📚 Mock Unit Test for OCR Pre-U Further Mathematics: Analysis & Solutions | OCR Pre-U 进阶数学模拟单元测试卷解析
This article provides a complete walkthrough of a mock unit test crafted for the OCR Pre-U Further Mathematics syllabus. Each question targets a core pure-mathematics topic, and the solutions are presented step by step with exam-style commentary. Use this resource to test your knowledge, identify weak areas, and refine your problem-solving technique before the real assessment.
本文为 OCR 预科进阶数学课程精心设计了一套模拟单元测试卷，并逐题给出详尽解答与应试点评。试卷覆盖纯数核心专题，通过逐步解析帮助你检测知识掌握程度、查漏补缺、提升解题能力，为正式考试做好充分准备。
1. How to Use This Mock Test | 如何使用本模拟卷
Treat this mock paper as you would a real examination: set aside a quiet block of time, attempt all questions without referring to notes, and then mark your work against the provided solutions. Pay close attention to the allocation of marks — it reflects the depth of working expected. After completing each question, read the analysis to understand alternative methods and common pitfalls.
请将本模拟卷视为真实考试：安排一段不受干扰的时间，不看笔记尝试完成所有题目，然后对照解答自行批改。注意题目的分值分配，它反映了答题步骤的详细程度要求。完成每题后，阅读解析以掌握替代解法和常见陷阱。
2. Question 1: Complex Numbers – Roots of Unity | 第1题：复数 – 单位根
Problem: Solve the equation z³ = 1, giving all roots in the form re^(iθ) where r > 0 and -π < θ ≤ π. Hence plot the roots on an Argand diagram. (6 marks)
题目：解方程 z³ = 1，将所有根表示为 re^(iθ) 的形式，其中 r > 0 且 -π < θ ≤ π。据此在阿尔冈图上标出这些根。（6分）
First, recognise that 1 can be written in polar form as 1 = e^(i·0). The three distinct cube roots are obtained by adding multiples of 2πi/3 to the argument:
首先，将 1 写成极坐标形式 1 = e^(i·0)。三个不同的立方根可通过在辐角上加上 2πi/3 的整数倍获得：
zₖ = e^(i(2kπ)/3) for k = 0, 1, 2.
Evaluating each:
逐个计算：
z₀ = e^(i·0) = 1 = cos 0 + i sin 0,
z₁ = e^(i·2π/3) = cos(2π/3) + i sin(2π/3) = -½ + i(√3/2),
z₂ = e^(i·4π/3) = cos(4π/3) + i sin(4π/3) = -½ – i(√3/2).
All three roots have modulus 1 and are equally spaced by 2π/3 radians on the unit circle. On an Argand diagram, plot the points (1,0), (-½, √3/2) and (-½, -√3/2). Note that the principal argument of z₂ can also be written as -2π/3 to satisfy -π < θ ≤ π.
三个根的模均为 1，且在单位圆上均匀分布，相邻辐角相差 2π/3。在阿尔冈图上标出点 (1,0)、(-½, √3/2) 和 (-½, -√3/2)。注意 z₂ 的主辐角可写作 -2π/3，以满足 -π < θ ≤ π 的要求。
3. Question 2: Matrices – Inverse and Eigenvalues | 第2题：矩阵 – 逆与特征值
Problem: Let A = [ [2, 1], [1, 2] ]. Find A⁻¹, the eigenvalues of A, and corresponding eigenvectors. (8 marks)
题目：设 A = [ [2, 1], [1, 2] ]。求 A⁻¹、A 的特征值及对应的特征向量。（8分）
The determinant of A is det(A) = (2)(2) – (1)(1) = 3. Hence the inverse is given by (1/det(A)) adj(A):
矩阵 A 的行列式为 det(A) = (2)(2) – (1)(1) = 3。因此逆矩阵为 (1/det(A)) adj(A)：
A⁻¹ = ⅓ [ [2, -1], [-1, 2] ] = [ [⅔, -⅓], [-⅓, ⅔] ].
For eigenvalues, solve det(A – λI) = 0:
对于特征值，解 det(A – λI) = 0：
|2-λ 1|
| 1 2-λ| = (2-λ)² – 1 = λ² – 4λ + 3 = (λ-1)(λ-3) = 0.
The eigenvalues are λ₁ = 1 and λ₂ = 3.
特征值为 λ₁ = 1 和 λ₂ = 3。
For λ = 1, solve (A – I)x = 0: [ [1,1], [1,1] ] x = 0 gives x₁ + x₂ = 0, so an eigenvector is (1, -1)ᵀ.
对于 λ = 1，解 (A – I)x = 0：矩阵方程为 x₁ + x₂ = 0，可取特征向量为 (1, -1)ᵀ。
For λ = 3, solve (A – 3I)x = 0: [ [-1,1], [1,-1] ] x = 0 gives -x₁ + x₂ = 0, so an eigenvector is (1, 1)ᵀ.
对于 λ = 3，解 (A – 3I)x = 0：-x₁ + x₂ = 0，可取特征向量为 (1, 1)ᵀ。
4. Question 3: Vectors – Line and Plane Intersection | 第3题：向量 – 直线与平面的交点
Problem: Find the point of intersection between the line r = (1,2,3) + t(2,-1,1) and the plane x + 3y – z = 5. Determine the acute angle between the line and the plane. (7 marks)
题目：求直线 r = (1,2,3) + t(2,-1,1) 与平面 x + 3y – z = 5 的交点，并求直线与平面所成的锐角。（7分）
Write the line in parametric form: x = 1+2t, y = 2-t, z = 3+t. Substitute into the plane equation:
写出直线的参数方程：x = 1+2t, y = 2-t, z = 3+t。代入平面方程：
(1+2t) + 3(2-t) – (3+t) = 5 → 1+2t + 6-3t -3 -t = 5 → (1+6-3) + (2t-3t-t) = 5 → 4 – 2t = 5 → -2t = 1 → t = -½.
Thus the intersection point is obtained by substituting t = -½: x = 1+2(-½)=0, y = 2 – (-½)=2.5, z = 3 + (-½)=2.5. So the point is (0, 2.5, 2.5).
代入 t = -½ 得交点：x = 0, y = 2.5, z = 2.5，即点 (0, 5/2, 5/2)。
The angle θ between the line and the plane is the complement of the angle between the direction vector d = (2,-1,1) and the normal n = (1,3,-1). Compute the sine of the angle between line and plane: sin θ = |d·n| / (|d||n|).
直线与平面夹角 θ 是直线的方向向量 d 与平面法向量 n 夹角的余角。计算 sin θ = |d·n| / (|d||n|)。
d·n = 2(1) + (-1)(3) + 1(-1) = 2 – 3 – 1 = -2, so |d·n| = 2. |d| = √(4+1+1) = √6, |n| = √(1+9+1) = √11. sin θ = 2 / (√6·√11) = 2/√66. Then θ = arcsin(2/√66) ≈ 14.2°. The acute angle is approximately 14.2°.
d·n = -2，绝对值2。|d| = √6，|n| = √11，sin θ = 2/√66，θ = arcsin(2/√66) ≈ 14.2°。锐角约为 14.2°。
5. Question 4: Polar Coordinates – Area of a Curve | 第4题：极坐标 – 曲线面积
Problem: A curve is defined by the polar equation r = 2 + 2 cos θ for 0 ≤ θ ≤ 2π. Sketch the curve and find the total area enclosed by it. (8 marks)
题目：曲线由极坐标方程 r = 2 + 2 cos θ（0 ≤ θ ≤ 2π）定义。画出曲线的草图，并求曲线所围成的总面积。（8分）
This is a cardioid. It is symmetrical about the initial line (θ = 0). The area can be found using A = ½ ∫₀²π r² dθ.
此曲线为心脏线，关于极轴对称。面积可用公式 A = ½ ∫₀²π r² dθ 计算。
Compute r² = (2+2cos θ)² = 4 + 8cos θ + 4cos²θ. Use cos²θ = (1+cos 2θ)/2 to integrate:
计算 r² = 4 + 8cos θ + 4cos²θ。利用 cos²θ = (1+cos 2θ)/2 进行积分：
r² = 4 + 8cos θ + 2(1+cos 2θ) = 6 + 8cos θ + 2cos 2θ.
Then A = ½ ∫₀²π (6 + 8cos θ + 2cos 2θ) dθ = ½ [6θ + 8sin θ + sin 2θ]₀²π.
因此 A = ½ [6θ + 8sin θ + sin 2θ]₀²π。
Evaluate: at 2π, 6(2π) + 0 + 0 = 12π; at 0, all terms are 0. So A = ½ (12π) = 6π.
计算：在 2π 处值为 12π，在 0 处为 0。因此 A = ½ × 12π = 6π。
The total enclosed area is 6π square units. The sketch should show a heart-shaped curve passing through (4,0) and having a cusp at the pole.
总面积为 6π 平方单位。草图中应展示一条经过 (4,0) 且在极点处有尖点的心脏形曲线。
6. Question 5: Hyperbolic Functions – Solving Equations | 第5题：双曲函数 – 解方程
Problem: Solve the equation 3 sinh x – 2 cosh x = 1, giving your answers in logarithmic form. (6 marks)
题目：解方程 3 sinh x – 2 cosh x = 1，答案用对数形式表示。（6分）
Express sinh x and cosh x in terms of exponentials: sinh x = (eˣ – e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2. Substitute:
用指数表示双曲函数：sinh x = (eˣ – e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2。代入：
3((eˣ – e⁻ˣ)/2) – 2((eˣ + e⁻ˣ)/2) = 1 → (3eˣ – 3e⁻ˣ – 2eˣ – 2e⁻ˣ)/2 = 1 → (eˣ – 5e⁻ˣ)/2 = 1.
Multiply both sides by 2: eˣ – 5e⁻ˣ = 2. Multiply through by eˣ: (eˣ)² – 5 = 2eˣ → (eˣ)² – 2eˣ – 5 = 0.
两边乘 2 得 eˣ – 5e⁻ˣ = 2。两边同乘 eˣ： (eˣ)² – 5 = 2eˣ → (eˣ)² – 2eˣ – 5 = 0。
This is a quadratic in eˣ. Solve: eˣ = [2 ± √(4 + 20)]/2 = [2 ± √24]/2 = [2 ± 2√6]/2 = 1 ± √6.
这是关于 eˣ 的二次方程。解：eˣ = 1 ± √6。
Since eˣ > 0, we discard the negative root 1 – √6 (which is negative). Hence eˣ = 1 + √6, giving x = ln(1 + √6).
因为 eˣ > 0，舍去负根 1 – √6。因此 eˣ = 1 + √6，x = ln(1 + √6)。
7. Question 6: Differential Equations – Integrating Factor | 第6题：微分方程 – 积分因子
Problem: Find the general solution of the differential equation dy/dx + 2y = e⁻ˣ. Hence find the particular solution satisfying y(0) = 3. (7 marks)
题目：求微分方程 dy/dx + 2y = e⁻ˣ 的通解，并求满足 y(0) = 3 的特解。（7分）
This is a first-order linear ODE. The integrating factor is I.F. = e^(∫2 dx) = e^(2x).
这是一阶线性微分方程。积分因子为 I.F. = e^(∫2 dx) = e^(2x)。
Multiply the ODE by e^(2x): e^(2x) dy/dx + 2e^(2x) y = e^(2x)·e⁻ˣ = eˣ. The left-hand side is d/dx (y e^(2x)).
方程两边同乘 e^(2x)：e^(2x) dy/dx + 2e^(2x) y = eˣ。左侧恰为 d/dx (y e^(2x))。
Integrate both sides: y e^(2x) = ∫ eˣ dx = eˣ + C. So y = e⁻²ˣ (eˣ + C) = e⁻ˣ + C e⁻²ˣ.
两边积分：y e^(2x) = eˣ + C。因此 y = e⁻²ˣ (eˣ + C) = e⁻ˣ + C e⁻²ˣ。
Apply initial condition y(0) = 3: 3 = e⁰ + C e⁰ = 1 + C ⇒ C = 2. The particular solution is y = e⁻ˣ + 2 e⁻²ˣ.
代入初始条件 y(0) = 3：3 = 1 + C，得 C = 2。特解为 y = e⁻ˣ + 2 e⁻²ˣ。
8. Question 7: Maclaurin Series – Expansion of a Function | 第7题：麦克劳林级数 – 函数展开
Problem: Find the Maclaurin series for f(x) = ln(1 + sin x) up to and including the term in x³. (6 marks)
题目：求 f(x) = ln(1 + sin x) 的麦克劳林级数，展开至含 x³ 项。（6分）
We need derivatives at x = 0. f(0) = ln(1+0) = 0.
需要计算在 x = 0 处的各阶导数。f(0) = 0。
f'(x) = cos x / (1+ sin x). f'(0) = 1/(1+0) = 1.
f'(x) = cos x / (1+ sin x)，f'(0) = 1。
f”(x): differentiate using quotient rule. Let u = cos x, v = 1+ sin x. u’ = -sin x, v’ = cos x. f” = (u’v – u v’)/v² = [ -sin x (1+ sin x) – cos x (cos x) ] / (1+ sin x)² = [ -sin x – sin²x – cos²x ] / (1+ sin x)² = [ -sin x – 1 ] / (1+ sin x)² = -(1+ sin x) / (1+ sin x)² = -1/(1+ sin x). So f”(0) = -1.
计算二阶导数：运用商法则得 f”(x) = -1/(1+ sin x)，故 f”(0) = -1。
f”'(x) = d/dx [-1/(1+ sin x)] = (cos x) / (1+ sin x)². f”'(0) = 1 / 1² = 1.
f”'(x) = cos x / (1+ sin x)²，f”'(0) = 1。
Thus the Maclaurin series is f(0) + f'(0) x + f”(0) x²/2! + f”'(0) x³/3! + … = 0 + 1·x + (-1)x²/2 + 1·x³/6 = x – ½ x² + ⅙ x³.
因此麦克劳林级数为 f(x) = x – ½ x² + ⅙ x³ + … 。
9. Question 8: Summation of Series – Method of Differences | 第8题：级数求和 – 差分法
Problem: Use the method of differences to evaluate ∑_{r=1}^{n} 1/(r(r+1)). Hence find the limit as n → ∞. (5 marks)
题目：用差分法求 ∑_{r=1}^{n} 1/(r(r+1))，并由此求 n → ∞ 时的极限。（5分）
Express the term using partial fractions: 1/(r(r+1)) = 1/r – 1/(r+1).
用部分分式表示：1/(r(r+1)) = 1/r – 1/(r+1)。
Write the sum: ∑_{r=1}^{n} (1/r – 1/(r+1)). This is a telescoping series:
写出和式：∑_{r=1}^{n} (1/r – 1/(r+1))。这是一个裂项相消级数：
(1/1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + … + (1/n – 1/(n+1)).
All intermediate terms cancel, leaving S_n = 1 – 1/(n+1) = n/(n+1).
所有中间项相消，得 S_n = 1 – 1/(n+1) = n/(n+1)。
As n → ∞, 1/(n+1) → 0, so the sum tends to 1.
当 n → ∞，1/(n+1) → 0，故级数和趋近于 1。
10. Common Mistakes and Key Takeaways | 常见错误与要点总结
Many candidates lose marks by not expressing complex roots in the required range for the argument; always check -π < θ ≤ π. In matrix questions,
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