OCR MEI Further Extra Pure 2020 November — Question 6 17 marks

Exam BoardOCR MEI
ModuleFurther Extra Pure (Further Extra Pure)
Year2020
SessionNovember
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind line of invariant points
DifficultyChallenging +1.2 This is a multi-part question on partial derivatives and 3D surfaces requiring standard techniques: finding stationary points via partial derivatives, factorising a quartic expression, sketching contours, finding tangent plane equations, and computing limits. While it involves several steps and some algebraic manipulation (especially the factorisation), each component uses routine A-level Further Maths methods without requiring novel insight or particularly challenging problem-solving.
Spec8.05e 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
6 A surface \(S\) is defined by \(z = \mathrm { f } ( x , y ) = 4 x ^ { 4 } + 4 y ^ { 4 } - 17 x ^ { 2 } y ^ { 2 }\).
    1. Show that there is only one stationary point on \(S\). The value of \(z\) at the stationary point is denoted by \(s\).
    2. State the value of \(s\).
    3. By factorising \(\mathrm { f } ( x , y )\), sketch the contour lines of the surface for \(z = s\).
    4. Hence explain whether the stationary point is a maximum point, a minimum point or a saddle point. C is a point on \(S\) with coordinates ( \(a , a , \mathrm { f } ( a , a )\) ) where \(a\) is a constant and \(a \neq 0\). \(\Pi\) is the tangent plane to \(S\) at C .
    1. Find the equation of \(\Pi\) in the form r.n \(= p\).
    2. The shortest distance from the origin to \(\Pi\) is denoted by \(d\). Show that \(\frac { d } { a } \rightarrow \frac { 3 \sqrt { 2 } } { 4 }\) as \(a \rightarrow \infty\).
    3. Explain whether the origin lies above or below \(\Pi\). \section*{END OF QUESTION PAPER}
Question 6:
AnswerMarks Guidance
6(a) (i)
=16x3 −34xy2
∂x
∂f
=16y3 −34x2y
∂y
16x3 – 34xy2 = 0 and 16y3 – 34x2y = 0
∂f
So x = 0 or 16x2 – 34y2 = 0 (or equivalent for =0)
∂y
But both x = 0 and y = 0 when
substituted into the othe2r equatio2n so
16𝑥𝑥 −34𝑦𝑦 =0 ⇒
AnswerMarks
x = 0 and y = 0 [is the only solution].B1
M1
M1
M1
A1
AnswerMarks
[5]1.1
1.1
1.1
1.1
AnswerMarks
1.1for both
both
For M0 here SC1 for
and subs x2 or
y2 int2o the oth2er equation
16𝑥𝑥 −34𝑦𝑦 =0
For M0 here SC1 for x = y = 0
only
AnswerMarks Guidance
6(a) (ii)
[1]1.1
6(a) (iii)
(2x – y)(2x + y)(x – 2y)(x + 2y) or
AnswerMarks
y = (+/-)2x and y = (+/-) ½xM1
A1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1Sketch of the four complete (ie
each line going across two
quadrants) lines y = ±2x and
y = ±½x. No scale necessary.
AnswerMarks Guidance
6(a) (iv)
‘wedges’ so it is not the case that z > 0 at all points
near P (the stationary point) so it is not a minimum and
similarly it is not the case that z < 0 at all points near P
so it is not a maximum.
AnswerMarks Guidance
So P must be a saddle point.B1 2.4
moving eg along x-axis, through
P, z is +ve, 0, +ve while along eg
y = x, through P, z is –ve, 0, –ve
or z is positive on the negative x-
axis and negative on the positive
AnswerMarks
branch of y = x etcSC1 No appeal to diagram
but correctly finding two z
coordinates, one positive
and one negative and
stating that there are no
other SPs
No FT for 0/3 in 6 (a) (iii)
AnswerMarks Guidance
6(b) (i)
   
n= 16a3 −34a2×a = −18a3
 
 −1   −1 
   
 a  −18a3
   
p=  a  . −18a3 
  4a4 +4a4 −17a2×a2     −1  
18a3
 
r.18a3  =27a4 oe
 
1
AnswerMarks
 M1
M1
AnswerMarks
A11.1
1.1
AnswerMarks
1.1or any non-zero multiple
their n
cao isw
AnswerMarks Guidance
6(b) (ii)
d = soi
n
n = 2 ( 18a3)2 +1≈18 2a3 for large a
d 1 27a4 3 2
∴ → × =
AnswerMarks
a a 18 2a3 4M1
M1
A1
AnswerMarks
[3]3.1a
3.1a
AnswerMarks
3.2aFT their (i)
FT their (i)
Or equivalent argument using
limits
AnswerMarks Guidance
AGNo limits discussion SC1
6(b) (iii)
If x = y = 0 in equation for Π then z = 27a4 > 0 so the
z-intercept is positive so the origin is below the
AnswerMarks Guidance
plane.E1
[1]2.4 Or in the equation
18a3
 
r.18a3  =27a4 the z component
 
1
 
of n is positive and so n is
pointing upward but p > 0 so O is
on the other side of the plane ie
below (or equivalent argument
AnswerMarks
with –ve signs).Needs more than z > 0
PPMMTT
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Shaftesbury Road
Cambridge
CB2 8EA
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Question 6:
6 | (a) | (i) | ∂f
=16x3 −34xy2
∂x
∂f
=16y3 −34x2y
∂y
16x3 – 34xy2 = 0 and 16y3 – 34x2y = 0
∂f
So x = 0 or 16x2 – 34y2 = 0 (or equivalent for =0)
∂y
But both x = 0 and y = 0 when
substituted into the othe2r equatio2n so
16𝑥𝑥 −34𝑦𝑦 =0 ⇒
x = 0 and y = 0 [is the only solution]. | B1
M1
M1
M1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | for both
both
For M0 here SC1 for
and subs x2 or
y2 int2o the oth2er equation
16𝑥𝑥 −34𝑦𝑦 =0
For M0 here SC1 for x = y = 0
only
6 | (a) | (ii) | (s =) 0 | B1
[1] | 1.1
6 | (a) | (iii) | 4x4 + 4y4 – 17x2y2 = (4x2 – y2)(x2 – 4y2)
(2x – y)(2x + y)(x – 2y)(x + 2y) or
y = (+/-)2x and y = (+/-) ½x | M1
A1
A1
[3] | 1.1
1.1
1.1 | Sketch of the four complete (ie
each line going across two
quadrants) lines y = ±2x and
y = ±½x. No scale necessary.
6 | (a) | (iv) | The z = 0 plane is divided into positive and negative
‘wedges’ so it is not the case that z > 0 at all points
near P (the stationary point) so it is not a minimum and
similarly it is not the case that z < 0 at all points near P
so it is not a maximum.
So P must be a saddle point. | B1 | 2.4 | or equivalent explanation eg
moving eg along x-axis, through
P, z is +ve, 0, +ve while along eg
y = x, through P, z is –ve, 0, –ve
or z is positive on the negative x-
axis and negative on the positive
branch of y = x etc | SC1 No appeal to diagram
but correctly finding two z
coordinates, one positive
and one negative and
stating that there are no
other SPs
No FT for 0/3 in 6 (a) (iii)
6 | (b) | (i) | 16a3 −34a×a2 −18a3
   
n= 16a3 −34a2×a = −18a3
 
 −1   −1 
   
 a  −18a3
   
p=  a  . −18a3 
  4a4 +4a4 −17a2×a2     −1  
18a3
 
r.18a3  =27a4 oe
 
1
  | M1
M1
A1 | 1.1
1.1
1.1 | or any non-zero multiple
their n
cao isw
6 | (b) | (ii) | 27a4
d = soi
n
n = 2 ( 18a3)2 +1≈18 2a3 for large a
d 1 27a4 3 2
∴ → × =
a a 18 2a3 4 | M1
M1
A1
[3] | 3.1a
3.1a
3.2a | FT their (i)
FT their (i)
Or equivalent argument using
limits
AG | No limits discussion SC1
6 | (b) | (iii) | Below.
If x = y = 0 in equation for Π then z = 27a4 > 0 so the
z-intercept is positive so the origin is below the
plane. | E1
[1] | 2.4 | Or in the equation
18a3
 
r.18a3  =27a4 the z component
 
1
 
of n is positive and so n is
pointing upward but p > 0 so O is
on the other side of the plane ie
below (or equivalent argument
with –ve signs). | Needs more than z > 0
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
6 A surface $S$ is defined by $z = \mathrm { f } ( x , y ) = 4 x ^ { 4 } + 4 y ^ { 4 } - 17 x ^ { 2 } y ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that there is only one stationary point on $S$.

The value of $z$ at the stationary point is denoted by $s$.
\item State the value of $s$.
\item By factorising $\mathrm { f } ( x , y )$, sketch the contour lines of the surface for $z = s$.
\item Hence explain whether the stationary point is a maximum point, a minimum point or a saddle point.

C is a point on $S$ with coordinates ( $a , a , \mathrm { f } ( a , a )$ ) where $a$ is a constant and $a \neq 0$. $\Pi$ is the tangent plane to $S$ at C .
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the equation of $\Pi$ in the form r.n $= p$.
\item The shortest distance from the origin to $\Pi$ is denoted by $d$. Show that $\frac { d } { a } \rightarrow \frac { 3 \sqrt { 2 } } { 4 }$ as $a \rightarrow \infty$.
\item Explain whether the origin lies above or below $\Pi$.

\section*{END OF QUESTION PAPER}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Extra Pure 2020 Q6 [17]}}
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