Year 13 OCR Maths: Mock Unit Test Analysis | 单元测试模拟卷解析

📚 Year 13 OCR Maths: Mock Unit Test Analysis | 单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test designed for Year 13 OCR Mathematics. The paper covers Pure, Mechanics and Statistics topics typically examined in the second year of A Level, including trigonometric identities, calculus techniques, 3D vectors, forces in equilibrium, and hypothesis testing. Each section presents a question, a step‑by‑step solution, and commentary on the underlying principles to help you master exam‑style problems.

本文为 Year 13 OCR 数学单元测试模拟卷的详细解析。试卷涵盖第二学年常见的纯数、力学和统计考点，如三角恒等式、微积分技巧、三维向量、力的平衡以及假设检验。每个小节展示一道题目、分步解答以及对核心原理的解读，帮助你掌握考试型问题。

1. Trigonometric Equation with Double Angle | 二倍角三角方程

Question: 3 cos 2θ + cos θ = 2 for 0° ≤ θ ≤ 360°. Give answers to 1 decimal place where appropriate.

题目： 解方程 3 cos 2θ + cos θ = 2，θ 范围 0° ≤ θ ≤ 360°。答案精确至小数点后 1 位。

Use the double‑ cos θ − 5 = 0. Factorise as (3 cos θ − 1)(2 cos θ + 5) = 0, yielding cos θ = 1/3 or cos θ = −5/2 (reject as cos θ must be between −1 and 1). For cos θ = 1/3, primary solution is θ = 70.5°. Since cosine is positive in the first and fourth quadrants, the second solution is 360° − 70.5° = 289.5°. Therefore θ = 70.5°, 289.5°.

使用二倍角恒等式 cos 2θ = 2 cos²θ − 1 将方程重写为仅含 cos θ 的形式。代入得 3(2 cos²θ − 1) + cos θ = 2 → 6 cos²θ − 3 + cos θ = 2 → 6 cos²θ + cos θ − 5 = 0。因式分解为 (3 cos θ − 1)(2 cos θ + 5) = 0，解得 cos θ = 1/3 或 cos θ = −5/2（舍去，因为 cos θ 的值域为 [−1, 1]）。由 cos θ = 1/3 得主解 θ = 70.5°；余弦在第一和第四象限为正，因此第二解为 360° − 70.5° = 289.5°。最终答案为 θ = 70.5°，289.5°。

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2. Chain Rule Differentiation | 链式法则求导

Question: Differentiate y = ln(sin(2x)).

题目： 求 y = ln(sin(2x)) 的导数。

le, dy/dx = dy/du × du/dx = (1 / sin(2x)) × (2 cos(2x)) = 2 cot(2x). This is the simplified derivative.

从外到内应用链式法则。设 u = sin(2x)，则 y = ln u，dy/du = 1/u。求 u 的导数：du/dx = cos(2x) × 2 = 2 cos(2x)。根据链式法则，dy/dx = dy/du × du/dx = (1 / sin(2x)) × (2 cos(2x)) = 2 cot(2x)。此为化简后的导数。

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3. Integration by Substitution | 换元积分法

Question: Find ∫ 3x (x² + 4)⁴ dx.

题目： 计算 ∫ 3x (x² + 4)⁴ dx。

u⁴ du. Integrate: (3/2) × (u⁵ / 5) = (3/10) u⁵ + C. Substitute back: (3/10) (x² + 4)⁵ + C.

作换元 u = x² + 4，则 du/dx = 2x，即 dx = du/(2x)。重写被积函数：3x u⁴ × (du/(2x)) = (3/2) u⁴ du。积分得 (3/2) × (u⁵ / 5) = (3/10) u⁵ + C。代回原变量：(3/10) (x² + 4)⁵ + C。

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4. Exponential Growth & Decay | 指数增长与衰减

Question: The number of bacteria P in a culture is modelled by P = 500 e^0.4t, where t is the time in hours. Find the time taken for the population to double.

题目： 培养液中细菌数量 P 满足模型 P = 500 e^0.4t，其中 t 为小时数。求细菌数量翻倍所需的时间。

Initial population P₀ = 500. Doubling means P = 1000. Set up: 1000 = 500 e^0.4t → 2 = e^0.4t nutes: 0.73275 × 60 ≈ 44 minutes. Thus doubling time is 1 hour 44 minutes (or 1.73 hours).

初始数量 P₀ = 500。翻倍意味着 P = 1000。建立方程：1000 = 500 e^0.4t → 2 = e^0.4t。取自然对数：ln 2 = 0.4t → t = (ln 2) / 0.4。计算：ln 2 ≈ 0.6931，故 t ≈ 1.73275 小时。将 0.73275 小时转换为分钟，得约 44 分钟。因此翻倍时间约为 1 小时 44 分钟（或 1.73 小时）。

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5. 3D Vector Geometry | 三维向量几何

Question: ave position vectors a = i + 2j − k and b = 3i − j + 2k. Find the vector AB and the length |AB|.

题目： 点 A 和点 B 的位置向量分别为 a = i + 2j − k，b = 3i − j + 2k。求向量 AB 及其模长 |AB|。

Vector AB = b − a = (3−1)i + (−1−2)j + (2−(−1))k = 2i − 3j + 3k. The magnitude |AB| = √(2² + (−3)² + 3²) = √4 + 9 + 9 = √22. So AB = 2i − 3j + 3k, length = √22 ≈ 4.69 (3 s.f.).

向量 AB = b − a = (3−1)i + (−1−2)j + (2−(−1))k = 2i − 3j + 3k。模长 |AB| = √(2² + (−3)² + 3²) = √4 + 9 + 9 = √22。因此 AB = 2i − 3j + 3k，长度约为 4.69（保留三位有效数字）。

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6. Hypothesis Testing for Binomial | 二项分布假设检验

Question: A coin is tossed 20 times and lands heads 7 times. Test at the 5% significance level whether the coin is biased against heads. Let p = P(head).

题目： 一枚硬币抛掷 20 次，出现 7 次正面。以 5% 显著性水平检验该硬币是否偏向反面。设 p = P(正面)。

Define hypotheses: H₀: p = 0.5; H₁: p < 0.5 (one‑tailed test because we test if coin is biased *against* heads). Under H₀, X ~ B(20, 0.5). We need the p‑ approx). Since 0.1316 > 0.05, we do not reject H₀. There is insufficient evidence at the 5% level to conclude the coin is biased against heads.

建立假设：H₀: p = 0.5；H₁: p < 0.5（单尾检验，因为检验硬币是否偏向反面）。在 H₀ 下，X ~ B(20, 0.5)。计算 p 值 = P(X ≤ 7)。通过二项分布表或计算器得 P(X ≤ 7) ≈ 0.1316。由于 0.1316 > 0.05，不拒绝原假设。在 5% 显著性水平下，没有足够证据表明硬币偏向反面。

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7. Forces in Equilibrium | 力的平衡

Question: A particle of weight 40 N is suspended by two light inextensible strings. One string is horizontal, the other makes an angle 30° with the vertical. Find the tensions in both strings.

题目： 一重量 40 N 的质点由两条轻质不可伸长绳悬挂。一条绳水平，另一条与竖直方向成 30° 角。求两绳的张力。

cal component of T₂). Resolve horizontally: T₁ = T₂ sin 30°. From vertical equation: T₂ = 40 / cos 30° = 40 / (√3/2) = 80/√3 ≈ 46.19 N. Then T₁ = (80/√3) × sin 30° = (80/√3) × 0.5 = 40/√3 ≈ 23.09 N. So tensions are approximately 46.2 N and 23.1 N.

设水平绳张力为 T₁，斜绳张力为 T₂。竖直方向分解：T₂ cos 30° = 40（重量由 T₂ 的竖直分量平衡）。水平方向分解：T₁ = T₂ sin 30°。由竖直方程得 T₂ = 40 / cos 30° = 40 / (√3/2) = 80/√3 ≈ 46.19 N。因此 T₁ = (80/√3) × sin 30° = (80/√3) × 0.5 = 40/√3 ≈ 23.09 N。张力近似为 46.2 N 和 23.1 N。

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8. Projectile Motion | 抛体运动

Question: izontal. Calculate the greatest height reached and the time of flight. Take g = 9.8 m s⁻².

题目： 一球从地面以 14 m s⁻¹ 的速率、与水平面成 25° 角踢出。计算到达的最大高度和飞行时间。取 g = 9.8 m s⁻²。

Initial vertical component: u_y : 0 = (5.9164)² − 2 × 9.8 × H → H = (5.9164²) / (19.6) ≈ 1.79 m. For time of flight, use s_y = u_y t − ½ g t² = 0. Factorising: t (u_y − 4.9t) = 0 → t = 0 (launch) or t = u_y .9 ≈ 5.9164 / 4.9 ≈ 1.207 s. So time of flight ≈ 1.21 s (3 s.f.). Maximum height ≈ 1.79 m.

竖直初速度分量：u_y = 14 sin 25° ≈ 14 × 0.4226 = 5.9164 m s⁻¹。在最高点，竖直速度为零。利用竖直方向公式 v² = u² + 2as：0 = (5.9164)² − 2 × 9.8 × H → H = (5.9164²) / 19.6 ≈ 1.79 m。飞行时间利用 s_y = u_y t − ½ g t² = 0。因式分解：t (u_y − 4.9t) = 0 → t = 0（抛出时刻）或 t = u_y / 4.9 ≈ 5.9164 / 4.9 ≈ 1.207 s。因此飞行时间 ≈ 1.21 s（三位有效数字），最大高度 ≈ 1.79 m。

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9. Implicit Differentiation & Tangent | 隐函数求导与切线

Question: quation x² + 3xy − y² = 5. Find dy/dx and the equation of the tangent at the point (1, 2).

题目： 曲线方程为 x² + 3xy − y² = 5。求 dy/dx 以及在点 (1, 2) 处的切线方程。

ct dy/dx terms: (3x − 2y) dy/dx = −2x − 3y. Thus dy/dx = (−2x − 3y) / (3x − 2y). At (1, 2): dy/dx = (−2 − 6) / (3 − 4) = (−8)/(−1) = 8. Equation of tangent: y − 2 = 8(x − 1) → y = 8x − 6.

两边同时对 x 求导：2x + 3(y + x dy/dx) − 2y dy/dx = 0（对 3xy 使用乘法法则）。整理得：2x + 3y + 3x dy/dx − 2y dy/dx = 0。合并含 dy/dx 的项：(3x − 2y) dy/dx = −2x − 3y。因此 dy/dx = (−2x − 3y) / (3x − 2y)。在点 (1, 2) 处：dy/dx = (−2 − 6) / (3 − 4) = (−8)/(−1) = 8。切线方程：y − 2 = 8(x − 1) → y = 8x − 6。

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10. Volume of Revolution | 旋转体体积

Question: The region bounded by y = √x, the x‑axis, and the lines x = 0 and x = 4 is rotated through 360° about the x‑axis. Find the volume of the solid generated.

题目： 由曲线 y = √x、x 轴以及直线 x = 0 和 x = 4 围成的区域绕 x 轴旋转 360°。求生成旋转体的体积。

The volume V = π ∫₀⁴ y² dx = π ∫₀⁴ (√x ½ x² ]₀⁴ = π × (½ × 16) = 8π. Therefore the volume is 8π cubic units (≈ 25.1 cubic units).

体积 V = π ∫₀⁴ y² dx = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx。计算：π [ ½ x² ]₀⁴ = π × (½ × 16) = 8π。因此体积为 8π 立方单位（约 25.1 立方单位）。

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