AQA A Level Maths June 2018 Pure Mathematics Mark Scheme Key Points | AQA A Level 数学 2018 年 6 月纯数学评分方案要点解析

📚 AQA A Level Maths June 2018 Pure Mathematics Mark Scheme Key Points | AQA A Level 数学 2018 年 6 月纯数学评分方案要点解析

The June 2018 AQA A Level Mathematics Pure paper is a core component of the linear 7357 specification. The mark scheme rewards both method and accuracy, so a partially correct approach can still earn valuable marks. This revision article summarises the main topic areas and marking principles that students often overlook when reviewing the official mark scheme.

2018 年 6 月 AQA A Level 数学纯数试卷是 7357 线性课程的核心部分。评分方案同时奖励方法与准确性，因此部分正确的步骤也能获得宝贵分数。本文总结学生在查阅官方评分方案时经常忽略的主要考点与评分原则。

1. Surds and Rationalising Denominators | 根式与分母有理化

Candidates were expected to simplify expressions such as (√5 + 1)/(√5 – 1). The mark scheme gives one M1 for multiplying numerator and denominator by the conjugate √5 + 1, and one A1 for the simplified form (3 + √5)/2.

考生需化简形如 (√5 + 1)/(√5 – 1) 的式子。评分方案对分子分母同乘共轭根式 √5 + 1 给一个 M1 分，对最简结果 (3 + √5)/2 给一个 A1 分。

A common error is to rationalise with √5 – 1 instead of √5 + 1, which changes the denominator to a negative value but can still lead to a correct final answer if simplification is completed fully.

常见错误是乘 √5 – 1 而不是 √5 + 1，这样分母会变为负数，但只要完整化简，最终答案仍可正确，因此同样能得分。

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2. Differentiation from First Principles | 从第一性原理求导

For a function such as f(x) = x², the mark scheme awards an M1 for writing the limit expression [f(x+h) – f(x)] / h, and an A1 for expanding x² + 2xh + h² – x² to 2xh + h².

对于 f(x) = x² 这类函数，评分方案对写出极限表达式 [f(x+h) – f(x)] / h 给 M1 分，对将 x² + 2xh + h² – x² 化简为 2xh + h² 给 A1 分。

The final answer f ‘(x) = 2x gains the last A1 only if the limit as h → 0 is clearly shown and no incorrect notation is used.

最终答案 f ‘(x) = 2x 只有在清楚写出 h → 0 的极限过程且没有错误记号时才能获得最后一个 A1 分。

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3. Binomial Expansion | 二项式展开

In expanding (1 + 3x)^4, the mark scheme usually gives B1 for the correct binomial coefficients 1, 4, 6, 4, 1, and M1 for substituting the correct powers of 3x.

在展开 (1 + 3x)^4 时，评分方案通常对正确的二项式系数 1, 4, 6, 4, 1 给 B1 分，对正确代入 3x 的各次幂给 M1 分。

The simplified terms 1 + 12x + 54x² + 108x³ + 81x⁴ are then awarded A1. If the question asks for the expansion up to x³, the term 81x⁴ must not be included.

化简后的各项 1 + 12x + 54x² + 108x³ + 81x⁴ 获得 A1 分。如果题目只要求展开到 x³，则不能包含 81x⁴ 这一项。

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4. Trigonometric Equations and Identities | 三角方程与恒等式

The June 2018 paper included a trigonometric equation such as 3 sin² x – sin x – 2 = 0. The mark scheme allows M1 for factorising as (3 sin x + 2)(sin x – 1) = 0.

2018 年 6 月试卷包含类似 3 sin² x – sin x – 2 = 0 的三角方程。评分方案允许对因式分解为 (3 sin x + 2)(sin x – 1) = 0 给 M1 分。

Correct solutions in a given interval require each value to be found from the CAST diagram or graph. A1 is given for each exact angle such as 90°, 221.8° and 318.2°.

在给定区间内的正确解需要借助 CAST 图或图像逐一求出。每个精确角度如 90°、221.8° 和 318.2° 得到相应的 A1 分。

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5. Exponentials and Logarithms | 指数与对数

A typical question asks to solve e^(2x) = 7. The mark scheme awards M1 for taking natural logarithms of both sides to get 2x = ln 7, and A1 for x = (1/2) ln 7.

典型题目要求解 e^(2x) = 7。评分方案对两边取自然对数得到 2x = ln 7 给 M1 分，对 x = (1/2) ln 7 给 A1 分。

If the equation is 5^(x+1) = 60, students should take log base 5 or use the change-of-base formula. The mark scheme accepts x = log₅ 60 – 1 or x = (ln 60 / ln 5) – 1.

如果方程是 5^(x+1) = 60，学生应取以 5 为底的对数或使用换底公式。评分方案接受 x = log₅ 60 – 1 或 x = (ln 60 / ln 5) – 1。

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6. Integration and Area Under a Curve | 积分与曲线下方面积

For ∫ x√(x² + 1) dx, the substitution u = x² + 1 gives du = 2x dx. The mark scheme grants M1 for the correct substitution and A1 for (1/3)(x² + 1)^(3/2) + C.

对于 ∫ x√(x² + 1) dx，令 u = x² + 1 得 du = 2x dx。评分方案对正确换元给 M1 分，对 (1/3)(x² + 1)^(3/2) + C 给 A1 分。

When finding an area between a curve and the x-axis, the mark scheme requires limits to be substituted correctly; omitting the lower limit loses the final A1.

求曲线与 x 轴之间的面积时，评分方案要求正确代入上下限；漏掉下限会失去最后一个 A1 分。

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7. Sequences and Sigma Notation | 数列与求和记号

Questions on sigma notation often ask for Σ (3r + 2) from r = 1 to n. The mark scheme gives M1 for splitting the sum into 3Σr + Σ2, and A1 for 3n(n+1)/2 + 2n.

求和符号题目常要求计算 Σ (3r + 2)（r 从 1 到 n）。评分方案对将求和拆分为 3Σr + Σ2 给 M1 分，对 3n(n+1)/2 + 2n 给 A1

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