OCR MEI Further Pure with Technology 2019 June — Question 3 20 marks

Exam BoardOCR MEI
ModuleFurther Pure with Technology (Further Pure with Technology)
Year2019
SessionJune
Marks20
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeIterative/numerical methods
DifficultyChallenging +1.2 This is a multi-part Further Maths question requiring isocline analysis, solving separable/linear ODEs, and implementing RK4 in a spreadsheet. While it covers several techniques and requires careful work, each individual part is relatively standard: isoclines are geometric interpretation, parts (b) and (c) involve routine integration, and RK4 implementation follows a given formula. The question is longer and more involved than typical A-level, justifying a positive score, but doesn't require novel insights or particularly deep problem-solving.
Spec1.09d Newton-Raphson method4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts
3 This question concerns the family of differential equations \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - x ^ { a } y \left( { } ^ { * } \right)\) where \(a\) is \(- 1,0\) or 1 .
  1. Determine and describe geometrically the isoclines of (\textit{) when
    1. \(a = - 1\),
    2. \(a = 0\),
    3. \(a = 1\).
  2. In this part of the question \(a = 0\).
    1. Write down the solution to \(( * )\) which passes through the point \(( 0 , b )\) where \(b \neq 1\).
    2. Write down the equation of the asymptote to this solution.
  3. In this part of the question \(a = - 1\).
    1. Write down the solution to \(( * )\) which passes through the point \(( c , d )\) where \(c \neq 0\).
    2. Describe the relationship between \(c\) and \(d\) when the solution in part (i) has a stationary point.
  4. In this part of the question \(a = 1\).
    1. The standard Runge-Kutta method of order 4 for the solution of the differential equation \(\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }\) is as follows. \(k _ { 1 } = h \mathrm { f } \left( x _ { n } , y _ { n } \right)\) \(k _ { 2 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 1 } } { 2 } \right)\) \(k _ { 3 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 2 } } { 2 } \right)\) \(k _ { 4 } = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 3 } \right)\) \(y _ { n + 1 } = y _ { n } + \frac { 1 } { 6 } \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right)\).
      Construct a spreadsheet to solve (}) in the case \(x _ { 0 } = 0\) and \(y _ { 0 } = 1.5\). State the formulae you have used in your spreadsheet.
    2. Use your spreadsheet with \(h = 0.05\) to find an approximation to the value of \(y\) when \(x = 1\).
    3. The solution to \(( * )\) in which \(x _ { 0 } = 0\) and \(y _ { 0 } = 1.5\) has a maximum point ( \(r , s\) ) with \(0 < r < 1\). Use your spreadsheet with suitable values of \(h\) to estimate \(r\) to two decimal places. Justify your answer.
Question 3:
AnswerMarks Guidance
3(a) i
y
1 m y(1m)x.
x
AnswerMarks
These are straight lines through the origin.M1
B1
AnswerMarks
[2]1.1b
1.2Substituting a = -1 and
rearranging into a suitable form.
Substituting a = 0 and rearranging
into a suitable form.
Substituting a = 1 and rearranging
into a suitable form.
Any reasonable geometrical
description allowed. Condone
lack of consideration of m = 1.
AnswerMarks
iiIf a = 0 the isoclines satisfy
1ymy1m.
AnswerMarks
These are horizontal lines.M1
B1
AnswerMarks
[2]1.1b
1.2
AnswerMarks
iiiIf a = 1 the isoclines satisfy
1m
1xym y .
x
1
This is an stretch/enlargement of y
x
except when m = 1 when it is the line y = 0)
OR
Hyperbolae with the x and y axes as
asymptotes (except when m = 1 when it is the
AnswerMarks
line y = 0).M1
B1
AnswerMarks
[2]1.1b
1.2
AnswerMarks Guidance
(b)i Solution is y1(b1)ex .
[1]1.1a
iiAsymptote is y = 1. B1
[1]1.2
(c)i x2 2cd c2
Solution is y
AnswerMarks
2xB1
[1]1.1a
iidy x2 c2 2cd
dx 2x2
dy
So 0 x2 c(2dc) o.e
dx
dy
0 has a solution x if and only if
dx
c(2dc)0.
This is if and only if either (c0 and 2d c)
AnswerMarks
or (c0 and 2d c). caoM1
M1
M1
A1
AnswerMarks
[4]1.1b
1.1b
3.1a
AnswerMarks
3.2a[x ≠ 0] Alternatively, using the
differential equation.
dy y
01 0 yx
dx x
Then substituting into y equation
gives x2 c(2dc)
Condone use of ≥.
Or equivalent description of
region in (c, d) plane. Must be
strict inequalities since x ≠ 0.
SC 1 for correct answer
unsupported.
Solution possible involving
x
identifying y as common
2
asymptote to solutions in (c) (i).
Conclusion then needs to be fully
explained. Identifying the
common asymptote with correct
AnswerMarks Guidance
conclusion only awarded SC 2.Need not be factorised.
(d)i A2 = 0, B2 = 1.5, H1 = 0.05
C2 = \(H\)1*(1-A2*B2)
D2 =\(H\)1*(1-(A2+\(H\)1/2)*(B2+C2/2))
E2 =\(H\)1*(1-(A2+\(H\)1/2)*(B2+D2/2))
F2=\(H\)1*(1-(A2+\(H\)1)*(B2+E2))
A3 =A2+\(H\)1
AnswerMarks
B3 = B2 + (1/6)(C2 + 2*D2+2*E2+F2)M1
M1
M1
M1
AnswerMarks
[4]1.1
3.1a
3.1a
AnswerMarks
2.5Give reasonable BOD on possible
transcription errors and consider a
correct answer to 3(d)(ii) as
evidence of correct formulae in
the spreadsheet.
Columns for x and y or equivalent
Columns for k and k or
1 2
equivalent
Columns for k and k or
3 4
equivalent
Formulae for x and y
n+1 n+1
AnswerMarks Guidance
iiSpreadsheet gives 1.63457442 (to 8 d.p.) B1
[1]1.1 Correct answer to at least 3 s.f.
Must for correct for the number of
significant figures given.
AnswerMarks
iiiSetting h = 0.005 in the spreadsheet gives
these estimates, x coordinates on the left,
corresponding y coordinates on the right
(extract from spreadsheet):
AnswerMarks
This gives an estimate of 0.56 (to 2 d.p.).M1
A1
AnswerMarks
[2]3.1a
3.2aMust see at least three suitable
(x, y) pairs. Or argument based
sign change in gradient
sufficiently close to 0.56.
SC 1 for 0.56 unsupported.
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
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programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
Question 3:
3 | (a) | i | If a = -1 the isoclines satisfy
y
1 m y(1m)x.
x
These are straight lines through the origin. | M1
B1
[2] | 1.1b
1.2 | Substituting a = -1 and
rearranging into a suitable form.
Substituting a = 0 and rearranging
into a suitable form.
Substituting a = 1 and rearranging
into a suitable form.
Any reasonable geometrical
description allowed. Condone
lack of consideration of m = 1.
ii | If a = 0 the isoclines satisfy
1ymy1m.
These are horizontal lines. | M1
B1
[2] | 1.1b
1.2
iii | If a = 1 the isoclines satisfy
1m
1xym y .
x
1
This is an stretch/enlargement of y
x
except when m = 1 when it is the line y = 0)
OR
Hyperbolae with the x and y axes as
asymptotes (except when m = 1 when it is the
line y = 0). | M1
B1
[2] | 1.1b
1.2
(b) | i | Solution is y1(b1)ex . | B1
[1] | 1.1a
ii | Asymptote is y = 1. | B1
[1] | 1.2
(c) | i | x2 2cd c2
Solution is y
2x | B1
[1] | 1.1a
ii | dy x2 c2 2cd

dx 2x2
dy
So 0 x2 c(2dc) o.e
dx
dy
0 has a solution x if and only if
dx
c(2dc)0.
This is if and only if either (c0 and 2d c)
or (c0 and 2d c). cao | M1
M1
M1
A1
[4] | 1.1b
1.1b
3.1a
3.2a | [x ≠ 0] Alternatively, using the
differential equation.
dy y
01 0 yx
dx x
Then substituting into y equation
gives x2 c(2dc)
Condone use of ≥.
Or equivalent description of
region in (c, d) plane. Must be
strict inequalities since x ≠ 0.
SC 1 for correct answer
unsupported.
Solution possible involving
x
identifying y as common
2
asymptote to solutions in (c) (i).
Conclusion then needs to be fully
explained. Identifying the
common asymptote with correct
conclusion only awarded SC 2. | Need not be factorised.
(d) | i | A2 = 0, B2 = 1.5, H1 = 0.05
C2 = $H$1*(1-A2*B2)
D2 =$H$1*(1-(A2+$H$1/2)*(B2+C2/2))
E2 =$H$1*(1-(A2+$H$1/2)*(B2+D2/2))
F2=$H$1*(1-(A2+$H$1)*(B2+E2))
A3 =A2+$H$1
B3 = B2 + (1/6)(C2 + 2*D2+2*E2+F2) | M1
M1
M1
M1
[4] | 1.1
3.1a
3.1a
2.5 | Give reasonable BOD on possible
transcription errors and consider a
correct answer to 3(d)(ii) as
evidence of correct formulae in
the spreadsheet.
Columns for x and y or equivalent
Columns for k and k or
1 2
equivalent
Columns for k and k or
3 4
equivalent
Formulae for x and y
n+1 n+1
ii | Spreadsheet gives 1.63457442 (to 8 d.p.) | B1
[1] | 1.1 | Correct answer to at least 3 s.f.
Must for correct for the number of
significant figures given.
iii | Setting h = 0.005 in the spreadsheet gives
these estimates, x coordinates on the left,
corresponding y coordinates on the right
(extract from spreadsheet):
This gives an estimate of 0.56 (to 2 d.p.). | M1
A1
[2] | 3.1a
3.2a | Must see at least three suitable
(x, y) pairs. Or argument based
sign change in gradient
sufficiently close to 0.56.
SC 1 for 0.56 unsupported.
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
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
3 This question concerns the family of differential equations $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - x ^ { a } y \left( { } ^ { * } \right)$\\
where $a$ is $- 1,0$ or 1 .
\begin{enumerate}[label=(\alph*)]
\item Determine and describe geometrically the isoclines of (\textit{) when
\begin{enumerate}[label=(\roman*)]
\item $a = - 1$,
\item $a = 0$,
\item $a = 1$.
\end{enumerate}\item In this part of the question $a = 0$.
\begin{enumerate}[label=(\roman*)]
\item Write down the solution to $( * )$ which passes through the point $( 0 , b )$ where $b \neq 1$.
\item Write down the equation of the asymptote to this solution.
\end{enumerate}\item In this part of the question $a = - 1$.
\begin{enumerate}[label=(\roman*)]
\item Write down the solution to $( * )$ which passes through the point $( c , d )$ where $c \neq 0$.
\item Describe the relationship between $c$ and $d$ when the solution in part (i) has a stationary point.
\end{enumerate}\item In this part of the question $a = 1$.
\begin{enumerate}[label=(\roman*)]
\item The standard Runge-Kutta method of order 4 for the solution of the differential equation $\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }$ is as follows.\\
$k _ { 1 } = h \mathrm { f } \left( x _ { n } , y _ { n } \right)$\\
$k _ { 2 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 1 } } { 2 } \right)$\\
$k _ { 3 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 2 } } { 2 } \right)$\\
$k _ { 4 } = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 3 } \right)$\\
$y _ { n + 1 } = y _ { n } + \frac { 1 } { 6 } \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right)$.\\
Construct a spreadsheet to solve (}) in the case $x _ { 0 } = 0$ and $y _ { 0 } = 1.5$. State the formulae you have used in your spreadsheet.
\item Use your spreadsheet with $h = 0.05$ to find an approximation to the value of $y$ when $x = 1$.
\item The solution to $( * )$ in which $x _ { 0 } = 0$ and $y _ { 0 } = 1.5$ has a maximum point ( $r , s$ ) with $0 < r < 1$. Use your spreadsheet with suitable values of $h$ to estimate $r$ to two decimal places. Justify your answer.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure with Technology 2019 Q3 [20]}}
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