| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiation from First Principles |
| Type | Numerical gradient deduction |
| Difficulty | Standard +0.8 This is a Further Maths numerical methods question requiring interpretation of spreadsheet data for derivative estimation, understanding convergence order from ratios, recognizing rounding error effects, and making justified accuracy decisions. While systematic, it demands sophisticated understanding of numerical analysis concepts beyond standard A-level calculus, placing it moderately above average difficulty. |
| Spec | 1.09f Trapezium rule: numerical integration |
| A | B | C | D | |
| 1 | h | estimate | difference | ratio |
| 2 | 0.1 | 6.3050005 | ||
| 3 | 0.01 | 6.0300512 | -0.274949 | |
| 4 | 0.001 | 6.0030018 | -0.027049 | 0.098379 |
| 5 | 0.0001 | 6.0003014 | -0.0027 | 0.099835 |
| 6 | 0.00001 | 6.0000314 | -0.00027 | 0.099983 |
| 7 | 0.000001 | 6.0000044 | \(- 2.7 \mathrm { E } - 05\) | 0.099994 |
| 8 | 1E-07 | 6.0000016 | \(- 2.71 \mathrm { E } - 06\) | 0.100352 |
| 9 | 1E-08 | 6.0000013 | \(- 3.02 \mathrm { E } - 07\) | 0.111457 |
| 10 | 1E-09 | 6.0000018 | \(4.885 \mathrm { E } - 07\) | -1.61765 |
| 11 | 1E-10 | 6.0000049 | \(3.109 \mathrm { E } - 06\) | 6.363636 |
| 12 | 1E-11 | 6.0000005 | \(- 4.44 \mathrm { E } - 06\) | -1.42857 |
| 13 | 1E-12 | 6.0005334 | 0.0005329 | -120 |
| 14 | 1E-13 | 5.9952043 | -0.005329 | -10 |
| 15 | 1E-14 | 6.1284311 | 0.1332268 | -25 |
| 16 | 1E-15 | 5.3290705 | -0.799361 | -6 |
| 17 | 1E-16 | 0 | -5.329071 | 6.666667 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | = B3 ‒ B2 |
| = C4/C3 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | in cell C3 |
| in cell D4 | if B0B0 allow SC1 for |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | these values are approximately equal / appear |
| Answer | Marks |
|---|---|
| which suggests 1st order | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 2.2b | B0 for eg just “it’s a first |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | [1×] 10‒9 or 0.000000001 |
| [1] | 1.1 | ignore cell references |
| 5 | (d) | the ratios are diverging oe |
| Answer | Marks |
|---|---|
| the bottom of the column oe | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (e) | 6.000001 seems likely or 6.00000 seems |
| Answer | Marks |
|---|---|
| approximation after 6.0000013 is larger oe | B1 |
| Answer | Marks |
|---|---|
| [2] | 2.2b |
| 2.4 | allow eg values agree to this |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (f) | [f(x) =] ln(√sinh(x3)) |
| [1] | 2.5 | or ½ln(sinhx3) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (g) | the calculation involves the subtraction of |
| Answer | Marks |
|---|---|
| to the precision the spreadsheet operates oe | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
Question 5:
5 | (a) | = B3 ‒ B2
= C4/C3 | B1
B1
[2] | 1.1
1.1 | in cell C3
in cell D4 | if B0B0 allow SC1 for
B3 ‒ B2 and C4/C3 seen
5 | (b) | these values are approximately equal / appear
to be tending to 0.1
which suggests 1st order | B1
B1
[2] | 2.4
2.2b | B0 for eg just “it’s a first
order method”
5 | (c) | [1×] 10‒9 or 0.000000001 | B1
[1] | 1.1 | ignore cell references
5 | (d) | the ratios are diverging oe
(which suggests that) the approximations to the
derivative are (becoming) less accurate towards
the bottom of the column oe | B1
B1
[2] | 2.4
2.2b
5 | (e) | 6.000001 seems likely or 6.00000 seems
certain
approximation after 6.0000013 is larger oe | B1
B1
[2] | 2.2b
2.4 | allow eg values agree to this
precision before they start to
diverge
5 | (f) | [f(x) =] ln(√sinh(x3)) | B1
[1] | 2.5 | or ½ln(sinhx3) | allow omission of inner
brackets;
B0 for modulus instead
of brackets
5 | (g) | the calculation involves the subtraction of
nearly equal numbers (in the numerator) oe
they are stored as the same (non-zero) number
to the precision the spreadsheet operates oe | B1
B1
[2] | 3.1b
3.2b5 Fig. 5 shows spreadsheet output concerning the estimation of the derivative of a function $\mathrm { f } ( x )$ at $x = 2$ using the forward difference method.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
& A & B & C & D \\
\hline
1 & h & estimate & difference & ratio \\
\hline
2 & 0.1 & 6.3050005 & & \\
\hline
3 & 0.01 & 6.0300512 & -0.274949 & \\
\hline
4 & 0.001 & 6.0030018 & -0.027049 & 0.098379 \\
\hline
5 & 0.0001 & 6.0003014 & -0.0027 & 0.099835 \\
\hline
6 & 0.00001 & 6.0000314 & -0.00027 & 0.099983 \\
\hline
7 & 0.000001 & 6.0000044 & $- 2.7 \mathrm { E } - 05$ & 0.099994 \\
\hline
8 & 1E-07 & 6.0000016 & $- 2.71 \mathrm { E } - 06$ & 0.100352 \\
\hline
9 & 1E-08 & 6.0000013 & $- 3.02 \mathrm { E } - 07$ & 0.111457 \\
\hline
10 & 1E-09 & 6.0000018 & $4.885 \mathrm { E } - 07$ & -1.61765 \\
\hline
11 & 1E-10 & 6.0000049 & $3.109 \mathrm { E } - 06$ & 6.363636 \\
\hline
12 & 1E-11 & 6.0000005 & $- 4.44 \mathrm { E } - 06$ & -1.42857 \\
\hline
13 & 1E-12 & 6.0005334 & 0.0005329 & -120 \\
\hline
14 & 1E-13 & 5.9952043 & -0.005329 & -10 \\
\hline
15 & 1E-14 & 6.1284311 & 0.1332268 & -25 \\
\hline
16 & 1E-15 & 5.3290705 & -0.799361 & -6 \\
\hline
17 & 1E-16 & 0 & -5.329071 & 6.666667 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Write down suitable cell formulae for
\begin{itemize}
\item cell C3,
\item cell D4.
\item Explain what the entries in cells D4 to D8 tell you about the order of the convergence of the forward difference method.
\item Write the entry in cell A10 in standard mathematical notation.
\item Explain what the values displayed in cells D10 to D17 suggest about the values in cells B10 to B16.
\item Write down the value of the derivative of $\mathrm { f } ( x )$ at $x = 2$ to an accuracy that seems justified, explaining your answer.
\end{itemize}
The formula in cell B2 is $\quad = ( \mathrm { LN } ( \mathrm { SQRT } ( \mathrm { SINH } ( ( 2 + \mathrm { A } 2 ) \wedge 3 ) ) ) - \mathrm { LN } ( \mathrm { SQRT } ( \mathrm { SINH } ( 2 \wedge 3 ) ) ) ) / \mathrm { A } 2$ and equivalent formulae are entered in cells B3 to B17.
\item Write $\mathrm { f } ( x )$ in standard mathematical notation.
The value displayed in cell B17 is zero, even though the calculation results in a non-zero answer.
\item Explain how this has arisen.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2019 Q5 [12]}}