| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Interval Bisection from Spreadsheet |
| Difficulty | Standard +0.3 This is a straightforward interval bisection question requiring students to identify errors in spreadsheet formulas (mixing up cell references), write correct formulas using IF statements, and apply the bisection method to find a root with error bounds. While it involves multiple parts, each step is routine application of standard numerical methods taught at this level, with no novel problem-solving required. |
| Spec | 1.09a Sign change methods: locate roots |
| A | B | C | D | E | F | |
| 1 | a | f(a) | b | f(b) | \(x _ { \text {new } }\) | \(\mathrm { f } \left( x _ { \text {new } } \right)\) |
| 2 | 0 | 0.5 | 0.5 | -1.51831 | 0.25 | -1.133 |
| 3 | 0 | 0.5 | 0.25 | -1.133 | 0.125 | -0.42 |
| 4 | 0 | 0.5 | 0.125 | -0.42001 | 0.0625 | 0.01873 |
| 5 | 0.0625 | 0.01873 | 0.125 | -0.42001 | 0.09375 | -0.2065 |
| A | B | C | D | |
| 1 | \(x\) | f(x) | difference | ratio |
| 2 | 0.4 | -1.5799 | ||
| 3 | 0.6 | -1.0504 | ||
| 4 | 0.99671 | 8.4245 | ||
| 5 | 0.64398 | -0.6809 | -0.35273 | |
| 6 | 0.67036 | -0.4026 | 0.026378 | -0.0748 |
| 7 | 0.70852 | 0.08386 | 0.038164 | 1.44682 |
| 8 | 0.70194 | -0.0075 | -0.00658 | -0.1724 |
| 9 | 0.70248 | -0.0001 | 0.00054 | -0.082 |
| 10 | 0.70249 | \(1.8 \mathrm { E } - 07\) | \(8.88 \mathrm { E } - 06\) | 0.01646 |
| Answer | Marks |
|---|---|
| 6 | 1.49682560956 |
| (7 | 1.49682560953 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | both f(0) and f(1) are positive oe |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | = EXP(3*A2) ‒ 11*A2 ‒ 0.5 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | =IF(F2>0,C2,E2) |
| or =IF(F2<0,E2,C2) | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (d) | = ] 0.078125 |
| Answer | Marks |
|---|---|
| 2 | B1 |
| B1 | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (e) | secant method |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (f) | the ratios of differences are decreasing in |
| Answer | Marks |
|---|---|
| so the convergence is faster than first order | B1 |
| B1 | 1.1 |
| 2.2b | if B0B0 allow SC1 for eg ratios not |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (g) | f(0.702485) and f(0.702495) calculated |
| Answer | Marks |
|---|---|
| so Liren is correct | M1 |
| A1 | 2.1 |
Question 6:
6 | 1.49682560956
(7 | 1.49682560953
)
6 | (a) | both f(0) and f(1) are positive oe | B1 | 2.4
[1]
6 | (b) | = EXP(3*A2) ‒ 11*A2 ‒ 0.5 | B1 | 1.1
[1]
6 | (c) | =IF(F2>0,C2,E2)
or =IF(F2<0,E2,C2) | B1 | 1.1
[1]
6 | (d) | = ] 0.078125
0.0625+0.09375
[mpe = 0.0156 25
2 | B1
B1 | 3.1a
1.1
[2]
6 | (e) | secant method | B1 | 1.2
[1]
6 | (f) | the ratios of differences are decreasing in
magnitude oe
so the convergence is faster than first order | B1
B1 | 1.1
2.2b | if B0B0 allow SC1 for eg ratios not
equal so convergence not 1st order
[2]
6 | (g) | f(0.702485) and f(0.702495) calculated
awrt ‒0.00006… and + 0.00008…sign change,
so Liren is correct | M1
A1 | 2.1
2.2a
[2]6 Fig. 6.1 shows the graph of $y = \mathrm { e } ^ { 3 x } - 11 x - 0.5$ for $- 0.5 \leqslant x \leqslant 1$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{87bb8eb7-b725-48b0-b32b-0bfce624cd91-08_576_881_315_333}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}
The equation $\mathrm { e } ^ { 3 x } - 11 x - 0.5 = 0$ has two roots, $\alpha$ and $\beta$, such that $\alpha < \beta$. Dennis is going to use the method of interval bisection with starting values denoted by $a$ and $b$.
\begin{enumerate}[label=(\alph*)]
\item Explain why the method of interval bisection starting with $a = 0$ and $b = 1$ may not be used to find either $\alpha$ or $\beta$.
Dennis uses the method of interval bisection starting with $a = 0$ and $b = 0.5$ to find $\alpha$. Some spreadsheet output is shown in Fig. 6.2.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
& A & B & C & D & E & F \\
\hline
1 & a & f(a) & b & f(b) & $x _ { \text {new } }$ & $\mathrm { f } \left( x _ { \text {new } } \right)$ \\
\hline
2 & 0 & 0.5 & 0.5 & -1.51831 & 0.25 & -1.133 \\
\hline
3 & 0 & 0.5 & 0.25 & -1.133 & 0.125 & -0.42 \\
\hline
4 & 0 & 0.5 & 0.125 & -0.42001 & 0.0625 & 0.01873 \\
\hline
5 & 0.0625 & 0.01873 & 0.125 & -0.42001 & 0.09375 & -0.2065 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{table}
Dennis states that the formula in cell B2 is
$$= \operatorname { EXP } \left( 3 ^ { * } \mathrm {~A} 1 \right) - 11 \mathrm {~A} 2 - 0.5$$
Dennis has made two errors.
\item Write a correct version of Dennis's formula for cell B2.
The formula in cell A3, which is correct, is\\
= IF(F2 > 0, E2, A2)
\item Write a suitable formula for cell C3.
\item Use the information in Fig. 6.2 to
\begin{itemize}
\item find the value of $\alpha$ as accurately as possible,
\item state the maximum possible error in this estimate.
\end{itemize}
Liren uses a different method to find a sequence of estimates of the value of $\beta$ using a spreadsheet. The output, together with some further analysis, is shown in Fig. 6.3.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
& A & B & C & D \\
\hline
1 & $x$ & f(x) & difference & ratio \\
\hline
2 & 0.4 & -1.5799 & & \\
\hline
3 & 0.6 & -1.0504 & & \\
\hline
4 & 0.99671 & 8.4245 & & \\
\hline
5 & 0.64398 & -0.6809 & -0.35273 & \\
\hline
6 & 0.67036 & -0.4026 & 0.026378 & -0.0748 \\
\hline
7 & 0.70852 & 0.08386 & 0.038164 & 1.44682 \\
\hline
8 & 0.70194 & -0.0075 & -0.00658 & -0.1724 \\
\hline
9 & 0.70248 & -0.0001 & 0.00054 & -0.082 \\
\hline
10 & 0.70249 & $1.8 \mathrm { E } - 07$ & $8.88 \mathrm { E } - 06$ & 0.01646 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.3}
\end{center}
\end{table}
The formula in cell A4 is
$$= ( \mathrm { A } 2 * \mathrm {~B} 3 - \mathrm { A } 3 * \mathrm {~B} 2 ) / ( \mathrm { B } 3 - \mathrm { B } 2 )$$
\item State the method being used.
\item Explain what the values in column D tell you about the order of convergence of this sequence of estimates.
Liren states that $\beta = 0.70249$ correct to 5 decimal places.
\item Determine whether Liren is correct.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2020 Q6 [10]}}