📚 Year 13 Edexcel Mock Unit Test: Pure Mathematics Walkthrough | Edexcel 十三年级数学模拟单元测试卷解析
This article provides a full worked solution and examiner-style commentary for a mock unit test designed for Year 13 Edexcel Pure Mathematics. The paper covers the key topics from the Pure 3 and Pure 4 specifications, including advanced algebra, trigonometry, calculus, vectors, numerical methods, proof by induction and parametric equations. By working through each question, you will reinforce essential techniques and develop confidence for the final A Level examination.
本文为一份针对十三年级（Year 13）Edexcel 纯数单元测试的模拟试卷提供完整解析和评分员视角的点评。试卷涵盖了纯数 3 和纯数 4 课程的核心主题，包括高等代数、三角学、微积分、向量、数值方法、归纳法证明以及参数方程。通过逐一攻克每道题目，你将巩固关键解题技巧，并为 A Level 最终考试树立信心。
1. Algebra and Functions: Composite and Inverse Functions | 代数与函数：复合函数与反函数
Question: Given f(x) = 2x + 3 and g(x) = x² – 1, find (a) fg(x) and (b) g⁻¹(x).
题目：已知 f(x) = 2x + 3，g(x) = x² – 1，求 (a) fg(x) 和 (b) g⁻¹(x)。
For part (a), we need to apply the function g first, then f. fg(x) = f(g(x)) = f(x² – 1). Substitute the expression for g into f: f(x² – 1) = 2(x² – 1) + 3 = 2x² – 2 + 3 = 2x² + 1. Always remember that the domain of the composite function may be restricted by the inner function, but here no further restrictions are required.
对于 (a) 部分，需要先代入 g 再代入 f。fg(x) = f(g(x)) = f(x² – 1)。将 g 的表达式代入 f：f(x² – 1) = 2(x² – 1) + 3 = 2x² – 2 + 3 = 2x² + 1。务必记住复合函数的定义域可能会受到内层函数的限制，不过本题无需额外限定。
For part (b), to find the inverse of g(x) = x² – 1, we first write y = x² – 1. Swap x and y to obtain x = y² – 1. Solve for y: y² = x + 1, so y = ±√(x + 1). However, an inverse must be a one-to-one function, so we need to restrict the domain of g. Typically, for x ≥ 0, the inverse is g⁻¹(x) = √(x + 1) with domain x ≥ -1. If no domain is given, state both branches with the appropriate domain restrictions.
对于 (b) 部分，求 g(x) = x² – 1 的反函数，首先令 y = x² – 1。交换 x 与 y 得 x = y² – 1。解出 y：y² = x + 1，故 y = ±√(x + 1)。然而反函数必须是一对一的，因此需要限制 g 的定义域。通常，若 x ≥ 0，反函数就是 g⁻¹(x) = √(x + 1)，定义域为 x ≥ -1。如果题目未给定定义域，则应同时写出两个分支并标明对应的定义域限制。
2. Trigonometry: Solving Trigonometric Equations | 三角学：求解三角方程
Question: Solve 2sin²θ – cosθ = 1 for 0 ≤ θ ≤ 2π, giving answers in terms of π.
题目：求解 2sin²θ – cosθ = 1，其中 0 ≤ θ ≤ 2π，答案用 π 表示。
Use the identity sin²θ = 1 – cos²θ to express everything in terms of cosθ. Substituting gives 2(1 – cos²θ) – cosθ = 1 → 2 – 2cos²θ – cosθ = 1. Rearrange: -2cos²θ – cosθ + 1 = 0, or multiplying by -1: 2cos²θ + cosθ – 1 = 0. This is a quadratic in cosθ.
利用恒等式 sin²θ = 1 – cos²θ，将所有项用 cosθ 表示。代入得 2(1 – cos²θ) – cosθ = 1 → 2 – 2cos²θ – cosθ = 1。整理得 -2cos²θ – cosθ + 1 = 0，或者乘以 -1：2cos²θ + cosθ – 1 = 0。这是关于 cosθ 的二次方程。
2cos²θ + cosθ – 1 = 0
Factorise: (2cosθ – 1)(cosθ + 1) = 0. Hence cosθ = ½ or cosθ = -1. Solve each case separately for 0 ≤ θ ≤ 2π. For cosθ = ½, base solutions are θ = π/3 and θ = 5π/3. For cosθ = -1, θ = π. The solution set is θ = π/3, π, 5π/3. Always check the interval and reject any solutions outside it.
分解因式：(2cosθ – 1)(cosθ + 1) = 0。因此 cosθ = ½ 或 cosθ = -1。分别在 0 ≤ θ ≤ 2π 上求解。对于 cosθ = ½，基本解为 θ = π/3 和 θ = 5π/3。对于 cosθ = -1，θ = π。故解集为 θ = π/3, π, 5π/3。务必检查区间范围，舍去区间外的解。
3. Exponentials and Logarithms: Solving Equations | 指数与对数：方程求解
Question: Solve log₂(x + 1) – log₂(x – 1) = 3.
题目：求解 log₂(x + 1) – log₂(x – 1) = 3。
Combine the logarithms using the quotient rule: log₂((x + 1)/(x – 1)) = 3. Rewrite in exponential form: (x + 1)/(x – 1) = 2³ = 8. Cross-multiply: x + 1 = 8(x – 1) → x + 1 = 8x – 8. Rearrange: 7x = 9 → x = 9/7. Check the domain: x – 1 > 0 and x + 1 > 0, so x > 1. Since 9/7 > 1, this is a valid solution.
运用商法则合并对数：log₂((x + 1)/(x – 1)) = 3。转化为指数形式：(x + 1)/(x – 1) = 2³ = 8。交叉相乘：x + 1 = 8(x – 1) → x + 1 = 8x – 8。移项得 7x = 9 → x = 9/7。检查定义域：需要 x – 1 > 0 且 x + 1 > 0，即 x > 1。因为 9/7 > 1，该解有效。
Always remember to verify solutions in logarithmic equations, as algebraic manipulation can introduce extraneous roots that fall outside the domain of the original logarithms.
要始终牢记核对对数方程的解，因为代数变形可能会引入原对数定义域之外的增根。
4. Differentiation: Product Rule | 微分：乘积法则
Question: Differentiate y = e^(3x) sin(2x) with respect to x.
题目：对 y = e^(3x) sin(2x) 关于 x 求导。
Identify u = e^(3x) and v = sin(2x). Then u’ = 3e^(3x) and v’ = 2cos(2x) (using the chain rule for each). Apply the product rule: dy/dx = u’v + uv’ = 3e^(3x) sin(2x) + e^(3x) · 2cos(2x). Factorise e^(3x): dy/dx = e^(3x) [3 sin(2x) + 2 cos(2x)].
令 u = e^(3x)，v = sin(2x)。那么 u’ = 3e^(3x)，v’ = 2cos(2x)（各自使用链式法则）。应用乘积法则：dy/dx = u’v + uv’ = 3e^(3x) sin(2x) + e^(3x) · 2cos(2x)。提公因式 e^(3x)：dy/dx = e^(3x) [3 sin(2x) + 2 cos(2x)]。
This simplified form is preferred in marking schemes. Do not forget that the derivative of e^(kx) is ke^(kx) and that the derivative of sin(kx) requires multiplying by k.
这种化简形式在评分方案中更受青睐。不要忘记 e^(kx) 的导数是 ke^(kx)，且对 sin(kx) 求导时需要乘以系数 k。
5. Implicit Differentiation | 隐函数微分
Question: The curve C has equation x² + xy + y² = 12. Find dy/dx in terms of x and y, and hence determine the coordinates of the stationary points on the curve.
题目：曲线 C 的方程为 x² + xy + y² = 12。求 dy/dx（用 x 和 y 表示），并据此求出曲线上的驻点坐标。
Differentiate both sides with respect to x. The derivative of x² is 2x. For xy, use the product rule: d/dx (xy) = (1)y + x(dy/dx) = y + x dy/dx. For y², use the chain rule: d/dx (y²) = 2y dy/dx. The right side differentiates to 0. Hence 2x + y + x dy/dx + 2y dy/dx = 0.
对方程两边关于 x 求导。x² 的导数为 2x。对于 xy，使用乘积法则：d/dx (xy) = (1)y + x(dy/dx) = y + x dy/dx。对于 y²，应用链式法则：d/dx (y²) = 2y dy/dx。右边常数的导数为 0。于是得到 2x + y + x dy/dx + 2y dy/dx = 0。
Collect dy/dx terms: (x + 2y) dy/dx = -2x – y. Therefore dy/dx = -(2x + y) / (x + 2y). For stationary points, set dy/dx = 0, giving -2x – y = 0, or y = -2x. Substitute into the original equation: x² + x(-2x) + (-2x)² = 12 → x² – 2x² + 4x² = 12 → 3x² = 12 → x² = 4 → x = ±2. Corresponding y values: when x = 2, y = -4; when x = -2, y = 4. So stationary points are (2, -4) and (-2, 4).
合并含 dy/dx 的项：(x + 2y) dy/dx = -2x – y。因此 dy/dx = -(2x + y) / (x + 2y)。令 dy/dx = 0 求驻点，得 -2x – y = 0，即 y = -2x。代入原方程：x² + x(-2x) + (-2x)² = 12 → x² – 2x² + 4x² = 12 → 3x² = 12 → x² = 4 → x = ±2。对应的 y 值：x = 2 时 y = -4；x = -2 时 y = 4。故驻点为 (2, -4) 和 (-2, 4)。
6. Integration: Integration by Parts | 积分：分部积分法
Question: Find ∫ x e^(2x) dx.
题目：计算 ∫ x e^(2x) dx。
Use the integration by parts formula: ∫ u dv = uv – ∫ v du. Choose u = x (which simplifies when differentiated) and dv = e^(2x) dx. Then du = dx and v = (1/2)e^(2x). Substitute: ∫ x e^(2x) dx = x · (1/2)e^(2x) – ∫ (1/2)e^(2x) dx = (1/2)x e^(2x) – (1/2) ∫ e^(2x) dx = (1/2)x e^(2x) – (1/2)·(1/2)e^(2x) + C = (1/2)x e^(2x) – (1/4)e^(2x) + C.
使用分部积分公式：∫ u dv = uv – ∫ v du。选择 u = x（求导后简化）以及 dv = e^(2x) dx。则 du = dx，v = (1/2)e^(2x)。代入：∫ x e^(2x) dx = x·(1/2)e^(2x) – ∫ (1/2)e^(2x) dx = (1/2)x e^(2x) – (1/2) ∫ e^(2x) dx = (1/2)x e^(2x) – (1/2)·(1/2)e^(2x) + C = (1/2)x e^(2x) – (1/4)e^(2x) + C。
∫ x e^(2x) dx = ½ x e^(2x) – ¼ e^(2x) + C
Don’t forget the constant of integration. To verify, differentiate your answer and confirm it gives the original integrand x e^(2x). This is a classic pattern: integration by parts with a polynomial factor and an exponential.
不要忘记积分常数 C。你可以通过求导验证结果是否等于原被积函数 x e^(2x)。这是一个经典模式：多项式因子与指数函数相乘时使用分部积分法。
7. Vectors: Dot Product and Lines in 3D | 向量：点积与三维直线
Question: Given points A(1, 2, 3) and B(4, 5, 6), find the vector AB and the angle between AB and the vector c = (1, 0, -1).
题目：已知点 A(1, 2, 3) 和 B(4, 5, 6)，求向量 AB 以及 AB 与向量 c = (1, 0, -1) 的夹角。
Vector AB = (4-1, 5-2, 6-3) = (3, 3, 3) = 3(1, 1, 1). Its magnitude |AB| = √(3² + 3² + 3²) = √27 = 3√3. Magnitude of c: |c| = √(1² + 0² + (-1)²) = √2.
向量 AB = (4-1, 5-2, 6-3) = (3, 3, 3) = 3(1, 1, 1)。其模长 |AB| = √(3² + 3² + 3²) = √27 = 3√3。向量 c 的模长：|c| = √(1² + 0² + (-1)²) = √2。
The dot product AB · c = 3×1 + 3×0 + 3×(-1) = 3 + 0 – 3 = 0. Using cosθ = (AB·c) / (|AB||c|), we have cosθ = 0 / (3√3 × √2) = 0. Thus θ = π/2 (90°). The vectors are perpendicular.
点积 AB·c = 3×1 + 3×0 + 3×(-1) = 3 + 0 – 3 = 0。利用公式 cosθ = (AB·c) / (|AB||c|)，得 cosθ = 0 / (3√3 × √2) = 0。因此 θ = π/2 (90°)，两向量相互垂直。
Recognising a zero dot product as an indicator of perpendicularity saves time. In exam questions, always show the full formula even if the result is obvious.
识别出点积为零即表明垂直，可以节省时间。考试答题时，即使结果显而易见，也要完整展现公式。
8. Numerical Methods: Fixed-Point Iteration | 数值方法：不动点迭代
Question: The equation x³ – 2x – 5 = 0 has a root α near 2. Show that the equation can be rearranged into the form x = ³√(2x + 5), and use the iterative formula x_(n+1) = ³√(2x_n + 5) with x₀ = 2 to find α correct to 2 decimal places.
题目：方程 x³ – 2x – 5 = 0 有一根 α 位于 2 附近。证明该方程可改写为 x = ³√(2x + 5)，并用迭代公式 x_(n+1) = ³√(2x_n + 5)，初始值 x₀ = 2，求 α 精确至 2 位小数。
Starting from x³ – 2x – 5 = 0, add 2x + 5 to both sides: x³ = 2x + 5. Take the cube root: x = ³√(2x + 5). This is the required rearrangement. Now apply iteration: x₁ = ³√(2×2 + 5) = ³√9 ≈ 2.0801; x₂ = ³√(2×2.0801 + 5) ≈ ³√9.1602 ≈ 2.0925; x₃ = ³√(2×2.0925 + 5) ≈ ³√9.185 ≈ 2.0942; x₄ ≈ ³√(2×2.0942+5) ≈ ³√9.1884 ≈ 2.0945; x₅ ≈ ³√9.189 ≈ 2.0946. The sequence converges to 2.09 (2 d.p.) as subsequent differences fall below 0.005.
从 x³ – 2x – 5 = 0 出发，两边同时加 2x + 5：x³ = 2x + 5。开立方得 x = ³√(2x + 5)，即所需的重写形式。进行迭代：x₁ = ³√(2×2 + 5) = ³√9 ≈ 2.0801；x₂ = ³√(2×2.0801 + 5) ≈ ³√9.1602 ≈ 2.0925；x₃ = ³√(2×2.0925 + 5) ≈ ³√9.185 ≈ 2.0942；x₄ ≈ ³√(2×2.0942+5) ≈ ³√9.1884 ≈ 2.0945；x₅ ≈ 2.0946。序列收敛于 2.09（两位小数），因为后续差值小于 0.005。
Always check that the rearrangement will lead to convergence. Here the derivative of the iteration function near the root ensures convergence. In the exam, present your workings clearly in a table and state the final approximation to the required accuracy.
应始终检查所选的迭代形式是否收敛。此处迭代函数在根附近的导数保证了收敛性。考试中，要在表格中清晰呈现计算过程，并按要求精度给出最终近似值。
9. Proof by Induction: Summation | 数学归纳法证明：求和公式
Question: Prove by mathematical induction that for all positive integers n, ∑ (r=1 to n) r(r+1) = n(n+1)(n+2)/3.
题目：用数学归纳法证明，对所有正整数 n，有 ∑ (r=1 to n) r(r+1) = n(n+1)(n+2)/3。
Base case (n=1): LHS = 1×2 = 2. RHS = 1×2×3/3 = 6/3 = 2. So the formula holds for n=1.
归纳奠基 (n=1)：左式 = 1×2 = 2。右式 = 1×2×3/3 = 6/3 = 2。因此 n=1 时公式成立。
Inductive step: Assume true for n=k: ∑ (r=1 to k) r(r+1) = k(k+1)(k+2)/3. For n=k+1, LHS = ∑ (r=1 to k) r(r+1) + (k+1)(k+2) = k(k+1)(k+2)/3 + (k+1)(k+2). Factor (k+1)(k+2): = (k+1)(k+2)[k/3 + 1] = (k+1)(k+2)[(k+3)/3] = (k+1)(k+2)(k+3)/3. This is RHS with n=k+1. Hence true for n=k+1. By induction, the statement holds for all n ∈ ℤ⁺.
归纳递推：假设 n=k 时命题成立：∑ (r=1 to k) r(r+1) = k(k+1)(k+2)/3。考虑 n=k+1，左式 = ∑ (r=1 to k) r(r+1) + (k+1)(k+2) = k(k+1)(k+2)/3 + (k+1)(k+2)。提取公因子 (k+1)(k+2)：= (k+1)(k+2)[k/3 + 1] = (k+1)(k+2)[(k+3)/3] = (k+1)(k+2)(k+3)/3。这正是 n=k+1 时的右式。因此 n=k+1 时命题也成立。由数学归纳法，所述命题对所有正整数 n 均成立。
∑ (r=1 to n) r(r+1) = n(n+1)(n+2)/3
Proof by induction requires a clear structure: base case, assumption, and the chain of algebra that transforms the (k+1) sum into the target expression. Examiners award method marks for correct factoring and logical flow.
数学归纳法证明需要结构清晰：奠基、归纳假设，以及将 k+1 情况下的求和转化为目标表达式的代数链。评分官会给正确的因式分解和逻辑流程赋予方法分。
10. Parametric Equations: Differentiation and Tangents | 参数方程：微分与切线
Published by TutorHao | Year 13 Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply