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

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

This article provides a complete, step-by-step walkthrough of a mock unit test designed for the Cambridge Pre-U Mathematics (CAIE) specification. The paper focuses on Pure Mathematics and is built to resemble a typical 90‑minute assessment. Each question is analysed in detail, with an emphasis on the key techniques, common pitfalls, and efficient problem‑solving strategies required at this level.

本文为针对剑桥 Pre‑U 数学 (CAIE) 大纲设计的模拟单元测试提供一份完整的逐题解析。试卷聚焦于纯数学部分，模拟一张典型的 90 分钟考核。我们将详细分析每一道题目，着重讲解核心技巧、常见误区以及在这个层次上所需的高效解题策略。

1. Overview of the Mock Test | 模拟卷概述

The mock test consists of 10 compulsory questions, totalling 75 marks. Students are advised to spend about 90 minutes on the paper. The topics covered include polynomial algebra, exponentials and logarithms, trigonometry, differentiation, integration, complex numbers, vectors, differential equations, arithmetic and geometric sequences, and integration by parts. A good command of algebraic manipulation and precise use of standard results is essential.

本模拟卷包含 10 道必答题，满分 75 分，建议用时 90 分钟。覆盖的专题包括多项式代数、指数与对数、三角、微分、积分、复数、向量、微分方程、等差与等比数列以及分部积分。扎实的代数运算功底以及对标准结论的准确运用是取得高分的关键。

The paper is designed to test both routine technique and the ability to combine several concepts in a single question. In this walkthrough, every solution is broken into logical steps, and alternative approaches are mentioned where relevant. The model solutions also show how to set out work clearly to gain full method marks.

试卷旨在既考查常规技巧，也考查在一道题中综合多个概念的能力。在本次解析中，每个解答都被拆分成逻辑步骤，并在相关之处提及替代方法。范例解答同时展示了如何清晰书写步骤以获得全部方法分。

2. Q1: Polynomial Factors & Remainder Theorem | 问题1：多项式因式与余式定理

The polynomial f(x) = 2x³ − 9x² + ax + b has a factor (x − 1) and leaves a remainder of −10 when divided by (x − 2). Find the values of a and b. Hence factorise f(x) completely.

已知多项式 f(x) = 2x³ − 9x² + ax + b 含有因式 (x − 1)，且除以 (x − 2) 时余数为 −10。求 a 和 b 的值，进而将 f(x) 完全分解因式。

By the Factor Theorem, (x − 1) being a factor implies f(1) = 0. Substituting x = 1 gives 2(1) − 9(1) + a + b = 0, so a + b = 7.

由因式定理，(x − 1) 是因式意味着 f(1) = 0。代入 x = 1 得到 2(1) − 9(1) + a + b = 0，即 a + b = 7。

The Remainder Theorem states that the remainder when f(x) is divided by (x − 2) is f(2). Since the remainder is −10, we have f(2) = 2(8) − 9(4) + 2a + b = 16 − 36 + 2a + b = 2a + b − 20. Setting this equal to −10 yields 2a + b = 10.

余式定理指出，f(x) 除以 (x − 2) 的余数为 f(2)。已知余数为 −10，故有 f(2) = 2(8) − 9(4) + 2a + b = 16 − 36 + 2a + b = 2a + b − 20。令其等于 −10，得到 2a + b = 10。

Now solve the simultaneous equations a + b = 7 and 2a + b = 10. Subtracting the first from the second gives a = 3. Substituting back gives b = 4. Thus f(x) = 2x³ − 9x² + 3x + 4.

现在解联立方程 a + b = 7 和 2a + b = 10。第二式减去第一式得到 a = 3。代回得 b = 4。因此 f(x) = 2x³ − 9x² + 3x + 4。

To factorise, divide f(x) by the known factor (x − 1). Using synthetic division with root 1 on coefficients 2, −9, 3, 4: bring down 2; 1×2 = 2, add to −9 gives −7; 1×(−7) = −7, add to 3 gives −4; 1×(−4) = −4, add to 4 gives 0. The quotient is 2x² − 7x − 4.

为了分解因式，用已知因式 (x − 1) 去除 f(x)。使用综合除法，以根 1 对系数 2, −9, 3, 4 运算：拉下 2；1×2 = 2，加至 −9 得 −7；1×(−7) = −7，加至 3 得 −4；1×(−4) = −4，加至 4 得 0。商式为 2x² − 7x − 4。

Factorising the quadratic 2x² − 7x − 4 requires two numbers that multiply to (2 × −4) = −8 and add to −7. These are −8 and 1. Split the middle term: 2x² − 8x + x − 4 = 2x(x − 4) + 1(x − 4) = (2x + 1)(x − 4). Hence the complete factorisation is f(x) = (x − 1)(2x + 1)(x − 4).

将二次式 2x² − 7x − 4 分解因式，需要两个数乘积为 (2 × −4) = −8，且和为 −7。这两个数是 −8 和 1。拆分中项：2x² − 8x + x − 4 = 2x(x − 4) + 1(x − 4) = (2x + 1)(x − 4)。因此完整的因式分解为 f(x) = (x − 1)(2x + 1)(x − 4)。

3. Q2: Exponential & Logarithmic Equations | 问题2：指数与对数方程

Solve the equation 3^2x+1 = 5^x−2, giving your answer in the form x = ln p / ln q where p and q are integers.

解方程 3^2x+1 = 5^x−2，并将答案写成 x = ln p / ln q 的形式，其中 p 和 q 为整数。

Take natural logarithms of both sides: ln(3^2x+1) = ln(5^x−2). Using the power rule, this becomes (2x + 1) ln 3 = (x − 2) ln 5.

两边取自然对数：ln(3^2x+1) = ln(5^x−2)。利用幂的对数性质，得 (2x + 1) ln 3 = (x − 2) ln 5。

Expand both sides: 2x ln 3 + ln 3 = x ln 5 − 2 ln 5. Now collect the terms containing x on one side and constant terms on the other: 2x ln 3 − x ln 5 = −2 ln 5 − ln 3.

展开两边：2x ln 3 + ln 3 = x ln 5 − 2 ln 5。接下来将含 x 的项与常数项分别移至等式两边：2x ln 3 − x ln 5 = −2 ln 5 − ln 3。

Factor out x on the left: x(2 ln 3 − ln 5) = −(2 ln 5 + ln 3). Using logarithm rules, 2 ln 3 = ln 9 and 2 ln 5 = ln 25, so x(ln 9 − ln 5) = −(ln 25 + ln 3) which simplifies to x(ln(9/5)) = −ln(75).

左边提取公因子 x：x(2 ln 3 − ln 5) = −(2 ln 5 + ln 3)。利用对数法则，2 ln 3 = ln 9 且 2 ln 5 = ln 25，因此 x(ln 9 − ln 5) = −(ln 25 + ln 3)，化简得 x(ln(9/5)) = −ln(75)。

Finally, x = −ln 75 / ln(9/5). This can be written as x = ln(1/75) / ln(9/5) = ln(75⁻¹) / ln(9/5). Recognising that −ln 75 = ln(1/75), we express the answer with positive integers in the log arguments as x = ln(1/75) / ln(9/5), or equivalently x = ln(75) / ln(5/9) with a negative sign absorbed, but the required form is often x = ln(5/9) / ln(75)? No, careful: We need x = ln p / ln q. From x = −ln 75 / ln(9/5), we can multiply numerator and denominator by −1 to get x = ln 75 / ln(5/9). So p = 75, q = 5/9, but q is not an integer. Better to rewrite as x = ln(75) / ln(5/9) = ln 75 / (ln 5 − ln 9) which is not ln p / ln q with integers p,q. Let’s find an integer form: Since x = (ln 75) / (ln 5/9) is valid but q = 5/9 is not integer. We can instead start from x(2 ln 3 − ln 5) = −2 ln 5 − ln 3, giving x = (−2 ln 5 − ln 3) / (2 ln 3 − ln 5). Multiply numerator and denominator by −1: x = (2 ln 5 + ln 3) / (ln 5 − 2 ln 3) = ln(25×3) / ln(5/9) = ln 75 / ln(5/9). Not integer p,q. However, the equation can be manipulated to get nice integers. How about writing 3^2x+1 / 5^x−2 = 1 => (3^2x+1 5^2−x) = 1 => 9^x·3 · 5²·5^−x = 1 => (9/5)^x · 75 = 1 => (9/5)^x = 1/75 => x = log_9/5(1/75) = −ln 75 / ln(9/5). So the answer can be given as x = ln(1/

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