WJEC Year 13 Further Maths: Unit Test Mock Exam Walkthrough | WJEC高年级进阶数学单元测试模拟卷解析

📚 WJEC Year 13 Further Maths: Unit Test Mock Exam Walkthrough | WJEC高年级进阶数学单元测试模拟卷解析

This mock exam walkthrough is designed for Year 13 students following the WJEC Further Mathematics specification. It covers a representative set of topics from the pure core of the course, including complex numbers, matrices, series, hyperbolic functions, polar coordinates, differential equations, vectors, Maclaurin series, and reduction formulae. Each question is worked through in detail, highlighting key methods and common pitfalls. Use this resource to consolidate your understanding and sharpen your exam technique.

本模拟卷解析面向遵循WJEC进阶数学大纲的高三年级学生。它覆盖了课程纯数核心中的一组代表性主题，包括复数、矩阵、级数、双曲函数、极坐标、微分方程、向量、麦克劳林级数以及归约公式。每道题都详细演算，重点突出关键方法和常见错误。利用这份资源巩固理解、提升应试技巧。

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1. Complex Numbers – de Moivre’s Theorem and Roots | 复数 – 棣莫弗定理与求根

The complex number z = 1 + i√3 is given. First, express z in modulus-argument form: the modulus r = √(1² + (√3)²) = 2, and the argument θ = arctan(√3/1) = π/3, so z = 2(cos(π/3) + i sin(π/3)).

已知复数 z = 1 + i√3。首先将 z 表示为模–辐角形式：模 r = √(1² + (√3)²) = 2，辐角 θ = arctan(√3/1) = π/3，因此 z = 2(cos(π/3) + i sin(π/3))。

e Moivre’s theorem: z⁵ = 2⁵ [cos(5 × π/3) + i sin(5 × π/3)] = 32(cos(5π/3) + i sin(5π/3)).

为了计算 z⁵，应用棣莫弗定理：z⁵ = 2⁵ [cos(5 × π/3) + i sin(5 × π/3)] = 32(cos(5π/3) + i sin(5π/3))。

Now cos(5π/3) = cos(2π − π/3) = cos(π/3) = 1/2, and sin(5π/3) = −sin(π/3) = −√3/2. Hence z⁵ = 32(½ − i√3/2) = 16 − 16i√3.

因为 cos(5π/3) = cos(2π − π/3) = cos(π/3) = 1/2，且 sin(5π/3) = −sin(π/3) = −√3/2。因此 z⁵ = 32(½ − i√3/2) = 16 − 16i√3。

Next, solve the equation w⁵ = 1 − i√3. The right-hand side has modulus √(1² + (√3)²) = 2 and argument arctan(−√3/1) = −π/3 (equivalently 5π/3). So the five fifth roots are given by w_k = 2^1/5 [cos((−π/3 + 2kπ)/5) + i sin((−π/3 + 2kπ)/5)] for k = 0, 1, 2, 3, 4.

接下来求解方程 w⁵ = 1 − i√3。右边模为 √(1² + (√3)²) = 2，辐角为 arctan(−√3/1) = −π/3（也可表示为 5π/3）。因此五个五次方根由 w_k = 2^1/5 [cos((−π/3 + 2kπ)/5) + i sin((−π/3 + 2kπ)/5)] 给出，其中 k = 0, 1, 2, 3, 4。

Writing the roots in exponential form gives very concise answers: w_k = 2^1/5 e^(i(6k−1)π/15), k = 0,1,2,3,4. For instance, when k=0: w₀ = 2^1/5 e^−iπ/15.

利用指数形式可以写出很简洁的答案：w_k = 2^1/5 e^(i(6k−1)π/15)，k = 0,1,2,3,4。例如当 k=0 时：w₀ = 2^1/5 e^−iπ/15。

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2. Matrices – Determinant, Inverse and Linear Equations | 矩阵 – 行列式、逆与线性方程组

Consider the matrix M =
[ 2 −1 0 ]
[ 3 1 2 ]
[ 1 1 1 ]. First evaluate det(M) by expanding along the first row:

考虑矩阵 M = [ 2 −1 0; 3 1 2; 1 1 1 ]。首先按第一行展开计算 det(M)：

×1) − (−1) × (3×1 − 2×1) + 0 = 2 × (1 − 2) + 1 × (3 − 2) = 2 × (−1) + 1 × 1 = −2 + 1 = −1.

2×1) − (−1) × (3×1 − 2×1) + 0 = 2 × (1 − 2) + 1 × (3 − 2) = 2 × (−1) + 1 × 1 = −2 + 1 = −1。

nd the cofactor matrix, then transpose to get the adjugate, and multiply by 1/det(M) = −1.

因为 det(M) ≠ 0，逆矩阵存在。求出余子式矩阵，转置得到伴随矩阵，再乘以 1/det(M) = −1。

The matrix of cofactors C is found by computing each (−1)ⁱ⁺ʲ minor. After calculation we obtain M⁻¹ = (−1) × C^T. The result is: M⁻¹ =
[ 1 1 2 ]
[ 1 −2 −4 ]
[ −2 3 5 ] (multiplied by the scalar −1 gives these entries – check with your own working).

余子式矩阵 C 通过计算每个 (−1)ⁱ⁺ʲ 子式得到。计算后可得 M⁻¹ = (−1) × C^T。结果为：M⁻¹ = [ 1 1 2; 1 −2 −4; −2 3 5 ]（乘以标量 −1 后得到这些元素——建议自行验算）。

To solve Mx = b with b = [1, 5, 3]ᵀ 1 −2×5 −4×3 = 1−10−12 = −21; x₃ = −2×1 + 3×5 + 5×3 = −2+15+15 = 28. Thus x = [12, −21, 28]ᵀ.

为了解 Mx = b，其中 b = [1, 5, 3]ᵀ，使用 x = M⁻¹ b。相乘得到：x₁ = 1×1 + 1×5 + 2×3 = 12；x₂ = 1×1 −2×5 −4×3 = −21；x₃ = −2×1 + 3×5 + 5×3 = 28。因此解为 x = [12, −21, 28]ᵀ。

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3. Series – Proof by Induction | 级数 – 归纳法证明

Prove by induction that Σ_r=1ⁿ r(r+1) = (1/3) n (n+1)(n+2). Base case n=1: LHS = 1×2 = 2; RHS = (1/3)×1×2×3 = 2. True.

用归纳法证明 Σ_r=1ⁿ r(r+1) = (1/3) n (n+1)(n+2)。基础情形 n=1：左边 = 1×2 = 2；右边 = (1/3)×1×2×3 = 2。成立。

Assume the formula holds for n = k, i.e. Σ_r=1ᵏ r(r+1) = (1/3) k (k+1)(k+2).

假设公式对 n = k 成立，即 Σ_r=1ᵏ r(r+1) = (1/3) k (k+1)(k+2)。

Now for n = k+1, the sum becomes Σ_r=1ᵏ r(r+1) + (k+1)(k+2). Substitute the assumption:

对于 n = k+1，求和式为 Σ_r=1ᵏ r(r+1) + (k+1)(k+2)。代入假设：

+2) + (k+1)(k+2) = (k+1)(k+2)[ (k/3) + 1 ] = (k+1)(k+2)[ (k+3)/3 ] = (1/3)(k+1)(k+2)(k+3).

左边 k+2) + (k+1)(k+2) = (k+1)(k+2)[ (k/3) + 1 ] = (k+1)(k+2)[ (k+3)/3 ] = (1/3)(k+1)(k+2)(k+3)。

ement is true for n = k+1. By induction, the formula is valid for all positive integers n.

这与 n = k+1 时公式的右边一致。因此命题对 n = k+1 成立。由归纳法，该公式对所有正整数 n 成立。

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4. Hyperbolic Functions – Identity and Differentiation | 双曲函数 – 恒等式与微分

Starting from the definitions cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ − e⁻ˣ)/2, verify the identity cosh²x − sinh²x = 1.

从定义 cosh x = (eˣ + e⁻ˣ)/2 以及 sinh x = (eˣ − e⁻ˣ)/2 出发，验证恒等式 cosh²x − sinh²x = 1。

Compute cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4, sinh²x = (e²ˣ − 2 + e⁻²ˣ)/4. Their difference is (4/4) = 1.

计算 cosh²x = (e²ˣ + 2 + e⁻²ˣ)/4，sinh²x = (e²ˣ − 2 + e⁻²ˣ)/4。两者之差为 (4/4) = 1。

Now differentiate y = arsinh x. Let y = arsinh x ⇒ sinh y = x. Differentiate implicitly with respect to x: cosh y (dy/dx) = 1 ⇒ dy/dx = 1 / cosh y.

现对 y = arsinh x 求导。令 y = arsinh x ⇒ sinh y = x。关于 x 隐式求导：cosh y (dy/dx) = 1 ⇒ dy/dx = 1 / cosh y。

Using the identity cosh²y − sinh²y = 1, we have cosh y = √1 + sinh²y = √1 + x². Hence dy/dx = 1/√1 + x².

利用恒等式 cosh²y − sinh²y = 1，得到 cosh y = √1 + sinh²y = √1 + x²。因此 dy/dx = 1/√1 + x²。

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5. Polar Coordinates – Area of a Cardioid | 极坐标 – 心形线面积

by one loop (the full cardioid) is given by A = ½ ∫₀²π r² dθ = ½ a² ∫₀²π (1 + cos θ)² dθ.

一条曲线定义为 r = a(1 + cos θ)，0 ≤ θ ≤ 2π。一圈（完整心形线）所围面积由 A = ½ ∫₀²π r² dθ = ½ a² ∫₀²π (1 + cos θ)² dθ 给出。

d: (1 + cos θ)² = 1 + 2cos θ + cos²θ. Use the double-angle identity cos²θ = ½(1 + cos 2θ).

将被积函数展开：(1 + cos θ)² = 1 + 2cos θ + cos²θ。利用倍角公式 cos²θ = ½(1 + cos 2θ)。

hen A = ½ a² ∫₀²π [1 + 2cos θ + ½ + ½ cos 2θ] dθ = ½ a² ∫₀²π [3/2 + 2cos θ + ½ cos 2θ] dθ.

于是 A = ½ a² ∫₀²π [1 + 2cos θ + ½ + ½ cos 2θ] dθ = ½ a² ∫₀²π [3/2 + 2cos θ + ½ cos 2θ] dθ。

2θ over a full period 0 to 2π are zero. So A = ½ a² × [3/2 × 2π] = ½ a² × 3π = (3π a²)/2.

逐项积分：cos θ 和 cos 2θ 在一个完整周期 0 到 2π 上的积分为零。因此 A = ½ a² × [3/2 × 2π] = ½ a² × 3π = (3π a²)/2。

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6. First-Order Linear Differential Equation – Integrating Factor | 一阶线性微分方程 – 积分因子

e initial value problem: dy/dx + 2y = x, with y(0) = 1. The integrating factor is μ(x) = e^{∫ 2 dx} = e^2x.

求解初值问题：dy/dx + 2y = x，其中 y(0) = 1。积分因子为 μ(x) = e^{∫ 2 dx} = e^2x。

Multiply both sides of the DE by μ: e^2x dy/dx + 2e^2x y = x e^2x. The left side is the derivative of y e^2x.

将方程两边乘以 μ：e^2x dy/dx + 2e^2x y = x e^2x。左边正是 y e^2x 的导数。

Integrate both sides with respect to x: y e^2x = ∫ x e^2x dx. Use integration by parts: let u = x, dv = e^2x dx.

两边对 x 积分：y e^2x = ∫ x e^2x dx。使用分部积分法：令 u = x，dv = e^2x dx。

Then ∫ x e^2x dx = (x e^2x)/2 − ∫ (e^2x)/2 dx = (x e^2x)/2 − e^2x/4 + C. So y e^2x = (2x − 1) e^2x/4 + C.

得 ∫ x e^2x dx = (x e^2x)/2 − ∫ (e^2x)/2 dx = (x e^2x)/2 − e^2x/4 + C。因此 y e^2x = (2x − 1) e^2x/4 + C。

Multiply through by e^−2x: y = (2x − 1)/4 + C e^−2x. Apply y(0)=1: 1 = (−1)/4 + C ⇒ C = 5/4. The solution: y = (2x − 1 + 5e^−2x)/4.

两边乘以 e^−2x：y = (2x − 1)/4 + C e^−2x。代入 y(0)=1：1 = (−1)/4 + C ⇒ C = 5/4。解为 y = (2x − 1 + 5e^−2x)/4。

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7. Second-Order ODE – Constant Coefficients | 二阶常微分方程 – 常系数非齐次

Find the general solution of d²y/dx² − 3 dy/dx + 2y = eˣ mentary function is m² − 3m + 2 = 0, which factors as (m−1)(m−2)=0, giving roots m=1, m=2.

求 d²y/dx² − 3 dy/dx + 2y = eˣ 的通解。首先，补函数的辅助方程为 m² − 3m + 2 = 0，分解为 (m−1)(m−2)=0，根为 m=1, m=2。

Hence the complementary function is y_c = A eˣ + B e^2x, where A and B are arbitrary constants.

因此补函数为 y_c = A eˣ + B e^2x，其中 A、B 为任意常数。

Since the right-hand side eˣ ots of the auxiliary equation (m=1), we must use a trial particular integral of the form y_p = C x eˣ.

由于右边 eˣ 对应辅助方程的一个根 (m=1)，我们必须使用试特解形式 y_p = C x eˣ。

Differentiate: y_p′ = C eˣ + C x eˣ, y_p″ = 2C eˣ + C x eˣ. Substitute into the ODE and collect terms: (2C eˣ + C x eˣ) − 3(C eˣ + C x eˣ) + 2C x eˣ = eˣ.

求导：y_p′ = C eˣ + C x eˣ，y_p″ = 2C eˣ + C x eˣ。代入方程并合并项：(2C eˣ + C x eˣ) − 3(C eˣ + C x eˣ) + 2C x eˣ = eˣ。

The x eˣ terms cancel: C − 3C + 2C = 0. The eˣ terms give 2C − 3C = −C = 1 ⇒ C = −1. So y_p = −x eˣ.

x eˣ 项相消：C − 3C + 2C = 0。eˣ 项给出 2C − 3C = −C = 1 ⇒ C = −1。因此 y_p = −x eˣ。

The general solution is y = y_c + y_p = A eˣ + B e^2x − x eˣ.

通解为 y = y_c + y_p = A eˣ + B e^2x − x eˣ。

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8. Vectors – Shortest Distance from a Point to a Line | 向量 – 点到直线的最短距离

. Let A be the fixed point on the line (2, −1, 3) and the direction vector b = (1, 2, −2).

求点 P(1, 2, −1) 到直线 L: r = (2i − j + 3k) + t(i + 2j − 2k) 的最短距离。令 A 为直线上定点 (2, −1, 3)，方向向量 b = (1, 2, −2)。

orm the vector AP = (1−2, 2−(−1), −1−3) = (−1, 3, −4). The shortest distance d is given by |AP × b| / |b|.

构造向量 AP = (1−2, 2−(−1), −1−3) = (−1, 3, −4)。最短距离 d 由 |AP × b| / |b| 给出。

× (1, 2, −2) = determinant: i(3×(−2) − (−4)×2) − j((−1)×(−2) − (−4)×1) + k((−1)×2 − 3×1).

计算叉积 AP × b = (−1, 3, −4) × (1, 2, −2) = 行列式：i(3×(−2) − (−4)×2) − j((−1)×(−2) − (−4)×1) + k((−1)×2 − 3×1)。

Simplify: i(−6 + 8) − j(2 + 4) + k(−2 − 3) = 2i − 6j − 5k = (2, −6, −5).

化简：i(−6 + 8) − j(2 + 4) + k(−2 − 3) = 2i − 6j − 5k = (2, −6, −5)。

Its magnitude |AP × b| = √(2² + (−6)² + (−5)²) = √4+36+25 = √65. The magnitude of b is |b| = √(1²+2²+(−2)²) = √9 = 3. Thus d = √65 / 3.

其模 |AP × b| = √(2² + (−6)² + (−5)²) = √4+36+25 = √65。b 的模 |b| = √(1²+2²+(−2)²) = √9 = 3。因此 d = √65 / 3。

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9. Maclaurin Series – Expansion of e^{sin x} | 麦克劳林级数 – e^{sin x} 展开

Expand f(x) = e^{sin x} as a Maclaurin series up to and including the term in x³. We need f(0), f′(0), f″(0) and f‴(0).

将 f(x) = e^{sin x} 展开为麦克劳林级数，直到含 x³ 的项。需要计算 f(0)、f′(0)、f″(0) 和 f‴(0)。

f(0) = e^{sin 0} = e⁰ = 1. Then f′(x) = e^{sin x} cos x ⇒ f′(0) = 1×1 = 1.

f(0) = e^{sin 0} = e⁰ = 1。然后 f′(x) = e^{sin x} cos x ⇒ f′(0) = 1×1 = 1。

Differentiate again: f″(x) = e^{sin x} cos² x − e^{sin x} sin x = e^{sin x}(cos² x − sin x). At x=0: cos²0=1, sin0=0, so f″(0)=1×(1−0)=1.

再次求导：f″(x) = e^{sin x} cos² x − e^{sin x} sin x = e^{sin x}(cos² x − sin x)。在 x=0 处：cos²0=1，sin0=0，故 f″(0)=1×(1−0)=1。

Third derivative: f‴(x) = f″(x) cos x − f′(x) sin x − f′(x) sin x − f(x) cos x … Better: differentiate f″(x) = e^{sin x}(cos² x − sin x). Let u = e^{sin x}, v = cos² x − sin x. Then f‴(x) = u′ v + u v′. u′ = e^{sin x} 2 cos x sin x − cos x = −cos x(2 sin x + 1). At x=0: u=1, u′=1, v=1, v′ = −1(0+1)=−1. So f‴(0) = 1×1 + 1×(−1) = 0.

三阶导数：f‴(x) = f″(x) cos x − f′(x) sin x − f′(x) sin x − f(x) cos x … 更好的做法：对 f″(x) = e^{sin x}(cos² x − sin x) 求导。令 u = e^{sin x}，v = cos² x − sin x。则 f‴(x) = u′ v + u v′。u′ = e^{sin x} cos x；v′ = −2 cos x sin x − cos x = −cos x(2 sin x + 1)。在 x=0 处：u=1, u′=1, v=1, v′ = −1(0+1)=−1。所以 f‴(0) = 1×1 + 1×(−1) = 0。

Thus the Maclaurin polynomial up to x³ is f(0) + f′(0)x + f″(0)x²/2! + f‴(0)x

Published by TutorHao | Year 13 进阶数学 Revision Series | aleveler.com

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