋转 90° 的矩阵为 R = [0 −1; 1 0]。每个点用列向量表示，左乘矩阵 R。

For A(2, 1): [0 −1; 1 0] × [2; 1] = [0×2 + (−1)×1 ; 1×2 + 0×1] = [−1; 2]. So A' = (−1, 2). For B(5, 1): gives (−1, 5). For C(3, 4): gives (−4, 3).

对于 A(2, 1)：[0 −1; 1 0] × [2; 1] = [−1; 2]。故 A' = (−1, 2)。B(5, 1) 得 (−1, 5)。C(3, 4) 得 (−4, 3)。

Plot both triangles to verify the transformation. This case study demonstrates how matrix operations provide an efficient way to carry out geometric transformations.

绘制两个三角形以验证变换。这一案例展示了矩阵运算如何高效实现几何变换。

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5. Coordinate Geometry: Equation of a Tangent to a Circle | 坐标几何：圆的切线方程

− 6x + 10y − 2 = 0. Find the equation of the tangent at the point P(7, −3) on the circle.

圆 C 的方程为 x² + y² − 6x + 10y − 2 = 0。求过圆上点 P(7, −3) 的切线方程。

First rewrite in centre‑radius form by completing the square: (x − 3)² − 9 + (y + 5)² − 25 − 2 = 0 ⇒ (x − 3)² + (y + 5)² = 36. Centre is (3, −5).

首先用配方法化为标准形式：(x − 3)² − 9 + (y + 5)² − 25 − 2 = 0 ⇒ (x − 3)² + (y + 5)² = 36。圆心为 (3, −5)。

The radius CP has gradient m_CP (−5)) / (7 − 3) = 2 / 4 = ½. The tangent is perpendicular to the radius, so its gradient m_tangent = −1 / (½) = −2.

半径 CP 的斜率 m_CP = (−3 − (−5)) / (7 − 3) = 2/4 = ½。切线与半径垂直，故切线斜率 m_tangent = −1/(½) = −2。

Using point‑ ): y + 3 = −2(x − 7) → y = −2x + 14 − 3 → y = −2x + 11. The final equation is 2x + y = 11.

利用点斜式代入 P(7, −3)：y + 3 = −2(x − 7) → y = −2x + 14 − 3 → y = −2x + 11。最终方程为 2x + y = 11。

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6. Algebraic Fractions: Simplifying Before Integration | 代数分式：积分前化简

s (2x² + 7x − 4) / (x + 4) in simplified form and then find ∫ (2x² + 7x − 4) / (x + 4) dx.

将 (2x² + 7x − 4) / (x + 4) 化简，并求 ∫ (2x² + 7x − 4) / (x + 4) dx。

(2x − 1)(x + 4). Thus the fraction becomes (2x − 1)(x + 4) / (x + 4) = 2x − 1, for x ≠ −4.

将分子因式分解：2x² + 7x − 4 = (2x − 1)(x + 4)。于是分式变为 (2x − 1)(x + 4)/(x + 4) = 2x − 1，x ≠ −4。

x = x² − x + C. Recognising common factors before integrating avoids complicated division.

然后积分变得简单：∫ (2x − 1) dx = x² − x + C。在积分前识别公因子，可避免繁琐的除法。

If the numerator were not factorable, polynomial long division would be needed; this case highlights the value of algebraic insight.

若分子无法分解，则需使用多项式长除法；本例突显了代数观察力的重要性。

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7. Trigonometric Equations: Solving for 0° ≤ θ ≤ 360° | 三角方程：在 0° ≤ θ ≤ 360° 内求解

Solve sin 2θ = 0.5 for 0° ≤ θ ≤ 360°, giving answers to one decimal place.

在 0° ≤ θ ≤ 360° 范围内求解 sin 2θ = 0.5，答案精确至 0.1°。

X = 2θ, then sin X = 0.5. The principal solution is X = 30°, and the second solution in 0°‑360° is 180° − 30° = 150°. Since X = 2θ, the range for X is 0° to 720°.

令 X = 2θ，则 sin X = 0.5。基本解为 X = 30°，在 0°‑360° 内的第二个解为 180° − 30° = 150°。由于 X = 2θ，X 的范围为 0° 至 720°。

150°, 390°, 510°. Then θ = X / 2: θ = 15°, 75°, 195°, 255°. All are within 0° ≤ θ ≤ 360°.

通过加 360° 生成全部 X 值：X = 30°, 150°, 390°, 510°。于是 θ = X/2：θ = 15°, 75°, 195°, 255°。均在要求范围内。

Always check the domain of the original variable after solving the transformed equation. The step of expanding the range for 2θ is crucial to capture all solutions.

解完变换后的方程后，务必核对原变量的定义域。扩展 2θ 的范围是捕获全部解的关键步骤。

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8. Calculus: Classifying Stationary Points | 微积分：驻点分类

Find the stationary points of the curve y = x³ − 6x² + 9x + 2 and determine their nature.

求曲线 y = x³ − 6x² + 9x + 2 的驻点并判断其类型。

12x + 9. Set dy/dx = 0 → 3(x² − 4x + 3) = 0 → 3(x − 1)(x − 3) = 0, giving x = 1 and x = 3.

求导：dy/dx = 3x² − 12x + 9。令 dy/dx = 0 → 3(x² − 4x + 3) = 0 → 3(x − 1)(x − 3) = 0，得 x = 1 和 x = 3。

6. At x = 3: d²y/dx² = 18 − 12 = 6 (> 0), so a local minimum. y(3) = 27 − 54 + 27 + 2 = 2.

求二阶导数：d²y/dx² = 6x − 12。在 x = 1 处：d²y/dx² = 6 − 12 = −6（小于 0），故为局部极大值点，y(1) = 6。在 x = 3 处：d²y/dx² = 18 − 12 = 6（大于 0），故为局部极小值点，y(3) = 2。

Thus the stationary points are maximum at (1, 6) and minimum at (3, 2). This case reinforces the second derivative test without needing a sign table.

因此驻点为 (1, 6) 极大值和 (3, 2) 极小值。本例强化了无需符号表的二阶导数判定法。

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9. Vector Geometry: Proving Collinearity | 向量几何：证明三点共线

b = 5i − 3j and c = 11i − 11j. Prove A, B and C are collinear and find the ratio AB : BC.

点 A、B、C 的位置向量分别为 a = 2i + j, b = 5i − 3j, c = 11i − 11j。证明 A、B、C 共线，并求比值 AB : BC。

B = b − a = (5−2)i + (−3−1)j = 3i − 4j. Find BC = c − b = (11−5)i + (−11−(−3))j = 6i − 8j.

求向量 AB = b − a = 3i − 4j。求 BC = c − b = 6i − 8j。

of AB, the vectors are parallel and share point B, so A, B, C are collinear. The magnitude |BC| = 2|AB|, therefore AB : BC = 1 : 2.

观察到 BC = 2 × (3i − 4j) = 2 AB。由于 BC 是 AB 的标量倍数，且两向量共点 B，故 A、B、C 共线。又 |BC| = 2|AB|，因此 AB : BC = 1 : 2。

This example shows how vector methods elegantly handle collinearity and section ratio problems without coordinate geometry.

此例展示了向量方法如何优雅地解决共线与分割比问题，无需坐标几何。

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10. Probability: Conditional Probability with Tree Diagrams | 概率：树状图与条件概率

A bag contains 4 red and 5 blue discs. Two discs are drawn without replacement. Given that the first disc drawn is red, find the probability that the second disc is also red. Then find the probability that both discs are the same colour.

一个袋子装有 4 个红色和 5 个蓝色圆盘。依次不放回抽取两个圆盘。已知第一个抽到的是红色，求第二个也是红色的概率。并求两个圆盘颜色相同的概率。

iagram: first pick: P(R) = 4/9, P(B) = 5/9. If red first, remaining: 3 red, 5 blue, so P(R|R) = 3/8. The conditional probability asked is simply P(second red | first red) = 3/8.

画出树状图：第一次抽到红的概率为 4/9，蓝为 5/9。若第一次为红，剩下 3 红 5 蓝，故 P(第二个红 | 第一个红) = 3/8。所要求的条件概率即为 3/8。

: (5/9) × (4/8) = 20/72 = 5/18. Total same colour = 1/6 + 5/18 = 3/18 + 5/18 = 8/18 = 4/9.

计算同色概率：情况 RR：(4/9)×(3/8)=12/72=1/6。情况 BB：(5/9)×(4/8)=20/72=5/18。同色总和 = 1/6 + 5/18 = 3/18+5/18 = 8/18 = 4/9。

Tree diagrams systematically expose the changing probabilities in without‑replacement scenarios and reinforce the multiplication rule for dependent events.

树状图能系统地展示不放回场景中概率的变化，强化了相依事件的乘法法则。

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