| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2020 |
| Session | November |
| Marks | 13 |
| 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 standard Further Maths numerical methods question covering routine techniques: central difference formula (straightforward substitution), rearranging an equation to obtain an iterative formula (algebraic manipulation), applying convergence conditions (|g'(x)| < 1), and performing iterations with a calculator. All parts follow textbook procedures with no novel problem-solving required. While it's a Further Maths topic, the execution is mechanical and well-practiced. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| \(x\) | 1.45 | 1.55 |
| \(g ( x )\) | 1.49468 | 1.49949 |
| \(n\) | \(x _ { n }\) |
| 0 | 4.5 |
| 1 | 4.81826433 |
| 2 | 6.27473453 |
| 3 | 23.2937196 |
| 4 | \(2.0654 \mathrm { E } + 10\) |
| 5 | \#NUM! |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 1.49949 ā1.49468 |
| Answer | Marks |
|---|---|
| 1.55 ā1.45 | M1 |
| A1 | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | eg |
| Answer | Marks |
|---|---|
| š„š„šš+1= ļæ½ 8 = 2 | M1 |
| A1 | 2.1 |
| 1.1 | constructive intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | (gā²(1.5) ā gā²(Ī±) and) ā1 < gā²(1.5) < 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | r xr |
| Answer | Marks |
|---|---|
| (7 ) | M1 |
| A1 | 1.1 |
| 1.1 | iterative formula used to obtain at |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (e) | 1.496 825 61 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (f) | the number in the cell is too large for the |
| spreadsheet to display | B1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| r | xr | |
| 5 | 1.49682561031 | |
| 5 | (g) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (h) | Ī» = ā 0.4 |
| Answer | Marks |
|---|---|
| so iteration diverges | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | use of relaxed iteration with either |
| Answer | Marks |
|---|---|
| converges to Ī² = 4.38(2) | must see at least 2 |
Question 5:
5 | (a) | 1.49949 ā1.49468
= 0.048 cao
1.55 ā1.45 | M1
A1 | 1.1a
1.1
[2]
5 | (b) | eg
3 š„š„
8š„š„ = š„š„ + 25
3 š„š„šš
š„š„šš
3 ļ潚„š„šš +25
š„š„šš +25
š„š„šš+1= ļæ½ 8 = 2 | M1
A1 | 2.1
1.1 | constructive intermediate step
all subscripts present and correct
[2]
5 | (c) | (gā²(1.5) ā gā²(Ī±) and) ā1 < gā²(1.5) < 1 | B1 | 2.4
[1]
5 | (d) | r xr
0 1.5
1 1.49697756786
2 1.49683286384
3 1.49682595579
4 1.49682562605
5 1.49682561031
6 1.49682560956
1.49682560953
(7 ) | M1
A1 | 1.1
1.1 | iterative formula used to obtain at
least
3 correct values
all correct to at least 5 dp
[2]
5 | (e) | 1.496 825 61 | B1 | 2.2a
[1]
5 | (f) | the number in the cell is too large for the
spreadsheet to display | B1 | 2.2a
[1]
r | xr
5 | 1.49682561031
5 | (g) | B1 | 2.4 | diverging staircase with values
approximately correct
[1]
5 | (h) | Ī» = ā 0.4
Ī» = 0.5
so iteration diverges | M1
A1
A1 | 1.1
1.1
1.1 | use of relaxed iteration with either
value
converges to Ī² = 4.38(2) | must see at least 2
iterates for M1
[3]5 You are given that\\
$g ( x ) = \frac { \sqrt [ 3 ] { x ^ { x } + 25 } } { 2 }$.
Fig. 5.1 shows two values of $x$ and the associated values of $\mathrm { g } ( x )$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
$x$ & 1.45 & 1.55 \\
\hline
$g ( x )$ & 1.49468 & 1.49949 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use the central difference method to calculate an estimate of $\mathrm { g } ^ { \prime } ( 1.5 )$, giving your answer correct to 3 decimal places.
The equation $x ^ { x } - 8 x ^ { 3 } + 25 = 0$ has two roots, $\alpha$ and $\beta$, such that $\alpha \approx 1.5$ and $\beta \approx 4.4$.
\item Obtain the iterative formula $x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }$.
\item Use your answer to part (a) to explain why it is possible that the iterative formula $x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }$ may be used to find $\alpha$.
\item Starting with $x _ { 0 } = 1.5$, use the iterative formula to find $x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }$, and $x _ { 6 }$.
\item Use your answer to part (d) to state the value of $\alpha$ correct to 8 decimal places.
Starting with $x _ { 0 } = 4.5$ the same iterative formula is used in an attempt to find $\beta$. The results are shown in Fig. 5.2.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | l | }
\hline
$n$ & \multicolumn{1}{|c|}{$x _ { n }$} \\
\hline
0 & 4.5 \\
\hline
1 & 4.81826433 \\
\hline
2 & 6.27473453 \\
\hline
3 & 23.2937196 \\
\hline
4 & $2.0654 \mathrm { E } + 10$ \\
\hline
5 & \#NUM! \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\end{center}
\end{table}
\item Explain why \#NUM! is displayed in the cell for $x _ { 5 }$.
\item On the diagram in the Printed Answer Booklet, starting with $x _ { 0 } = 4.5$, illustrate how the iterative formula works to find $x _ { 1 }$ and $x _ { 2 }$.
\item Determine what happens when the relaxed iteration $x _ { n + 1 } = ( 1 - \lambda ) x _ { n } + \lambda g \left( x _ { n } \right)$ is used to try to find $\beta$ with $x _ { 0 } = 4.5$, in each of the following cases.
\begin{itemize}
\item $\lambda = 0.5$
\item $\lambda = - 0.4$
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2020 Q5 [13]}}