Answers to Exercises – Further Pure 2 | 进阶纯数 2 练习答案

📚 Answers to Exercises – Further Pure 2 | 进阶纯数 2 练习答案

This revision set provides fully worked answers for common AQA Further Pure 2 exercises, covering complex numbers, hyperbolic functions, matrices, polar coordinates, differential equations, series, reduction formulae and numerical methods.

本复习资料为 AQA 进阶纯数 2 的常见练习提供完整解答，涵盖复数、双曲函数、矩阵、极坐标、微分方程、级数、递推公式和数值方法。

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1. Complex Numbers: Roots of Unity and Geometry | 复数：单位根与几何

Exercise 1 asks for the fifth roots of unity. Write 1 in exponential form as 1 = e^2kπi, where k is an integer.

练习 1 要求求五次单位根。将 1 写成指数形式 1 = e^2kπi，其中 k 为整数。

Raise both sides to the power 1/5 to obtain the five distinct roots.

两边同时开五次方，得到五个不同的根。

z = e^2kπi/5, k = 0, 1, 2, 3, 4

The roots are therefore 1, e^2πi/5, e^4πi/5, e^6πi/5 and e^8πi/5.

因此五个根为 1、e^2πi/5、e^4πi/5、e^6πi/5 和 e^8πi/5。

All roots lie on the unit circle with equal angular spacing of 2π/5, so they form a regular pentagon.

所有根都位于单位圆上，角间距相等为 2π/5，因此它们构成正五边形。

The sum of the roots is 0 because the coefficient of z⁴ in z⁵ - 1 = 0 is zero.

根的和为 0，因为方程 z⁵ - 1 = 0 中 z⁴ 的系数为零。

the roots is 1, since for the monic polynomial z⁵ - 1 = 0 the product is (-1)⁵ × (-1) = 1.

根的积为 1，因为对于首一多项式 z⁵ - 1 = 0，根的积为 (-1)⁵ × (-1) = 1。

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2. Hyperbolic Functions: Identities and Equations | 双曲函数：恒等式与方程

Exercise 2 requires solving 5 sinh x - 3 cosh x = 2. Use the definitions sinh x = (eˣ - e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2.

练习 2 要求解方程 5 sinh x - 3 cosh x = 2。使用定义 sinh x = (eˣ - e⁻ˣ)/2 和 cosh x = (eˣ + e⁻ˣ)/2。

Substitute into the equation and simplify the fractions.

代入方程并化简分式。

5(eˣ - e⁻ˣ)/2 - 3(eˣ + e⁻ˣ)/2 = 2

Multiplying by 2 gives 5eˣ - 5e⁻ˣ - 3eˣ - 3e⁻ˣ = 4, which simplifies to 2eˣ - 8e⁻ˣ = 4.

两边乘以 2 得 5eˣ - 5e⁻ˣ - 3eˣ - 3e⁻ˣ = 4，化简为 2eˣ - 8e⁻ˣ = 4。

Divide by 2 and multiply through by eˣ to obtain a quadratic in eˣ.

除以 2 后两边乘以 eˣ，得到关于 eˣ 的二次方程。

e²ˣ - 2eˣ - 4 = 0

Let y = eˣ, so y² - 2y - 4 = 0. Solving gives y = 1 ± √5.

设 y = eˣ，则 y² - 2y - 4 = 0。解得 y = 1 ± √5。

Since eˣ > 0, we take the positive root eˣ = 1 + √5. Hence x = ln(1 + √5).

因为 eˣ > 0，取正根 eˣ = 1 + √5。因此 x = ln(1 + √5)。

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3. Matrices: Eigenvalues and Diagonalisation | 矩阵：特征值与对角化

ercise 3 uses the matrix M = [[4, 1], [2, 3]]. To find eigenvalues, solve det(M - λI) = 0.

练习 3 使用矩阵 M = [[4, 1], [2, 3]]。求特征值需解 det(M - λI) = 0。

(4 - λ)(3 - λ) - 2 = 0

Expanding gives λ² - 7λ + 10 = 0, so λ = 2 or λ = 5.

展开得 λ² - 7λ + 10 = 0，所以 λ = 2 或 λ = 5。

2I)v = 0: the matrix [[2, 1], [2, 1]] gives 2x + y = 0, so an eigenvector is v₁ = (1, -2).

对于 λ = 2，解 (M - 2I)v = 0：矩阵 [[2, 1], [2, 1]] 给出 2x + y = 0，因此一个特征向量为 v₁ = (1, -2)。

I)v = 0: the matrix [[-1, 1], [2, -2]] gives -x + y = 0, so an eigenvector is v₂ = (1, 1).

对于 λ = 5，解 (M - 5I)v = 0：矩阵 [[-1, 1], [2, -2]] 给出 -x + y = 0，因此一个特征向量为 v₂ = (1, 1)。

Let P = [[1, 1], [-2, 1]] and D = [[2, 0], [0, 5]]. Then M = PDP⁻¹.

令 P = [[1, 1], [-2, 1]]，D = [[2, 0], [0, 5]]，则 M = PDP⁻¹。

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4. Polar Coordinates: Area Enclosed by a Polar Curve | 极坐标：极曲线围成的面积

Exercise 4 asks for the area enclosed by r = a(1 + cos θ) for 0 ≤ θ < 2π.

练习 4 要求求极曲线 r = a(1 + cos θ) 在 0 ≤ θ < 2π 内围成的

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