📚 Case Study Hands-on Practice for Year 12 AQA Further Mathematics | Year 12 AQA 进阶数学：案例分析实战演练
In the AQA Year 12 Further Mathematics course, students encounter a wide range of advanced topics — from complex numbers and matrices to vectors, proof by induction, polar coordinates, and hyperbolic functions. Applying these concepts to real-world or exam-style scenarios is essential for deep understanding. This article presents a series of case studies, each designed as a focused practice problem that bridges theory and technique. Work through each example to sharpen your problem-solving skills and develop the logical thinking required for top grades.
在 AQA 12 年级进阶数学课程中，学生会接触到复数、矩阵、向量、数学归纳法证明、极坐标以及双曲函数等一系列高阶主题。将这些概念应用于实际问题或考试风格的场景对深入理解至关重要。本文提供一系列案例分析，每个案例都设计为一个聚焦的实战问题，将理论与解题技巧相连。请逐一完成每一道例题，以提升你的解题能力，并锻炼取得高分所需的逻辑思维。
1. Complex Numbers in Exponential Form and Circular Motion | 复数指数形式与圆周运动
moves in the complex plane such that its position at time t seconds is given by z(t) = 2eiπt/3 e the motion and find the position when t = 2, giving your answer in exact Cartesian form.
一粒子在复平面运动，t 秒时位置为 z(t) = 2eiπt/3。试描述运动情况，并求 t = 2 时的位置，结果以精确的笛卡尔形式表示。
The modulus |z| tant rate of π/3 rad/s, indicating anticlockwise uniform circular motion. At t = 2, z = 2ei·2π/3 = 2(cos(2π/3) + i sin(2π/3)). Since cos(2π/3) = −1/2 and sin(2π/3) = √3/2, the position is 2(−1/2 + i√3/2) = −1 + i√3.
模长 |z| = 2 恒定，因此粒子在以原点为圆心、半径为 2 的圆上运动。辐角 θ = πt/3 以恒定速率 π/3 rad/s 增加，表明为逆时针匀速圆周运动。t = 2 时，z = 2ei·2π/3 = 2(cos(2π/3) + i sin(2π/3))。cos(2π/3) = −1/2，sin(2π/3) = √3/2，故位置为 −1 + i√3。
2. Deriving Trigonometric Identities Using de Moivre’s Theorem | 用棣莫弗定理推导三角恒等式
terms of cos θ, and hence find an exact value for cos 36° given that cos 108° = −cos 72°.
利用棣莫弗定理将 cos 3θ 用 cos θ 表示，并由此已知 cos 108° = −cos 72°，求 cos 36° 的精确值。
By de Moivre, (cos θ + i sin θ)3 = cos 3θ + i sin 3θ. Expanding the left side using the binomial theorem gives cos3θ + 3i cos2θ sin θ − 3 cos θ sin2θ − i sin3θ. Equating real parts: cos 3θ = cos3θ − 3 cos θ sin2θ. Substitute sin2θ = 1 − cos2θ: cos 3θ = 4 cos3θ − 3 cos θ. Now let θ = 36°. Then cos 108° = 4 cos3 (180° − 72°) = −cos 72°. Also cos 72° = sin 18°, and cos 36° is linked: use cos 2θ = 2 cos2θ − 1. To find cos 36° explicitly, we can set x = cos 36° and note cos 72° = 2x2 − 1. Then cos 108° = −(2x2 − 1) = 1 − 2x2. Equate to 4x3 − 3x: 4x3 − 3x = 1 − 2x2 → 4x3 + 2x2 − 3x − 1 = 0. Factorising (x + 1)(4x2 − 2x − 1) = 0. Since x = cos 36° > 0 and not −1, solve 4x2 − 2x − 1 = 0, giving x = (2 ± √4 + 16)/8 = (2 ± √20)/8 = (1 ± √5)/4. cos 36° is positive, so cos 36° = (1 + √5)/4.
由棣莫弗定理，(cos θ + i sin θ)3 = cos 3θ + i sin 3θ。左端用二项式定理展开得 cos3θ + 3i cos2θ sin θ − 3 cos θ sin2θ − i sin3θ。比较实部：cos 3θ = cos3θ − 3 cos θ sin2θ。代入 sin2θ = 1 − cos2θ 得 cos 3θ = 4 cos3θ − 3 cos θ。令 θ = 36°，则 cos 108° = 4 cos336° − 3 cos 36°。而 cos 108° = cos(180° − 72°) = −cos 72°。又 cos 72° = sin 18°，可设 x = cos 36°，cos 72° = 2x2 − 1，于是 cos 108° = 1 − 2x2。令等式 4x3 − 3x = 1 − 2x2，整理得 4x3 + 2x2 − 3x − 1 = 0。因式分解为 (x + 1)(4x2 − 2x − 1) = 0，由于 x = cos 36° > 0 且不等于 −1，解 4x2 − 2x − 1 = 0 得 x = (1 ± √5)/4，正值给出 cos 36° = (1 + √5)/4。
3. Matrix Encryption and Decryption | 矩阵编码与解码
number (A=1, B=2, ..., Z=26) and then multiplying each pair of numbers by a 2×2 matrix M = [3 2] [1 4]. The encoded pairs are (26, 20), (45, 38) and (30, 22). Decode the original message.
一条消息通过将每个字母表示为数字 (A=1, B=2, ..., Z=26) 后，以 2×2 矩阵 M = [3 2] [1 4] 乘以每对数字进行编码。已编码的对为 (26, 20), (45, 38) 和 (30, 22)。请解码出原始消息。
be column vectors p. Encoding means M p = encoded vector. To decode, compute the inverse M−1. det M = 3×4 − 2×1 = 10. M−1 = (1/10) [4 −2] [−1 3]. Apply to each encoded pair. First: (1/10)[4 −2; −1 3](26, 20)T = (6.4, 3.4). This suggests the numbers must be integers 1-26; maybe I need to work mod 26? rhaps the matrix is just invertible over reals and we expect integer original numbers. Let' b=26 → 10b=34 → b=3.4 not integer. Indeed a realistic cipher would use mod 26. The problem likely expects mod 26 arithmetic. Let’s reinterpret: the encoded numbers might be given in plain, but we should work modulo 26. However the example states (26,20) etc. Could be that the plaintext letters were numbers 1-26, and the matrix product is taken without modulo, giving these numbers. But we got non-integers. Perhaps the matrix is different? Let' x+4y=20. Solve exactly: multiply second by 3: 3x+12y=60, subtract first: 10y=34, y=3.4, x=20−13.6=6.4, not integers. So maybe the encoding was done modulo 26, and the numbers given (26,20) are the residues mod 26, where 26 is actually 0 mod 26. Then we work mod 26. Let's adopt that approach. Then (26,20) mod 26 is (0,20). Similarly (45,38) ≡ (45−26=19, 38−26=12) = (19,12). (30,22) ≡ find M inverse mod 26: det=10, which is not coprime to 26 (gcd(10,26)=2), so inverse doesn' use M = [3 2; 1 5], det=15−2=13, gcd(13,26)=13, still not 1. Let’s change to M = [3 2; 1 3], det=9−2=7, gcd(7,26)=1, good. So I will adjust the case study accordingly and provide a consistent example. I'll redo the entire case with a proper scenario.
Let’s restructure: The matrix M = [5 2] [3 7]? det=35−6=29, gcd(29,26)=1. That works. To keep numbers small, use M = [4 1] [1 3], det=11. I'll craft: message "CODE" +1*5, 1*4+3*5) = (16+5,4+15) = (21,19). So encoded pairs (1,22) and (21,19). Provide these, and ask to decode. However the problem statement in the article should be fixed. I will rewrite this section with valid numbers and clear explanation, including modular arithmetic.
给定编码矩阵 M = [4 1] [1 3]，在模 26 下工作。已知两段编码对为 (1, 22) 和 (21, 19)。试求原始字母。
First, find M−1 3 − 1×1 = 11. Since gcd(11,26)=1, the inverse exists mod 26. Need a number k such that 11k ≡ 1 mod 26. 11×19=209, 209 mod 26 = 209 − 8×26 = 209−208=1, so k=19. M−1 = 19 × [3 −1] [−1 4] mod 26 = [57 −19] [−19 76] d 26. 57 mod 26 = 57−2×26=5; −19 mod 26 = 7 (since −19+26=7); 76 mod 26 = 76−2×26=24. So M−1 = [5 7] [7 24] mod 26. Apply to first pair (1,22): [5 7] [7 24] × (1,22)T d pair (21,19): (5×21+7×19, 7×21+24×19) = (105+133, 147+456) = (238,603) mod 26. 238 mod 26: 26×9=234, remainder 4; 603 mod 26: 26×23=598, remainder 5. Plain pair = (4,5) → D, E. Message: CODE.
首先求 M−1 在模 26 下。det M = 11，由于 gcd(11,26)=1，逆存在。需要满足 11k ≡ 1 mod 26，11×19=209，209 mod 26 = 1，故 k=19。M−1 = 19 × [3 −1] [−1 4] mod 26 = [5 7] [7 24] mod 26。作用于第一个对 (1,22) 得 (3,15)，对应 C 和 O。第二个对 (21,19) 得 (4,5)，对应 D 和 E。原消息为 CODE。
4. Invariant Lines Under a Matrix Transformation | 矩阵变换下的不变直线
A linear transformation T is defined by the matrix M = [2 1] [1 2]. Find any lines through the origin that are invariant under T, and state the scaling factor along each line.
线性变换 T 由矩阵 M = [2 1] [1 2] 定义。求所有过原点且在该变换下不变的直线，并指出每条直线上的缩放因子。
ge of a general point (x, mx) must lie on the same line, meaning its coordinates satisfy y' = m x' where (x', y') = M (x, mx)T. Compute: x' = 2x + mx = (2+m)x, y' = 1·x + 2·mx = (1+2m)x. The condition is y' = m x' → (1+2m)x = m(2+m)x. For x ≠ 0, 1+2m = 2m + m² → m² = 1 → m = ±1. For m = 1, line y = x: x' = 3x, y' = 3x, scaling factor 3. For m = −1, line y = −x: x' = (2−1)x = x, y' = (1−2)x = −x, scaling factor 1 (actually points are fixed? , -x) → (2x - x, x - 2x) = (x, -x), so factor 1). So invariant lines are y = x and y = −x.
设过原点的直线为 y = mx。不变性要求一般点 (x, mx) 的像仍在同一直线上，即变换后的坐标满足 y' = m x'，其中 (x', y') = M (x, mx)T。计算得 x' = (2+m)x，y' = (1+2m)x。代入 y' = m x' 得 1+2m = 2m + m² → m² = 1 → m = ±1。对 m=1，直线 y=x，缩放因子 3；对 m=−1，直线 y=−x，像点为 (x, −x)，缩放因子 1。故不变直线为 y=x 和 y=−x。
5. Shortest Distance from a Point to a Plane Using Vectors | 用向量求点到平面的最短距离
Find the shortest distance from the point P(2, −1, 3) to the plane Π: r ⋅ (i + 2j − 2k) = 5, and determine the coordinates of the foot of the perpendicular.
求点 P(2, −1, 3) 到平面 Π: r ⋅ (i + 2j − 2k) = 5 的最短距离，并求出垂足的坐标。
The plane has normal vector n = (1, 2, −2). Write P as position vector p. The distance d = |p⋅n − 5| / |n|. Compute p⋅n = 2×1 + (−1)×2 + 3×(−2) = 2 − 2 − 6 = −6. |n| = √(1²+2²+(−2)²) = √9 = 3. So d = |−6 − 5| / 3 = |−11| tor PQ is parallel to n. So Q = P + t n, where t chosen so that Q lies on plane: (p + t n)⋅n = 5 → p⋅n + t |n| hus Q = (2, −1, 3) + (11/9)(1,2,−2) = (2 + 11/9, −1 + 22/9, 3 − 22/9) = (29/9, 13/9, 5/9).
平面法向量 n = (1, 2, −2)。P 的位置向量为 p。距离 d = |p⋅n − 5| / |n|。计算 p⋅n = 2×1 + (−1)×2 + 3×(−2) = −6，|n| = 3，得 d = |−6−5|/3 = 11/3。垂足 Q 满足 PQ 平行于 n，设 Q = P + t n，代入平面方程 −6 + 9t = 5，解得 t = 11/9，从而 Q = (29/9, 13/9, 5/9)。
6. Proof by Induction: Summation of a Series | 数学归纳法证明：级数求和
Prove by induction that for all positive integers n, Σr=1n r(r+1) = (1/3)n(n+1)(n+2).
用数学归纳法证明对所有正整数 n，Σr=1n r(r+1) = (1/3)n(n+1)(n+2)。
LHS = 1×2 = 2. RHS = (1/3)×1×2×3 = 2. True. Inductive step: assume true for n = k, i.e., Σr=1k r(r+1) = (1/3)k(k+1)(k+2). For n = k+1, LHS = Σr=1k 1)(k+2)(k+3), which matches RHS for n=k+1. Hence by induction, the formula holds for all n ∈ ℕ.
奠基：n=1 时，左边 = 1×2 = 2，右边 = (1/3)×1×2×3 = 2，成立。归纳步骤：假设 n=k 时成立，即 Σr=1k r(r+1) = (1/3)k(k+1)(k+2)。则 n=k+1 时，左边加一项 (k+1)(k+2)，提出公因子得 (k+1)(k+2)(k/3 + 1) = (1/3)(k+1)(k+2)(k+3)，恰为 n=k+1 时的右边。因此由归纳法，公式对所有正整数 n 成立。
7. Area Enclosed by a Polar Curve | 极坐标曲线围成的面积
ve C is defined by r = 2 + cos θ for 0 ≤ θ ≤ 2π. Find the area of the region bounded by C.
极坐标曲线 C 由方程 r = 2 + cos θ，0 ≤ θ ≤ 2π 定义。求由 C 围成区域的面积。
Area A = (1/2) ∫02π r² dθ = (1/2) ∫02π om 0 to 2π: ∫ 9/2 dθ = (9/2)θ; ∫ 4cos θ dθ = 4 sin θ; ∫ (1/2)cos 2θ dθ = (1/4) sin 2θ. Evaluate between limits: sin terms vanish at multiples of 2π. So A = (1/2)[(9/2)×2π] = (1/2)×9π = (9π)/2.
面积 A = (1/2) ∫02π r² dθ = (1/2)∫ (4 + 4cos θ + cos²θ) dθ。利用 cos²θ = (1+cos 2θ)/2，被积函数化为 9/2 + 4cos θ + (1/2)cos 2θ。从 0 到 2π 积分，正弦项为零，得 A = (1/2) × (9/2)×(2π) = 9π/2。
8. Hyperbolic Functions: Solving Equations and Integration | 双曲函数：解方程与积分
= 4, giving your answers in logarithmic form. Additionally, evaluate ∫ sinh 2x cosh x dx.
解方程 5 cosh x − 3 sinh x = 4，答案用对数形式表示。并计算积分 ∫ sinh 2x cosh x dx。
First, express cosh x and sinh x in exponentials: cosh x = (ex + e−x)/2, sinh x = (ex − e−x)/2. Then 5(ex+e−x)/2 − 3(ex−e−x)/2 = 4 → (1/2)[5ex+5e−x−3ex+3e−x] = 4 → (1/2)(2ex+8e−x) = 4 → ex + 4e−x = 4. Multiply by ex: e2x − 4ex + 4 = 0 → (ex − 2)2 = 0 → ex ∫ u² du = (2/3)u³ + C = (2/3) cosh³ x + C. Alternatively, one can use hyperbolic identities.
先用指数式表达双曲函数，方程化为 (1/2)(2ex+8e−x) = 4 → ex + 4e−x = 4，乘以 ex 得 (ex − 2)² = 0，故 ex = 2，x = ln 2。积分部分，sinh 2x = 2 sinh x cosh x，故 ∫ 2 sinh x cosh²x dx，令 u = cosh x，得 2∫ u² du = (2/3)u³ + C = (2/3) cosh³ x + C。
9. Vector Equation of a Line and Intersection with a Plane | 直线的向量方程及其与平面的交点
. Find its Cartesian equations and the point of intersection with the plane x + y − z = 4.
一直线 L 过点 A(1, 2, −1)，方向向量 d = 2i − j + 3k。求其笛卡尔方程以及与平面 x + y − z = 4 的交点。
x+y−z=4: (1+2λ) + (2−λ) − (−1+3λ) = 4 → 1+2λ+2−λ+1−3λ = 4 → (4−2λ)=4 → −2λ=0 → λ=0. Thus intersection point is at A itself: (1,2,−
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