📚 AQA International AS Further Maths 9665/FM02 June 2018 Examiner Report Analysis | AQA国际AS进阶数学9665/FM02 2018年6月考官报告分析
The June 2018 examiner report for AQA International AS Further Mathematics 9665/FM02 gives a clear picture of candidate performance across Pure, Statistics and Mechanics. This article explains the key findings for each section, highlights common mistakes, and shows how to use the examiner's comments to sharpen exam technique.
2018年6月AQA国际AS进阶数学9665/FM02考官报告清晰呈现了考生在纯数、统计和力学各部分的表现。本文解读各部分的主要发现，指出常见错误，并说明如何利用考官评语改进答题技巧。
1. Overview of the June 2018 FM02 Paper | 2018年6月FM02试卷概览
The FM02 Unit 1 paper tested Pure, Statistics and Mechanics within a single AS Further Mathematics unit. Examiners noted that well-prepared candidates could access the majority of marks, but time management and notation were frequent weaknesses.
FM02第一单元试卷在同一个AS进阶数学单元中考查了纯数、统计和力学。考官指出，准备充分的考生能够拿到大部分分数，但时间管理和符号书写是常见弱点。
The pure mathematics questions were answered most consistently, while the mechanics section produced the widest spread of marks. Many candidates lost marks through premature rounding and omitted units or stages of working.
纯数题目的作答最稳定，而力学部分的分数差距最大。许多考生因过早取整、遗漏单位或省略解题步骤而失分。
Overall, the report emphasised that questions requiring standard routines, such as finding a determinant or evaluating a Poisson probability, were completed well, but multi-step modelling problems exposed gaps in reasoning.
总体而言，报告强调，像求行列式或计算泊松概率这类标准套路题完成得较好，但多步建模问题暴露了推理上的不足。
2. Complex Numbers and Argand Diagrams | 复数与阿尔冈图
d candidates to manipulate complex numbers in the form z = a + bi and to interpret modulus |z| and argument arg z. The report noted that most candidates could calculate |z| = √a² + b² but made sign errors when finding arg z in the second or third quadrant.
第1题要求考生化简形如z = a + bi的复数，并解释模|z|和辐角arg z。报告指出，大多数考生能计算|z| = √a² + b²，但在求第二或第三象限的arg z时出现符号错误。
z = a + bi, |z| = √a² + b², arg z = tan⁻¹(b/a) adjusted by quadrant
ct argument is π − tan⁻¹(4/3), but many candidates stopped at tan⁻¹(4/3) or gave a negative angle without adding π.
例如当z = −3 + 4i时，正确的辐角是π − tan⁻¹(4/3)，但许多考生只写到tan⁻¹(4/3)，或者给出负角而没有加上π。
n form too early. Examiners recommended using polar form for products and quotients, since |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg z₁ + arg z₂.
在复数的乘除运算中，较弱的作答过早地展开成笛卡尔形式。考官建议对乘除使用极坐标形式，因为|z₁z₂| = |z₁||z₂|且arg(z₁z₂) = arg z₁ + arg z₂。
3. Roots of Polynomials | 多项式根的关系
Questions on roots of quadratics and cubics tested Vieta' c ax² + bx + c = 0 with roots α and β, candidates should recall α + β = −b/a and αβ = c/a.
有关二次和三次多项式根的问题考查了韦达关系。对于有根α和β的二次方程ax² + bx + c = 0，考生应记得α + β = −b/a和αβ = c/a。
α + β = −b/a, αβ = c/a
ecially when coefficients were negative. Examiners also observed that many candidates could not form a new quadratic from transformed roots such as 2α and 2β.
2018年6月的一个常见错误是漏掉负号，把α + β写成b/a，尤其是在系数为负时。考官还发现许多考生不会从变换后的根（如2α和2β）构造新的二次方程。
, the report reminded candidates that α + β + γ = −b/a, αβ + βγ + γα = c/a and αβγ = −d/a.
对于有根α、β和γ的三次方程ax³ + bx² + cx + d = 0，报告提醒考生α + β + γ = −b/a，αβ + βγ + γα = c/a，αβγ = −d/a。
The key skill is to express symmetric functions such as α² + β² as (α + β)² − 2αβ before substituting numerical values, which many candidates forgot to do.
关键技巧是先利用恒等式把对称函数如α² + β²写成(α + β)² − 2αβ，再代入数值，但许多考生忘记这样做。
4. Matrices and Linear Transformations | 矩阵与线性变换
sually computed correctly, but some candidates forgot to swap a and d and change signs of b and c in the adjugate.
矩阵题要求考生求一个2×2矩阵的逆矩阵，并解释旋转或反射。行列式det M = ad − bc通常计算正确，但部分考生在伴随矩阵中忘记交换a和d并改变b和c的符号。
M⁻¹ = 1/(ad − bc) × [[d, −b], [−c, a]]
M⁻¹ = I. Those who did this detected sign errors before moving to the transformation part.
报告强调，考生应通过检验M M⁻¹ = I来验证逆矩阵。这样做的考生能够在进入变换部分之前发现符号错误。
For transformation geometry, examiners required candidates to state both the type of transformation and its parameters, for example 'rotation 90° anticlockwise about the origin' rather than just 'rotation'.
对于变换几何，考官要求考生同时说明变换类型及其参数，例如“绕原点逆时针旋转90°”，而不只是写“旋转”。
Several responses confused clockwise and anticlockwise rotations when identifying the angle from the matrix, so drawing a simple unit square was recommended.
有部分作答在从矩阵判断角度时混淆了顺时针和逆时针旋转，因此建议画一个简单的单位正方形辅助判断。
5. Series and Proof by Induction | 级数与归纳法证明
ere needed in the series question. The report noted arithmetic slips when expanding and simplifying (n + 1)² or (n + 1)³.
级数题需要用到∑r、∑r²和∑r³的标准结果。报告指出展开和化简(n + 1)²或(n + 1)³时常出现算术错误。
∑r = n(n + 1)/2, ∑r² = n(n + 1)(2n + 1)/6, ∑r³ = n²(n + 1)²/4
jority of candidates could handle the basis step, but weaker proofs omitted the conclusion 'hence true for n = k + 1 by induction' or failed to link the assumption to the inductive step.
在归纳法证明中，大多数考生能处理基础步骤，但较弱的证明遗漏了“因此由归纳法可知对n = k + 1成立”的结论，或者没有把假设与归纳步骤联系起来。
for n = k, assume it true, prove for n = k + 1, then write the closing induction sentence.
考官建议使用清晰的模板：写出n = k时的命题，假设其成立，证明n = k + 1时成立，然后写出归纳结论句。
A frequent error was treating the assumption as a formula to rearrange without showing why the next case follows, which lost the final A1 mark.
一个常见错误是把假设当作公式进行变形，却没有说明为什么下一情况成立，因此丢掉了最后的A1分。
6. Discrete Random Variables and Expectation | 离散随机变量与期望
ion table. Candidates should use E(X) = Σ x P(X = x) and Var(X) = Σ x² P(X = x) − [E(X)]².
报告强调了一道要求从概率分布表求E(X)和Var(X)的题目。考生应使用E(X) = Σ x P(X = x)和Var(X) = Σ x² P(X = x) − [E(X)]²。
E(X) = Σ x P(X = x), Var(X) = Σ x² P(X = x) − [E(X)]²
A frequent mistake was subtracting E(X²) instead of [E(X)]², or computing E(X²) as [E(X)]². Examiners also found probabilities not summing to 1 left uncorrected.
一个常见错误是减去了E(X²)而不是[E(X)]²，或者把E(X²)算成[E(X)]²。考官还发现概率之和不为1时考生没有修正。
. The report noted that many used aVar(X) for the transformed variance, losing easy marks.
对于线性函数，考生应知道E(aX + b) = aE(X) + b以及Var(aX + b) = a²Var(X)。报告指出许多人把变换后的方差写成aVar(X)，丢掉了容易得到的分数。
P(X = x), xP(X = x) and x²P(X = x) was strongly recommended to minimise numerical errors.
报告强烈建议清晰列出x、P(X = x)、xP(X = x)和x²P(X = x)的表格，以尽量减少数值错误。
7. Poisson Distribution and Conditions | 泊松分布及其
Published by TutorHao | AS Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图