| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Applied context requiring Newton-Raphson |
| Difficulty | Standard +0.3 This is a standard Newton-Raphson application with straightforward setup. Part (a) requires basic circle geometry (segment area = sector - triangle), parts (b) and (c) are routine application of the Newton-Raphson formula with simple derivatives. The geometry is A-level standard and the iteration is mechanical, making this slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Sector area \(= \frac{1}{2}r^2x\) and Triangle \(= \frac{1}{2}r^2\sin x\) | M1, A1 | Both areas seen; segment area found |
| Area of segment \(= \frac{1}{2}r^2(x - \sin x)\) | B1 | \(0.05\pi r^2\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(x - \sin x = 2 \times 0.05 \times \pi \Rightarrow x - \sin x - \frac{1}{10}\pi = 0\) | A1 | AG Must be fully shown and correct rearrangement |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_{n+1} = x_n - \dfrac{x_n - \sin x_n - \frac{1}{10}\pi}{1 - \cos x_n}\) | B1 | \(x_{n+1}\) must be seen; algebraic form must be seen; derivative must be worked out. Condone \(x\) used instead of \(x_n\) in fraction part. |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_3 = 1.26894...\) | M1 | 3 iterations recorded |
| A1 | The first 3 iterations correct | |
| So root is \(1.269\) to 3 dp | A1 | Allow 1.27, 1.269 or more decimal places if correct. Root is 1.268947865 to 9 dp |
## Question 14:
**Part (a):**
Sector area $= \frac{1}{2}r^2x$ and Triangle $= \frac{1}{2}r^2\sin x$ | M1, A1 | Both areas seen; segment area found
Area of segment $= \frac{1}{2}r^2(x - \sin x)$ | B1 | $0.05\pi r^2$ seen
$\frac{1}{2}r^2(x-\sin x) = 0.05\pi r^2$
$x - \sin x = 2 \times 0.05 \times \pi \Rightarrow x - \sin x - \frac{1}{10}\pi = 0$ | A1 | AG Must be fully shown and correct rearrangement
**Part (b):**
$x_{n+1} = x_n - \dfrac{x_n - \sin x_n - \frac{1}{10}\pi}{1 - \cos x_n}$ | B1 | $x_{n+1}$ must be seen; algebraic form must be seen; derivative must be worked out. Condone $x$ used instead of $x_n$ in fraction part.
**Part (c):**
$x_0 = 1.2$
$x_1 = 1.27245...$
$x_2 = 1.26895...$
$x_3 = 1.26894...$ | M1 | 3 iterations recorded
| A1 | The first 3 iterations correct
So root is $1.269$ to 3 dp | A1 | Allow 1.27, 1.269 or more decimal places if correct. Root is 1.268947865 to 9 dp
---14 Fig. 14 shows a circle with centre O and radius $r \mathrm {~cm}$. The chord AB is such that angle $\mathrm { AOB } = x$ radians. The area of the shaded segment formed by AB is $5 \%$ of the area of the circle.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{59e924e6-8fa9-4035-9173-705fce487bd9-7_497_496_356_251}
\captionsetup{labelformat=empty}
\caption{Fig. 14}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that $x - \sin x - \frac { 1 } { 10 } \pi = 0$.
The Newton-Raphson method is to be used to find $x$.
\item Write down the iterative formula to be used for the equation in part (a).
\item Use three iterations of the Newton-Raphson method with $x _ { 0 } = 1.2$ to find the value of $x$ to a suitable degree of accuracy.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q14 [8]}}