Year 13 AQA Maths: Unit Test Mock Paper Walkthrough | Year 13 AQA 数学：单元测试模拟卷解析

📚 Year 13 AQA Maths: Unit Test Mock Paper Walkthrough | Year 13 AQA 数学：单元测试模拟卷解析

This mock paper walkthrough is designed for Year 13 students preparing for the AQA A-Level Mathematics exam. It covers key topics from pure mathematics, statistics, and mechanics, including differentiation, integration, numerical methods, vectors, normal distribution, hypothesis testing, and projectile motion. Each question is fully worked through, with step‑by‑step reasoning and clear links to the specification. Use this resource to test your understanding and to refine your problem‑solving skills.

本模拟卷解析专为备战 AQA A-Level 数学考试的 Year 13 学生设计，涵盖纯数、统计和力学中的核心专题：微分、积分、数值方法、向量、正态分布、假设检验以及抛体运动。每一道题目都给出了完整的演算过程，并附有分步推理和考纲链接。利用这份资源检验自己的理解，并打磨解题技巧。

1. Product Rule Differentiation | 乘积法则求导

The first question requires the derivative of f(x) = e³ˣ an exponential function and a trigonometric function, so the product rule must be applied.

第一题要求对 f(x) = e³ˣ sin(2x) 求导。这是一个指数函数与三角函数的乘积，因此必须使用乘积法则。

Let u = e³ˣ and v = sin(2x). Then u' = 3e³ˣ (using the chain rule) and v' = 2cos(2x).

令 u = e³ˣ，v = sin(2x)。则 u' = 3e³ˣ（使用链式法则），v' = 2cos(2x)。

Product rule states: f'(x) = u'v + uv'. Substituting gives f'(x) = 3e³ˣ sin(2x) + e³ˣ · 2cos(2x).

乘积法则指出：f'(x) = u'v + uv'。代入可得 f'(x) = 3e³ˣ sin(2x) + e³ˣ · 2cos(2x)。

Factor out e³ˣ to obtain the final simplified form: f'(x) = e³ˣ (3 sin(2x) + 2 cos(2x)).

提取公因子 e³ˣ，得到最终的简化形式：f'(x) = e³ˣ (3 sin(2x) + 2 cos(2x))。

Always check if further factorisation is possible; here the trigonometric expression cannot be simplified further.

务必检查是否可以进一步因式分解；这里三角表达式已无法再化简。

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2. Integration by Parts | 分部积分法

a product of a polynomial and a logarithmic function, making integration by parts the appropriate technique.

计算 ∫ x² ln x dx。被积函数是多项式与对数函数的乘积，适合使用分部积分法。

= uv - ∫ v du. Choose u = ln x (because it simplifies when differentiated) and dv = x² dx.

分部积分公式为 ∫ u dv = uv - ∫ v du。选择 u = ln x（因为求导后可简化），dv = x² dx。

Then du = (1/x) dx, and v = ∫ x² dx = x³/3.

那么 du = (1/x) dx，v = ∫ x² dx = x³/3。

he formula: ∫ x² ln x dx = (ln x)(x³/3) - ∫ (x³/3) · (1/x) dx = (x³ ln x)/3 - ∫ (x²/3) dx.

代入公式：∫ x² ln x dx = (ln x)(x³/3) - ∫ (x³/3) · (1/x) dx = (x³ ln x)/3 - ∫ (x²/3) dx。

Complete the integration: ∫ (x²/3) dx = x³/9. Thus the result is ∫ x² ln x dx = (x³ ln x)/3 - x³/9 + C, where C is the constant of integration.

完成积分：∫ (x²/3) dx = x³/9。因此结果为 ∫ x² ln x dx = (x³ ln x)/3 - x³/9 + C，C 为积分常数。

It is good practice to differentiate the antiderivative mentally to confirm the original integrand is recovered.

好的习惯是在心里对原函数求导，验证是否还原为被积函数。

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3. Newton‑Raphson Iteration | 牛顿‑拉夫森迭代

- 2x - 5 = 0 has a root near 2. Starting with x₀ = 2, perform two iterations of the Newton‑Raphson method to find x₁ and x₂.

方程 x³ - 2x - 5 = 0 在 2 附近有一个根。以 x₀ = 2 为初始值，运用牛顿‑拉夫森法进行两次迭代，求 x₁ 和 x₂。

The iterative formula is xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ), where f(x) = x³ - 2x - 5 and f'(x) = 3x² - 2.

迭代公式为 xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)，其中 f(x) = x³ - 2x - 5，f'(x) = 3x² - 2。

First iteration: f(2) = 8 - 4 - 5 = -1; f'(2) = 3(4) - 2 = 10. So x₁ = 2 - (-1)/10 = 2 + 0.1 = 2.1.

第一次迭代：f(2) = 8 - 4 - 5 = -1；f'(2) = 3(4) - 2 = 10。故 x₁ = 2 - (-1)/10 = 2 + 0.1 = 2.1。

Second iteration: f(2.1) = 2.1³ - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061; f' 2 = 13.23 - 2 = 11.23. Then x₂ = 2.1 - 0.061/11.23 ≈ 2.1 - 0.00543 = 2.09457 (to 5 d.p.).

第二次迭代：f(2.1) = 2.1³ - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061；f'(2.1) = 3(4.41) - 2 = 13.23 - 2 = 11.23。于是 x₂ = 2.1 - 0.061/11.23 ≈ 2.1 - 0.00543 = 2.09457（保留 5 位小数）。

The root is converging rapidly. In an exam, show all substitutions clearly and round appropriately.

根正在快速收敛。考试中要清晰地展示所有代入过程，并适当取整。

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4. Vector Operations and Triangle Area | 向量运算与三角形面积

Given points A(1,2,3), B(4,0,5) and C(2,1,4), find the vectors AB and AC, and hence the area of triangle ABC.

已知点 A(1,2,3)、B(4,0,5) 和 C(2,1,4)，求向量 AB 和 AC，并由此计算三角形 ABC 的面积。

= B - A = (4-1, 0-2, 5-3) = (3, -2, 2). Vector AC = C - A = (2-1, 1-2, 4-3) = (1, -1, 1).

向量 AB = B - A = (3, -2, 2)；向量 AC = C - A = (1, -1, 1)。

e is half the magnitude of the cross product AB × AC. Compute the cross product: AB × AC = |i j k; 3 -2 2; 1 -1 1| = i((-2)(1) - (2)(-1)) - j((3)(1) - (2)(1)) + k((3)(-1) - (-2)(1)).

三角形面积为 AB × AC 的模长的一半。计算叉积：AB × AC = |i j k; 3 -2 2; 1 -1 1| = i((-2)(1) - (2)(-1)) - j((3)(1) - (2)(1)) + k((3)(-1) - (-2)(1))。

Simplify: i(-2 + 2) - j(3 - 2) + k(-3 + 2) = (0, -1, -1).

化简得：(0, -1, -1)。

Magnitude of this vector is √(0² + (-1)² + (-1)²) = √2. Therefore area = (1/2)√2 square units.

该向量的模为 √(0² + (-1)² + (-1)²) = √2。因此面积为 (1/2)√2 平方单位。

Remember that the cross product yields a vector perpendicular to the plane; its length gives the parallelogram area, so halving it gives the triangle area.

记住叉积得到的向量垂直于该平面；其长度表示平行四边形面积，除以二即得三角形面积。

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5. Normal Distribution Probability | 正态分布概率

0 cm and standard deviation σ = 10 cm. Find the proportion of students taller than 185 cm.

学生身高服从正态分布，均值 μ = 170 cm，标准差 σ = 10 cm。求身高高于 185 cm 的学生比例。

Standardise: Z = (X - μ)/σ. For X = 185, Z = (185 - 170)/10 = 1.5.

标准化：Z = (X - μ)/σ。当 X = 185，Z = (185 - 170)/10 = 1.5。

re P(X > 185) = P(Z > 1.5) = 1 - Φ(1.5). Using the standard normal table, Φ(1.5) = 0.9332.

需要求 P(X > 185) = P(Z > 1.5) = 1 - Φ(1.5)。查标准正态分布表，Φ(1.5) = 0.9332。

Thus the proportion is 1 - 0.9332 = 0.0668, or 6.68%.

因此比例为 1 - 0.9332 = 0.0668，即 6.68%。

Always sketch a bell curve to confirm the tail area you are finding.

始终画一个钟形曲线草图，以确认你正在求的尾部面积。

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6. Hypothesis Test for a Population Mean | 总体均值的假设检验

an height of 174 cm. Using the same population standard deviation σ = 10 cm, test at the 5% cance level whether there is evidence that the true mean height has increased from 170 cm.

某样本包含 25 名学生，其平均身高为 174 cm。已知总体标准差 σ = 10 cm，在 5% 显著性水平下检验是否有证据表明真实平均身高已从 170 cm 上升。

State hypotheses: H₀: μ = 170; H₁: μ > 170 (one‑tailed test).

陈述假设：H₀: μ = 170；H₁: μ > 170（单侧检验）。

The test statistic is Z = (x̄ - μ) / (σ/√n) = (174 - 170) / (10/5) = 4 / 2 = 2.

检验统计量为 Z = (x̄ - μ) / (σ/√n) = (174 - 170) / (10/5) = 4 / 2 = 2。

For a one‑tailed test at the 5% level, the critical value from the normal distribution is 1.6449. Since 2 > 1.6449, the test statistic falls in the rejection region.

在 5% 水平的单侧检验中，正态分布的临界值为 1.6449。因为 2 > 1.6449，检验统计量落入拒绝域。

Therefore we reject H₀. There is sufficient evidence at the 5% level to conclude that the mean height has increased.

因此拒绝 H₀。在 5% 的显著性水平下有充分证据表明平均身高有所增加。

Always interpret the result in context and mention the significance level.

始终结合具体情境解释结果，并提及显著性水平。

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7. Projectile Motion: Time of Flight | 抛体运动：飞行时间

20 m/s at an angle θ = 30° to the horizontal. Take g = 9.8 m/s². Find the time of flight.

一个质点从水平地面以初速度 u = 20 m/s、与水平方向夹角 θ = 30° 斜抛。取 g = 9.8 m/s²。求飞行时间。

The vertical component of initial velocity is u_y = u sin θ = 20 sin 30° = 20 × 0.5 = 10 m/s.

初始速度的竖直分量为 u_y = u sin θ = 20 sin 30° = 20 × 0.5 = 10 m/s。

Using s = u_y - (1/2)g t² and setting vertical displacement s = 0 (returns to ground): 0 = 10t - 4.9t².

运用 s = u_y t - (1/2)g t² 并令竖直位移 s = 0（返回地面）：0 = 10t - 4.9t²。

Factor: t(10 - 4.9t) = 0. The non‑zero solution gives the time of flight: t = 10 / 4.9 ≈ 2.04 seconds (to 3 s.f.).

因式分解：t(10 - 4.9t) = 0。非零解即为飞行时间：t = 10 / 4.9 ≈ 2.04 秒（保留 3 位有效数字）。

ich yields the same result: t = (2 × 20 × sin 30°)/9.8 = 40 × 0.5 / 9.8 = 20/9.8 ≈ 2.04 s.

也可以使用公式 t = (2u sin θ)/g，结果相同：t = (2 × 20 × sin 30°)/9.8 = 40 × 0.5 / 9.8 = 20/9.8 ≈ 2.04 s。

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8. Projectile Motion: Range and Maximum Height | 抛体运动：射程与最大高度

20 m/s, θ = 30°, g = 9.8 m/s²), find the horizontal range and the maximum height reached.

在相同发射条件下（u = 20 m/s，θ = 30°，g = 9.8 m/s²），求水平射程和到达的最大高度。

Horizontal component of velocity is constant: u_x = u cos θ = 20 cos 30° = 20 × (√3/2) = 10√3 ≈ 17.32 m/s.

速度的水平分量守恒：u_x = u cos θ = 20 cos 30° = 20 × (√3/2) = 10√3 ≈ 17.32 m/s。

Range = u_x unrounded time). More precisely, t = 20/9.8, so range = (20 cos 30°) × (20/9.8) = (400 × (√3/2))/9.8 = (200√3)/9.8 ≈ 35.3 m.

射程 = u_x × 飞行时间 = 17.32 × 2.04 ≈ 35.3 m（使用精确时间）。更精确地，t = 20/9.8，因此射程 = (20 cos 30°) × (20/9.8) = (400 × (√3/2))/9.8 = (200√3)/9.8 ≈ 35.3 m。

Maximum height occurs when vertical velocity = 0. Use v_y² = u_y² - 2gH: 0 = (10)² - 2 × 9.8 × H → H = 100 / 19.6 ≈ 5.10 m.

最大高度发生在竖直速度为零时。使用 v_y² = u_y² - 2gH：0 = (10)² - 2 × 9.8 × H → H = 100 / 19.6 ≈ 5.10 m。

Always check that the time to maximum height is half the total time of flight, confirming the symmetry of the parabolic path (1.02 s vs 2.04 s).

始终验证到达最大高度的时间是否为总飞行时间的一半，这印证了抛物线轨迹的对称性（1.02 s 对 2.04 s）。

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