| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Challenging +1.2 This is a standard Further Maths numerical methods question requiring mechanical application of difference tables and Newton's forward difference formula. While it's a Further Maths topic (making it inherently harder than core A-level), the question follows a routine algorithmic procedure with no problem-solving insight required. The steps are clearly signposted across three parts, making it easier than average Further Maths questions but harder than typical core A-level work. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { f } ( x )\) | - 0.65 | - 0.35 | 1.77 | 5.71 | 11.47 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | x |
| A1 | 1.1 | |
| 1.1 | finds 4 Δ values, allow one error |
| Answer | Marks |
|---|---|
| 1 | ‒0.65 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | ‒0.35 | 1.82 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 1.77 | 1.82 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | 5.71 | 1.82 |
| Answer | Marks |
|---|---|
| 5 | 11.47 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | the second differences are constant oe |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (c) | ‒0.65 + 0.3(x ‒ 1) + 1.82× |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | must be correct form; allow 1 |
Question 2:
2 | (a) | x | f(x) | Δ | Δ² | M1
A1 | 1.1
1.1 | finds 4 Δ values, allow one error
all correct
1 | ‒0.65
0.3
2 | ‒0.35 | 1.82
2.12
3 | 1.77 | 1.82
3.94
4 | 5.71 | 1.82
5.76
5 | 11.47
[2]
2 | (b) | the second differences are constant oe | B1 | 1.1 | allow the 3rd differences are zero
[1]
2 | (c) | ‒0.65 + 0.3(x ‒ 1) + 1.82×
(𝑥𝑥−1)(𝑥𝑥−2)
2!
[P (x) =] 0.91x² ‒ 2.43x + 0.87
2 | M1
A1
A1 | 1.1
1.1
1.1 | must be correct form; allow 1
substitution error
two of three terms correct
all correct
[3]2 The table shows some values of $x$ and the associated values of $\mathrm { f } ( x )$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { f } ( x )$ & - 0.65 & - 0.35 & 1.77 & 5.71 & 11.47 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the difference table in the Printed Answer Booklet.
\item Explain why the data may be interpolated by a polynomial of degree 2.
\item Use Newton's forward difference interpolation formula to obtain a polynomial of degree 2 for the data.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2021 Q2 [6]}}