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Year 13 AQA Maths: Summer Bridging & Preparation Course | AQA 高三数学暑期预习与衔接课程

2026-07-22 17:17 · 本文部分内容由AI辅助创作

📚 Year 13 AQA Maths: Summer Bridging & Preparation Course | AQA 高三数学暑期预习与衔接课程

Summer is the perfect time to bridge the gap between Year 12 and the demanding Year 13 AQA Mathematics syllabus. This course will revisit essential AS topics and introduce the core A2 concepts, ensuring you start the final year with confidence and a solid foundation. We'll cover advanced pure maths, mechanics, and statistics techniques that build directly on your prior knowledge.

暑假是弥合 Year 12 与高要求 Year 13 AQA 数学课程之间差距的最佳时机。该衔接课程将重温关键的 AS 知识点并引入核心的 A2 概念,确保你在最后一年开始时充满信心,拥有扎实的基础。我们将涵盖高等纯数学、力学和统计技巧,这些内容直接建立在你的已有知识之上。

1. Bridging from Year 12: Key AS Topics Review | 从 AS 到 A2:关键 AS 知识点回顾

Before diving into Year 13 topics, it's crucial to consolidate your understanding of the following AS concepts: differentiation from first principles, basic integration, solving quadratic equations, coordinate geometry, and the laws of logarithms. Without fluency in these, advanced topics like implicit differentiation and integration by substitution become far more challenging.

在深入 Year 13 内容前,巩固以下 AS 概念至关重要:导数定义、基本积分、解二次方程、坐标几何以及对数运算法则。如果对这些不熟练,隐函数微分和换元积分等高等主题将变得异常困难。

r attention to trigonometric identities, such as tan θ = sin θ / cos θ and sin² θ + cos² θ ≡ 1, as they are used extensively in A2 calculus.

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花时间复习 Year 12 笔记,并做过去的 AS 试卷,重点放在纯数学上。特别注意三角恒等式,如 tan θ = sin θ / cos θ 和 sin² θ + cos² θ ≡ 1,它们在 A2 微积分中被广泛使用。

Also, ensure you are confident with handling functions, including domain and range, composite functions, and inverse functions. These underpin many topics in Year 13, such as parametric equations and modulus functions.

另外,确保你熟练掌握函数处理,包括定义域和值域、复合函数与反函数。这将支撑 Year 13 的许多主题,如参数方程和模函数。


2. Advanced Algebra: Partial Fractions & Binomial Expansion | 高等代数:部分分式与二项展开式

In Year 13, you'll extend your algebraic toolkit. Partial fractions are essential for integrating rational functions. For distinct linear factors, an expression like 1/[(x+1)(x-2)] can be decomposed into A/(x+1) + B/(x-2). For repeated factors, e.g., 1/[(x+2)²(x-1)], you need terms A/(x+2) + B/(x+2)² + C/(x-1). Make sure you can solve for constants using substitution or equating coefficients.

在 Year 13,你将扩展代数工具。部分分式对积分有理函数至关重要。对于不同线性因子,如 1/[(x+1)(x-2)] 可分解为 A/(x+1) + B/(x-2)。对于重根,如 1/[(x+2)²(x-1)],您需要 A/(x+2) + B/(x

Published by TutorHao | Year 13 Mathematics Revision Series | aleveler.com

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