📚 TMUA Mock Questions and Problem-Solving Strategies | TMUA 真题模拟与解题思路
This article offers a set of original TMUA-style questions with full solutions. The TMUA is used by universities such as Cambridge and LSE to test mathematical thinking, so the questions below are designed to reflect both the content and the reasoning style of the real test.
本文提供一套原创的TMUA风格题目及完整解析。TMUA被剑桥大学、伦敦政治经济学院等院校用于考查数学思维，因此下面的题目在内容和推理方式上都力求贴近真实考试。
1. Understand the TMUA Format | 理解TMUA考试结构
The TMUA consists of two 75-minute papers. Paper 1, Mathematical Thinking, has 20 multiple-choice questions on pure mathematics content: algebra, functions, calculus, trigonometry, sequences, and logarithms. Paper 2, Mathematical Reasoning, also has 20 multiple-choice questions and focuses on logical arguments, proof, and interpreting mathematical statements.
TMUA由两场各75分钟的试卷组成。第一卷“数学思维”包含20道选择题，覆盖纯数内容：代数、函数、微积分、三角、数列与对数。第二卷“数学推理”同样有20道选择题，重点考查逻辑论证、证明和对数学命题的理解。
No calculator or formula booklet is allowed. This means numerical fluency and algebraic accuracy are essential under time pressure.
考试不允许使用计算器或公式册，因此数字敏感度和代数准确性在时间压力下非常重要。
- Paper 1: 20 questions, 75 minutes, standard A-level techniques. | 第一卷：20题，75分钟，ALevel标准技巧。
- Paper 2: 20 questions, 75 minutes, reasoning and logic. | 第二卷：20题，75分钟，推理与逻辑。
2. Algebraic Manipulation | 代数变形
Sample question: Given that x² + 3x – 10 = 0, find the value of x² + 3x + 1.
模拟题：已知 x² + 3x – 10 = 0，求 x² + 3x + 1 的值。
Do not solve the quadratic unless you have no other option. The original equation can be rewritten as x² + 3x = 10. Therefore x² + 3x + 1 = 10 + 1 = 11.
不要一上来就去解二次方程，除非别无选择。原方程可改写为 x² + 3x = 10，因此 x² + 3x + 1 = 10 + 1 = 11。
This question illustrates the TMUA principle: use what you are given. Factoring x² + 3x – 10 = (x – 2)(x + 5) would also give x = 2 or x = -5, but that is slower.
这道题体现了TMUA的重要原则：利用已知条件。若分解 x² + 3x – 10 = (x – 2)(x + 5)，可得 x = 2 或 x = -5，但这样更慢。
Always look for a substitution or a rearrangement before expanding.
在做展开前，先考虑能否换元或整体恒等变形。
3. Functions and Graphs | 函数与图像
Sample question: The functions f and g are defined by f(x) = x² – 4x + 3 and g(x) = 2x – 1. What is the minimum value of f(g(x))?
模拟题：已知函数 f(x) = x² – 4x + 3，g(x) = 2x – 1，求 f(g(x)) 的最小值。
First rewrite f(x) in completed-square form:
f(x) = (x – 2)² – 1
This shows that f(x) has minimum value -1 at x = 2. Because g(x) = 2x – 1 is a linear function and can take every real value, there is some x for which g(x) = 2. Therefore the minimum of f(g(x)) is -1.
首先将 f(x) 配方：f(x) = (x – 2)² – 1。这说明 f(x) 在 x = 2 时取最小值 -1。由于 g(x) = 2x – 1 是一次函数，能取到所有实数，所以一定存在某个 x 使 g(x) = 2。因此 f(g(x)) 的最小值为 -1。
Graphically, the vertex of y = f(x) is at (2, -1). A linear input does not change the possible range of this quadratic, because its range is already all real numbers.
从图像上看，y = f(x) 的顶点在 (2, -1)。线性输入不会改变这个二次函数的取值范围，因为一次函数的取值范围本就是全体实数。
4. Differentiation and Integration | 微分与积分
Sample question: A curve has equation y = x³ – 6x² + 9x. Which x-coordinate gives a local maximum?
模拟题：曲线 y = x³ – 6x² + 9x 在哪个 x 坐标处取得局部极大值？
Differentiate to find stationary points:
dy/dx = 3x² – 12x + 9 = 3(x – 1)(x – 3)
Setting dy/dx = 0 gives x = 1 or x = 3. To distinguish a maximum from a minimum, use the second derivative:
d²y/dx² = 6x – 12
At x = 1, d²y/dx² = -6 < 0, so this is a local maximum. At x = 3, d²y/dx² = 6 > 0, so this is a local minimum.
求导找驻点：dy/dx = 3x² – 12x + 9 = 3(x – 1)(x – 3)。令 dy/dx = 0 得 x = 1 或 x = 3。用二阶导数判断极大或极小：d²y/dx² = 6x – 12。当 x = 1 时，d²y/dx² = -6 < 0，因此是局部极大值；当 x = 3 时，d²y/dx² = 6 > 0，因此是局部极小值。
Integration also appears frequently. For example, if dy/dx = 6x² – 12x + 9 and y = 6 when x = 1, then y = 2x³ – 6x² + 9x + 1.
积分也经常出现。例如，若 dy/dx = 6x² – 12x + 9，且 x = 1 时 y = 6，则 y = 2x³ – 6x² + 9x + 1。
5. Logarithms and Exponentials | 对数与指数
Sample question: Solve log₂ x + log₂ (x – 2) = 3.
模拟题：解方程 log₂ x + log₂ (x – 2) = 3。
Use the logarithm law for sums:
log₂ x + log₂ (x – 2) = log₂ [x(x – 2)]
Therefore:
x(x – 2) = 2³ = 8
Expanding gives x² – 2x – 8 = 0, so (x – 4)(x + 2) = 0, hence x = 4 or x = -2. However, the logarithm domain requires x > 0 and x – 2 > 0, so x > 2. The only valid solution is x = 4.
利用对数加法法则：log₂ x + log₂ (x – 2) = log₂ [x(x – 2)]。因此 x(x – 2) = 2³ = 8。展开得 x² – 2x – 8 = 0，即 (x – 4)(x + 2) = 0，所以 x = 4 或 x = -2。但对数的定义域要求 x > 0 且 x – 2 > 0，即 x > 2，所以唯一有效解为 x = 4。
This is a classic TMUA trap: a quadratic may produce extra roots that fail the domain condition. Always check.
这是TMUA的经典陷阱：二次方程可能产生不满足定义域的增根，务必代入检验。
6. Series and Sequences | 级数与数列
Sample question: The sum of the first n terms of an arithmetic sequence is Sₙ = n² + 3n. Find the first term and the common difference.
模拟题：某等差数列的前 n 项和为 Sₙ = n² + 3n，求首项和公差。
The first term is S₁ = 1² + 3 × 1 = 4. The sum of the first two terms is S₂ = 2² + 3 × 2 = 10. Since S₂
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