| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Recurrence relation asymptotic behaviour |
| Difficulty | Standard +0.3 This is a standard application of Newton's forward difference method requiring routine table completion, polynomial fitting, and basic interpretation. While it involves multiple parts, each step follows textbook procedures with no novel insight required. The asymptotic behaviour question in part (e) requires recognizing that a cubic eventually becomes negative, which is straightforward reasoning for Further Maths students. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.05a Roots and coefficients: symmetric functions |
| \(n\) | 1 | 2 | 3 | 4 | 5 |
| \(y\) | 9 | 32 | 63 | 96 | 125 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | 3 columns of differences seen |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | Δ³ column has same values or Δ4 is zero |
| [1] | 2.3 | |
| 3 | (c) | 8 |
| Answer | Marks |
|---|---|
| FT their cubic [y] in terms of n | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | all four terms |
| Answer | Marks |
|---|---|
| additional terms | allow up to M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (d) | f(6) = 144 good fit |
| f(7) = 147 good fit | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 3.4 | allow SC1 for 144 and 147 seen |
| Answer | Marks |
|---|---|
| comment | FT their 10n2 ‒ n3 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (e) | 10n2 ‒ n3 = 0 gives n = 10 (or 0) so maximum |
| Answer | Marks | Guidance |
|---|---|---|
| predicted after week 10 | B1 | |
| [1] | 3.5b | n = 10 unsupported is insufficient |
Question 3:
3 | (a) | M1
A1
A1
[3] | 1.1
1.1
1.1 | 3 columns of differences seen
for Δ column correct
Δ² and Δ³ correct FT
3 | (b) | Δ³ column has same values or Δ4 is zero | B1
[1] | 2.3
3 | (c) | 8
9(n1)23(n1)(n2)
2!
6
(n1)(n2)(n3)
3!
10n2 ‒ n3 may be eg in terms of x
FT their cubic [y] in terms of n | M1
A1
A1
A1
[4] | 3.3
1.1
1.1
1.1 | all four terms
all correctly placed, may be implied
by correct answer
both terms correct with no
additional terms | allow up to M1A1A1
for polynomial in x
3 | (d) | f(6) = 144 good fit
f(7) = 147 good fit | B1
B1
[2] | 3.4
3.4 | allow SC1 for 144 and 147 seen
and no comment or incorrect
comment | FT their 10n2 ‒ n3
for each mark
3 | (e) | 10n2 ‒ n3 = 0 gives n = 10 (or 0) so maximum
possible number of weeks it could be viable is
10, since a negative number of cases is
predicted after week 10 | B1
[1] | 3.5b | n = 10 unsupported is insufficient | FT their cubic3 In the first week of an outbreak of influenza, 9 patients were diagnosed with the virus at a medical practice in Pencaster. Records were kept of $y$, the total number of patients diagnosed with influenza in week $n$. The data are shown in Fig. 3.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$n$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$y$ & 9 & 32 & 63 & 96 & 125 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Complete the difference table in the Printed Answer Booklet.
\item Explain why a cubic model is appropriate for the data.
\item Use Newton's method to find the interpolating polynomial of degree 3 for these data.
In both week 6 and week 7 there were 145 patients in total diagnosed with influenza at the medical practice.
\item Determine whether the model is a good fit for these data.
\item Determine the maximum number of weeks for which the model could possibly be valid.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2019 Q3 [11]}}