AQA Paper 1 Specimen — Question 9 8 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeIncreasing/decreasing intervals
DifficultyStandard +0.3 This is a standard quotient rule differentiation followed by routine algebraic manipulation and solving a quadratic inequality. While it requires multiple steps (differentiate, simplify, determine sign conditions, solve inequality), each component is a textbook technique with no novel insight required. The 8 marks reflect length rather than conceptual difficulty, making it slightly above average but well within typical A-level scope.
Spec1.02g Inequalities: linear and quadratic in single variable1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
A curve has equation \(y = \frac{2x + 3}{4x^2 + 7}\)
    1. Find \(\frac{dy}{dx}\) [2 marks]
    2. Hence show that \(y\) is increasing when \(4x^2 + 12x - 7 < 0\) [4 marks]
  2. Find the values of \(x\) for which \(y\) is increasing. [2 marks]
Question 9:
AnswerMarks
9(a)(i)Selects an appropriate
routine procedure; evidence
of quotient rule or product
AnswerMarks Guidance
ruleAO1.1a M1
dx (4x2 7)2
Obtains correct derivative
AnswerMarks Guidance
(no need for simplification)AO1.1b A1
(a)(ii)States clearly that
dy
0 y is increasing
AnswerMarks Guidance
dxAO2.4 R1
y is increasing  0
dx
2(4x2 7)8x(2x3)
0
(4x2 7)2
(4x2 7)2 0for all x
2(4x2 7)8x(2x3)0
8x2 1416x2 24x0
4x212x70 (AG)
Forms inequality from ‘their’
dy
0
AnswerMarks Guidance
dxAO3.1a B1
Deduces numerator must be
AnswerMarks Guidance
positiveAO2.2a R1
Considers denominator alone
and sets out clear argument
to justify given inequality AG
Only award this mark if they
have a completely correct
solution, which is clear, easy
to follow and contains no
AnswerMarks Guidance
slipsAO2.1 R1
(b)Solves the correct quadratic
inequality
(accept evidence of
factorising, completing the
square, use of formula, or
AnswerMarks Guidance
correct critical values stated)AO1.1a M1
7 1
x ,
2 2
7 1
 <x<
2 2
 7 1  7 1
Or x   ,  Or x  x:  x 
 2 2  2 2
Obtains fully correct answer,
given as an inequality or
AnswerMarks Guidance
using set notationAO1.1b A1
Total8
QMarking Instructions AO 
Question 9:
--- 9(a)(i) ---
9(a)(i) | Selects an appropriate
routine procedure; evidence
of quotient rule or product
rule | AO1.1a | M1 | dy 2(4x2 7)8x(2x3)

dx (4x2 7)2
Obtains correct derivative
(no need for simplification) | AO1.1b | A1
(a)(ii) | States clearly that
dy
0 y is increasing
dx | AO2.4 | R1 | dy
y is increasing  0
dx
2(4x2 7)8x(2x3)
0
(4x2 7)2
(4x2 7)2 0for all x
2(4x2 7)8x(2x3)0
8x2 1416x2 24x0
4x212x70 (AG)
Forms inequality from ‘their’
dy
0
dx | AO3.1a | B1
Deduces numerator must be
positive | AO2.2a | R1
Considers denominator alone
and sets out clear argument
to justify given inequality AG
Only award this mark if they
have a completely correct
solution, which is clear, easy
to follow and contains no
slips | AO2.1 | R1
(b) | Solves the correct quadratic
inequality
(accept evidence of
factorising, completing the
square, use of formula, or
correct critical values stated) | AO1.1a | M1 | (2x + 7)(2x – 1)
7 1
x ,
2 2
7 1
 <x<
2 2
 7 1  7 1
Or x   ,  Or x  x:  x 
 2 2  2 2
Obtains fully correct answer,
given as an inequality or
using set notation | AO1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
A curve has equation $y = \frac{2x + 3}{4x^2 + 7}$

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$
[2 marks]
\item Hence show that $y$ is increasing when $4x^2 + 12x - 7 < 0$
[4 marks]
\end{enumerate}
\item Find the values of $x$ for which $y$ is increasing.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1  Q9 [8]}}
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