Edexcel PURE 2024 October — Question 2

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicParametric differentiation
TypeTangent parallel to axis condition
DifficultyModerate -0.8 This is a straightforward parametric differentiation question requiring only the reciprocal rule (dy/dx = 1/(dx/dy)) and recognizing that a vertical tangent means dx/dy = 0. Both parts involve standard techniques with minimal problem-solving, making it easier than average.
Spec1.07s Parametric and implicit differentiation
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b9472037-c143-4b68-86e2-801f71029773-04_761_758_251_657} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of the curve with equation $$x = 2 y ^ { 2 } + 5 y - 6$$
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\). The point \(P\) lies on the curve and is shown in Figure 1.
    Given that the tangent to the curve at \(P\) is parallel to the \(y\)-axis,
  2. find the coordinates of \(P\).
Question 2:
AnswerMarks
2 (a)x = 2 y 2 + 5 y − 6
dx dy 1
=4y+5 =
AnswerMarks
dy dx 4y+5M1, A1
(2)
AnswerMarks
(b)5
Sets their 4 y + 5 = 0  y = −
AnswerMarks
4M1
5
Substitutes their y = − into x = 2 y 2 + 5 y − 6 to find x
AnswerMarks
4dM1
 7 3 5 
− , −
AnswerMarks
8 4A1
(3)

Total 5

Further guidance to highlighted sentence.
If (a) is incorrect or incomplete follow the following guidance.
Incorrect (a): If they incorrectly state
d
d
y
x
= 4 y + 5 o.e. and go on to use 4 y + 5 = 0  y = ... and write
down a correct set of coordinates for P they will score NO MARKS in (b)
Incomplete (a): If they state
d
d
x
y
= 4 y + 5 in (a) and go on to use 4 y + 5 = 0  y = ... in (b) they will score
no marks in (a) but could potentially score full marks in (b)
 73 5
You can award all 3 marks for  − ,−  following either
 8 4
d
d
x
y
= 4 y + 5 or
d
d
y
x
=
4 y
1
+ 5
PMT
………………………………………………………………………………………………
Alternative approach: Part (a) may well be blank or incorrect. Part(b) done via completion of square
It is possible to do part (b) by not doing any differentiating.
2
 5 73
x = 2y2 +5y−6 x = 2 y+  −
 4 8
M1: An allowable attempt to complete the square followed by a valid attempt at either x or y from their
attempt
A1: For a correct x or y
A1: For a correct x and y
………………………………………………………………………………………………………..
Question 2:
--- 2 (a) ---
2 (a) | x = 2 y 2 + 5 y − 6
dx dy 1
=4y+5 =
dy dx 4y+5 | M1, A1
(2)
(b) | 5
Sets their 4 y + 5 = 0  y = −
4 | M1
5
Substitutes their y = − into x = 2 y 2 + 5 y − 6 to find x
4 | dM1
 7 3 5 
− , −
8 4 | A1
(3)
Total 5
Further guidance to highlighted sentence.
If (a) is incorrect or incomplete follow the following guidance.
Incorrect (a): If they incorrectly state
d
d
y
x
= 4 y + 5 o.e. and go on to use 4 y + 5 = 0  y = ... and write
down a correct set of coordinates for P they will score NO MARKS in (b)
Incomplete (a): If they state
d
d
x
y
= 4 y + 5 in (a) and go on to use 4 y + 5 = 0  y = ... in (b) they will score
no marks in (a) but could potentially score full marks in (b)
 73 5
You can award all 3 marks for  − ,−  following either
 8 4
d
d
x
y
= 4 y + 5 or
d
d
y
x
=
4 y
1
+ 5
PMT
………………………………………………………………………………………………
Alternative approach: Part (a) may well be blank or incorrect. Part(b) done via completion of square
It is possible to do part (b) by not doing any differentiating.
2
 5 73
x = 2y2 +5y−6 x = 2 y+  −
 4 8
M1: An allowable attempt to complete the square followed by a valid attempt at either x or y from their
attempt
A1: For a correct x or y
A1: For a correct x and y
………………………………………………………………………………………………………..
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b9472037-c143-4b68-86e2-801f71029773-04_761_758_251_657}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of the curve with equation

$$x = 2 y ^ { 2 } + 5 y - 6$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$.

The point $P$ lies on the curve and is shown in Figure 1.\\
Given that the tangent to the curve at $P$ is parallel to the $y$-axis,
\item find the coordinates of $P$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel PURE 2024 Q2}}
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