📚 Mock Unit Test Walkthrough: Year 12 Edexcel Mathematics | 模拟单元测试解析：Edexcel Year 12 数学
This walkthrough breaks down a mock unit test designed for the Year 12 Edexcel AS Mathematics specification, covering topics from Pure Mathematics 1. The test is structured to mimic a typical 90-minute in‑class assessment, with questions that target algebraic manipulation, coordinate geometry, calculus, trigonometry, and the binomial expansion. Each section below highlights a representative question, explains the method step by step, and draws attention to common pitfalls.
本文详细解析一套针对 Year 12 Edexcel AS 数学课程设计的模拟单元测试，内容涵盖纯数1的核心考点。试卷模拟校内90分钟单元测验，题目涉及代数运算、坐标几何、微积分、三角学和二项式展开。以下每个小节选取一道典型考题，逐步展示解题方法，并聚焦常见错误。
1. Introduction & Test Overview | 引言与考试概览
The mock paper contains eight questions totalling 50 marks, reflecting the balance expected in an Edexcel AS Pure test. All working must be shown, and answers should be given in exact form unless otherwise stated. A solid understanding of index laws, quadratic theory, functions, calculus, and trigonometry is essential for success.
模拟卷共8道题，满分50分，体现Edexcel AS纯数测试的题型分布。所有解题过程必须呈现，除非特别说明，答案应以准确形式给出。成功完成这套试卷需要扎实掌握指数法则、二次函数理论、函数、微积分和三角学基础。
2. Question 1: Indices and Surds | 问题1：指数与根式
Question: Simplify (5⁻² × 25³)1/2, writing your answer as 5ⁿ. Hence, rationalise the denominator of 12 / (√75 − √27) and express it in the form p√3.
题目：化简 (5⁻² × 25³)1/2，答案写成5ⁿ。据此将 12 / (√75 − √27) 的分母有理化，并将结果表示为 p√3 的形式。
First, rewrite 25 as 5². Then 25³ = (5²)³ = 5⁶. The product inside the parentheses becomes 5⁻² × 5⁶ = 5⁴. Taking the square root gives (5⁴)1/2 = 5², so the answer is 5².
首先，把25写成5²，于是25³ = (5²)³ = 5⁶。括号内的乘积为5⁻² × 5⁶ = 5⁴。开平方后得(5⁴)1/2 = 5²，答案为5²。
For the surds, simplify √75 and √27: √75 = √(25×3) = 5√3, and √27 = √(9×3) = 3√3. The denominator becomes 5√3 − 3√3 = 2√3. The expression is then 12 / (2√3) = 6/√3.
接着处理根式：√75 = √(25×3) = 5√3，√27 = √(9×3) = 3√3。分母变为5√3 − 3√3 = 2√3。式子即为12 / (2√3) = 6/√3。
Rationalising gives 6/√3 × √3/√3 = 6√3/3 = 2√3, so p = 2. The step of scaling to the simplest surd form avoids common sign errors.
有理化：6/√3 × √3/√3 = 6√3/3 = 2√3，故p = 2。将根式化至最简形式可避免常见的符号错误。
3. Question 2: Quadratic Equations and the Discriminant | 问题2：二次方程与判别式
Question: The quadratic equation x² − 2kx + (k + 2) = 0 has two distinct real roots. Find the range of possible values for k.
题目：二次方程 x² − 2kx + (k + 2) = 0 有两个不相等的实数根。求k的取值范围。
For two distinct real roots the discriminant Δ must be positive: b² − 4ac > 0. Here a = 1, b = −2k, c = k + 2. So Δ = (−2k)² − 4(1)(k + 2) = 4k² − 4k − 8. Set the inequality: 4k² − 4k − 8 > 0. Divide by 4 to obtain k² − k − 2 > 0.
有两个不同实根时判别式Δ必须大于零：b² − 4ac > 0。此方程中a = 1, b = −2k, c = k + 2。于是Δ = (−2k)² − 4(1)(k + 2) = 4k² − 4k − 8。建立不等式：4k² − 4k − 8 > 0，除以4得 k² − k − 2 > 0。
Factorising the quadratic: (k − 2)(k + 1) > 0. Critical values are k = −1 and k = 2. A sign diagram or test values show the inequality holds when k < −1 or k > 2. Neither critical value is included because the inequality is strict.
分解因式：(k − 2)(k + 1) > 0。临界值为 k = −1 和 k = 2。通过符号表或取值检验可得不等式在 k < −1 或 k > 2 时成立。不等式不带等号，故临界值不包含在内。
Thus the solution set is k < −1 ∪ k > 2. Always restate the range in set notation to secure full marks on Edexcel papers.
因此解集为 k < −1 ∪ k > 2。在Edexcel考试中，用集合表示法写出最终范围有助于拿到全部分数。
4. Question 3: Composite Functions and Range | 问题3：复合函数与值域
Question: Given f(x) = 2x + 1 and g(x) = x² − 4, find fg(x) and gf(x). Determine the range of gf(x) for x ∈ ℝ.
题目：已知 f(x) = 2x + 1，g(x) = x² − 4。求 fg(x) 和 gf(x)，并确定当 x ∈ ℝ 时 gf(x) 的值域。
fg(x) means apply g first then f: fg(x) = f(g(x)) = f(x² − 4) = 2(x² − 4) + 1 = 2x² − 8 + 1 = 2x² − 7.
fg(x) 表示先 g 后 f：fg(x) = f(g(x)) = f(x² − 4) = 2(x² − 4) + 1 = 2x² − 8 + 1 = 2x² − 7。
gf(x) applies f first then g: gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)² − 4 = 4x² + 4x + 1 − 4 = 4x² + 4x − 3.
gf(x) 先 f 后 g：gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)² − 4 = 4x² + 4x + 1 − 4 = 4x² + 4x − 3。
To find the range, complete the square: 4x² + 4x − 3 = 4(x² + x) − 3 = 4[(x + ½)² − ¼] − 3 = 4(x + ½)² − 1 − 3 = 4(x + ½)² − 4. Since (x + ½)² ≥ 0, the minimum value is −4 (when x = −½). As x → ±∞, the expression → +∞. Hence the range is y ≥ −4, or [−4, ∞).
为求值域，使用配方法：4x² + 4x − 3 = 4(x² + x) − 3 = 4[(x + ½)² − ¼] − 3 = 4(x + ½)² − 1 − 3 = 4(x + ½)² − 4。因为 (x + ½)² ≥ 0，最小值是 −4（当 x = −½）。x → ±∞ 时函数值趋于 +∞。所以值域为 y ≥ −4，即 [−4, ∞)。
Many students mistakenly swap the order of composition; always read fg as f after g. Exploring the order helps avoid losing marks on simple substitution.
许多学生容易混淆复合顺序，务必记住 fg 表示先 g 后 f。分清次序能避免在简单代入上失分。
5. Question 4: Coordinate Geometry – Circle and Tangent | 问题4：坐标几何——圆与切线
Question: A circle has equation (x − 3)² + (y + 1)² = 25. Find the centre and radius. The line y = 2x + k is tangent to the circle. Determine the possible values of k.
题目：圆的方程为 (x − 3)² + (y + 1)² = 25。求圆心与半径。直线 y = 2x + k 与圆相切，求k的可能取值。
Centre C(3, −1), radius r = 5. For tangency, the perpendicular distance from the centre to the line must equal the radius. Write the line in general form: 2x − y + k = 0. Distance d = |2(3) − (−1) + k| / √(2² + (−1)²) = |6 + 1 + k| / √5 = |k + 7|/√5.
圆心 C(3, −1)，半径 r = 5。直线与圆相切的条件是圆心到直线的垂直距离等于半径。将直线方程化为一般式：2x − y + k = 0。距离 d = |2(3) − (−1) + k| / √(2² + (−1)²) = |7 + k|/√5。
Set d = r: |k + 7|/√5 = 5 → |k + 7| = 5√5. Therefore k + 7 = ±5√5, giving k = −7 ± 5√5. So the two values are k = −7 + 5√5 and k = −7 − 5√5.
令 d = r：|k + 7|/√5 = 5 → |k + 7| = 5√5。因此 k + 7 = ±5√5，解得 k = −7 ± 5√5。两个可能值分别为 k = −7 + 5√5 和 k = −7 − 5√5。
An alternative method substitutes the line into the circle to create a quadratic and sets its discriminant to zero. Both approaches are valid; the distance method is often quicker.
另一种方法是将直线表达式代入圆的方程，得到关于x的二次方程并令判别式为零。两种方法均可，距离法通常更快捷。
6. Question 5: Differentiation – Tangents and Stationary Points | 问题5：微分——切线与驻点
Question: The curve C has equation y = x³ − 5x² + 3x + 4. Find the equation of the tangent to C at x = 2. Hence find the coordinates of the stationary points and determine their nature.
题目：曲线C的方程为 y = x³ − 5x² + 3x + 4。求曲线在 x = 2 处的切线方程，并求出驻点坐标且判断其性质。
Differentiate: dy/dx = 3x² − 10x + 3. At x = 2, gradient m = 3(2)² − 10(2) + 3 = 12 − 20 + 3 = −5. The y‑coordinate: y(2) = 8 − 20 + 6 + 4 = −2. Using y − y₁ = m(x − x₁): y − (−2) = −5(x − 2) → y + 2 = −5x + 10 → y = −5x + 8.
求导：dy/dx = 3x² − 10x + 3。代入 x = 2 得斜率 m = 3(2)² − 10(2) + 3 = 12 − 20 + 3 = −5。此时 y 坐标为：y(2) = 8 − 20 + 6 + 4 = −2。由点斜式 y − y₁ = m(x − x₁)：y + 2 = −5(x − 2) → y = −5x + 8。
For stationary points set dy/dx = 0: 3x² − 10x + 3 = 0 → (3x − 1)(x − 3) = 0 → x = 1/3, x = 3. Substitute back: x = 1/3 gives y = (1/27) − 5/9 + 1 + 4 = (1 − 15 + 27 + 108)/27 = 121/27; x = 3 gives y = 27 − 45 + 9 + 4 = −5. So points are (1/3, 121/27) and (3, −5).
令 dy/dx = 0 求驻点：3x² − 10x + 3 = 0 → (3x − 1)(x − 3) = 0 → x = 1/3, x = 3。回代：x = 1/3 得 y = 1/27 − 5/9 + 1 + 4 = (1 − 15 + 27 + 108)/27 = 121/27；x = 3 得 y = 27 − 45 + 9 + 4 = −5。两驻点为 (1/3, 121/27) 和 (3, −5)。
Nature: second derivative d²y/dx² = 6x − 10. At x = 1/3, f” = 2 − 10 = −8 < 0 → local maximum. At x = 3, f'' = 18 − 10 = 8 > 0 → local minimum.
性质判别：二阶导数 d²y/dx² = 6x − 10。在 x = 1/3 处，f” = 2 − 10 = −8 < 0 → 局部极大值。在 x = 3 处，f'' = 18 − 10 = 8 > 0 → 局部极小值。
Sketching a sign diagram for dy/dx is equally accepted. The question tests solid differentiation technique and interpretation.
采用一阶导数符号表判别同样可行。此题重在考查扎实的求导能力和几何意义的理解。
7. Question 6: Integration and Area Under a Curve | 问题6：积分与曲线下方面积
Question: Find the area enclosed between the curve y = 4x − x² and the x‑axis. The region is bounded by the curve and the lines x = 1 and x = 3.
题目：求由曲线 y = 4x − x² 与 x 轴围成的区域面积，该区域由曲线和直线 x = 1、x = 3 所界定。
First confirm the curve lies above the x‑axis on [1,3]. At x = 1, y = 3; at x = 2, y = 4; at x = 3, y = 3. The quadratic opens downward, vertex at x = 2, y = 4 – clearly positive. So area = ∫₍₁₎³ (4x − x²) dx.
首先确认曲线在区间 [1,3] 上位于 x 轴上方。x=1时 y=3；x=2时 y=4；x=3时 y=3。该二次函数开口向下，顶点 (2,4)，显然为正。因此面积 = ∫₁³ (4x − x²) dx。
Integrate term by term: ∫4x dx = 2x², ∫x² dx = x³/3. So the antiderivative is 2x² − x³/3. Evaluate from 1 to 3: [2(3)² − (27)/3] − [2(1)² − 1/3] = [18 − 9] − [2 − 1/3] = 9 − (5/3) = 22/3 square units.
逐项积分：∫4x dx = 2x²，∫x² dx = x³/3。原函数为 2x² − x³/3。从1到3求值：[2(9) − 27/3] − [2 − 1/3] = [18 − 9] − [2 − 1/3] = 9 − (5/3) = 22/3 平方单位。
Many students forget to subtract the lower limit properly; writing 2(1)² − 1/3 as 2 − 1/3 = 5/3 is essential. The answer is left as an exact fraction, which is preferred in Edexcel markschemes.
不少同学会忽略正确代入下限，务必准确计算 2(1)² − 1/3 = 5/3。答案保留准确分数形式，这符合Edexcel评分要求。
8. Question 7: Binomial Expansion | 问题7：二项式展开
Question: Expand (2 − 3x)⁴ up to the term in x³. Hence find the coefficient of x³ in the expansion of (1 + x)(2 − 3x)⁴.
题目：将 (2 − 3x)⁴ 展开至 x³ 项，并据此求出 (1 + x)(2 − 3x)⁴ 展开式中 x³ 项的系数。
Using the binomial theorem: (a + b)ⁿ = Σ [nCr] aⁿ⁻ʳ bʳ. With a = 2, b = −3x, n = 4. Terms: r = 0: ⁴C₀ (2)⁴(−3x)⁰ = 16. r = 1: ⁴C₁ (2)³ (−3x) = 4 × 8 × (−3x) = −96x. r = 2: ⁴C₂ (2)² (−3x)² = 6 × 4 × 9x² = 216x². r = 3: ⁴C₃ (2)¹ (−3x)³ = 4 × 2 × (−27x³) = −216x³. So (2 − 3x)⁴ = 16 − 96x + 216x² − 216x³ + …
利用二项式定理：(a + b)ⁿ = Σ [nCr] aⁿ⁻ʳ bʳ。令 a = 2, b = −3x, n = 4。逐项展开：r = 0: ⁴C₀·2⁴·(−3x)⁰ = 16。r = 1: ⁴C₁·2³·(−3x) = 4×8×(−3x) = −96x。r = 2: ⁴C₂·2²·(−3x)² = 6×4×9x² = 216x²。r = 3: ⁴C₃·2¹·(−3x)³ = 4×2×(−27x³) = −216x³。因此 (2 − 3x)⁴ = 16 − 96x + 216x² − 216x³ + …
Now multiply by (1 + x). The x³ coefficient arises from 1 × (−216x³) and x × (216x²) = 216x³. Sum = −216 + 216 = 0. Therefore the coefficient of x³ in the product is 0.
再乘以 (1 + x)。x³ 项来自 1 × (−216x³) 和 x × (216x²) = 216x³。总和为 −216 + 216 = 0。所以乘积中 x³ 的系数为0。
This neat cancellation is a classic exam twist. Always try to extend the original expansion one step beyond the requested term to avoid missing contributions from higher-degree products.
这种巧妙抵消是常见的考题策略。务必在展开时多求一项，避免遗漏更高次乘积的贡献。
9. Question 8: Trigonometric Equations | 问题8：三角方程
Question: Solve 2 sin² θ − cos θ = 1 for 0° ≤ θ ≤ 360°. Give your answers to 1 decimal place where necessary.
题目：解方程 2 sin² θ − cos θ = 1，θ 的范围为 0° ≤ θ ≤ 360°。若需要，答案精确至小数点后一位。
Use sin² θ = 1 − cos² θ. Substitute: 2(1 − cos² θ) − cos θ = 1 → 2 − 2 cos² θ − cos θ = 1 → −2 cos² θ − cos θ + 1 = 0. Multiply by −1: 2 cos² θ + cos θ − 1 = 0.
利用恒等式 sin² θ = 1 − cos² θ 代入：2(1 − cos² θ) − cos θ = 1 → 2 − 2 cos² θ −
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