📚 Worked Solutions for AQA A-Level Further Pure 2 | AQA A-Level 进阶纯数 2 习题精解
Further Pure 2 (FP2) is a demanding module that pushes your algebraic manipulation and problem-solving skills to a new level. Mastering it requires more than just memorising formulas; you need to see how they are applied in complex, multi-step scenarios. This guide provides fully worked solutions to representative FP2 exercise types, breaking down the methods so you can replicate them in your exam.
进阶纯数 2 (FP2) 是一门要求很高的模块，极其考验你的代数运算和问题解决能力。掌握它不仅需要记住公式，更需要理解如何在复杂、多步骤的题目中灵活运用。本指南提供了 FP2 典型习题的完整解答，逐步拆解方法，助你在考试中举一反三。
1. Complex Numbers: De Moivre’s Theorem | 复数：棣莫弗定理
Exercise: Simplify (cos 3θ + i sin 3θ)⁴ using De Moivre’s theorem.
题目：使用棣莫弗定理化简 (cos 3θ + i sin 3θ)⁴。
De Moivre’s theorem is a fundamental tool for dealing with powers and roots of complex numbers. It states that for any integer n, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. The key is to treat the expression inside the brackets as a single complex number r(cos α + i sin α) and multiply the angle by the power.
棣莫弗定理是处理复数幂与根的基本工具。它指出对于任何整数 n，有 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。关键在于将括号内的表达式视为一个整体，即 r(cos α + i sin α) 的形式，然后将角度乘以幂指数。
(cos 3θ + i sin 3θ)⁴ = cos(4 × 3θ) + i sin(4 × 3θ) = cos 12θ + i sin 12θ
Thus, the expression simplifies to cos 12θ + i sin 12θ. Be careful to distribute the power correctly; a common mistake is to forget to multiply the angle by the power.
因此，原式化简为 cos 12θ + i sin 12θ。注意要将指数正确分配，一个常见错误是忘记将角度与幂相乘。
2. Roots of Polynomial Equations | 多项式方程的根
Exercise: Given that α, β, γ are the roots of x³ – 3x² + 4 = 0, find the value of α² + β² + γ².
题目：已知 α, β, γ 是方程 x³ – 3x² + 4 = 0 的根，求 α² + β² + γ² 的值。
For a cubic equation of the form ax³ + bx² + cx + d = 0, the relationships between the roots and coefficients are: Σα = -b/a, Σαβ = c/a, and αβγ = -d/a. Here, a = 1, b = -3, c = 0, and d = 4. The identity α² + β² + γ² = (α + β + γ)² – 2(αβ + βγ + γα) connects the sum of squares to the elementary symmetric sums.
对于形如 ax³ + bx² + cx + d = 0 的三次方程，根与系数之间存在如下关系：Σα = -b/a，Σαβ = c/a，以及 αβγ = -d/a。在此题中，a = 1, b = -3, c = 0, d = 4。恒等式 α² + β² + γ² = (α + β + γ)² – 2(αβ + βγ + γα) 将平方和与基本对称和联系起来。
Σα = 3, Σαβ = 0, αβγ = -4
α² + β² + γ² = (α + β + γ)² – 2(αβ + βγ + γα) = 3² – 2(0) = 9
Notice that because the x term is missing, the sum of the pairwise products Σαβ is zero, which simplifies the calculation significantly.
请注意，由于原方程不含 x 项，所以两两乘积之和 Σαβ 为零，这极大地简化了计算。
3. Summation of Series: Method of Differences | 级数求和：差分法
Exercise: Use the method of differences to find the sum of the series Σᵣ₌₁ⁿ 1/(r(r+1)).
题目：使用差分法求级数 Σᵣ₌₁ⁿ 1/(r(r+1)) 的和。
The method of differences, or telescoping, relies on expressing a term as the difference of two consecutive terms of another sequence. This causes most terms in the sum to cancel out. We first decompose the general term 1/(r(r+1)) using partial fractions.
差分法，又称裂项相消法，其核心思想是将通项拆分成另一个序列中相邻两项之差，使得求和时大部分项相互抵消。我们首先对通项 1/(r(r+1)) 进行部分分式分解。
1/(r(r+1)) = 1/r – 1/(r+1)
Writing out the sum: (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1)). All intermediate terms cancel, leaving the first and last terms.
将求和式展开： (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1))。所有中间项均相互抵消，仅剩首项和末项。
Σᵣ₌₁ⁿ 1/(r(r+1)) = 1 – 1/(n+1) = n/(n+1)
This technique is highly efficient for rational functions where the denominator factors into consecutive linear terms.
对于分母可分解为连续线性因子的有理函数，此方法尤为高效。
4. Maclaurin Series Expansions | 麦克劳林级数展开
Exercise: Find the Maclaurin series for eˣ sin x up to and including the x³ term.
题目：求 eˣ sin x 的麦克劳林级数，展开到含 x³ 的项。
The Maclaurin series for eˣ and sin x are standard and can be multiplied together. We only need terms up to x³, so we ignore any terms of degree 4 or higher. The series are: eˣ = 1 + x + x²/2 + x³/6 + … and sin x = x – x³/6 + …
eˣ 和 sin x 的麦克劳林级数都是标准形式，可直接相乘。我们只需要最高三次项，因此忽略四次及更高次的项。具体级数为：eˣ = 1 + x + x²/2 + x³/6 + … 和 sin x = x – x³/6 + …
eˣ sin x = (1 + x + x²/2 + x³/6)(x – x³/6)
Multiplying out and collecting the relevant terms: The x term is x. The x² term is x². The x³ term comes from 1 × (-x³/6) plus (x²/2) × x, which equals -x³/6 + x³/2 = x³/3.
展开并合并同类项：一次项为 x。二次项为 x²。三次项来自 1 × (-x³/6) 加上 (x²/2) × x，即 -x³/6 + x³/2 = x³/3。
eˣ sin x = x + x² + x³/3 + …
Always be meticulous when multiplying series; it is easy to miss cross-terms that contribute to a specific power of x.
在级数相乘时要格外细致，很容易遗漏那些会贡献到特定 x 幂次的交叉项。
5. Polar Coordinates: Area Enclosed | 极坐标：求封闭区域面积
Exercise: Find the area enclosed by the curve r = a(1 + cos θ).
题目：求曲线 r = a(1 + cos θ) 所围成的面积。
This curve is a cardioid, and the formula for the area enclosed by a polar curve is A = ½ ∫ₐᵇ r² dθ. Because the curve is symmetrical about the initial line, we can integrate from 0 to 2π directly, or from 0 to π and double. Here, integrating from 0 to 2π is straightforward.
该曲线是心形线，极坐标曲线所围面积的公式为 A = ½ ∫ₐᵇ r² dθ。由于该曲线关于极轴对称，我们可以直接积分从 0 到 2π，或者从 0 到 π 再乘以 2。此处，直接积分 0 到 2π 即可。
A = ½ ∫₀²π a²(1 + cos θ)² dθ = ½ a² ∫₀²π (1 + 2cos θ + cos²θ) dθ
We use the identity cos²θ = (1 + cos 2θ)/2 to integrate cos²θ. The integral of 1 over 0 to 2π is 2π. The integral of 2cos θ is 0, and the integral of cos²θ is π.
我们使用恒等式 cos²θ = (1 + cos 2θ)/2 来对 cos²θ 进行积分。1 在 0 到 2π 上的积分为 2π。2cos θ 的积分为 0，而 cos²θ 的积分为 π。
A = ½ a² (2π + 0 + π) = 3πa²/2
Hence, the area of the cardioid is 3πa²/2 square units. Remember to square the function r, not just substitute it directly.
因此，心形线的面积为 3πa²/2 平方单位。切记要对函数 r 进行平方，而不是直接代入。
6. Polar Coordinates: Tangents at the Pole | 极坐标：极点处的切线
Exercise: Find the equation of the tangent to the curve r = 2a cos θ
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