AS OCR Further Mathematics: Case Study Practical Exercises | AS OCR 进阶数学：案例分析实战演练

📚 AS OCR Further Mathematics: Case Study Practical Exercises | AS OCR 进阶数学：案例分析实战演练

In AS OCR Further Mathematics, applying theoretical concepts to real-world case studies deepens understanding and prepares students for complex problem-solving. This article presents a series of practical exercises covering core topics such as complex numbers, matrices, vectors, polynomials, inequalities, sequences, and proof by induction. Each case study demonstrates how abstract mathematical techniques are used to model and solve genuine challenges, from electrical engineering to finance and cryptography. By working through these examples, students will strengthen their analytical skills and exam readiness.

在 AS OCR 进阶数学中，将理论概念应用于实际案例分析能加深理解，并为解决复杂问题做好准备。本文展示一系列实战演练，涵盖复数、矩阵、向量、多项式、不等式、数列和归纳证明等核心主题。每个案例都展示了如何运用抽象数学技巧来建模并解决现实挑战，涉及电气工程、金融和密码学等领域。通过这些示例，学生将提升分析能力和考试应对水平。

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1. Complex Numbers in AC Circuit Analysis | 交流电路分析中的复数

ing complex numbers. Impedance, the AC analogue of resistance, is expressed as Z = R + i(X_L − X_C), where X_L = ωL and X_C = 1/(ωC). Phasor voltages and currents are then related by Ohm' ing magnitude and phase calculations. This approach avoids solving differential equations.

使用复数可以优雅地分析含有电阻、电感和电容的交流电路。阻抗是交流中的电阻类比，表示为 Z = R + i(X_L − X_C)，其中 X_L = ωL，X_C = 1/(ωC)。相量电压与电流通过欧姆定律 V = IZ 联系，可进行幅值和相位计算，避免了求解微分方程。

Case Study: pply of v(t) = 100 sin(100πt) volts. Determine the complex impedance and the peak current.

案例：一个串联 RLC 电路，R = 10 Ω，L = 5 mH，C = 100 μF，接在电压 v(t) = 100 sin(100πt) 伏的电源上。求复数阻抗和峰值电流。

Angular frequency ω = 100π rad/s.
X_L = ωL = 100π × 5×10⁻³ = 0.5π ≈ 1.571 Ω.
X_C = 1/(ωC) = 1/(100π × 100×10⁻⁶) = 1/(0.01π) = 100/π ≈ 31.831 Ω.
Thus Z = 10 + i(1.571 − 31.831) = 10 − i30.26 Ω.

角频率 ω = 100π rad/s。感抗 X_L = ωL = 100π × 5×10⁻³ = 0.5π ≈ 1.571 Ω；容抗 X_C = 1/(ωC) = 1/(100π × 100×10⁻⁶) = 1/(0.01π) = 100/π ≈ 31.831 Ω。因此 Z = 10 + i(1.571 − 31.831) = 10 − i30.26 Ω。

Impedance magnitude |Z| = √10² + 30.26² ≈ √100 + 915.7 = √1015.7 ≈ 31.87 Ω.
The peak current I₀ = V₀ / |Z| 00 / 31.87 ≈ 3.14 A. The current lags the voltage by phase φ = arctan(−30.26/10) ≈ −71.8°.

阻抗模 |Z| = √10² + 30.26² ≈ √100 + 915.7 = √1015.7 ≈ 31.87 Ω。峰值电流 I₀ = V₀ / |Z| = 100 / 31.87 ≈ 3.14 A。电流滞后电压角度 φ = arctan(−30.26/10) ≈ −71.8°。

Complex numbers turn a messy differential equation into simple algebra—perfect for handling frequency-dependent components.

复数将复杂的微分方程转化为简单的代数运算，是处理频率相关元件的理想工具。

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2. Matrix Encryption and Decryption | 矩阵加密与解密

o encrypt messages via linear transformations. Each letter is assigned a number (A=1, B=2, … encryption matrix yields the ciphertext. Decryption requires the inverse matrix modulo 26.

矩阵可通过线性变换加密消息。每个字母分配一个数字（A=1, B=2, …, Z=26），并分组为向量。乘以选定的加密矩阵得到密文。解密需要该矩阵的模26逆矩阵。

Case Study: Use the matrix M =

3	2
1	5

to encrypt the word "HELP". Then find the decryption matrix and recover the plaintext.

案例：使用矩阵 M =

3	2
1	5

加密单词 "HELP"，再求解密矩阵并恢复明文。

s vectors (8,5) and (12,16). Compute ciphertext: (8,5) × M = (8×3+5×1, 8×2+5×5) = (29, 41) ≡ (3,15) mod 26 → C,O. Second vector: (12×3+16×1, 12×2+16×5) = (52, 104) ≡ (0,0) mod 26? 6 = 0, giving (0,0). This suggests we need to adjust the scheme; typical mapping uses A=0,…,Z=25. Let's adopt A=0. Then HELP → 7,4,11,15. (7,4)·M = (7×3+4×1, 7×2+4×5) = (25, 34) ≡ (25,8) mod 26 → Z,I. (11,15)·M = (11×3+15×1, 11×2+15×5) = (48, 97) ≡ (22,19) mod 26 → W,T. Ciphertext: ZIWT.

字母转换：采用 A=0，则 H=7, E=4, L=11, P=15。分组向量 (7,4) 与 (11,15)。计算密文：(7,4)·M = (7×3+4×1, 7×2+4×5) = (25,34) ≡ (25,8) mod 26 → Z,I。(11,15)·M = (11×3+15×1, 11×2+15×5) = (48,97) ≡ (22,19) mod 26 → W,T。密文为 ZIWT。

Determinant of M: det = 3×5 − 2×1 = 13. Since 13 has inverse mod 26 only if gcd(13,26)=1? must have determinant coprime with 26. For demonstration, use a different matrix, say N =

5	3
2	7

, det = 5×7−3×2=29 ≡ 3 mod 26, and gcd(3,26)=1. Its inverse mod 26 is (3×9=27≡ he key lesson: matrices enable secure communication, but constraints on determinant apply.

M 的行列式为 13，与 26 不互质，没有模逆。正确的加密矩阵行列式应与 26 互质。例如改用 N =

5	3
2	7

，行列式=29≡3 mod 26，其逆为 9 乘伴随矩阵模26。该案例说明矩阵在安全通信中的应用及行列式约束的重要性。

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3. Vectors in Static Equilibrium | 静力平衡中的向量

When an object is in static equilibrium, the vector sum of all forces acting on it equals zero. Expressing forces in component form (i, j, k) allows unknown forces to be found algebraically.

物体处于静力平衡时，所有作用力的矢量和为零。将力用分量（i, j, k）表示，可通过代数方法求解未知力。

Case Study: A particle is acted upon by three forces: F₁ = 10i N, F₂ = −8i + 6j N. Find the third force F₃ required for equilibrium, and its magnitude.

案例：一质点受三个力作用：F₁ = 10i N，F₂ = −8i + 6j N。求平衡所需的第三个力 F₃ 及其大小。

Equilibrium condition: F₁ + F₂ + F₃ = 0 ⇒ F₃ = −(F₁ + F₂) = −(10i −8i + 6j) = −(2i + 6j) = −2i − 6j N.

平衡条件：F₁ + F₂ + F₃ = 0 ⇒ F₃ = −(F₁ + F₂) = −(10i −8i + 6j) = −(2i + 6j) = −2i − 6j N。

Magnitude |F₃| = √((−2)² + (−6)²) = √4+36 = √40 = 2√10 = arctan(3) ≈ 71.6° below the negative x-axis, i.e., 180°+71.6° = 251.6° from positive x.

大小为 |F₃| = √((−2)² + (−6)²) = √4+36 = √40 = 2√10 ≈ 6.32 N。方向角 θ = arctan((−6)/(−2)) = arctan(3) ≈ 71.6°，位于第三象限，与正x轴夹角 180°+71.6°=251.6°。

Vector equilibrium is fundamental in structural engineering, ensuring bridges and cranes remain stationary.

向量平衡在结构工程中至关重要，确保桥梁和起重机保持静止。

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4. Polynomial Roots in Break-Even Analysis | 盈亏平衡分析中的多项式根

Quadratic functions often model profit in business. The roots of the profit polynomial give the break-even points where revenue equals cost. Using Vieta's formulas, we can quickly link the sum and product of the roots to the coefficients.

二次函数常用来模拟商业利润。利润多项式的根给出了收入等于成本的盈亏平衡点。利用韦达定理，可以快速

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