AQA Paper 3 2019 June — Question 9 15 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2019
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeShow no stationary points exist
DifficultyChallenging +1.2 This is a solid implicit differentiation question with multiple parts requiring careful algebraic manipulation. Part (a) is straightforward substitution. Part (b)(i) requires standard implicit differentiation technique with product rule. Parts (b)(ii) and (b)(iii) involve analyzing the derivative and finding a specific tangent line. While it requires multiple techniques and careful algebra, these are all standard A-level procedures without requiring novel insight or particularly complex reasoning. The question is moderately harder than average due to the algebraic complexity and multi-step nature, but remains within typical A-level scope.
Spec1.07s Parametric and implicit differentiation
A curve has equation $$x^2y^2 + xy^4 = 12$$
  1. Prove that the curve does not intersect the coordinate axes. [2 marks]
    1. Show that \(\frac{dy}{dx} = -\frac{2xy + y^3}{2x^2 + 4xy^2}\) [5 marks]
    2. Prove that the curve has no stationary points. [4 marks]
    3. In the case when \(x > 0\), find the equation of the tangent to the curve when \(y = 1\) [4 marks]
Question 9:
AnswerMarks
9(a)Demonstrates by substitution that
x=0or y=0 leads to value on
AnswerMarks Guidance
the LHS = 02.4 E1
2 2 4
When y = 00 𝑦𝑦 + 0𝑦𝑦 = 0
2 2 4
This is a cπ‘₯π‘₯on0tra+dic π‘₯π‘₯ti0on =be0cause
so the curve
do2es2 not in4tersect either axis
Completes rigorous argument to
AnswerMarks Guidance
show required result2.1 R1
(b)(i) ---
9
AnswerMarks Guidance
(b)(i)Uses implicit differentiation 3.1a
2xy2 +2x2y + y4 +4xy3 =0
dx dx
dy 2xy2 + y4
=βˆ’
dx 2x2y+4xy3
3
𝑦𝑦 (2π‘₯π‘₯𝑦𝑦+𝑦𝑦 )
2xy+2y3 2
== βˆ’βˆ’π‘¦π‘¦(2π‘₯π‘₯ +4π‘₯π‘₯,𝑦𝑦 )
2x2 +4xy2
Product rule used LHS (at least
AnswerMarks Guidance
one pair of terms correct)1.1a M1
Differentiates equation of curve
AnswerMarks Guidance
fully correctly1.1b A1
dy
Collects their terms in an
dx
AnswerMarks Guidance
equation and factorises3.1a M1
Completes convincing argument
to obtain required result by
factorising then simplifying y
AnswerMarks Guidance
AG2.1 R1
(b)(ii) ---
9
AnswerMarks
(b)(ii)Begins argument by setting
dy
=0 to form an equation for
dx
x and y
AnswerMarks Guidance
PI by 2xy+ y3 =02.1 M1
dy
=0
dx
β‡’2xy+ y3 =0
β‡’ y2 =βˆ’2x
β‡’ x2y2 +x (βˆ’2x ) y2 =12
β‡’βˆ’x2y2 =12
βˆ’x2y2
Since < 0 there can be no
stationary points.
Obtains y2 =βˆ’2x or y =
or x = βˆšβˆ’2π‘₯π‘₯
AnswerMarks Guidance
21.1b A1
βˆ’ 𝑦𝑦
Substitutes y2 =βˆ’2x or
2
x = into equation for curve
AnswerMarks Guidance
21.1a M1
βˆ’ 𝑦𝑦
Completes convincing argument
2
AnswerMarks Guidance
to deduce the required result2.2a R1
(b)(iii) ---
9
AnswerMarks
(b)(iii)Substitutes y = 1 into equation of
curve to obtain correct quadratic
AnswerMarks Guidance
ACF3.1a M1
β‡’ x=3 ( x>0 )
dy 7
β‡’ =βˆ’
dx 30
7
yβˆ’1=βˆ’ ( xβˆ’3 )
30
Deduces x = 3
PI by substituting their x in their
AnswerMarks Guidance
dy/dx2.2a R1
Substitutes their x and y = 1 in
AnswerMarks Guidance
their dy/dx1.1a M1
Obtains correct equation of
tangent
ACF
AnswerMarks Guidance
ISW1.1b A1
Total15
QMarking Instructions AO 
Question 9:
--- 9(a) ---
9(a) | Demonstrates by substitution that
x=0or y=0 leads to value on
the LHS = 0 | 2.4 | E1 | When x = 0
2 2 4
When y = 00 𝑦𝑦 + 0𝑦𝑦 = 0
2 2 4
This is a cπ‘₯π‘₯on0tra+dic π‘₯π‘₯ti0on =be0cause
so the curve
do2es2 not in4tersect either axis
Completes rigorous argument to
show required result | 2.1 | R1
--- 9
(b)(i) ---
9
(b)(i) | Uses implicit differentiation | 3.1a | M1 | π‘₯π‘₯ 𝑦𝑦 + π‘₯π‘₯𝑦𝑦 =dy12 dy
2xy2 +2x2y + y4 +4xy3 =0
dx dx
dy 2xy2 + y4
=βˆ’
dx 2x2y+4xy3
3
𝑦𝑦 (2π‘₯π‘₯𝑦𝑦+𝑦𝑦 )
2xy+2y3 2
== βˆ’βˆ’π‘¦π‘¦(2π‘₯π‘₯ +4π‘₯π‘₯,𝑦𝑦 )
2x2 +4xy2
Product rule used LHS (at least
one pair of terms correct) | 1.1a | M1
Differentiates equation of curve
fully correctly | 1.1b | A1
dy
Collects their terms in an
dx
equation and factorises | 3.1a | M1
Completes convincing argument
to obtain required result by
factorising then simplifying y
AG | 2.1 | R1
--- 9
(b)(ii) ---
9
(b)(ii) | Begins argument by setting
dy
=0 to form an equation for
dx
x and y
PI by 2xy+ y3 =0 | 2.1 | M1 | For stationary points
dy
=0
dx
β‡’2xy+ y3 =0
β‡’ y2 =βˆ’2x
β‡’ x2y2 +x (βˆ’2x ) y2 =12
β‡’βˆ’x2y2 =12
βˆ’x2y2
Since < 0 there can be no
stationary points.
Obtains y2 =βˆ’2x or y =
or x = βˆšβˆ’2π‘₯π‘₯
2 | 1.1b | A1
βˆ’ 𝑦𝑦
Substitutes y2 =βˆ’2x or
2
x = into equation for curve
2 | 1.1a | M1
βˆ’ 𝑦𝑦
Completes convincing argument
2
to deduce the required result | 2.2a | R1
--- 9
(b)(iii) ---
9
(b)(iii) | Substitutes y = 1 into equation of
curve to obtain correct quadratic
ACF | 3.1a | M1 | y =1β‡’ x2 +xβˆ’12=0
β‡’ x=3 ( x>0 )
dy 7
β‡’ =βˆ’
dx 30
7
yβˆ’1=βˆ’ ( xβˆ’3 )
30
Deduces x = 3
PI by substituting their x in their
dy/dx | 2.2a | R1
Substitutes their x and y = 1 in
their dy/dx | 1.1a | M1
Obtains correct equation of
tangent
ACF
ISW | 1.1b | A1
Total | 15
Q | Marking Instructions | AO | Mark | Typical Solution
A curve has equation

$$x^2y^2 + xy^4 = 12$$

\begin{enumerate}[label=(\alph*)]
\item Prove that the curve does not intersect the coordinate axes. [2 marks]

\item \begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dy}{dx} = -\frac{2xy + y^3}{2x^2 + 4xy^2}$ [5 marks]

\item Prove that the curve has no stationary points. [4 marks]

\item In the case when $x > 0$, find the equation of the tangent to the curve when $y = 1$ [4 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2019 Q9 [15]}}
This paper (17 questions)
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Q1 1 Q2 1 Q3 1 Q4 3 Q5 5 Q6 4 Q7 8 Q8 12 Q9 15 Q10 1 Q11 1 Q12 6 Q13 10 Q14 7 Q15 3 Q16 10 Q17 12