| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure with Technology (Further Pure with Technology) |
| Year | 2022 |
| Session | June |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Iterative/numerical methods |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question involving qualitative analysis of differential equations, including isoclines, tangent fields, and solving a specific case. Part (a)(ii) requires solving a first-order DE (likely separable or with integrating factor), which is standard Further Maths content. The qualitative analysis (sketching regions, interpreting tangent fields) requires understanding but not deep insight. The turning point analysis adds some challenge. Overall, this is a solid Further Maths question requiring multiple techniques but following predictable patterns for this specification. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | (i) |
| Answer | Marks |
|---|---|
| [3] | 1.2 |
| Answer | Marks |
|---|---|
| 1.1a | For sketch of y = x + 1 only |
| Answer | Marks |
|---|---|
| d x | Need general shape correct |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | ( ) ( ) | |
| Solution is y = b 2 − 2 x 2 + 2 b 2 − 1 x + b 2 | B1 | |
| [1] | 1.2 | oe |
| (iii) | With y as in (ii) |
| Answer | Marks |
|---|---|
| 2 − b 2 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1b | Could also use y 2 = x + 1 direct |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | (i) | Around a = 0. |
| [1] | 1.1b | Allow − 0 .5 a 0 .5 |
| (ii) | Around a = 0.6 | B1 |
| [1] | 1.1b | Allow 0 .5 a 1 (has to be |
| Answer | Marks |
|---|---|
| (iii) | ) |
| = | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1b |
| 1.1b | Needs to have a maximum and |
| Answer | Marks | Guidance |
|---|---|---|
| (iv) | One curve has a turning point (local max between | |
| 0 and 5), the) other doesn’t. | B1 | |
| [1] | 1.2 | Also allow ‘not strictly increasing’ |
| Answer | Marks | Guidance |
|---|---|---|
| ‘stationary point’. | Not ‘asymptote’. | |
| (c) | (i) | Insert 0 and 2 resp into cells A1 and B1. |
| Answer | Marks |
|---|---|
| And copy down. | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1a |
| Answer | Marks |
|---|---|
| 2.1 | Give reasonable BOD on |
| Answer | Marks |
|---|---|
| (ii) | Spreadsheet gives that y is approximately |
| Answer | Marks | Guidance |
|---|---|---|
| = 3. | B1 | |
| [1] | 1.1b | Allow any value between 2.2 and |
| Answer | Marks |
|---|---|
| (iii) | Using the spreadsheet with h = 0.01 and a = -0.2 |
| Answer | Marks |
|---|---|
| numerical methods confirms this) | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1a |
| Answer | Marks |
|---|---|
| 2.2b | For stating existence of increasing |
| Answer | Marks |
|---|---|
| For correct answer of 0.8. | Or stating existence of |
Question 3:
3 | (a) | (i) | B1
B1
B1
[3] | 1.2
1.1a
1.1a | For sketch of y = x + 1 only
needed for x 0 .
d y
For region 0 y x + 1
d x
d y
For region 0 y x + 1
d x | Need general shape correct
(decreasing positive gradient
and includes (0,1) and other
coordinates approximately
correct). Must be correct in
upper right quadrant. Ignore
anything in other quadrants.
Only award if general shape
correct (see above)
Only award if general shape
correct (see above)
(ii) | ( ) ( )
Solution is y = b 2 − 2 x 2 + 2 b 2 − 1 x + b 2 | B1
[1] | 1.2 | oe
(iii) | With y as in (ii)
) ( b 2 − 2 ) x + b 2 − 1
d y
=
d x ( b 2 − 2 ) x 2 + 2 ( b 2 − 1 ) x + b 2
d y 1 − b 2
So = 0 x =
d x b 2 − 2
This has a solution with x 0 if and only if
1 b 2
1−b2
When x= in the solution in (ii)
b2−2
1
y =
2 − b 2 | M1
A1
A1
B1
[4] | 1.1a
1.1b
3.2a
1.1b | Could also use y 2 = x + 1 direct
from (*).
Numerator = 0 or equivalent.
dy
For x in terms of b from =0
dx
FT using their x
Allow anything equivalent to this y
and FT using their x
(b) | (i) | Around a = 0. | B1
[1] | 1.1b | Allow − 0 .5 a 0 .5
(ii) | Around a = 0.6 | B1
[1] | 1.1b | Allow 0 .5 a 1 (has to be
different to answer to (b)(i) to be
awarded).
(iii) | )
= | B1
B1
[2] | 1.1b
1.1b | Needs to have a maximum and
negative second derivative
throughout. Must extend to at least
x = 4.5.
No turning point and negative
second derivative throughout. Must
extend to at least x = 4.5.
(iv) | One curve has a turning point (local max between
0 and 5), the) other doesn’t. | B1
[1] | 1.2 | Also allow ‘not strictly increasing’
or ‘intersects the x -axis’ or
‘stationary point’. | Not ‘asymptote’.
(c) | (i) | Insert 0 and 2 resp into cells A1 and B1.
Using cells I1 and I2 for the values of h and a
respectively other formulae then could be
Cell C1: ==I$1*((B1^I$2)/(A1+1)-1/B1)
Cell A2: =A1+I$1
Cell B2: =B1+C1
And copy down. | B1
B1
B1
[3] | 1.1a
3.1a
2.1 | Give reasonable BOD on
possible transcription errors and
consider correct answers to
3(c)(ii), 3(c)(iii) as evidence of
correct formulae in the
spreadsheet.
Cols for x and y and initialised x
and y values.
Allows for a and h to be varied
(doesn’t need cell reference, can be
implied)
Formulae for x and y
n+1 n+1.
(ii) | Spreadsheet gives that y is approximately
2.222487439 (to no. of dec places shown) when x
= 3. | B1
[1] | 1.1b | Allow any value between 2.2 and
2.23 inclusive.
(iii) | Using the spreadsheet with h = 0.01 and a = -0.2
gives (layout as in c(i), rows 81 to 86 shown
below)
This suggest that the local maximum has x-
coordinate between 0.82 and 0.84 and so would
be a value which rounds to 0.8 to one decimal
place. Smaller values of h (and use of other
numerical methods confirms this) | B1
M1
A1
[3] | 1.1a
3.1a
2.2b | For stating existence of increasing
then decreasing y values (as x
increases).
For analysis with values of h less
than or equal to 0.05.
For correct answer of 0.8. | Or stating existence of
positive then negative values
d y
of (as x increases).
d x
Note that h = 0.01 to be sure
of accuracy of correct
answer (but not required to
state this to award A1).
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.3 In this question you are required to consider the family of differential equations $\frac { d y } { d x } = \frac { y ^ { a } } { x + 1 } - \frac { 1 } { y } ( * )$\\
and its solutions. The parameter $a$ is a real number.
You should assume that $x \geqslant 0$ and $y > 0$ throughout this question.
\begin{enumerate}[label=(\alph*)]
\item In this part of the question $a = 1$.
\begin{enumerate}[label=(\roman*)]
\item On the axes in the Printed Answer Booklet
\begin{itemize}
\item Sketch the isocline defined by $\frac { d y } { d x } = 0$.
\item Shade and label the region in which $\frac { \mathrm { dy } } { \mathrm { dx } } > 0$.
\item Shade and label the region in which $\frac { \mathrm { dy } } { \mathrm { dx } } < 0$.
\item For $b > 0$, find, in terms of $b$, the solution to $( * )$ which passes through the point $( 0 , b )$.
\item Determine
\item The values of $b > 0$ for which the solution in (ii) has a turning point.
\item The corresponding maximum value of $y$.
\item Fig. 3.1 and Fig. 3.2 show tangent fields for two distinct but unspecified values of $a$. In each case a sketch of the solution curve $y = \mathrm { g } ( x )$ which passes through $( 0,2 )$ is shown for $0 \leqslant x \leqslant 0.5$.
\end{itemize}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{43fdb360-0f80-4794-917c-f28b04181fa4-4_656_648_1777_301}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{43fdb360-0f80-4794-917c-f28b04181fa4-4_656_652_1777_1117}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item For the case in Fig. 3.1 suggest a possible value of $a$.
\item For the case in Fig. 3.2 suggest a possible value of $a$.
\item In each case, continue the sketch of the solution curves for $0.5 \leqslant x \leqslant 5$ in the Printed Answer Booklet.
\item State a feature which is present in one of the curves in part (iii) for $0.5 \leqslant x \leqslant 5$ but not in the other.
\end{enumerate}
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item The Euler method for the solution of the differential equation $\frac { \mathrm { dy } } { \mathrm { dx } } = \mathrm { f } ( x , y )$ is as follows
$$y _ { n + 1 } = y _ { n } + h f \left( x _ { n } , y _ { n } \right)$$
It is given that $x _ { 0 } = 0$ and $y _ { 0 } = 2$.
\begin{itemize}
\item Construct a spreadsheet to solve (*) using the Euler method so that the value of $a$ and the value of $h$ can be varied, in the case $x _ { 0 } = 0$ and $y _ { 0 } = 2$.
\item State the formulae you have used in your spreadsheet.\\[0pt]
[3]
\item In this part of the question $a = 0.1$.
\end{itemize}
Use your spreadsheet with $h = 0.1$ to approximate the value of $y$ when $x = 3$ for the solution to (*) in which $y = 2$ when $x = 0$.
\item In this part of the question $a = - 0.2$.
Use your spreadsheet to approximate, to $\mathbf { 1 }$ decimal place, the $x$-coordinate of the local maximum for the solution to (*) in which $y = 2$ when $x = 0$.
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure with Technology 2022 Q3 [20]}}