AQA Paper 1 2024 June — Question 19 7 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2024
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind constant from gradient condition
DifficultyChallenging +1.2 This question requires implicit differentiation of a moderately complex expression, finding where dy/dx = 0, and using the constraint that x = 0 at the stationary point. While it involves multiple steps and careful algebraic manipulation, it follows a standard implicit differentiation procedure without requiring novel insight. The 7 marks reflect the working required rather than exceptional conceptual difficulty, placing it slightly above average.
Spec1.07s Parametric and implicit differentiation

A curve has equation $$y^3e^{2x} + 2y - 16x = k$$ where \(k\) is a constant. The curve has a stationary point on the \(y\)-axis. Determine the value of \(k\) [7 marks]

Question 19:
AnswerMarks
19Uses implicit differentiation, with
dy d y
Ay2 or 2 seen.
AnswerMarks Guidance
dx d x3.1a M1
d y d y
3 2 2 y e x + 2 y 3 2 e x + 2 − 1 6 = 0
d x d x
d y
= 0 x , = 0
d x
2 3 y − 1 6 = 0
y = 2
2 3 0 + e 4 − 1 6  0 = k
k = 1 2
Uses product rule to differentiate
y3e x
and obtains
dy
Ay2e2x +By3e2x
AnswerMarks Guidance
dx3.1a M1
Obtains correctly
dy dy
3y2e2x +2y3e2x +2 −16=0
AnswerMarks Guidance
dx dx1.1b A1
dy
Substitutes =0,x=0into
dx
their differentiated equation or
rearranged equation to obtain a
value for y.
Their equation needs to have
contained either
dy d y
Ay2 or 2
dx d x
AnswerMarks Guidance
and involve e 2 x3.1a M1
Obtains y = 2
Must have achieved
M1M1A1M1 so far.
2
PI substituting y = and x=0
x 23
e
AnswerMarks Guidance
into y 3 e 2 x + 2 y − 1 6 x1.1b A1
Substitutes x=0and their y = 2
into y 3 e 2 x + 2 y − 1 6 x to obtain a
AnswerMarks Guidance
value of k3.1a M1
Deduces k =12
Must have achieved all previous
AnswerMarks Guidance
marks2.2a R1
Question 19 Total7
QMarking instructions AO
Question 19:
19 | Uses implicit differentiation, with
dy d y
Ay2 or 2 seen.
dx d x | 3.1a | M1 | y x 3 2 e + 2 y − 1 6 x = k
d y d y
3 2 2 y e x + 2 y 3 2 e x + 2 − 1 6 = 0
d x d x
d y
= 0 x , = 0
d x
2 3 y − 1 6 = 0
y = 2
2 3 0 + e 4 − 1 6  0 = k
k = 1 2
Uses product rule to differentiate
y3e x
and obtains
dy
Ay2e2x +By3e2x
dx | 3.1a | M1
Obtains correctly
dy dy
3y2e2x +2y3e2x +2 −16=0
dx dx | 1.1b | A1
dy
Substitutes =0,x=0into
dx
their differentiated equation or
rearranged equation to obtain a
value for y.
Their equation needs to have
contained either
dy d y
Ay2 or 2
dx d x
and involve e 2 x | 3.1a | M1
Obtains y = 2
Must have achieved
M1M1A1M1 so far.
2
PI substituting y = and x=0
x 23
e
into y 3 e 2 x + 2 y − 1 6 x | 1.1b | A1
Substitutes x=0and their y = 2
into y 3 e 2 x + 2 y − 1 6 x to obtain a
value of k | 3.1a | M1
Deduces k =12
Must have achieved all previous
marks | 2.2a | R1
Question 19 Total | 7
Q | Marking instructions | AO | Marks | Typical solution
A curve has equation
$$y^3e^{2x} + 2y - 16x = k$$
where $k$ is a constant.

The curve has a stationary point on the $y$-axis.

Determine the value of $k$
[7 marks]

\hfill \mbox{\textit{AQA Paper 1 2024 Q19 [7]}}