AQA Paper 3 Specimen — Question 6 8 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
SessionSpecimen
Marks8
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Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypeBasic partial fractions then integrate
DifficultyChallenging +1.2 This is an 8-mark integration question requiring partial fractions decomposition and logarithmic integration. While it involves multiple steps (factorizing the denominator, splitting into partial fractions, integrating, and applying limits), these are all standard A-level techniques with no novel insight required. The algebraic manipulation is moderately involved but routine for Paper 3, placing it above average difficulty but not exceptionally challenging.
Spec1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions
Find the value of \(\int_1^2 \frac{6x + 1}{6x^2 - 7x + 2} dx\), expressing your answer in the form \(m\ln 2 + n\ln 3\), where \(m\) and \(n\) are integers. [8 marks]
Question 6:
AnswerMarks
6Uses partial fractions with linear
denominators
6x+1 A B
= +
AnswerMarks Guidance
6x2 −7x+2 ax+b cx+dAO3.1a M1
= +
6x2 −7x+2 3x−2 2x−1
A(2x−1)+B(3x−2)=6x+1
2 1
x= , A =5 so A=15
3 3
( )
x = 1 , B − 1 = 4 so B = −8
2 2
2 15 8
∫
− dx
1 3x − 2 2x −1
= [ 5ln(3x − 2) − 4ln(2x −1) ]2
1
=5ln(4)−4ln(3)−(5ln(1)4ln(1))
=10ln(2)−4ln(3)
AnswerMarks Guidance
Obtains correct linear denominatorsAO1.1b B1
Obtains at least one numerator
correct
(using any valid method,
eg equating coefficients or
AnswerMarks Guidance
substitution of values)AO1.1b A1
Obtains partial fractions completely
AnswerMarks Guidance
correctAO1.1b A1
Integrates ‘their’ partial fractions,
must include logs
AnswerMarks Guidance
pln(ax+b)+qln(cx+d)AO1.1a M1
‘Their’ integral correct (ignore limits)AO1.1b A1F
Substitutes limits into ‘their’ integralAO1.1a M1
Correct final answer in correct form
AnswerMarks Guidance
CAOAO1.1b A1
Total8
QMarking Instructions AO 
Question 6:
6 | Uses partial fractions with linear
denominators
6x+1 A B
= +
6x2 −7x+2 ax+b cx+d | AO3.1a | M1 | 6x+1 A B
= +
6x2 −7x+2 3x−2 2x−1
A(2x−1)+B(3x−2)=6x+1
2 1
x= , A =5 so A=15
3 3
( )
x = 1 , B − 1 = 4 so B = −8
2 2
2 15 8
∫
− dx
1 3x − 2 2x −1
= [ 5ln(3x − 2) − 4ln(2x −1) ]2
1
=5ln(4)−4ln(3)−(5ln(1)4ln(1))
=10ln(2)−4ln(3)
Obtains correct linear denominators | AO1.1b | B1
Obtains at least one numerator
correct
(using any valid method,
eg equating coefficients or
substitution of values) | AO1.1b | A1
Obtains partial fractions completely
correct | AO1.1b | A1
Integrates ‘their’ partial fractions,
must include logs
pln(ax+b)+qln(cx+d) | AO1.1a | M1
‘Their’ integral correct (ignore limits) | AO1.1b | A1F
Substitutes limits into ‘their’ integral | AO1.1a | M1
Correct final answer in correct form
CAO | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
Find the value of $\int_1^2 \frac{6x + 1}{6x^2 - 7x + 2} dx$, expressing your answer in the form $m\ln 2 + n\ln 3$, where $m$ and $n$ are integers. [8 marks]

\hfill \mbox{\textit{AQA Paper 3  Q6 [8]}}
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Q1 1 Q2 6 Q3 13 Q4 5 Q5 11 Q6 8 Q7 12 Q8 2 Q9 3 Q10 7 Q11 3 Q12 10 Q13 8 Q14 11 Q15 6