Case Study Practical Exercises for CCEA Year 12 Further Mathematics | CCEA 进阶数学案例分析实战演练

📚 Case Study Practical Exercises for CCEA Year 12 Further Mathematics | CCEA 进阶数学案例分析实战演练

Case study exercises provide a powerful way to link abstract further mathematics topics with realistic, exam-style contexts. In this article, we work through ten carefully chosen problems that mirror the CCEA AS Further Mathematics specification. Each case breaks down a concept into manageable steps, pairing clear English reasoning with Chinese translations to reinforce understanding. By practising these model answers, you will build both procedural fluency and the analytical skills needed to tackle unfamiliar questions under timed conditions.

案例分析实战演练是将抽象的进阶数学主题与真实考试情境联系起来的有效方法。本文精选了十个与 CCEA AS 进阶数学大纲高度匹配的典型问题。每个案例将一个概念分解为可操作的步骤，用清晰的英语推理解释并配以汉语翻译，以加深理解。通过练习这些标准答案，你不仅能提高解题熟练度，还能培养在限时条件下应对陌生题目所需的分析能力。

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1. Complex Loci on the Argand Diagram | 阿根图中的复数轨迹

A point P represents a complex number z such that |z − (3 + 4i)| e in the Argand diagram. The centre of the circle is the point representing the complex number 3 + 4i, and the radius is 5 units.

点 P 表示复数 z，满足条件 |z − (3 + 4i)| = 5。这一条件在阿根图中表示一个圆，圆心为表示复数 3 + 4i 的点，半径为 5 个单位。

To find the Cartesian equation of the locus, we let z = x + iy. Then |(x + iy) − (3 + 4i)| = 5 gives |(x − 3) + i(y − 4)| = 5. Squaring both sides yields (x − 3)² + (y − 4)² = 25.

为求出轨迹的笛卡尔方程，设 z = x + iy。则 |(x + iy) − (3 + 4i)| = 5 化为 |(x − 3) + i(y − 4)| = 5。两边平方得到 (x − 3)² + (y − 4)² = 25。

ne from the origin making an angle of 45° with the positive real axis, so y = x for x > 0. Substituting y = x into the circle equation gives (x − 3)² + (x − 4)² = 25. Expanding and simplifying leads to 2x² − 14x = 0, so x = 0 or x = 7. Since x must be positive, the intersection point is z = 7 + 7i.

如果同时附加条件 arg(z) = π/4，我们可以求出交点。arg(z) = π/4 表示 z 位于从原点出发、与正实轴成 45° 角的射线上，因此对于 x > 0 有 y = x。将 y = x 代入圆的方程得到 (x − 3)² + (x − 4)² = 25。展开并化简得 2x² − 14x = 0，因此 x = 0 或 x = 7。由于 x 必须为正，交点为 z = 7 + 7i。

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2. Matrix Transformations Representing a Reflection | 表示反射的矩阵变换

nd j = (0, 1). Under reflection in y = x, i becomes (0, 1) and j becomes (1, 0). Thus the matrix is composed of these image vectors as columns, giving M = [0 1; 1 0].

线性变换 T 将点关于直线 y = x 进行反射。为求出表示该变换的 2×2 矩阵，我们考虑单位向量 i = (1, 0) 和 j = (0, 1) 的像。在关于 y = x 的反射下，i 变为 (0, 1)，j 变为 (1, 0)。因此，矩阵的列由这些像向量构成，即 M = [0 1; 1 0]。

tes. For example, A′ = M × [1; 2] = [2; 1]. The images are A′(2, 1), B′(2, 3), C′(5, 3), D′(5, 1). The original rectangle has been reflected into a congruent rectangle across the line y = x.

现在将 T 应用于顶点为 A(1, 2)、B(3, 2)、C(3, 5) 和 D(1, 5) 的四边形。将每个位置向量与矩阵 M 相乘，相当于交换 x 和 y 坐标。例如，A′ = M × [1; 2] = [2; 1]。像点依次为 A′(2, 1)、B′(2, 3)、C′(5, 3)、D′(5, 1)。原矩形被反射为关于直线 y = x 对称的全等矩形。

We can also verify that the determinant of M is −1, confirming that the transformation is an indirect isometry (a reflection). The area of the shape remains unchanged under reflection.

我们还能验证矩阵 M 的行列式为 −1，证实该变换是一个间接等距变换（反射）。图形面积在反射下保持不变。

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3. Cubic Equation with Complex Conjugate Roots | 具有共轭复根的三次方程

other two roots. Since the coefficients are real, complex roots occur in conjugate pairs. We can divide the cubic by (x − 2) to obtain the remaining quadratic factor.

考虑三次方程 x³ − 4x² + 6x − 4 = 0。已知一个实数根为 x = 2，求另外两个根。由于系数均为实数，复数根一定以共轭对的形式出现。我们可以用因式 (x − 2) 去除三次式，得到剩余的二次因式。

s quadratic equal to zero, x² − 2x + 2 = 0, and using the quadratic formula gives x = [2 ± √4 − 8] / 2 = 1 ± i. So the three roots are 2, 1 + i, and 1 − i.

进行除法运算：(x³ − 4x² + 6x − 4) ÷ (x − 2) = x² − 2x + 2。令此二次式等于零，x² − 2x + 2 = 0，利用二次公式求得 x = [2 ± √4 − 8] / 2 = 1 ± i。因此三个根为 2、1 + i 和 1 − i。

ls −(−4)/1 = 4. Indeed, 2 + (1+i) + (1−i) = 4, which matches the coefficient relationship.

验证根的和等于 −(−4)/1 = 4 是有益的。事实上，2 + (1+i) + (1−i) = 4，这与系数的关系一致。

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4. Summation of a Finite Series Using Standard Results | 利用标准结果求有限级数的和

Evaluate the sum Σ_r=1ⁿ ing the standard formulae Σr = n(n+1)/2 and Σr² = n(n+1)(2n+1)/6, we obtain a closed form.

计算和式 Σ_r=1ⁿ r(r+1)。该表达式可展开为 Σ(r² + r) = Σr² + Σr。利用标准公式 Σr = n(n+1)/2 和 Σr² = n(n+1)(2n+1)/6，我们可以得到封闭形式。

Thus Σ_r=1ⁿ +1)/2. Factorising n(n+1)/6 gives (n(n+1)/6)[(2n+1) + 3] = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3.

因此 Σ_r=1ⁿ r(r+1) = n(n+1)(2n+1)/6 + n(n+1)/2。提取因子 n(n+1)/6 得到 (n(n+1)/6)[(2n+1) + 3] = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3。

11 × 12 / 3 = 440. This technique is often needed in proof by induction problems as well.

当 n = 10 时，其和为 10 × 11 × 12 / 3 = 440。这种方法在归纳法证明题中也经常用到。

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5. Maclaurin Series Expansion of a Rational Function | 有理函数的麦克劳林级数展开

³. We can start with the geometric series expansion: 1/(1 + x) = 1 − x + x² − x³ + ... for |x| g both sides gives −1/(1 + x)² = −1 + 2x − 3x² + ..., so f(x) = 1 − 2x + 3x² − 4x³ + ... .

求函数 f(x) = 1/(1+x)² 在 x³ 项以内的麦克劳林级数。我们可以从几何级数展开开始：对于 |x| < 1，有 1/(1+x) = 1 − x + x² − x³ + ...。两边求导得到 −1/(1+x)² = −1 + 2x − 3x² + ...，因此 f(x) = 1 − 2x + 3x² − 4x³ + ... 。

e the binomial expansion for negative integer powers: (1 + x)⁻² = 1 + (−2)x + (−2)(−3)x²/2! + (−2)(−3)(−4)x³/3! + ... which simplifies to 1 − 2x + 3x² − 4x³. The series converges for |x| < 1.

另一种方法是利用负整数指数的二项式展开：(1+x)⁻² = 1 + (−2)x + (−2)(−3)x²/2! + (−2)(−3)(−4)x³/3! + ...，化简后得到 1 − 2x + 3x² − 4x³。该级数在 |x| < 1 时收敛。

.1, f(0.1) ≈ 1 − 0.2 + 0.03 − 0.004 = 0.826, which is close to the exact value 0.826446...

此展开式可用于小量 x 时的函数近似；例如，当 x = 0.1 时，f(0.1) ≈ 1 − 0.2 + 0.03 − 0.004 = 0.826，与精确值 0.826446... 非常接近。

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6. Solving Equations with Hyperbolic Functions | 求解含双曲函数的方程

x + 4 cosh x = 5. We can rewrite the hyperbolic functions in exponential form: sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. Substituting these yields 3(eˣ − e⁻ˣ)/2 + 4(eˣ + e⁻ˣ)/2 = 5, which simplifies to (7eˣ + e⁻ˣ)/2 = 5.

解方程 3 sinh x + 4 cosh x = 5。我们可以将双曲函数写成指数形式：sinh x = (eˣ − e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2。代入后得到 3(eˣ − e⁻ˣ)/2 + 4(eˣ + e⁻ˣ)/2 = 5，化简得 (7eˣ + e⁻ˣ)/2 = 5。

Multiplying through by 2 and rearranging: 7eˣ + e⁻ˣ = 10. Multiply by eˣ: 7e²ˣ + 1 = 10eˣ, giving the quadratic 7e²ˣ − 10eˣ + 1 = 0. Let y = eˣ, then 7y² − 10y + 1 = 0. Solving, y = (10 ± √100 − 28)/14 = (10 ± √72)/14 = (10 ± 6√2)/14 = (5 ± 3√2)/7.

两边乘以 2 并整理：7eˣ + e⁻ˣ = 10。两边同乘 eˣ：7e²ˣ + 1 = 10eˣ，得到二次方程 7e²ˣ − 10eˣ + 1 = 0。设 y = eˣ，则 7y² − 10y + 1 = 0。解得 y = (10 ± √100 − 28)/14 = (10 ± √72)/14 = (10 ± 6√2)/14 = (5 ± 3√2)/7。

Since eˣ > 0, we take the positive root: y = (5 + 3√2)/7. Hence x = ln((5 + 3√2)/7). The other root (5 − 3√2)/7 is positive but smaller; however, only one solution yields a real x that satisfies the original equation after checking domain.

由于 eˣ > 0，我们取正根：y = (5+3√2)/7。因此 x = ln((5+3√2)/7)。另一根 (5−3√2)/7 虽为正数但较小，不过经检验定义域后仅有一个解满足原方程。

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7. First-Order Linear Differential Equation with Integrating Factor | 一阶线性微分方程与积分因子

Solve dy/dx + 2xy = x, given y(0) = 1. The integrating factor is e^{∫2x dx} = e^x². Multiply the entire equation by this factor: e^x² dy/dx + 2x e^x² y = x e^x². The left-hand side is the derivative of (y e^x²), so d/dx (y e^x²) = x e^x².

求解 dy/dx + 2xy = x，已知 y(0) = 1。积分因子为 e^{∫2x dx} = e^x²。用该因子乘以整个方程：e^x² dy/dx + 2x e^x² y = x e^x²。左边是 (y e^x²) 的导数，因此 d/dx (y e^x²) = x e^x²。

Integrate both sides with respect to x: y e^x² = ∫ x e^x² dx. Let u = x², du = 2x dx, so ∫ x e^x² dx = ½ ∫ e^u du = ½ e^x² + C. Thus y e^x² = ½ e^x² + C, giving y = ½ + C e^−x².

两边对 x 积分：y e^x² = ∫ x e^x² dx。令 u = x²，du = 2x dx，则 ∫ x e^x² dx = ½ ∫ e^u du = ½ e^x² + C。因此 y e^x² = ½ e^x² + C，得到 y = ½ + C e^−x²。

the initial condition: y(0) = ½ + C = 1, so C = ½. The particular solution is y = ½ + ½ e^−x². This function tends to ½ as x → ∞.

应用初始条件：y(0) = ½ + C = 1，故 C = ½。特解为 y = ½ + ½ e^−x²。当 x → ∞ 时，该函数趋近于 ½。

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8. Area Enclosed by a Polar Curve | 极坐标曲线围成的面积

ver a full cycle. Because of symmetry, we can integrate from 0 to π and double the result.

求心脏线 r = 2(1 + cos θ) 所围成的面积。极坐标下的面积公式为 A = ½ ∫ r² dθ，在一个完整周期上积分。由对称性，我们可以从 0 到 π 积分再将结果加倍。

ave r² = 4(1 + 2cos θ + ½ + ½ cos 2θ) = 4(3/2 + 2cos θ + ½ cos 2θ) = 6 + 8cos θ + 2cos 2θ.

计算 r² = 4(1 + cos θ)² = 4(1 + 2cos θ + cos²θ)。利用恒等式 cos²θ = (1 + cos 2θ)/2，得到 r² = 4(1 + 2cos θ + ½ + ½ cos 2θ) = 4(3/2 + 2cos θ + ½ cos 2θ) = 6 + 8cos θ + 2cos 2θ。

× 2 × ∫₀^π r² dθ = ∫₀^π r² dθ = 6π. Wait, careful: A = ½ ∫₀^(2π) r² dθ = ½ × 12π = 6π. Yes, the enclosed area is 6π square units.

从 0 到 π 积分：∫₀^π (6 + 8cos θ + 2cos 2θ) dθ = [6θ + 8sin θ + sin 2θ]₀^π = 6π。那么总面积为 A = ½ ∫₀^(2π) r² dθ = ½ × 12π = 6π。因此所围面积为 6π 平方单位。

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9. Vector Equation of a Plane and Shortest Distance | 平面的向量方程与最短距离

The points A(1, 2, −1), B(3, 0, 2) and C(0, −1, 4) lie in a plane Π. Find a vector equation of the plane and the shortest distance from the point P(5, 1, 3) to Π.

点 A(1, 2, −1)、B(3, 0, 2) 和 C(0, −1, 4) 位于平面 Π 内。求该平面的向量方程以及点 P(5, 1, 3) 到 Π 的最短距离。

(−2)(−1) ) = i(−10 + 9) − j(10 + 3) + k(−6 − 2) = (−1, −13, −8). We can use n = (1, 13, 8) by multiplying by −1 for simplicity.

平面内的方向向量为 AB = (2, −2, 3) 和 AC = (−1, −3, 5)。法向量 n = AB × AC = i( (−2)(5) − (3)(−3) ) − j( (2)(5) − (3)(−1) ) + k( (2)(−3) − (−2)(−1) ) = i

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