Answers to Exercises – Further Pure 3 | 习题答案：进阶纯数 3

📚 Answers to Exercises – Further Pure 3 | 习题答案：进阶纯数 3

This article provides worked solutions to typical exercises from the AQA A-Level Further Pure 3 syllabus. Each topic is summarised with a model question and a step-by-step answer, so you can check your methods and improve your exam technique.

本文精选 AQA A-Level 进阶纯数 3 的典型习题，给出完整解答与关键步骤。通过对照答案，你可以检查自己的解题方法，提升考试技巧。

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1. Complex Numbers – de Moivre’s Theorem | 复数：德莫弗定理

Exercise: Find all three cube roots of 8i in polar form.

习题：求 8i 的三个三次方根，并写出极坐标形式。

First write 8i in modulus-argument form. The modulus is 8 and the argument is π/2, so we use de Moivre’s theorem for roots.

首先将 8i 写为模辐角形式。模为 8，辐角为 π/2，然后利用德莫弗定理求根。

z = 2( cos(π/6 + 2kπ/3) + i sin(π/6 + 2kπ/3) ), k = 0, 1, 2

• k = 0: z = 2( cos π/6 + i sin π/6 ) = √3 + i

  k = 0：z = 2( cos π/6 + i sin π/6 ) = √3 + i

• k = 1: z = 2( cos 5π/6 + i sin 5π/6 ) = −√3 + i

  k = 1：z = 2( cos 5π/6 + i sin 5π/6 ) = −√3 + i

• k = 2: z = 2( cos 3π/2 + i sin 3π/2 ) = −2i

  k = 2：z = 2( cos 3π/2 + i sin 3π/2 ) = −2i

The three roots are equally spaced around a circle of radius 2, separated by an angle of 2π/3.

三个根均匀分布在半径为 2 的圆上，相邻根之间的夹角为 2π/3。

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2. Hyperbolic Functions – Identities and Inverses | 双曲函数：恒等式与反函数

Exercise: Prove that cosh²x − sinh²x = 1 and hence show that arcosh x = ln( x + √x² − 1 ).

习题：证明 cosh²x − sinh²x = 1，并由此推出 arcosh x = ln( x + √x² − 1 )。

Using the definitions cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ − e⁻ˣ)/2, we expand:

根据定义 cosh x = (eˣ + e⁻ˣ)/2 和 sinh x = (eˣ − e⁻ˣ)/2，展开得：

cosh²x − sinh²x = (e²ˣ + 2 + e⁻²ˣ − e²ˣ + 2 − e⁻²ˣ)/4 = 4/4 = 1

For the inverse, let x = cosh y. Then eʸ satisfies e²ʸ − 2x eʸ + 1 = 0, so eʸ = x ± √x² − 1. The principal branch takes the positive root, giving y = ln( x + √x² − 1 ).

对反函数，令 x = cosh y。则 eʸ 满足 e²ʸ − 2x eʸ + 1 = 0，因此 eʸ = x ± √x² − 1。主值分支取正号，所以 y = ln( x + √x² − 1 )。

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3. Maclaurin Series – Expansions and Limits | 麦克劳林级数：展开与极限

Exercise: Find the Maclaurin series of eˣ sin x up to and including the x³ term.

习题：求 eˣ sin x 的麦克劳林级数，展开到 x³ 项为止。

We recall the standard series: eˣ = 1 + x + x²/2 + x³/6 + … and sin x = x − x³/6 + … Multiply and collect terms with degree at most 3.

回顾标准展开：eˣ = 1 + x + x²/2 + x³/6 + …，sin x = x − x³/6 + …。相乘后合并次数不超过 3 的项。

eˣ sin x = x + x² + x³/3 + …

For an alternative method, write eˣ sin x as Im( eˣ⁺ᶦˣ ) = Im( e⁽¹⁺ⁱ⁾ˣ ) and expand the complex exponential. This quickly gives the same coefficients.

另一种方法是利用 eˣ sin x = Im( eˣ⁺ᶦˣ ) = Im( e⁽¹⁺ⁱ⁾ˣ )，展开复指数后取虚部，可以快速得到相同系数。

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4. Improper Integrals – Convergence and Evaluation | 反常积分：收敛性与计算

Exercise: Evaluate ∫₀^∞ x e⁻ˣ dx.

习题：计算 ∫₀^∞ x e⁻ˣ dx。

Use integration by parts with u = x and dv = e⁻ˣ dx. Then du = dx and v = −e⁻ˣ.

使用分部积分法，令 u = x，dv = e⁻ˣ dx。则 du = dx，v = −e⁻ˣ。

∫₀^∞ x e⁻ˣ dx = [ −x e⁻ˣ ]₀^∞ + ∫₀^∞ e⁻ˣ dx = 0 + 1 = 1

The boundary term vanishes because x e⁻ˣ → 0 as x → ∞, confirming convergence.

边界项在 x → ∞ 时趋于 0，因为 x e⁻ˣ → 0，因此反常积分收敛。

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5. First-Order Differential Equations – Integrating Factors | 一阶微分方程：积分因子

Exercise: Solve x dy/dx + 2y = x³ with y(1) = 0.

习题：解微分方程 x dy/dx + 2y = x³，初值条件为 y(1) = 0。

Rewrite in standard form dy/dx + (2/x)y = x². The integrating factor is exp( ∫ (2/x) dx ) = x².

将方程改写为标准形式 dy/dx + (2/x)y = x²。积分因子为 exp( ∫ (2/x) dx ) = x²。

d/dx ( x² y ) = x⁴

Integrate both sides: x² y = x⁵/5 + C. Use y(1) = 0 to get C = −1/5, so the particular solution is y = (x⁵ − 1)/(5x²).

两边积分得 x² y = x⁵/5 + C。由 y(1) = 0 得 C = −1/5，所以特解为 y = (x⁵ − 1)/(5x²)。

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6. Second-Order Differential Equations – Particular Integrals | 二阶微分方程：特解

Exercise: Find the general solution of y″ − 3y′ + 2y = eˣ.

习题：求微分方程 y″ − 3y′ + 2y = eˣ 的通解。

The auxiliary equation is m² − 3m + 2 = 0, with roots m = 1 and m = 2. The complementary function is y_c = A eˣ + B e²ˣ.

辅助方程为 m² − 3m + 2 = 0，根为 m = 1 和 m = 2。互补函数为 y_c = A eˣ + B e²ˣ。

Since eˣ already appears in the complementary function, try a particular integral of the form y_p = k x eˣ.

由于 eˣ 已出现在互补函数中，设特解为 y_p = k x eˣ。

y_p = −x eˣ, y = A eˣ + B e²ˣ − x eˣ

The constant k is found to be −1 after substituting y_p and its derivatives into the original equation.

代入原方程后求得 k = −1，因此特解为 y_p = −x eˣ。

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7. Polar Coordinates – Area and Tangents | 极坐标：面积与切线

Exercise: Show that the area enclosed by r = a(1 + cos θ), where a > 0, is 3πa²/2.

习题：证明曲线 r = a(1 + cos θ)（a > 0）围成的面积为 3πa²/2。

The area is given by the polar formula A = ½ ∫₀^2π r² dθ.

极坐标面积公式为 A = ½ ∫₀^2π r² dθ。

A = (a²/2) ∫₀^2π (1 + 2 cos θ + cos²θ) dθ = (a²/2)(2π + 0 + π) = 3πa²/2

The cross term 2cos θ integrates to zero over a full period, while cos²θ integrates to π.

交叉项 2cos θ 在一个完整周期内积分为零，而 cos²θ 的积分为 π。

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8. Vectors in 3D – Lines and Planes | 三维向量：直线与平面

Exercise: Find the intersection of the line r = (1, 2, 3) + t(2, −1, 1) with the plane x + 2y − z = 6.

习题：求直线 r = (1, 2, 3) + t(2, −1, 1) 与平面 x + 2y − z = 6 的交点。

Substitute the line components into the plane equation:

将直线的分量代入平面方程：

(1 + 2t) + 2(2 − t) − (3 + t) = 6

Simplify to 2 − t = 6, so t = −4. The intersection point is (1 − 8, 2 + 4, 3 − 4) = (−7, 6, −1).

化简得 2 − t = 6，所以 t = −4。交点为 (1 − 8, 2 + 4, 3 − 4) = (−7, 6, −1)。

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9. Scalar Triple Product – Coplanarity Test | 三重标量积：共面判定

Exercise: Decide whether the points A(1,0,1), B(2,1,3), C(3,−1,2) and D(4,2,5) are coplanar.

习题：判断点 A(1,0,1)、B(2,1,3)、C(3,−1,2)、D(4,2,5) 是否共面。

Form vectors AB = (1,1,2), AC = (2,−1,1), AD = (3,2,4). Compute the scalar triple product AB · (AC × AD).

构造向量 AB = (1,1,2)，AC = (2,−1,1)，AD = (3,2,4)。计算三重标量积 AB · (AC × AD)。

det = 1(−1×4 − 1×2) − 1(2×4 − 1×3) + 2(2×2 + 1×3) = −6 − 5 + 14 = 3

Since the determinant is not zero, the four points are not coplanar.

行列式不为零，所以四点不共面。

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10. Exam-Style Mixed Exercise | 考试风格综合题

Question: A curve has polar equation r = 2 cos θ. Find its Cartesian equation and the area swept out as θ varies from 0 to π/4.

题目：曲线极坐标方程为 r = 2 cos θ。求其直角坐标方程，并求 θ 从 0 到 π/4 时扫过的面积。

Multiply by r: r² = 2r cos θ. Since r² = x² + y² and r cos θ = x, the Cartesian equation is x² + y² = 2x, or (x − 1)² + y² = 1.

两边乘以 r：r² = 2r cos θ。因 r² = x² + y²，r cos θ = x，直角坐标方程为 x² + y² = 2x，即 (x − 1)² + y² = 1。

Area = ½ ∫₀^π/4 (2 cos θ)² dθ = 2 ∫₀^π/4 cos²θ dθ = (π/4) + 1/2

Use cos²θ = (1 + cos 2θ)/2 to integrate easily. This question combines polar coordinates, Cartesian conversion and integration, which are all essential FP3 skills.

利用 cos²θ = (1 + cos 2θ)/2 可轻松积分。本题综合了极坐标、直角坐标转换与积分，是进阶纯数 3 的重要考点。

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