📚 Year 13 CCEA Maths: Unit Test Mock Paper Walkthrough | 英国高考 CCEA 数学：单元测试模拟卷解析
This article walks you through a typical CCEA Year 13 unit test mock paper, providing detailed solutions and key techniques for pure mathematics topics including differentiation, integration, trigonometry, exponentials, sequences, vectors, and differential equations.
本文为你逐题解析一份典型的 CCEA 13 年级单元测试模拟卷，涵盖纯数主题的详细解法与关键技巧，包括微分、积分、三角、指数、数列、向量和微分方程。
1. Differentiation and Tangent Line | 微分与切线
Question: Find the equation of the tangent to the curve y = x³ − 4x² + 5x − 2 at the point where x = 2.
题目：求曲线 y = x³ − 4x² + 5x − 2 在 x = 2 处的切线方程。
Differentiate the function: dy/dx = 3x² − 8x + 5. Substituting x = 2 gives the gradient m = 3(2)² − 8(2) + 5 = 12 − 16 + 5 = 1.
对函数求导：dy/dx = 3x² − 8x + 5。代入 x = 2 得梯度 m = 3(2)² − 8(2) + 5 = 12 − 16 + 5 = 1。
The y-coordinate at x = 2 is y(2) = 2³ − 4×2² + 5×2 − 2 = 8 − 16 + 10 − 2 = 0. The point is (2, 0).
x = 2 处的 y 坐标为 y(2) = 2³ − 4×2² + 5×2 − 2 = 8 − 16 + 10 − 2 = 0，所求点为 (2, 0)。
Using the point–gradient form, the tangent equation is y − 0 = 1(x − 2), hence y = x − 2.
利用点斜式，切线方程为 y − 0 = 1(x − 2)，即 y = x − 2。
2. Definite Integration and Area | 定积分与面积
Question: Find the area bounded by the curve y = 6 − x − x² and the x-axis.
题目：求曲线 y = 6 − x − x² 与 x 轴围成的面积。
The curve meets the x-axis when 6 − x − x² = 0, which factors to (3 − x)(2 + x) = 0, giving intersection points at x = −3 and x = 2.
曲线与 x 轴相交于 6 − x − x² = 0，因式分解为 (3 − x)(2 + x) = 0，得交点 x = −3 与 x = 2。
The required area is the definite integral ∫₋₃² (6 − x − x²) dx. Integrate term by term to get [6x − ½x² − ⅓x³]₋₃².
所求面积为定积分 ∫₋₃² (6 − x − x²) dx。逐项积分得 [6x − ½x² − ⅓x³]₋₃²。
Evaluating at the upper limit x = 2: 6(2) − ½(4) − ⅓(8) = 12 − 2 − 8/3 = 22/3. At the lower limit x = −3: 6(−3) − ½(9) − ⅓(−27) = −18 − 4.5 + 9 = −13.5. Subtract: (22/3) − (−13.5) = 22/3 + 27/2 = (44+81)/6 = 125/6.
代入上限 x = 2：6(2) − ½(4) − ⅓(8) = 12 − 2 − 8/3 = 22/3；代入下限 x = −3：6(−3) − ½(9) − ⅓(−27) = −18 − 4.5 + 9 = −13.5。相减得 (22/3) − (−13.5) = 22/3 + 27/2 = (44+81)/6 = 125/6 平方单位。
3. Trigonometric Equation | 三角方程
Question: Solve 2 sin²θ − 3 sin θ + 1 = 0 for 0° ≤ θ ≤ 360°.
题目：在 0° ≤ θ ≤ 360° 范围内解方程 2 sin²θ − 3 sin θ + 1 = 0。
Treat the equation as a quadratic in sin θ. Factorise: (2 sin θ − 1)(sin θ − 1) = 0.
将方程视为关于 sin θ 的二次方程，因式分解得 (2 sin θ − 1)(sin θ − 1) = 0。
Thus sin θ = ½ or sin θ = 1. For sin θ = ½, the principal solutions are θ = 30° and θ = 150°. For sin θ = 1, the solution is θ = 90°.
因此 sin θ = ½ 或 sin θ = 1。由 sin θ = ½ 得基本解 θ = 30° 与 θ = 150°；由 sin θ = 1 得 θ = 90°。
The complete solution set is θ = 30°, 90°, 150°.
完整解集为 θ = 30°, 90°, 150°。
4. Exponential and Logarithmic Equations | 指数与对数方程
Question: Solve e²ˣ − 5eˣ + 6 = 0.
题目：解方程 e²ˣ − 5eˣ + 6 = 0。
Let u = eˣ. The equation becomes u² − 5u + 6 = 0, which factorises to (u − 2)(u − 3) = 0.
设 u = eˣ，原方程化为 u² − 5u + 6 = 0，因式分解为 (u − 2)(u − 3) = 0。
Hence u = 2 or u = 3. Replacing u gives eˣ = 2 ⇒ x = ln 2, and eˣ = 3 ⇒ x = ln 3.
所以 u = 2 或 u = 3。代回得 eˣ = 2 ⇒ x = ln 2，以及 eˣ = 3 ⇒ x = ln 3。
5. Implicit Differentiation | 隐函数求导
Question: Find the equation of the tangent to the circle x² + y² = 25 at the point (3, 4).
题目：求圆 x² + y² = 25 在点 (3, 4) 处的切线方程。
Differentiate both sides with respect to x: 2x + 2y (dy/dx) = 0. Solve for dy/dx to obtain dy/dx = −x / y.
两边对 x 求导：2x + 2y (dy/dx) = 0，整理得 dy/dx = −x / y。
At the point (3, 4), the gradient is m = −3 / 4. Using y − 4 = m(x − 3) gives the tangent line y − 4 = −¾ (x − 3), or 4y − 16 = −3x + 9 ⇒ 3x + 4y = 25.
在点 (3, 4) 处，梯度为 m = −3 / 4。代入点斜式 y − 4 = m(x − 3) 得切线方程 y − 4 = −¾ (x − 3)，化简为 3x + 4y = 25。
6. Binomial Expansion | 二项式展开
Question: In the expansion of (2 − x)⁴, find the coefficient of the x³ term.
题目：求 (2 − x)⁴ 的展开式中 x³ 项的系数。
The general term is given by Tₖ₊₁ = C(4, k) × 2⁴⁻ᵏ × (−x)ᵏ. For x³, we need k = 3.
通项公式为 Tₖ₊₁ = C(4, k) × 2⁴⁻ᵏ × (−x)ᵏ。要得到 x³，取 k = 3。
The term is C(4, 3) × 2¹ × (−x)³ = 4 × 2 × (−x³) = −8 x³. Hence the coefficient of x³ is −8.
该项为 C(4, 3) × 2¹ × (−x)³ = 4 × 2 × (−x³) = −8 x³。因此 x³ 的系数为 −8。
7. Arithmetic and Geometric Sequences | 等差数列与等比数列
Question: In an arithmetic sequence, the fourth term is 11 and the tenth term is 29. Find the first term and common difference, and then find the sum of the first 20 terms.
题目：在一个等差数列中，第四项为 11，第十项为 29。求首项和公差，并计算前 20 项的和。
Using uₙ =
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