AQA AS Paper 2 2018 June — Question 1 1 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2018
SessionJune
Marks1
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeBasic indefinite integration
DifficultyEasy -1.8 This is a trivial integration question requiring only the power rule: rewrite as x^(-2)/6, integrate to get x^(-1)/(-6) = -1/(6x). It's a 1-mark multiple choice question testing basic recall with no problem-solving element, making it significantly easier than average A-level questions.
Spec1.08b Integrate x^n: where n != -1 and sums
Given that \(\frac{dy}{dx} = \frac{1}{6x^2}\), find \(y\). Circle your answer. \(\frac{-1}{3x^3} + c\) \quad \(\frac{1}{2x^3} + c\) \quad \(\frac{-1}{6x} + c\) \quad \(\frac{-1}{3x} + c\) [1 mark]
Question 1:
AnswerMarks Guidance
1Circles correct answer AO1.1b
+ c
6𝑥
AnswerMarks Guidance
Total1
QMarking Instructions AO 
Question 1:
1 | Circles correct answer | AO1.1b | B1 | −1
+ c
6𝑥
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
Given that $\frac{dy}{dx} = \frac{1}{6x^2}$, find $y$.

Circle your answer.

$\frac{-1}{3x^3} + c$ \quad $\frac{1}{2x^3} + c$ \quad $\frac{-1}{6x} + c$ \quad $\frac{-1}{3x} + c$

[1 mark]

\hfill \mbox{\textit{AQA AS Paper 2 2018 Q1 [1]}}
This paper (19 questions)
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Q1 1 Q2 1 Q3 2 Q4 4 Q5 4 Q6 6 Q7 6 Q8 4 Q9 3 Q10 5 Q11 9 Q12 8 Q13 1 Q14 1 Q15 6 Q16 4 Q17 2 Q18 6 Q19 7