| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Secant Method or False Position |
| Difficulty | Standard +0.3 This is a standard Further Maths numerical methods question testing routine application of the secant method. Parts (a)-(d) require basic recall of the secant formula and spreadsheet implementation. Part (e) involves straightforward error checking using the given iterates. Part (f) requires minimal insight about choosing starting values. While it's a Further Maths topic, the question is entirely procedural with no novel problem-solving required, making it slightly easier than average overall. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| I | J | K | L | M | |
| 2 | \(r\) | \(\mathrm { x } _ { \mathrm { r } }\) | f(x) | \(\mathrm { X } _ { \mathrm { r } + 1 }\) | \(\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } + 1 } \right)\) |
| 3 | 0 | 1 | -0.0914 | 2 | -0.0476 |
| 4 | 1 | 2 | -0.0476 | 3.08529 | 0.95784 |
| 5 | 2 | 3.08529 | 0.95784 | 2.05134 | -0.0298 |
| 6 | 3 | 2.05134 | -0.0298 | 2.08259 | -0.0181 |
| 7 | 4 | 2.08259 | -0.0181 | 2.13042 | 0.00155 |
| 8 | 5 | 2.13042 | 0.00155 | 2.12664 | \(- 7 \mathrm { E } - 05\) |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (b) | = L3 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (c) | = (J3*M3 ‒ K3*L3)/(M3 ‒ K3) |
| or = (K3*L3 ‒ J3*M3)/(K3 ‒ M3) oe | M1 | |
| A1 | 1.1 | |
| 1.1 | use of secant formula; allow spreadsheet syntax errors but not |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (d) | 2.12664 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (e) | f(2.126635) = awrt ‒0.00007 |
| Answer | Marks |
|---|---|
| no sign change so not correct to 5 dp | M1 |
| A1 | 2.1 |
| 2.2a | both calculated and one correct to 1 sf |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (f) | 1 < a ≤ 1.74 |
Question 8:
8 | (a) | B1 | 1.1 | chord from curve at x = 1 and through the curve at x = 2
which joins the x-axis to the right of the root isw
[1]
8 | (b) | = L3 | B1 | 1.1
[1]
8 | (c) | = (J3*M3 ‒ K3*L3)/(M3 ‒ K3)
or = (K3*L3 ‒ J3*M3)/(K3 ‒ M3) oe | M1
A1 | 1.1
1.1 | use of secant formula; allow spreadsheet syntax errors but not
error in formula
NB or = (J3*K4 ‒ K3*L3)/(K4 ‒ K3)
or = (K3*L3 ‒ J3*K4)/(K3 ‒ K4)
or (J3*K4 ‒ K3*J4)/(K4 ‒ K3)
or = (K3*J4 ‒ J3*K4)/(K3 ‒ K4)
all correct
[2]
8 | (d) | 2.12664 | B1 | 2.2a
[1]
8 | (e) | f(2.126635) = awrt ‒0.00007
f(2.126645) = awrt ‒0.00007
no sign change so not correct to 5 dp | M1
A1 | 2.1
2.2a | both calculated and one correct to 1 sf
NB ‒0.000070377215 and ‒0.000066109812
M1FT 2.13042
awrt 0.002 in each case
NB 0.0015487 and 0.001553
both values correct to 1 sf
A1 FT awrt 0.002 and 0.002 so no sign change so not
correct to 5 dp
[2]
8 | (f) | 1 < a ≤ 1.74 | B1 | 3.1a
[1]8 The graph of $\mathrm { y } = 0.2 \cosh \mathrm { x } - 0.4 \mathrm { x }$ for values of $x$ from 0 to 3.32 is shown on the graph below.\\
\includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-08_988_1561_312_244}
The equation $0.2 \cosh x - 0.4 x = 0$ has two roots, $\alpha$ and $\beta$ where $\alpha < \beta$, in the interval $0 < x < 3$. The secant method with $x _ { 0 } = 1$ and $x _ { 1 } = 2$ is to be used to find $\beta$.
\begin{enumerate}[label=(\alph*)]
\item On the copy of the graph in the Printed Answer Booklet, show how the secant method works with these two values of $x$ to obtain an improved approximation to $\beta$.
The spreadsheet output in the table below shows the result of applying the secant method with $x _ { 0 } = 1$ and $x _ { 1 } = 2$.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
& I & J & K & L & M \\
\hline
2 & $r$ & $\mathrm { x } _ { \mathrm { r } }$ & f(x) & $\mathrm { X } _ { \mathrm { r } + 1 }$ & $\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } + 1 } \right)$ \\
\hline
3 & 0 & 1 & -0.0914 & 2 & -0.0476 \\
\hline
4 & 1 & 2 & -0.0476 & 3.08529 & 0.95784 \\
\hline
5 & 2 & 3.08529 & 0.95784 & 2.05134 & -0.0298 \\
\hline
6 & 3 & 2.05134 & -0.0298 & 2.08259 & -0.0181 \\
\hline
7 & 4 & 2.08259 & -0.0181 & 2.13042 & 0.00155 \\
\hline
8 & 5 & 2.13042 & 0.00155 & 2.12664 & $- 7 \mathrm { E } - 05$ \\
\hline
\end{tabular}
\end{center}
\item Write down a suitable cell formula for cell J4.
\item Write down a suitable cell formula for cell L4.
\item Write down the most accurate approximation to $\beta$ which is displayed in the table.
\item Determine whether your answer to part (d) is correct to 5 decimal places. You should not calculate any more iterates.
\item It is decided to use the secant method with starting values $x _ { 0 } = 1$ and $\mathrm { x } _ { 1 } = \mathrm { a }$, where $a > 1$, to find $\alpha$. State a suitable value for $a$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2023 Q8 [8]}}