| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| 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 Simpson's rule with clearly specified intervals (n=6) to find an area, then multiply by thickness for volume. The function is given explicitly, requires only substitution of x-values and arithmetic. While it's Further Maths (FP1), Simpson's rule is a standard numerical method requiring careful calculation but no problem-solving insight or conceptual difficulty beyond following the formula. |
| Spec | 1.09f Trapezium rule: numerical integration4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Step \(\frac{1}{3}\) | B1 | Correct strip width for method chosen \(\frac{1}{3}\) for interval \([-1,1]\) |
| Table of values: \(x\): \(-1, -\frac{2}{3}, -\frac{1}{3}, 0, \frac{1}{3}, \frac{2}{3}, 1\); \(y\): \(0, 2.2981, 2.9544, 3, 2.9544, 2.2981, 0\) | M1 | Uses model to find appropriate values. At least two correct values to 4 s.f. needed. |
| \(y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6 = \) "42.203" \(\{0 + 4(2.2981) + 3 + 2.2981) + 2(2.9544 + 2.9544) + 0\}\) | M1 | Applies Simpson's rule bracket \(y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6\). Coefficients must be correct. |
| \(= 42.203 \left(= 24\cos\!\left(\frac{2\pi}{9}\right) + 12\cos\!\left(\frac{\pi}{18}\right) + 12\right)\) | A1 | Correct value for the bracket. If not explicit, may be implied by awrt 4.689 as cross-section area value. |
| Volume \(\approx \frac{85}{1000} \times \frac{1}{3} \times\) "42.203" | M1 | |
| \(= \text{awrt } 0.3986 \text{ m}^3\) | A1 | |
| (6 marks) |
## Question 2:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Step $\frac{1}{3}$ | **B1** | Correct strip width for method chosen $\frac{1}{3}$ for interval $[-1,1]$ |
| Table of values: $x$: $-1, -\frac{2}{3}, -\frac{1}{3}, 0, \frac{1}{3}, \frac{2}{3}, 1$; $y$: $0, 2.2981, 2.9544, 3, 2.9544, 2.2981, 0$ | **M1** | Uses model to find appropriate values. At least two correct values to 4 s.f. needed. |
| $y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6 = $ "42.203" $\{0 + 4(2.2981) + 3 + 2.2981) + 2(2.9544 + 2.9544) + 0\}$ | **M1** | Applies Simpson's rule bracket $y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6$. Coefficients must be correct. |
| $= 42.203 \left(= 24\cos\!\left(\frac{2\pi}{9}\right) + 12\cos\!\left(\frac{\pi}{18}\right) + 12\right)$ | **A1** | Correct value for the bracket. If not explicit, may be implied by awrt 4.689 as cross-section area value. |
| Volume $\approx \frac{85}{1000} \times \frac{1}{3} \times$ "42.203" | **M1** | |
| $= \text{awrt } 0.3986 \text{ m}^3$ | **A1** | |
| **(6 marks)** | | |2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9f127ab1-0e03-4f9f-87c2-01c553c54ee9-04_807_649_251_708}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the vertical cross-section of the entrance to a tunnel. The width at the base of the tunnel entrance is 2 metres and its maximum height is 3 metres.
The shape of the cross-section can be modelled by the curve with equation $y = \mathrm { f } ( x )$ where
$$f ( x ) = 3 \cos \left( \frac { \pi } { 2 } x ^ { 2 } \right) \quad x \in [ - 1,1 ]$$
A wooden door of uniform thickness 85 mm is to be made to seal the tunnel entrance.\\
Use Simpson's rule with 6 intervals to estimate the volume of wood required for this door, giving your answer in $\mathrm { m } ^ { 3 }$ to 4 significant figures.
\hfill \mbox{\textit{Edexcel FP1 2020 Q2 [6]}}