📚 Year 11 CIE Maths: Mock Unit Test Paper Analysis | CIE 数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a Year 11 CIE Mathematics mock unit test, covering Number, Algebra, Functions, Coordinate Geometry, Circle Theorems, Trigonometry, Statistics and Probability. Each question is broken down step by step to help you understand the key concepts and exam techniques.
本文详细解析了一套 Year 11 CIE 数学单元测试模拟卷，涵盖数、代数、函数、坐标几何、圆定理、三角学、统计与概率。每个问题都逐步分解，帮助您掌握关键概念与应试技巧。
1. Introduction to Mock Test | 模拟卷介绍
Mock tests are essential for identifying strengths and weaknesses before the actual examination. This paper simulates a typical 45-minute unit test with 8 questions totaling 50 marks. Attempting it under timed conditions will help you manage time effectively.
模拟测试对于在正式考试前识别强项与薄弱环节至关重要。本试卷模拟一份典型的45分钟单元测试，包含8道题，总分50分。在限时条件下完成测试有助于你有效管理时间。
We recommend you try the paper first, then review the solutions and commentary below to deepen your understanding of each topic.
我们建议你先独立完成试卷，然后对照下面的解析和评注，以加深对每个主题的理解。
2. Paper Structure and Topics | 试卷结构与考察主题
The table below outlines the question distribution and mark allocation. Understanding the structure can help you allocate revision time effectively.
下表概述了题目分布和分值安排。了解试卷结构有助于你合理分配复习时间。
| Question | Topic | Marks |
|---|---|---|
| 1 | Surds and Indices | 6 |
| 2 | Solving Quadratic Equations | 6 |
| 3 | Composite Functions | 5 |
| 4 | Coordinate Geometry | 7 |
| 5 | Circle Theorems | 6 |
| 6 | Trigonometry - Sine Rule | 7 |
| 7 | Statistics from Frequency Table | 6 |
| 8 | Probability Tree Diagram | 7 |
The total of 50 marks covers a balanced mix of pure and applied mathematics. Focus on method marks, as clear working can earn most of the marks even if the final answer is slightly off.
总分50分涵盖了纯数学与应用数学的均衡组合。要注重方法分，因为即使最终答案稍有偏差，清晰的解题过程也能拿到大部分分数。
3. Question 1: Surds and Indices | 题目1：根式与指数
Question: Simplify √48 + √75 − √12, giving your answer in the form k√3, where k is an integer. (6 marks)
题目：化简 √48 + √75 − √12，将答案表示为 k√3 的形式，其中 k 为整数。（6分）
We start by breaking down each surd into a product of a perfect square and another factor: √48 = √16×3 = 4√3. Similarly, √75 = √25×3 = 5√3, and √12 = √4×3 = 2√3.
我们首先将每个根式分解为一个完全平方数与另一个因式的乘积：√48 = √16×3 = 4√3，同理 √75 = √25×3 = 5√3，以及 √12 = √4×3 = 2√3。
Now combine the terms: 4√3 + 5√3 − 2√3 = (4 + 5 − 2)√3 = 7√3. Therefore, k = 7.
现在合并各项：4√3 + 5√3 − 2√3 = (4 + 5 − 2)√3 = 7√3，因此 k = 7。
Many students forget to fully factor out the largest perfect square, which leads to more complicated expressions. Always check if the number under the square root can be divided by 4, 9, 16, 25, etc.
许多学生忘记完全提取最大的完全平方数，导致表达式变复杂。一定要检查根号下的数字能否被 4、9、16、25 等整除。
In index form, this operation uses the rule √ab = √a × √b. Mastering this makes simplifying surds straightforward.
用指数形式表示，此运算使用了规则 √ab = √a × √b。掌握这一点能让根式化简变得简单明了。
4. Question 2: Solving Quadratic Equations | 题目2：二次方程求解
Question: Solve x² − 7x + 10 = 0. (6 marks)
题目：解方程 x² − 7x + 10 = 0。（6分）
This quadratic can be factorised. We look for two numbers that multiply to +10 and add up to −7. Those numbers are −5 and −2.
该二次式可以因式分解。我们寻找两个相乘得 +10、相加得 −7 的数，它们是 −5 和 −2。
7x + 10 = (x − 5)(x − 2). Setting each factor equal to zero gives x − 5 = 0 or x − 2 = 0.
因此，x² − 7x + 10 = (x − 5)(x − 2)。令每个因式等于零得到 x − 5 = 0 或 x − 2 = 0。
are x = 5 and x = 2. Always write the final answer clearly, separating the two roots with 'or'.
因此解为 x = 5 和 x = 2。始终清晰地写出最终答案，用“或”分隔两个根。
the quadratic cannot be factorised, remember you can use the quadratic formula: x = [−b ± √b² − 4ac] / 2a. For this equation a=1, b=−7, c=10, the discriminant is 49−40=9, giving √9 = 3, which leads to x = (7 ± 3)/2, i.e. x=5 or x=2.
如果二次式无法因式分解，记住可以使用求根公式：x = [−b ± √b² − 4ac] / 2a。对于此方程 a=1, b=−7, c=10，判别式为 49−40=9，得 √9 = 3，从而 x = (7 ± 3)/2，即 x=5 或 x=2。
5. Question 3: Composite Functions | 题目3：复合函数
Question: f(x) = 3x − 2 and g(x) = x² + 1. Find (a) fg(2), (b) gf(x). (5 marks)
题目：f(x) = 3x − 2，g(x) = x² + 1。求 (a) fg(2)，(b) gf(x)。（5分）
uate g(2): g(2) = (2)² + 1 = 4 + 1 = 5. Then f(5) = 3(5) − 2 = 15 − 2 = 13. So fg(2) = 13.
对于 (a) 部分，fg(2) 表示 f(g(2))。先计算 g(2)：g(2) = (2)² + 1 = 4 + 1 = 5，然后 f(5) = 3(5) − 2 = 15 − 2 = 13。因此 fg(2) = 13。
Expand the square: (3x − 2)² = 9x² − 12x + 4. Then add 1 to obtain gf(x) = 9x² − 12x + 5.
对于 (b) 部分，gf(x) 表示 g(f(x))。将 f(x) 代入 g： gf(x) = (3x − 2)² + 1。展开平方：(3x − 2)² = 9x² − 12x + 4，再加 1 得到 gf(x) = 9x² − 12x + 5。
Remember the order: fg(2) means apply g first, then f. A common mistake is to reverse the order or to misinterpret the notation.
请记住顺序：fg(2) 表示先应用 g，再应用 f。常见错误是颠倒顺序或误解符号。
6. Question 4: Coordinate Geometry – Midpoint and Distance | 题目4：坐标几何——中点与距离
Question: Points A(1, 2) and B(5, 8) are given. (a) Find the midpoint M of AB. (b) Calculate the exact length of AB in surd form. (7 marks)
题目：已知点 A(1, 2) 和 B(5, 8)。(a) 求线段 AB 的中点 M。(b) 以根式形式计算 AB 的精确长度。（7分）
(x₁+x₂)/2, (y₁+y₂)/2). Substituting the coordinates gives M = ((1+5)/2, (2+8)/2) = (3, 5).
对于中点，使用公式 ((x₁+x₂)/2, (y₁+y₂)/2)。代入坐标得 M = ((1+5)/2, (2+8)/2) = (3, 5)。
e AB, apply the distance formula: d = √[(x₂−x₁)² + (y₂−y₁)²]. So AB = √[(5−1)² + (8−2)²] = √4² + 6² = √16 + 36 = √52.
对于距离 AB，使用距离公式：d = √[(x₂−x₁)² + (y₂−y₁)²]。因此 AB = √[(5−1)² + (8−2)²] = √4² + 6² = √16 + 36 = √52。
Simplify √52 by extracting the largest square factor: √52 = √4×13 = 2√13. This is the exact length.
化简 √52，提取最大平方因子：√52 = √4×13 = 2√13。这就是精确长度。
Leaving the answer as 2√13 is preferred over a decimal approximation unless the question specifies otherwise. This shows precision and good surd handling.
除非题目另有说明，答案最好保留为 2√13 而非小数近似值。这体现了精确度和良好的根式处理能力。
7. Question 5: Circle Theorems – Angle Calculation | 题目5：圆定理——角度计算
Question: = 110°. Points A, B, and C lie on the circumference. Find the size of angle ACB. (6 marks)
题目：O 为圆心，∠AOB = 110°。点 A、B、C 均在圆周上。求 ∠ACB 的大小。（6分）
mference when both are subtended by the same arc AB. Therefore, angle ACB = ½ × angle AOB.
当圆心角和圆周角对应同一条弧 AB 时，圆心角是圆周角的两倍。因此 ∠ACB = ½ × ∠AOB。
Substituting gives angle ACB = 110°/2 = 55°. Always quote the reason 'angle at centre = 2 × angle at circumference' to secure method marks.
代入得 ∠ACB = 110°/2 = 55°。务必注明理由“圆心角等于圆周角的两倍”以获得方法分。
This theorem is one of the most frequently tested circle theorems. Practice recognising the 'arrowhead' shape formed by the centre and points on the circumference.
该定理是考试中最常出现的圆定理之一。请练习识别由圆心和圆周点构成的“箭头”形状。
If the question had asked about the reflex angle AOB (250°), then angle ACB on the opposite arc would be 125°. Pay close attention to which arc the angle subtends.
如果题目问的是优角 AOB (250°)，那么对应另一弧上的 ∠ACB 将是 125°。要密切注意角所对应的弧是哪个。
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