| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sign Change & Interval Methods |
| Type | Interval Bisection from Spreadsheet |
| Difficulty | Moderate -0.8 This is a straightforward spreadsheet-based interval bisection question requiring only basic understanding of the algorithm. Parts (a)-(d) involve reading formulas and completing a table with simple arithmetic, while parts (e)-(f) require minimal calculation or reasoning about convergence rates. No novel problem-solving or deep mathematical insight is neededβjust mechanical application of a standard numerical method. |
| Spec | 1.09a Sign change methods: locate roots |
| 1 | A | B | C | D | E | F |
| 1 | a | f(a) | \(b\) | f(b) | c | \(\mathrm { f } ( c )\) |
| 2 | 2 | -0.6109 | 3 | 6.08554 | 2.5 | 1.43249 |
| 3 | 2 | -0.6109 | 2.5 | 1.43249 | 2.25 | 0.17524 |
| 4 | 2 | -0.6109 | 2.25 | 0.17524 | 2.125 | -0.2677 |
| 5 | 2.125 | -0.2677 | 2.25 | 0.17524 | 2.1875 | -0.0598 |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | eπ₯ βπ₯2βπ₯β2 |
| eπ₯ βπ₯2βπ₯β2 = 0 | M1 | |
| A1 | 1.1 | |
| 1.1 | allow one sign error |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | =(A2 + C2)/2 oe |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | =IF(F2>0,E2,C2) or =IF(F2<0,C2,E2) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | x-values (column A, C, E) |
| y-values (column B, D, F) | B1 | |
| B1 | 1.1 | |
| 1.1 | 2.1875 -0.0598 2.25 0.17524 2.21875 0.0542[3] |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (e) | 2.1875 < root < 2.21875 so root = 2.2 (because |
| Answer | Marks | Guidance |
|---|---|---|
| 2.2) | B1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 2.1875 | -0.0598 | 2.25 |
| 5 | (f) | intervalΓ0.5π |
| Answer | Marks |
|---|---|
| allow[ π > 6.97] so π = 7 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 3.2a | interval may be 1 or an integer power of 0.5; |
| Answer | Marks |
|---|---|
| π = 6 or π = 7 | two bisections of maximum possible error; may see eg 0.5, 0.25, |
Question 5:
5 | (a) | eπ₯ βπ₯2βπ₯β2
eπ₯ βπ₯2βπ₯β2 = 0 | M1
A1 | 1.1
1.1 | allow one sign error
all correct
[2]
5 | (b) | =(A2 + C2)/2 oe | B1 | 1.1
[1]
5 | (c) | =IF(F2>0,E2,C2) or =IF(F2<0,C2,E2) | B1 | 1.1
[1]
5 | (d) | x-values (column A, C, E)
y-values (column B, D, F) | B1
B1 | 1.1
1.1 | 2.1875 -0.0598 2.25 0.17524 2.21875 0.0542[3]
[2]
5 | (e) | 2.1875 < root < 2.21875 so root = 2.2 (because
both endpoints round to 2.2, so any
approximation in this interval will also round to
2.2) | B1 | 2.2a | allow 2.2 because
eg 2.1875 and 2.21875 both agree to 1 dp
eg the last two approximations agree to 1 dp oe
eg the best 2 approximations agree to 2 sf oe
eg A6 and E6 agree to 1dp oe
do not allow 2.2 because
eg A6 and C6 agree to 1 dp
[1]
2.1875 | -0.0598 | 2.25 | 0.17524 | 2.21875 | 0.0542[3]
5 | (f) | intervalΓ0.5π
ln(interval Γ0.5π) < ln0.0005 oe
[π > 5.96] so π = 6
allow[ π > 6.97] so π = 7 | M1
M1
A1 | 3.1a
2.1
3.2a | interval may be 1 or an integer power of 0.5;
may see 5.96
may see 11 β 5 = 6 or 11 β 4 = 7
[3]
Alternatively
eg [0.03125], 0.015625, 0.0078125, 0.003906,
0.001953, 0.0009765, 0.0004883
0.0009765 and 0.0004883 seen
π = 6 or π = 7 | two bisections of maximum possible error; may see eg 0.5, 0.25,
0.125, 0.0625, 0.03125, β¦
this mark is dependent on the award of one M mark;
0.03125
if M0M0 allow SC2 for = 0.000488 π¨π so π = 6
26
0.0625
or SC2 for = 0.000488 π¨π so n = 7
27
1
or SC2 for = 0.000488 π¨π so π = 6 or n = 7
211
M1
M1
A1
[3]5 The root of the equation $\mathrm { f } ( x ) = 0$ is being found using the method of interval bisection. Some of the associated spreadsheet output is shown in the table below.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
1 & A & B & C & D & E & F \\
\hline
1 & a & f(a) & $b$ & f(b) & c & $\mathrm { f } ( c )$ \\
\hline
2 & 2 & -0.6109 & 3 & 6.08554 & 2.5 & 1.43249 \\
\hline
3 & 2 & -0.6109 & 2.5 & 1.43249 & 2.25 & 0.17524 \\
\hline
4 & 2 & -0.6109 & 2.25 & 0.17524 & 2.125 & -0.2677 \\
\hline
5 & 2.125 & -0.2677 & 2.25 & 0.17524 & 2.1875 & -0.0598 \\
\hline
6 & & & & & & \\
\hline
\end{tabular}
\end{center}
The formula in cell B2 is $\quad = \mathrm { EXP } ( \mathrm { A } 2 ) - \mathrm { A } 2 ^ { \wedge } 2 - \mathrm { A } 2 - 2$.
\begin{enumerate}[label=(\alph*)]
\item Write down the equation whose root is being found.
\item Write down a suitable formula for cell E2.
The formula in cell A3 is
$$= \mathrm { IF } ( \mathrm {~F} 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$$
.
\item Write down a similar formula for cell C3.
\item Complete row 6 of the table on the copy in the Printed Answer Booklet.
\item Without doing any calculations, write down the value of the root correct to the number of decimal places which seems justified. You must explain the precision quoted.
\item Determine how many more applications of the bisection method are needed such that the interval which contains the root is less than 0.0005 .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2024 Q5 [10]}}