| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Standard +0.3 This is a straightforward Further Maths numerical methods question testing standard concepts: recognizing when central differences can't be used (part a), applying forward difference formula (part b), interpreting ratio patterns to identify method order (part c), and using Richardson extrapolation or ratio analysis for improved accuracy (part d). All techniques are routine applications of textbook methods with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 1.5 | 1 | 2 |
| \(\mathrm { f } ( x )\) | 0.84089 | 1 | 1.18921 |
| \(h\) | 0.8 | 0.4 | 0.2 | 0.1 | 0.05 | 0.025 | 0.0125 | 0.00625 |
| approximation | 0.130452 | 0.138647 | 0.143381 | 0.145942 | 0.147277 | 0.147959 | 0.148304 | 0.148477 |
| difference | 0.008195 | 0.004734 | 0.002561 | 0.001335 | 0.000682 | 0.000345 | 0.000173 | |
| ratio | 0.577633 | 0.541099 | 0.521186 | 0.510762 | 0.505424 | 0.502723 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | because the other two x-values are both greater |
| than 1 | B1 | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | 1.18921−1 0.84089−1 |
| Answer | Marks |
|---|---|
| or ‒ 0.31822 or ‒ 0.3182 or ‒ 0.318 | M1 |
| A1 | 1.1 |
| 1.1 | B2 for either correct answer unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | (i) |
| Answer | Marks |
|---|---|
| halved in successive approximations) | B1 |
| B1 | 2.2b |
| 2.4 | allow |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | (ii) | this is to be expected since forward difference is |
| usually a first order method | B1 | 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (d) | 𝑟 |
| Answer | Marks |
|---|---|
| extrapolation greatly increases accuracy oe | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 3.2a | 𝑟 |
Question 6:
6 | (a) | because the other two x-values are both greater
than 1 | B1 | 2.4 | allow because
eg we don’t have a pair of values of x and y (which are equally
spaced) either side of x = 1
eg we need f(0.5)
eg we need (0,f(0))
eg we need a value for x = 0 (or 0.5)
[1]
6 | (b) | 1.18921−1 0.84089−1
or soi
2−1 1.5−1
0.18921 or 0.1892 or 0.189
or ‒ 0.31822 or ‒ 0.3182 or ‒ 0.318 | M1
A1 | 1.1
1.1 | B2 for either correct answer unsupported
[2]
Alternatively
eg [0.03125], 0.015625, 0.0078125, 0.003906,
0.001953, 0.0009765, 0.0004883
0.0009765 and 0.0004883 seen
𝑛 = 6 or 𝑛 = 7
two bisections of maximum possible error; may see eg 0.5, 0.25,
0.125, 0.0625, 0.03125, …
this mark is dependent on the award of one M mark;
0.03125
if M0M0 allow SC2 for = 0.000488 𝐨𝐞 so 𝑛 = 6
26
0.0625
or SC2 for = 0.000488 𝐨𝐞 so n = 7
27
1
or SC2 for = 0.000488 𝐨𝐞 so 𝑛 = 6 or n = 7
211
6 | (c) | (i) | ratio of differences is converging to 0.5 oe
this suggests order of method is 1 (since h is
halved in successive approximations) | B1
B1 | 2.2b
2.4 | allow
eg the order of convergence of the method is 1
eg first order (convergence)
eg method has first order convergence
[2]
(c) | (ii) | this is to be expected since forward difference is
usually a first order method | B1 | 1.2
[1]
6 | (d) | 𝑟
𝑋+𝐷×
1−𝑟
𝑟
0.148477+0.000173×
1−𝑟
1
or 0.148304+0.000173×
1−r
0.14865 to 0.1486521
accept 0.14865 or 0.1487 or 0.149 since
extrapolation greatly increases accuracy oe | M1
M1
A1
A1 | 3.1a
2.1
1.1
3.2a | 𝑟
allow 𝑋±𝐷×
1−𝑟
X is an approximation from the table, D is the associated difference
and r = 0.5 or one of the (possibly rounded) ratios from the table
1
may see 0.148304+0.000345×
1−r
𝑟
allow 0.148477±0.000173×
1−𝑟
1
or 0.148304±0.000173×
1−r
allow 0.5 ≤ 𝑟 ≤ 0.503
M1M1 may be implied by sight of correct value in range with
evidence of use of correct formula
1
NB may see may see 0.148304+0.000173× with correct
1−r
value of r
if M0M0 allow SC1 for partial extrapolation:
0.148477+0.000173×𝑟 = 0.1485635 to 0.1485639
then SC1 for accept 0.149 since extrapolation greatly increases
accuracy oe
[4]6 Table 6.1 shows some values of $x$ and the associated values of a function, $y = f ( x )$.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 6.1}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 1.5 & 1 & 2 \\
\hline
$\mathrm { f } ( x )$ & 0.84089 & 1 & 1.18921 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Explain why it is not possible to use the central difference method to calculate an estimate of $\frac { \mathrm { dy } } { \mathrm { dx } }$ when $x = 1$.
\item Use the forward difference method to calculate an estimate of $\frac { \mathrm { dy } } { \mathrm { dx } }$ when $x = 1$.
A student uses the forward difference method to calculate a series of approximations to $\frac { \mathrm { dy } } { \mathrm { dx } }$ when $x = 2$ with different values of the step length, $h$.
These approximations are shown in Table 6.2, together with some further analysis.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 6.2}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
$h$ & 0.8 & 0.4 & 0.2 & 0.1 & 0.05 & 0.025 & 0.0125 & 0.00625 \\
\hline
approximation & 0.130452 & 0.138647 & 0.143381 & 0.145942 & 0.147277 & 0.147959 & 0.148304 & 0.148477 \\
\hline
difference & & 0.008195 & 0.004734 & 0.002561 & 0.001335 & 0.000682 & 0.000345 & 0.000173 \\
\hline
ratio & & & 0.577633 & 0.541099 & 0.521186 & 0.510762 & 0.505424 & 0.502723 \\
\hline
\end{tabular}
\end{center}
\end{table}
\item \begin{enumerate}[label=(\roman*)]
\item Explain what the ratios of differences tell you about the order of the method in this case.
\item Comment on whether this is unusual.
\end{enumerate}\item Determine the value of $\frac { \mathrm { dy } } { \mathrm { dx } }$ when $x = 2$ as accurately as possible. You must justify the precision quoted.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2024 Q6 [10]}}