| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Challenging +1.2 This is a standard Further Maths numerical methods question requiring application of forward difference formula, analysis of convergence order through ratio examination, and Richardson extrapolation. While it involves multiple parts and interpretation of spreadsheet data, the techniques are routine for this specification with no novel problem-solving required. The conceptual demand is moderate—understanding convergence order and extrapolation—but execution is straightforward. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 3 | 3.5 |
| \(\mathrm { f } ( x )\) | 6.082763 | 4.596194 |
| \(x\) | 3 | 3.00001 | 3.0001 | 3.001 | 3.01 | 3.1 |
| \(\mathrm { f } ( x )\) | 6.082763 | 6.08274 | 6.082541 | 6.08054 | 6.060454 | 5.848846 |
| \(h\) | 0.1 | 0.01 | 0.001 | 0.0001 | 0.00001 |
| estimate | -2.339165 | -2.230883 | -2.220532 | -2.219501 | -2.219398 |
| difference | 0.1082815 | 0.010352 | 0.0010307 | 0.000103 | |
| ratio | 0.095602 | 0.099567 | 0.0999568 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | soi |
| Answer | Marks |
|---|---|
| ‒ 2.97301.5 cao | M1 |
| A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | (Tommy) should consider the ratio of |
| Answer | Marks | Guidance |
|---|---|---|
| Tommy is wrong | E1 | |
| E1 | 1.1 | |
| 2.2b | must mention ratio | must state that Tommy |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (c) | extrapolation used with estimate and difference |
| Answer | Marks |
|---|---|
| ‒2.2194 is certain or ‒2.21939 is probable | M1 |
| Answer | Marks |
|---|---|
| A1 | 2.1 |
| Answer | Marks |
|---|---|
| 2.2b | 0.1 |
| Answer | Marks |
|---|---|
| accept either answer | allow up to M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (d) | 0.02 × their f′(3) |
| ‒0.0443878 to ‒0.044388 | M1 | |
| A1 | 3.1a | |
| 1.1 | FT from part (c) correct to 4 sf or | |
| more | allow positive answer in |
Question 7:
7 | (a) | soi
4.596194 ‒6.082763
‒ 2.97301.5 cao | M1
A1 | 1.1
1.1
[2]
7 | (b) | (Tommy) should consider the ratio of
differences
which suggests 1st order convergence, since the
ratio appears to be converging (to 0.1) so
Tommy is wrong | E1
E1 | 1.1
2.2b | must mention ratio | must state that Tommy
is not correct for both
marks
[2]
7 | (c) | extrapolation used with estimate and difference
used from Fig 7.3
‒2.219398 and 0.000103 used
0.09995 ≤ r ≤ 0.1 used
‒2.21938666 to ‒ 2.21938655 inclusive
‒2.2194 is certain or ‒2.21939 is probable | M1
A1
A1
M1
A1 | 2.1
3.1a
3.1a
1.1
2.2b | 0.1
‒ 2.219398+ 0.000103 × 1 ‒0.1
accept either answer | allow up to M1A1A1
for partial extrapolation
eg ‒2.2193877 from
‒2.219398+0.000103 × 0.1
final M1A1 only
available for
extrapolation to infinity
[5]
7 | (d) | 0.02 × their f′(3)
‒0.0443878 to ‒0.044388 | M1
A1 | 3.1a
1.1 | FT from part (c) correct to 4 sf or
more | allow positive answer in
range
[2]
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recorded or monitored7 Fig. 7.1 shows two values of $x$ and the associated values of $\mathrm { f } ( x )$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
$x$ & 3 & 3.5 \\
\hline
$\mathrm { f } ( x )$ & 6.082763 & 4.596194 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use the forward difference method to calculate an estimate of the gradient of $\mathrm { f } ( x )$ at $x = 3$, giving your answer correct to 4 decimal places.
Fig. 7.2 shows some spreadsheet output with additional values of $x$ and the associated values of $\mathrm { f } ( x )$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 3 & 3.00001 & 3.0001 & 3.001 & 3.01 & 3.1 \\
\hline
$\mathrm { f } ( x )$ & 6.082763 & 6.08274 & 6.082541 & 6.08054 & 6.060454 & 5.848846 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 7.2}
\end{center}
\end{table}
These values have been used to produce a sequence of estimates of the gradient of $\mathrm { f } ( x )$ at $x = 3$, together with some further analysis. This is shown in the spreadsheet output in Fig. 7.3.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$h$ & 0.1 & 0.01 & 0.001 & 0.0001 & 0.00001 \\
\hline
estimate & -2.339165 & -2.230883 & -2.220532 & -2.219501 & -2.219398 \\
\hline
difference & 0.1082815 & 0.010352 & 0.0010307 & 0.000103 & \\
\hline
ratio & 0.095602 & 0.099567 & 0.0999568 & & \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 7.3}
\end{center}
\end{table}
Tommy states that the differences between successive estimates is decreasing so rapidly that the order of convergence of this sequence of estimates is much faster than first order.
\item Explain whether or not Tommy is correct.
\item Use extrapolation to determine the value of the gradient of $\mathrm { f } ( x )$ at $x = 3$ as accurately as possible, justifying the precision quoted.
\item Calculate an estimate of the absolute error when $\mathrm { f } ( 3 )$ is used as an approximation to $\mathrm { f } ( 3.02 )$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2020 Q7 [11]}}