AQA FM05 Further Maths 18 Jan 2023: Paper Analysis & Revision Guide | AQA FM05 进阶数学 2023年1月18日 试卷解析与复习指南

📚 AQA FM05 Further Maths 18 Jan 2023: Paper Analysis & Revision Guide | AQA FM05 进阶数学 2023年1月18日 试卷解析与复习指南

The AQA FM05 Discrete Mathematics paper, sat on 18 January 2023 in the international session, tested a broad range of decision mathematics topics. This article breaks down every major content area, provides worked examples in exam style, and highlights the techniques that earn full marks.

2023 年 1 月 18 日国际考次举行的 AQA FM05 离散数学试卷，全面考查了决策数学的各大知识板块。本文逐一拆解所有重点内容，提供考试风格的典型例题，并讲解如何通过规范步骤获得满分。

1. Paper Structure & Assessment Objectives | 试卷结构与评估目标

The FM05 paper is a 2-hour written examination worth 80 marks, making up 25% of the A-level Further Mathematics grade. It typically contains between 7 and 9 multi-part questions, with each question focusing on a major discrete mathematics topic from the specification.

FM05 试卷为 2 小时笔试，满分 80 分，占进阶数学 A-level 总成绩的 25%。试卷通常包含 7 至 9 道多小问大题，每一道题对应考纲中的一个主要离散数学主题。

Assessment objectives are weighted as follows:

• AO1: Recall and use of mathematical techniques – approximately 40%
• AO2: Construct and interpret mathematical proofs and arguments – approximately 40%
• AO3: Solve problems using mathematical reasoning – approximately 20%

• AO1：回忆与运用数学技巧——约占 40%
• AO2：构建并解释数学证明与论证——约占 40%
• AO3：运用数学推理解决实际问题——约占 20%

The 18 January 2023 paper rewarded candidates who wrote down each iteration of an algorithm clearly. Partial marks are plentiful, so never skip a pass in a sorting problem.

2023 年 1 月 18 日的试卷对清晰写出算法每一步迭代的考生格外有利。过程分很多，所以在排序题中切勿跳步。

2. Algorithms: Sorting & Searching | 算法：排序与查找

Sorting algorithms are a guaranteed starting point on FM05. The bubble sort, quick sort and insertion sort each have their own step-by-step routines, and marks are awarded for the working as much as the final answer.

排序算法是 FM05 的必考起点。冒泡排序、快速排序和插入排序各有其固定的分步流程，过程与最终答案同等给分。

Bubble sort on a list of n items requires at most n − 1 passes. After each pass, the largest remaining unsorted element ‘bubbles’ into its final position at the right-hand end. Let us sort 5, 3, 8, 1.

对含 n 个元素的列表进行冒泡排序，最多需要 n − 1 轮。每一轮结束后，剩余未排序元素中的最大值都会“冒泡”至右端最终位置。下面我们对 5, 3, 8, 1 排序：

Pass 1: compare 5 and 3 → swap → 3, 5, 8, 1; compare 5 and 8 → no swap; compare 8 and 1 → swap → 3, 5, 1, 8.

第 1 轮：比较 5 和 3 → 交换 → 3, 5, 8, 1；比较 5 和 8 → 不交换；比较 8 和 1 → 交换 → 3, 5, 1, 8。

Pass 2: compare 3 and 5 → no swap; compare 5 and 1 → swap → 3, 1, 5, 8; compare 5 and 8 → no swap.

第 2 轮：比较 3 和 5 → 不交换；比较 5 和 1 → 交换 → 3, 1, 5, 8；比较 5 和 8 → 不交换。

Pass 3: compare 3 and 1 → swap → 1, 3, 5, 8; compare 3 and 5 → no swap; compare 5 and 8 → no swap. The list is now sorted: 1, 3, 5, 8.

第 3 轮：比较 3 和 1 → 交换 → 1, 3, 5, 8；比较 3 和 5 → 不交换；比较 5 和 8 → 不交换。列表已排序：1, 3, 5, 8。

Quick sort chooses a pivot, partitions the list into elements less than the pivot and elements greater than the pivot, then recursively sorts each sub-list. The pivot is often chosen as the middle element in AQA papers.

快速排序选择一个基准元素（AQA 试卷中通常取中间元素），将列表分为小于基准和大于基准两部分，然后对每个子列表递归排序。

Consider 5, 3, 8, 1, 9, 2. The middle element of this six-item list is the 4th item, 1, in positional terms, but AQA convention often selects the (n + 1)/2-th item when using 1-indexing; for this six-item list, that is item 3, which is 8. Using 8 as pivot: left sub-list = 5, 3, 1, 2; right sub-list = 9. Repeating with the middle of 5, 3, 1, 2 gives pivot 1: left = empty, right = 5, 3, 2. Continuing eventually yields 1, 2, 3, 5, 8, 9.

以 5, 3, 8, 1, 9, 2 为例。对六元素列表，AQA 通常约定取第 (n + 1)/2 项为基准，即第 3 项 8。以 8 为基准：左侧子列表 = 5, 3, 1, 2；右侧 = 9。再取 5, 3, 1, 2 的中间项 1 为基准：左侧为空，右侧 = 5, 3, 2。继续递归最终得到 1, 2, 3, 5, 8, 9。

Binary search is the standard searching algorithm: repeatedly halve the list, comparing the target with the middle element. It requires the list to be pre-sorted.

折半查找是标准查找算法：反复将列表对半分割，将目标值与中间元素比较。使用前提是列表已经排好序。

3. Bin Packing Problems | 装箱问题

Bin packing asks how to fit items of given sizes into bins of fixed capacity using the smallest number of bins. AQA tests three algorithms: first-fit, first-fit decreasing, and occasionally full-bin packing.

装箱问题要求将给定尺寸的物品装入固定容量的箱子中，使所用箱子数最少。AQA 考查三种算法：首次适应、降序首次适应，偶尔考查完全装箱。

Example: pack items of sizes 3, 4, 2, 5, 2 into bins of capacity 7.

例：将尺寸为 3, 4, 2, 5, 2 的物品装入容量为 7 的箱子。

First-fit: place 3 in bin A; 4 in bin B; 2 in bin A (A now has 3 + 2 = 5); 5 in bin C; 2 in bin B

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