📚 Pre-U Cambridge Mathematics: Unit Test Mock Paper Analysis | Pre-U剑桥数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Pre-U Cambridge Mathematics, covering essential topics such as polynomials, binomial expansion, sequences, calculus, trigonometry, vectors, complex numbers and differential equations. Each question is followed by step-by-step solutions in both English and Chinese, designed to reinforce understanding and exam technique.
本文详细解析 Pre-U 剑桥数学单元测试模拟卷，涵盖多项式、二项式展开、数列、微积分、三角函数、向量、复数与微分方程等核心主题。每道题目均配有中英双语逐步解答，旨在巩固理解并提升应试技巧。
1. Paper Overview and Mark Distribution | 试卷概览与分值分布
The mock paper is written in the style of Cambridge Pre-U Mathematics (9794) Paper 1: a mixture of short technique questions and longer multi-part questions, with no calculators allowed in the first section and a graphics calculator permitted afterwards. Total marks are 80 and the time allowance is 2 hours, so candidates have roughly 1.5 minutes per mark.
本模拟卷依照 Cambridge Pre-U Mathematics (9794) Paper 1 的风格编写：前半部分为技巧型短题，后半部分为多小问长题；第一部分不允许使用计算器，之后可使用图形计算器。全卷满分 80 分，考试时间 2 小时，平均每分约 1.5 分钟。
| Topic | 主题 | Marks | 分值 | Weighting | 占比 |
| Polynomials | 多项式 | 10 | 12.5% |
| Binomial expansion | 二项式展开 | 8 | 10% |
| Sequences and series | 数列与级数 | 8 | 10% |
| Calculus | 微积分 | 24 | 30% |
| Trigonometry | 三角学 | 9 | 11.25% |
| Vectors, complex numbers, DEs | 向量、复数、微分方程 | 21 | 26.25% |
The table makes the revision priority obvious: calculus plus the three pure topics at the end account for well over half of the paper, so time spent on integration technique and complex-number algebra always pays back.
这张表让复习重点一目了然：微积分加上最后三个纯数专题合计超过全卷一半的分值，因此投入在积分技巧与复数运算上的时间回报最高。
2. Polynomial Division | 多项式除法
Question: Divide 4x³ - 3x² + 5x - 2 by (x - 2). State the quotient and remainder.
题目：用 (x - 2) 除 4x³ - 3x² + 5x - 2，给出商和余数。
t up the long division. Divide the leading term: 4x³ ÷ x = 4x². Write 4x² in the quotient.
步骤1：设立长除法。首项相除：4x³ ÷ x = 4x²，将 4x² 写入商中。
mial: (4x³ - 3x²) - (4x³ - 8x²) = 5x². Then bring down the next term +5x to form 5x² + 5x.
步骤2：将 (x - 2) 乘以 4x² 得 4x³ - 8x²，从原多项式中减去：(4x³ - 3x²) - (4x³ - 8x²) = 5x²。随后落下下一项 +5x，得到 5x² + 5x。
et 5x² - 10x, and subtract: (5x² + 5x) - (5x² - 10x) = 15x. Bring down -2 to form 15x - 2.
步骤3：5x² ÷ x = 5x，将 +5x 写入商中。将 (x - 2) 乘以 5x 得 5x² - 10x，相减：(5x² + 5x) - (5x² - 10x) = 15x。落下 -2，得到 15x - 2。
subtract: (15x - 2) - (15x - 30) = 28. The remainder is constant, so the algorithm stops.
步骤4：15x ÷ x = 15，将 +15 写入商中。将 (x - 2) 乘以 15 得 15x - 30，相减：(15x - 2) - (15x - 30) = 28。余数为常数，除法结束。
Quotient = 4x² + 5x + 15, Remainder = 28
商 = 4x² + 5x + 15，余数 = 28
3. Remainder and Factor Theorem Check | 余数定理与因式定理验证
2 = 32 - 12 + 10 - 2 = 28. This matches the long division exactly, which is a free mark check.
务必用余数定理检验除法结果：设 f(x) = 4x³ - 3x² + 5x - 2，则除以 (x - 2) 的余数等于 f(2) = 4(8) - 3(4) + 5(2) - 2 = 32 - 12 + 10 - 2 = 28，与长除法完全一致，这是一次免费的验算。
the remainder. If f(a) = 0, then (x - a) is a factor of f(x), which is the factor theorem.
考试中必须书写的公式是 f(x) = (x - a)Q(x) + f(a)，其中 Q(x) 为商，f(a) 为余数。若 f(a) = 0，则 (x - a) 是 f(x) 的因式，这就是因式定理。
-2 + k + 7 - 6 = 0, giving k = 1. Synthetic division then factorises the cubic completely.
常见延伸题是求未知系数：若 (x + 1) 是 2x³ + kx² - 7x - 6 的因式，则 f(-1) = -2 + k + 7 - 6 = 0，得 k = 1。随后用综合除法即可把三次式完全分解。
4. Binomial Expansion | 二项式展开
Question: Expand (2 - 3x)⁶ up to and including the term in x³, and state the coefficient of x³.
题目：将 (2 - 3x)⁶ 展开至含 x³ 项为止，并写出 x³ 的系数。
, 2, 3 the binomial coefficients are 1, 6, 15, 20, so the expansion is built term by term.
使用通项 C(6, k)·2⁶⁻ᵏ·(-3x)ᵏ。当 k = 0, 1, 2, 3 时二项式系数分别为 1, 6, 15, 20，逐项构造展开式。
(2 - 3x)⁶ = 64 - 576x + 2160x² - 4320x³ + …
2⁴ × 9 = 2160; the x³ term is 20 × 2³ × (-27) = -4320. So the coefficient of x³ is -4320.
检验：常数项为 2⁶ = 64；x 项为 6 × 2⁵ × (-3) = -576；x² 项为 15 × 2⁴ × 9 = 2160；x³ 项为 20 × 2³ × (-27) = -4320。因此 x³ 的系数为 -4320。
Pre-U also tests the negative and fractional index form. For (1 + x/2)⁻², valid for |x| ; 2, the first four terms are 1 - x + (3/4)x² - (1/2)x³, obtained from 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! with n = -2. Always state the range of validity - it carries a mark on its own.
Pre-U 还会考查负指数与分数指数形式。对于 (1 + x/2)⁻²，在 |x| < 2 时成立，其前四项为 1 - x + (3/4)x² - (1/2)x³，由 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! 取 n = -2 得到。务必写出成立范围，这一句本身就有分数。
5. Sequences and Series | 数列与级数
t the expression and use the two standard results, then combine over a common denominator.
典型题目要求计算 r(r + 1) 从 r = 1 到 n 的和。把表达式拆开后使用两个标准结果，再通分合并即可。
Σ r(r + 1) = Σ r² + Σ r = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(n + 2)/3
1) is a common factor: n(n + 1)[(2n + 1) + 3]/6 = n(n + 1)(2n + 4)/6 = n(n + 1)(n + 2)/3.
代数化简十分整齐，因为 n(n + 1) 是公因式：n(n + 1)[(2n + 1) + 3]/6 = n(n + 1)(2n + 4)/6 = n(n + 1)(n + 2)/3。
arⁿ⁻¹ and Sₙ = a(1 - rⁿ)/(1 - r) for r ≠ 1. The sum to infinity a/(1 - r) exists only when |r| ies such as Σ n·2⁻ⁿ is handled by multiplying the sum by the common ratio and subtracting.
等差与等比部分需牢记 uₙ = a + (n - 1)d、Sₙ = n/2[2a + (n - 1)d]、uₙ = arⁿ⁻¹ 以及 r ≠ 1 时的 Sₙ = a(1 - rⁿ)/(1 - r)。无穷和 a/(1 - r) 仅在 |r| < 1 时存在；而像 Σ n·2⁻ⁿ 这样的等差乘等比级数，则用"乘以公比后相减"的方法处理。
6. Differentiation Techniques | 微分技巧
1)⁵ ln x. This is a product of a composite power and a logarithm, so the product rule together with the chain rule is required.
题目 (a)：求 y = (3x + 1)⁵ ln x 的导数。该式为复合幂函数与对数的乘积，需同时使用乘积法则与链式法则。
dy/dx = 15(3x + 1)⁴ ln x + (3x + 1)⁵ / x
g out (3x + 1)⁴ gives (3x + 1)⁴[15 ln x + (3x + 1)/x], which is the form examiners prefer.
第一部分来自对内层括号求导：5(3x + 1)⁴ × 3 = 15(3x + 1)⁴；第二部分来自 d(ln x)/dx = 1/x。提取公因式 (3x + 1)⁴ 后得 (3x + 1)⁴[15 ln x + (3x + 1)/x]，这是考官偏好的形式。
y³ = 10. Differentiate each term with respect to x, remembering that y is a function of x.
题目 (b)：对隐函数曲线 3x² + 4xy - y³ = 10 求 dy/dx。逐项对 x 求导，切记 y 是 x 的函数。
6x + 4y + 4x·dy/dx - 3y²·dy/dx = 0, so dy/dx = (6x + 4y)/(3y² - 4x)
l step is simply collecting the dy/dx terms on one side and factorising. Never omit the dy/dx factor when differentiating a y term - this is the single most common lost mark in this topic.
中间项需用乘积法则：d(4xy)/dx = 4y + 4x·dy/dx。最后一步只是把含 dy/dx 的项移到一侧并提取公因式。对含 y 的项求导时切勿漏写 dy/dx，这是本专题最常见的失分点。
7. Integration and Area Under a Curve | 积分与曲线下面积
to e. Integration by parts with u = ln x and dv/dx = x gives the antiderivative directly.
题目 (a)：计算 x ln x 在 1 到 e 上的定积分。取 u = ln x、dv/dx = x 进行分部积分，可直接得到原函数。
∫ x ln x dx = (x²/2) ln x - x²/4 + c
t form is safer, since examiners usually accept either but exact values never lose accuracy marks.
代入上下限：[(e²/2)(1) - e²/4] - [0 - 1/4] = e²/4 + 1/4 = (e² + 1)/4 ≈ 2.097。保留精确形式更稳妥，虽然考官通常两种都接受，但精确值绝不会因取整而失分。
Question (b): find the area between y = x² + 1 and y = 2
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