OCR MEI Further Extra Pure 2022 June — Question 5 16 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2022
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeDescribe geometric transformation from matrix
DifficultyChallenging +1.8 This is a substantial multi-part question requiring partial differentiation, stationary point analysis, geometric transformations of graphs, classification using second derivatives/contours, and tangent plane equations. While each individual technique is A-level Further Maths standard, the question demands sustained reasoning across 6 parts with non-trivial algebraic manipulation (especially the exponential function) and geometric insight to connect 2D/3D representations. The transformation sequence and classification require more problem-solving than routine application.
Spec8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero8.05f Nature of stationary points: classify using Hessian matrix8.05g Tangent planes: equation at a given point on surface
5 A surface \(S\) is defined by \(z = f ( x , y )\), where \(f ( x , y ) = y e ^ { - \left( x ^ { 2 } + 2 x + 2 \right) y }\).
    1. Find \(\frac { \partial f } { \partial x }\).
    2. Show that \(\frac { \partial f } { \partial y } = - \left( x ^ { 2 } y + 2 x y + 2 y - 1 \right) e ^ { - \left( x ^ { 2 } + 2 x + 2 \right) y }\).
    3. Determine the coordinates of any stationary points on \(S\). Fig. 5.1 shows the graph of \(z = e ^ { - x ^ { 2 } }\) and Fig. 5.2 shows the contour of \(S\) defined by \(z = 0.25\). \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{76f3559a-f3b3-4a21-878f-adb261dd1236-5_478_686_822_244} \captionsetup{labelformat=empty} \caption{Fig. 5.1}
      \end{figure} \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{76f3559a-f3b3-4a21-878f-adb261dd1236-5_478_437_822_1105} \captionsetup{labelformat=empty} \caption{Fig. 5.2}
      \end{figure}
  2. Specify a sequence of transformations which transforms the graph of \(\mathrm { z } = \mathrm { e } ^ { - \mathrm { x } ^ { 2 } }\) onto the graph of the section defined by \(z = f ( x , 1 )\).
  3. Hence, or otherwise, sketch the section defined by \(z = f ( x , 1 )\).
  4. Using Fig. 5.2 and your answer to part (c), classify any stationary points on \(S\), justifying your answer. You are given that \(P\) is a point on \(S\) where \(z = 0\).
  5. Find, in vector form, the equation of the tangent plane to \(S\) at \(P\). The tangent plane found in part (e) intersects \(S\) in a straight line, \(L\).
  6. Write down, in vector form, the equation of \(L\).
Question 5:
AnswerMarks Guidance
5(a) (i)
∂xB1
[1]1.1 −2y2(x+1)e −(x2+2x+2)y
5(a) (ii)
=e
∂y
=(1−y(x2+2x+2))e −(x2+2x+2)y
AnswerMarks Guidance
=−(x2y+2xy+2y−1)e −(x2+2x+2)yB1
[1]2.1 Use of product and chain rules
must be clear.
AG Intermediate working must
be shown.
AnswerMarks Guidance
5(a) (iii)
=0⇒ y=0 or x=−1
∂x
∂f
y=0⇒ =1 so no SP
∂y
∂f
x=−1 and =0⇒(1− y)e −y =0⇒ y=1
∂y
AnswerMarks
(–1 , 1, e–1)M1
A1
M1
A1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks
1.1Solving f = 0 to produce at least
x
one numerical solution
Eliminating y = 0
Using x-value and f = 0
y
condition to find a y-value. This
mark can be awarded even with
‘rogue’ SPs or y = 0 not properly
considered.
Must be the only SP
AnswerMarks Guidance
5(b) Translation, 1 unit in the negative x-direction
and stretch, parallel to the z-axis, scale factor e–
AnswerMarks
1B1
B1
AnswerMarks
[2]3.1a
1.1or –1 unit in the x-direction or
defined by vector (may be 2D).
Condone stretch “in y-direction”
or “vertical” if intention clear.
AnswerMarks
Must be exact value.f(x,1)=e −(x2+2x+2) =e −1e −(x+1)2
The transformations can be given in
either order.
AnswerMarks Guidance
5(c) B1FT
[1]1.1 Sketch of given graph with peak
marked at (–1, e–1). Condone
awrt 0.37. FT their
transformation if properly defined
and stretch and translation only.
AnswerMarks Guidance
5(d) The SP is a maximum point...
...since e–1 > 0.25 and if moving from the SP to
the contour line in any direction we must go
AnswerMarks
down since the contour line is closed.B1
B1
AnswerMarks
[2]3.2a
2.4Or sketch of section shows a
maximum and a closed contour
line around a single SP can only
contain a maximum or a
minimum
z = 0 ⇒y = 0
 −(x2+2x+2)y×(−y(2x+2)) 
ye
 
∇f =  e −(x2+2x+2)y−y(x 2+2x+2)e −(x2+2x+2)y
 
 −1 
 
 
 0 
 
y = 0 => n=∇f = 1 cwo
 
−1
 0  x
   
1 . 0 =0for any x so equation of tangent
   
 −1 0
 0  x
   
plane is 1 . y =0 oe
   
AnswerMarks
 −1 zB1
M1
A1
AnswerMarks
A1FT3.1a
1.1
1.1
AnswerMarks
3.2aCould be seen at any stage
Use of ∂f ∂f
∇f= i+ j−k
∂x ∂y
Could be derivative from (a).
Could be used once values found.
Must be from the correct ∇f.
FT their normal with numerical
components.
Normal could be changed to eg
 0 
 
−1
 
AnswerMarks
 1  −2y2(x+1)e −(x2+2x+2)y 
 
 ( 1−y(x2+2x+2) ) e −(x2+2x+2)y
 
 −1 
 
Alternative method for last 3 marks:
∂f ∂f
y=0⇒ =0, =1
∂x ∂y
AnswerMarks
∴z−0=0(x−a)+1(y−0)M1
Calculating the values of the
partial derivatives (possibly
from(a)) and using
∂f ∂f
z−c= (x−a)+ (y−b)
∂x ∂y
AnswerMarks
∴Cartesian equation of tangent plane is z = yA1
Alternative method for last 3 marks:
∂f ∂f
y=0⇒ =0, =1
∂x ∂y
∴z−0=0(x−a)+1(y−0)
M1
 0  x
   
So vector equation of plane is 1 . y =0
   
AnswerMarks Guidance
 −1 zA1  0  x
   
or eg −1 . y =0
   
 1  z
[4]
AnswerMarks Guidance
5(f) 1
 
Eqn of line is r=λ 0
 
AnswerMarks Guidance
0B1
[1]3.2a a 1
   
Or r= 0 +λ 0 for any
   
0 0
AnswerMarks
numerical a.No justification for correct equation
is necessary but must be in correct
vector form with “r =” oe.
A1
AnswerMarks
ResponseMark 
Question 5:
5 | (a) | (i) | ∂f = ye −(x2+2x+2)y×(−y(2x+2) )oe
∂x | B1
[1] | 1.1 | −2y2(x+1)e −(x2+2x+2)y
5 | (a) | (ii) | ∂f −(x2+2x+2)y−y(x2+2x+2)e −(x2+2x+2)y
=e
∂y
=(1−y(x2+2x+2))e −(x2+2x+2)y
=−(x2y+2xy+2y−1)e −(x2+2x+2)y | B1
[1] | 2.1 | Use of product and chain rules
must be clear.
AG Intermediate working must
be shown.
5 | (a) | (iii) | ∂f
=0⇒ y=0 or x=−1
∂x
∂f
y=0⇒ =1 so no SP
∂y
∂f
x=−1 and =0⇒(1− y)e −y =0⇒ y=1
∂y
(–1 , 1, e–1) | M1
A1
M1
A1
[4] | 1.1
1.1
1.1
1.1 | Solving f = 0 to produce at least
x
one numerical solution
Eliminating y = 0
Using x-value and f = 0
y
condition to find a y-value. This
mark can be awarded even with
‘rogue’ SPs or y = 0 not properly
considered.
Must be the only SP
5 | (b) | Translation, 1 unit in the negative x-direction
and stretch, parallel to the z-axis, scale factor e–
1 | B1
B1
[2] | 3.1a
1.1 | or –1 unit in the x-direction or
defined by vector (may be 2D).
Condone stretch “in y-direction”
or “vertical” if intention clear.
Must be exact value. | f(x,1)=e −(x2+2x+2) =e −1e −(x+1)2
The transformations can be given in
either order.
5 | (c) | B1FT
[1] | 1.1 | Sketch of given graph with peak
marked at (–1, e–1). Condone
awrt 0.37. FT their
transformation if properly defined
and stretch and translation only.
5 | (d) | The SP is a maximum point...
...since e–1 > 0.25 and if moving from the SP to
the contour line in any direction we must go
down since the contour line is closed. | B1
B1
[2] | 3.2a
2.4 | Or sketch of section shows a
maximum and a closed contour
line around a single SP can only
contain a maximum or a
minimum
z = 0 ⇒y = 0
 −(x2+2x+2)y×(−y(2x+2)) 
ye
 
∇f =  e −(x2+2x+2)y−y(x 2+2x+2)e −(x2+2x+2)y
 
 −1 
 
 
 0 
 
y = 0 => n=∇f = 1 cwo
 
−1

 0  x
   
1 . 0 =0for any x so equation of tangent
   
 −1 0
 0  x
   
plane is 1 . y =0 oe
   
 −1 z | B1
M1
A1
A1FT | 3.1a
1.1
1.1
3.2a | Could be seen at any stage
Use of ∂f ∂f
∇f= i+ j−k
∂x ∂y
Could be derivative from (a).
Could be used once values found.
Must be from the correct ∇f.
FT their normal with numerical
components.
Normal could be changed to eg
 0 
 
−1
 
 1  |  −2y2(x+1)e −(x2+2x+2)y 
 
 ( 1−y(x2+2x+2) ) e −(x2+2x+2)y
 
 −1 
 
Alternative method for last 3 marks:
∂f ∂f
y=0⇒ =0, =1
∂x ∂y
∴z−0=0(x−a)+1(y−0) | M1
Calculating the values of the
partial derivatives (possibly
from(a)) and using
∂f ∂f
z−c= (x−a)+ (y−b)
∂x ∂y
∴Cartesian equation of tangent plane is z = y | A1
Alternative method for last 3 marks:
∂f ∂f
y=0⇒ =0, =1
∂x ∂y
∴z−0=0(x−a)+1(y−0)
M1
 0  x
   
So vector equation of plane is 1 . y =0
   
 −1 z | A1 |  0  x
   
or eg −1 . y =0
   
 1  z
[4]
5 | (f) | 1
 
Eqn of line is r=λ 0
 
0 | B1
[1] | 3.2a | a 1
   
Or r= 0 +λ 0 for any
   
0 0
numerical a. | No justification for correct equation
is necessary but must be in correct
vector form with “r =” oe.
A1
Response | Mark
5 A surface $S$ is defined by $z = f ( x , y )$, where $f ( x , y ) = y e ^ { - \left( x ^ { 2 } + 2 x + 2 \right) y }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \partial f } { \partial x }$.
\item Show that $\frac { \partial f } { \partial y } = - \left( x ^ { 2 } y + 2 x y + 2 y - 1 \right) e ^ { - \left( x ^ { 2 } + 2 x + 2 \right) y }$.
\item Determine the coordinates of any stationary points on $S$.

Fig. 5.1 shows the graph of $z = e ^ { - x ^ { 2 } }$ and Fig. 5.2 shows the contour of $S$ defined by $z = 0.25$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{76f3559a-f3b3-4a21-878f-adb261dd1236-5_478_686_822_244}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{76f3559a-f3b3-4a21-878f-adb261dd1236-5_478_437_822_1105}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\end{center}
\end{figure}
\end{enumerate}\item Specify a sequence of transformations which transforms the graph of $\mathrm { z } = \mathrm { e } ^ { - \mathrm { x } ^ { 2 } }$ onto the graph of the section defined by $z = f ( x , 1 )$.
\item Hence, or otherwise, sketch the section defined by $z = f ( x , 1 )$.
\item Using Fig. 5.2 and your answer to part (c), classify any stationary points on $S$, justifying your answer.

You are given that $P$ is a point on $S$ where $z = 0$.
\item Find, in vector form, the equation of the tangent plane to $S$ at $P$.

The tangent plane found in part (e) intersects $S$ in a straight line, $L$.
\item Write down, in vector form, the equation of $L$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2022 Q5 [16]}}
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