| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Standard +0.3 This is a straightforward application of the secant method (linear interpolation) with clear spreadsheet context. Parts (a)-(c) require recognizing the formula and performing routine calculations. Parts (d)-(e) test basic understanding of rounding and scientific notation. While it involves Further Maths content, the question is highly structured with given values, requiring minimal problem-solving or insight—just careful arithmetic and standard numerical methods knowledge. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| C | D | E | |
| 4 | \(r\) | \(\mathrm { x } _ { \mathrm { r } }\) | \(\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } } \right)\) |
| 5 | 0 | - 1 | - 0.4707963 |
| 6 | 1 | - 0.8 | - 0.0272952 |
| 7 | 2 | - 0.787691 | - 0.0193610 |
| 8 | 3 | - 0.7576546 | - 0.0020574 |
| 9 | 4 | - 0.7540834 | - 0.0001740 |
| 10 | 5 | ||
| 11 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | Secant (method) |
| [1] | 2.2a | |
| 3 | (b) | −0.7576546×−0.000174−(−0.7540834)×(−0.0020574) |
| Answer | Marks |
|---|---|
| ‒0.7537502 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | allow sign errors; may be implied by equivalent cell formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | ‒0.75375 cao since the last two |
| Answer | Marks | Guidance |
|---|---|---|
| is faster than 1st order oe | B1 | |
| [1] | 2.2b | must see value and justification |
| 3 | (d) | the spreadsheet uses the numbers to a higher |
| Answer | Marks | Guidance |
|---|---|---|
| uses the displayed values oe | B1 | |
| [1] | 2.4 | B0 if answer spoiled by eg spreadsheet stores exact values |
| 3 | (e) | ‒1.4629×10‒9 |
| or ‒ 0.000 000 001 462 9 | B1 | |
| [1] | 1.1 |
Question 3:
3 | (a) | Secant (method) | B1
[1] | 2.2a
3 | (b) | −0.7576546×−0.000174−(−0.7540834)×(−0.0020574)
(−0.000174−(−0.0020574))
oe
‒0.7537535
‒0.7537502 | M1
A1
A1
[3] | 1.1
1.1
1.1 | allow sign errors; may be implied by equivalent cell formulae
or by correct answers to 3 or more dp
allow ‒0.7537501 from working with values rounded to 7 dp
3 | (c) | ‒0.75375 cao since the last two
approximations agree to this precision
or ‒0.753750 is possible since convergence
is faster than 1st order oe | B1
[1] | 2.2b | must see value and justification
3 | (d) | the spreadsheet uses the numbers to a higher
precision than is displayed; the calculator
uses the displayed values oe | B1
[1] | 2.4 | B0 if answer spoiled by eg spreadsheet stores exact values
3 | (e) | ‒1.4629×10‒9
or ‒ 0.000 000 001 462 9 | B1
[1] | 1.13 The equation $\mathrm { f } ( x ) = \sin ^ { - 1 } ( x ) - x + 0.1 = 0$ has a root $\alpha$ such that $- 1 < \alpha < 0$.\\
Alex uses an iterative method to find a sequence of approximations to $\alpha$. Some of the associated spreadsheet output is shown in the table.
\begin{center}
\begin{tabular}{ | c | l | l | c | }
\hline
& C & \multicolumn{1}{|c|}{D} & E \\
\hline
4 & $r$ & \multicolumn{1}{c|}{$\mathrm { x } _ { \mathrm { r } }$} & $\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } } \right)$ \\
\hline
5 & 0 & - 1 & - 0.4707963 \\
\hline
6 & 1 & - 0.8 & - 0.0272952 \\
\hline
7 & 2 & - 0.787691 & - 0.0193610 \\
\hline
8 & 3 & - 0.7576546 & - 0.0020574 \\
\hline
9 & 4 & - 0.7540834 & - 0.0001740 \\
\hline
10 & 5 & & \\
\hline
11 & 6 & & \\
\hline
\end{tabular}
\end{center}
The formula in cell D7 is
$$= ( \mathrm { D } 5 * \mathrm { E } 6 - \mathrm { D } 6 * \mathrm { E } 5 ) / ( \mathrm { E } 6 - \mathrm { E } 5 )$$
and equivalent formulae are in cells D8 and D9.
\begin{enumerate}[label=(\alph*)]
\item State the method being used.
\item Use the values in the spreadsheet to calculate $x _ { 5 }$ and $x _ { 6 }$, giving your answers correct to 7 decimal places.
\item State the value of $\alpha$ as accurately as you can, justifying the precision quoted.
Alex uses a calculator to check the value in cell D9, his result is - 0.7540832686 .
\item Explain why this is different to the value displayed in cell D9.
The value displayed in cell E11 in Alex's spreadsheet is $- 1.4629 \mathrm { E } - 09$.
\item Write this value in standard mathematical notation.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2022 Q3 [7]}}