Pre-U WJEC Mathematics: Unit Test Mock Paper Walkthrough | Pre-U WJEC 数学：单元测试模拟卷解析

📚 Pre-U WJEC Mathematics: Unit Test Mock Paper Walkthrough | Pre-U WJEC 数学：单元测试模拟卷解析

Welcome to this mock paper walkthrough for the Pre-U WJEC Mathematics unit test. This resource provides detailed, step-by-step solutions to a selection of typical exam-style questions, covering key topics such as binomial expansion, functions, trigonometry, calculus, vectors, and series. Use this to consolidate your understanding and improve your problem-solving technique.

欢迎来到这份 Pre-U WJEC 数学单元测试模拟卷解析。本资源对一系列典型考题进行逐步详解，涵盖二项展开、函数、三角学、微积分、向量和级数等核心主题。请通过此解析巩固理解并提高解题技巧。

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1. Binomial Expansion and Approximation | 二项展开与近似计算

Question: Use the binomial theorem to expand (1+3x)⁴ in ascending powers of x up to and including the term in x³. Hence, use your expansion to find an approximation for (1.03)⁴, giving your answer to 4 decimal places.

题目：利用二项式定理将 (1+3x)⁴ 按 x 的升幂展开，直到含 x³ 项为止。然后，利用展开式求出 (1.03)⁴ 的近似值，结果精确至4位小数。

The binomial expansion of (1+ax)ⁿ is 1 + nax + [n(n−1)/2!](ax)² + [n(n−1)(n−2)/3!](ax)³ + ... . For (1+3x)⁴, a=3 and n=4.

(1+ax)ⁿ 的二项展开式为 1 + nax + [n(n−1)/2!](ax)² + [n(n−1)(n−2)/3!](ax)³ + ...。对于 (1+3x)⁴，有 a=3，n=4。

6 × 9 = 54 → term: 54x². Coefficient of x³: [4×3×2/6] × (3)³ = 4 × 27 = 108 → term: 108x³.

x 项系数：4 × 3 = 12 → 项：12x。x² 项系数：[4×3/2] × (3)² = 6 × 9 = 54 → 项：54x²。x³ 项系数：[4×3×2/6] × (3)³ = 4 × 27 = 108 → 项：108x³。

Thus, (1+3x)⁴ ≈ 1 + 12x + 54x² + 108x³ for small x.

因此，(1+3x)⁴ ≈ 1 + 12x + 54x² + 108x³（当 x 很小时）。

108(0.000001) = 1 + 0.12 + 0.0054 + 0.000108 = 1.125508. Rounding to 4 d.p. gives 1.1255.

要近似计算 (1.03)⁴，将其写作 (1 + 3×0.01)⁴，因此 x = 0.01。代入 x = 0.01：1 + 12(0.01) + 54(0.0001) + 108(0.000001) = 1 + 0.12 + 0.0054 + 0.000108 = 1.125508。四舍五入至4位小数得 1.1255。

Answer: (1.03)⁴ ≈ 1.1255

答案：(1.03)⁴ ≈ 1.1255

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2. Functions and Inverse Functions | 函数与反函数

f(x) = (2x + 1)/(x − 3) for x ≠ 3, find the inverse function f⁻¹(x) and state its domain.

题目：给定 f(x) = (2x + 1)/(x − 3)，x ≠ 3，求其反函数 f⁻¹(x) 并说明其定义域。

2x + 1)/(x − 3). To find the inverse, swap x and y: x = (2y + 1)/(y − 3). Now solve for y.

设 y = (2x + 1)/(x − 3)。为求反函数，交换 x 与 y：x = (2y + 1)/(y − 3)。现在解出 y。

1. Collect y terms: xy − 2y = 3x + 1 → y(x − 2) = 3x + 1. Therefore, y = (3x + 1)/(x − 2).

两边同乘 (y − 3)：x(y − 3) = 2y + 1 → xy − 3x = 2y + 1。将含 y 的项集中：xy − 2y = 3x + 1 → y(x − 2) = 3x + 1。因此，y = (3x + 1)/(x − 2)。

main of f⁻¹ is all real numbers except where the denominator is zero, i.e. x ≠ 2. This corresponds to the range of f.

因此，f⁻¹(x) = (3x + 1)/(x − 2)。f⁻¹ 的定义域为所有实数，除去分母为零之处，即 x ≠ 2。这与 f 的值域相对应。

Answer: f⁻¹(x) = (3x + 1)/(x − 2); domain: x ∈ ℝ, x ≠ 2.

答案：f⁻¹(x) = (3x + 1)/(x − 2)；定义域：x ∈ ℝ, x ≠ 2。

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3. Trigonometric Identities and Equations | 三角恒等式与方程

t (sin θ + cos θ)² = 1 + sin 2θ. (b) Hence solve the equation sin 2θ = 1/2 for 0 ≤ θ ≤ 2π.

题目：(a) 证明 (sin θ + cos θ)² = 1 + sin 2θ。(b) 由此解方程 sin 2θ = 1/2，其中 0 ≤ θ ≤ 2π。

dentities sin²θ + cos²θ = 1 and 2 sin θ cos θ = sin 2θ, we obtain 1 + sin 2θ. Thus proved.

展开左边：(sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ。利用恒等式 sin²θ + cos²θ = 1 以及 2 sin θ cos θ = sin 2θ，得到 1 + sin 2θ。得证。

θ in [0, 4π] are 2θ = π/6, 5π/6, 13π/6, 17π/6. Therefore, θ = π/12, 5π/12, 13π/12, 17π/12.

对 (b) 部分，sin 2θ = 1/2。2θ 在 [0, 4π] 上的主值为 2θ = π/6, 5π/6, 13π/6, 17π/6。因此，θ = π/12, 5π/12, 13π/12, 17π/12。

All four solutions lie within the required interval 0 ≤ θ ≤ 2π.

所有四个解均落在指定区间 0 ≤ θ ≤ 2π 内。

Answer: θ = π/12, 5π/12, 13π/12, 17π/12.

答案：θ = π/12, 5π/12, 13π/12, 17π/12。

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4. Differentiation: Product Rule and Tangent | 微分：乘积法则与切线

pect to x. Hence find the equation of the tangent to the curve at the point where x = π/2.

题目：对 y = x² sin(2x) 求关于 x 的导数。继而求出曲线在 x = π/2 处的切线方程。

Use the product rule: let u = x², v = sin(2x). Then u' = 2x, v' = 2 cos(2x). dy/dx = u'v + uv' = 2x sin(2x) + x²·2 cos(2x) = 2x sin(2x) + 2x² cos(2x).

使用乘积法则：设 u = x²，v = sin(2x)，则 u' = 2x，v' = 2 cos(2x)。dy/dx = u'v + uv' = 2x sin(2x) + x²·2 cos(2x) = 2x sin(2x) + 2x² cos(2x)。

: y = (π/2)² sin(π) = 0. Tangent equation: y − 0 = (−π²/2)(x − π/2) → y = (−π²/2)x + π³/4.

当 x = π/2 时：sin(π) = 0，cos(π) = −1。因此 dy/dx = 2(π/2)(0) + 2(π²/4)(−1) = −π²/2。y 坐标：y = (π/2)² sin(π) = 0。切线方程：y − 0 = (−π²/2)(x − π/2) → y = (−π²/2)x + π³/4。

Answer: dy/dx = 2x sin(2x) + 2x² cos(2x); tangent: y = (−π²/2)x + π³/4.

答案：dy/dx = 2x sin(2x) + 2x² cos(2x)；切线：y = (−π²/2)x + π³/4。

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5. Integration by Parts | 分部积分

Question: Evaluate the definite integral ∫₀¹ x eˣ dx.

题目：计算定积分 ∫₀¹ x eˣ dx。

Use integration by parts. Let u = x, so du/dx = 1. Let dv/dx = eˣ, so v = eˣ. The formula ∫ u dv = uv − ∫ v du gives: ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx.

使用分部积分法。设 u = x，则 du/dx = 1。设 dv/dx = eˣ，则 v = eˣ。公式 ∫ u dv = uv − ∫ v du 给出：∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx。

Evaluate the first term: [x eˣ]₀¹ = (1·e¹) − (0·e⁰) = e. The remaining integral: ∫₀¹ eˣ dx = [eˣ]₀¹ = e − 1. Hence the result is e − (e − 1) = 1.

计算第一项：[x eˣ]₀¹ = (1·e¹) − (0·e⁰) = e。剩余积分：∫₀¹ eˣ dx = [eˣ]₀¹ = e − 1。因此结果为 e − (e − 1) = 1。

Answer: 1

答案：1

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6. Exponential and Logarithmic Equations | 指数与对数方程

Question: Solve the equation 2e²ˣ − 5eˣ + 2 = 0, giving your answers in exact logarithmic form.

题目：解方程 2e²ˣ − 5eˣ + 2 = 0，以精确对数形式给出答案。

Let y = eˣ equation becomes 2y² − 5y + 2 = 0. Factorise: (2y − 1)(y − 2) = 0. Hence y = 1/2 or y = 2.

设 y = eˣ。则方程化为 2y² − 5y + 2 = 0。因式分解：(2y − 1)(y − 2) = 0。因此 y = 1/2 或 y = 2。

Substituting back: eˣ = 1/2 ⇒ x = ln(1/2) = −ln 2. eˣ = 2 ⇒ x = ln 2. Both are valid.

代回原变量：eˣ = 1/2 ⇒ x = ln(1/2) = −ln 2。eˣ = 2 ⇒ x = ln 2。两者均有效。

Answer: x = ln 2 or x = −ln 2.

答案：x = ln 2 或 x = −ln 2。

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7. Vectors: Magnitude and Angle | 向量：模与夹角

Question: Points A and B have coordinates A(1, 2, −3) and B(4, −1, 5). (a) Find the vector AB. (b) Calculate the angle between AB and the vector i + 2j − 2k.

题目：点 A 和点 B 的坐标分别为 A(1, 2, −3) 和 B(4, −1, 5)。(a) 求向量 AB。(b) 计算 AB 与向量 i + 2j − 2k 之间的夹角。

Vector AB = (4 − 1)i + (−1 − 2)j + (

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