AQA AS Paper 1 Specimen — Question 11 7 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeIncreasing/decreasing intervals
DifficultyStandard +0.3 This question requires differentiating two polynomials, setting up an inequality to test the claim, and solving a quadratic inequality. While it involves multiple steps (differentiate, form inequality, solve quadratic, interpret), each individual technique is standard AS-level content with no novel insight required. The 7 marks reflect the working needed rather than conceptual difficulty. Slightly easier than average due to straightforward algebraic manipulation.
Spec1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx
Chris claims that, "for any given value of \(x\), the gradient of the curve \(y = 2x^3 + 6x^2 - 12x + 3\) is always greater than the gradient of the curve \(y = 1 + 60x - 6x^2\)". Show that Chris is wrong by finding all the values of \(x\) for which his claim is not true. [7 marks]
Question 11:
AnswerMarks
11dy
Obtains
dx
for both the given curves – at least
one term must be correct for each
AnswerMarks Guidance
curveAO3.1a M1
6x2 12x12
dx
dy
6012x
dx
Chris’s claim is incorrect when
6x2 12x126012x
2x2 + 8x – 24  0
x2 + 4x – 12  0
(x + 6) (x – 2) 0
Critical values are x = –6 and 2
region x < –6 –6 < x < 2 x > 2
sign + – +
–6  x  2
Chris’s claim is incorrect for values of x
in the range –6  x 2, so he is wrong
AnswerMarks Guidance
States both derivatives correctlyAO1.1b A1
Translates problem into an
AnswerMarks Guidance
inequalityAO3.1a M1
States a correct quadratic
inequality
dy
FT from an incorrect provided
dx
AnswerMarks Guidance
both M1 marks have been awardedAO1.1b A1
Determines a solution to ‘their’
AnswerMarks Guidance
inequalityAO1.1a M1
Obtains correct range of values for
x
Must be correctly written with both
AnswerMarks Guidance
inequality signs correctAO1.1b A1
Interprets final solution in context of
the original question, must refer to
AnswerMarks Guidance
Chris’s claimAO3.2a R1
Total7
regionx < –6 –6 < x < 2
sign+ –
QMarking Instructions AO 
Question 11:
11 | dy
Obtains
dx
for both the given curves – at least
one term must be correct for each
curve | AO3.1a | M1 | dy
6x2 12x12
dx
dy
6012x
dx
Chris’s claim is incorrect when
6x2 12x126012x
2x2 + 8x – 24  0
x2 + 4x – 12  0
(x + 6) (x – 2) 0
Critical values are x = –6 and 2
region x < –6 –6 < x < 2 x > 2
sign + – +
–6  x  2
Chris’s claim is incorrect for values of x
in the range –6  x 2, so he is wrong
States both derivatives correctly | AO1.1b | A1
Translates problem into an
inequality | AO3.1a | M1
States a correct quadratic
inequality
dy
FT from an incorrect provided
dx
both M1 marks have been awarded | AO1.1b | A1
Determines a solution to ‘their’
inequality | AO1.1a | M1
Obtains correct range of values for
x
Must be correctly written with both
inequality signs correct | AO1.1b | A1
Interprets final solution in context of
the original question, must refer to
Chris’s claim | AO3.2a | R1
Total | 7
region | x < –6 | –6 < x < 2 | x > 2
sign | + | – | +
Q | Marking Instructions | AO | Marks | Typical Solution
Chris claims that, "for any given value of $x$, the gradient of the curve $y = 2x^3 + 6x^2 - 12x + 3$ is always greater than the gradient of the curve $y = 1 + 60x - 6x^2$".

Show that Chris is wrong by finding all the values of $x$ for which his claim is not true.
[7 marks]

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