📚 Case Study Analysis and Hands-on Practice | 案例分析实战演练
In Further Mathematics, applying concepts to real-world or abstract problems is essential for deep understanding. This article presents a series of mini case studies that walk you through key topics in the Eduqas Year 10 syllabus, showing step-by-step solutions to build your problem-solving skills.
在进阶数学中，将概念应用于实际或抽象问题是深入理解的关键。本文通过一系列小型案例分析，带你梳理 Eduqas 十年级考纲中的重要主题，逐步展示解题过程，提升你的解题能力。
1. Combining Matrix Transformations | 矩阵变换的组合
A triangle has vertices at A(1, 0), B(0, 2), and C(−1, 1). First, it is reflected in the y-axis, then stretched by a factor of 2 parallel to the x-axis. Find the coordinates of the final image and the single transformation matrix that would directly map the original triangle to the final image.
一个三角形的顶点为 A(1, 0)、B(0, 2) 和 C(−1, 1)。首先将三角形关于 y 轴反射，然后沿 x 轴方向作倍数为 2 的拉伸。求最终像的坐标，并写出将原三角形直接映射到最终像的单个变换矩阵。
Reflection in the y-axis is given by M = [-1 0; 0 1]. Stretch parallel to the x-axis by factor 2 is N = [2 0; 0 1]. Since we apply the reflection first then the stretch, the combined matrix is P = NM = [2 0; 0 1] × [-1 0; 0 1] = [-2 0; 0 1].
关于 y 轴的反射变换矩阵为 M = [-1 0; 0 1]。沿 x 轴方向倍数为 2 的拉伸为 N = [2 0; 0 1]。由于先反射再拉伸，复合矩阵为 P = NM = [2 0; 0 1] × [-1 0; 0 1] = [-2 0; 0 1]。
Now apply P to each vertex vector. For A(1,0): [-2 0; 0 1] [1; 0] = [-2; 0], so A′(−2, 0). For B(0,2): product gives [0; 2], so B′(0, 2). For C(−1,1): [-2 0; 0 1] × [-1; 1] = [2; 1], so C′(2, 1). The single matrix for the overall transformation is P = [-2 0; 0 1].
现在将 P 作用于每个顶点向量。对于 A(1,0)：[-2 0; 0 1] [1; 0] = [-2; 0]，得 A′(−2, 0)。对于 B(0,2)：相乘得 [0; 2]，得 B′(0, 2)。对于 C(−1,1)：[-2 0; 0 1] × [-1; 1] = [2; 1]，得 C′(2, 1)。实现整个变换的单个矩阵是 P = [-2 0; 0 1]。
2. Factor and Remainder Theorems | 因式定理与余数定理
2). Then, for g(x) = x³ + ax² − 4x − 12, if (x + 2) is a factor, determine the value of a.
已知多项式 f(x) = 2x³ − 5x² + 3x + 7，求其除以 (x − 2) 时的余数。另外，对于 g(x) = x³ + ax² − 4x − 12，若 (x + 2) 是其因式，求 a 的值。
f(2). Compute f(2) = 2(8) − 5(4) + 3(2) + 7 = 16 − 20 + 6 + 7 = 9. So the remainder is 9.
由余数定理，余数等于 f(2)。计算 f(2) = 2(8) − 5(4) + 3(2) + 7 = 16 − 20 + 6 + 7 = 9。因此余数为 9。
bstitute x = −2: (−2)³ + a(−2)² − 4(−2) − 12 = −8 + 4a + 8 − 12 = 4a − 12. Set 4a − 12 = 0 ⇒ a = 3. The polynomial becomes g(x) = x³ + 3x² − 4x − 12.
因为 (x + 2) 是因式，根据因式定理有 g(−2) = 0。代入 x = −2： (−2)³ + a(−2)² − 4(−2) − 12 = −8 + 4a + 8 − 12 = 4a − 12。令 4a − 12 = 0 ⇒ a = 3。此时多项式为 g(x) = x³ + 3x² − 4x − 12。
3. Proving Trigonometric Identities | 证明三角恒等式
Prove that (sin θ + cos θ)² + (sin θ − cos θ)² ≡ 2. Then, deduce the value of (sin 30° + cos 30°)² + (sin 30° − cos 30°)² without using a calculator.
证明 (sin θ + cos θ)² + (sin θ − cos θ)² ≡ 2。然后不用计算器，推算出 (sin 30° + cos 30°)² + (sin 30° − cos 30°)² 的值。
θ)² = sin²θ + 2 sin θ cos θ + cos²θ, and (sin θ − cos θ)² = sin²θ − 2 sin θ cos θ + cos²θ.
将每个平方展开： (sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ，以及 (sin θ − cos θ)² = sin²θ − 2 sin θ cos θ + cos²θ。
sin²θ + cos²θ = 1 for any angle θ, the sum simplifies to 2. Hence the identity is proved.
将两式相加： 2 sin θ cos θ 项抵消，剩下 2 sin²θ + 2 cos²θ = 2(sin²θ + cos²θ)。由于对任意角 θ 有 sin²θ + cos²θ = 1，总和化简为 2。因此该恒等式得证。
on automatically equals 2, irrespective of the sine and cosine values. So the result is 2.
当 θ = 30° 时，该表达式自动等于 2，与正弦和余弦的具体值无关。因此结果为 2。
4. Recurrence Relations and Sequences | 递推关系与数列
ₙ₊₁ = 2uₙ − 3. Find the value of u₅, and describe the long-term behaviour of the sequence.
一个数列由 u₁ = 4 和 uₙ₊₁ = 2uₙ − 3 定义。求 u₅ 的值，并描述该数列的长期变化趋势。
5 − 3 = 10 − 3 = 7. u₄ = 2×7 − 3 = 14 − 3 = 11. u₅ = 2×11 − 3 = 22 − 3 = 19. Thus u₅ = 19.
我们逐项计算： u₁ = 4。u₂ = 2×4 − 3 = 8 − 3 = 5。u₃ = 2×5 − 3 = 10 − 3 = 7。u₄ = 2×7 − 3 = 14 − 3 = 11。u₅ = 2×11 − 3 = 22 − 3 = 19。因此 u₅ = 19。
Looking at the pattern, each term increases and the differences between successive terms (1, 2, 4, 8) double each time. As n grows, uₙ grows without bound, so the sequence diverges to infinity.
观察模式，每一项都在增大，且相邻项的差（1、2、4、8）每次翻倍。随着 n 增大，uₙ 无限增大，因此该数列发散至无穷。
5. Finding Gradients and Tangents with Differentiation | 用导数求斜率和切线
− 7. Find the gradient of C at x = 1, and hence the equation of the tangent at that point.
曲线 C 的方程为 y = 4x³ − 2x² + 5x − 7。求曲线在 x = 1 处的斜率，并由此写出该点处的切线方程。
e is y = 4(1)³ − 2(1)² + 5(1) − 7 = 4 − 2 + 5 − 7 = 0. So the point of tangency is (1, 0).
求导得： dy/dx = 12x² − 4x + 5。在 x = 1 处，斜率 m = 12(1)² − 4(1) + 5 = 12 − 4 + 5 = 13。y 坐标为 y = 4(1)³ − 2(1)² + 5(1) − 7 = 4 − 2 + 5 − 7 = 0。因此切点为 (1, 0)。
radient form, the tangent equation is y − 0 = 13(x − 1), which simplifies to y = 13x − 13.
利用点斜式，切线方程为 y − 0 = 13(x − 1)，化简得 y = 13x − 13。
6. Vector Proofs: Collinearity and Ratios | 向量证明：共线与比例
Three points are P(2, 3), Q(5, 9) and R(8, 15). Using vectors, prove that P, Q and R are collinear, and find the ratio PQ : QR.
有三个点 P(2, 3)、Q(5, 9) 和 R(8, 15)。利用向量证明 P、Q、R 三点共线，并求比值 PQ : QR。
6). Both vectors are identical, meaning they are parallel and they share the point Q, so P, Q and R lie on the same straight line.
构造方向向量： PQ = (5−2, 9−3) = (3, 6)，QR = (8−5, 15−9) = (3, 6)。两个向量完全相同，说明它们平行且共享点 Q，因此 P、Q、R 三点共线。
Since PQ = QR, the distances between consecutive points are equal. Hence PQ : QR = 1 : 1.
由于 PQ = QR，相邻两点间的距离相等。因此 PQ : QR = 1 : 1。
7. Binomial Expansion and Approximation | 二项式展开与近似计算
Expand (1 + 2x)⁴ in ascending powers of x up to and including the term in x⁴. Use your expansion to estimate (1.02)⁴.
将 (1 + 2x)⁴ 按 x 的升幂展开，写出直到 x⁴ 项的展开式。利用展开式估算 (1.02)⁴。
⁴C₃(2x)³ + ⁴C₄(2x)⁴ = 1 + 4(2x) + 6(4x²) + 4(8x³) + 1(16x⁴) = 1 + 8x + 24x² + 32x³ + 16x⁴.
利用二项式定理， ⁴C₃(2x)³ + ⁴C₄(2x)⁴ = 1 + 4(2x) + 6(4x²) + 4(8x³) + 1(16x⁴) = 1 + 8x + 24x² + 32x³ + 16x⁴。
To approximate (1.02)⁴, set 1 + 2x = 1.02, so 2x = 0.02 ⇒ 32 + 0.00000016 = 1.08243216. Thus (1.02)⁴ ≈ 1.082432, which matches the calculator value.
要估算 (1.02)⁴，令 1 + 2x = 1.02，得 2x = 0.02 ⇒ x = 0.01。代入展开式： ) + 32(0.000001) + 16(0.00000001) = 1 + 0.08 + 0.0024 + 0.000032 + 0.00000016 = 1.08243216。因此 (1.02)⁴ ≈ 1.082432，与计算器结果一致。
8. Solving Trigonometric Equations | 解三角方程
Published by TutorHao | Year 10 进阶数学 Revision Series | aleveler.com
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