| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining basic geometry (sector area formula, triangle area) with standard A-level numerical methods. Part (a) requires simple area formulas and algebraic manipulation, part (b) is routine Newton-Raphson application, and part (c) is a basic percentage error calculation. All techniques are standard textbook exercises with no novel insight required. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}r \times \frac{r}{2}\sin\theta = \frac{1}{4}\left(\frac{1}{2}r^2\theta\right)\) | B1 | Uses \(A = \frac{1}{2}ab\sin C\) for triangle \(OAC\) or \(OAB\); PI by equation |
| \(\Rightarrow \frac{r^2}{4}\sin\theta = \frac{1}{8}r^2\theta\) | M1 | Forms equation relating area of \(OAC\) and \(ABC\) in form \(Ar^2\sin\theta = Br^2\theta\) |
| \(\Rightarrow 2r^2\sin\theta = r^2\theta\) | A1 | Obtains fully correct equation ACF |
| \(\Rightarrow 2\sin\theta = \theta\) AG | R1 | Simplifies to obtain required equation; only award if all working correct with rigorous argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(\theta) = \theta - 2\sin\theta = 0\) | M1 | Rearranges to form \(f(\theta) = 0\); PI by correct \(\theta_2\) or \(\theta_3\) |
| \(\theta_{n+1} = \theta_n - \frac{\theta_n - 2\sin\theta_n}{1 - 2\cos\theta_n}\) | A1 | Differentiates their \(f(\theta)\) or uses calculator; PI correct \(\theta_2\) or \(\theta_3\) |
| \(\theta_2 = 2.094395\ldots\) | A1 | |
| \(\theta_3 = 1.913222\ldots\) | ||
| \(\theta_3 = 1.91322\) (5 d.p.) | A1 | Obtains correct \(\theta_3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.935\%\) | B1 | Obtains percentage error for \(\theta_3\); AWRT \(0.94\%\) |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}r \times \frac{r}{2}\sin\theta = \frac{1}{4}\left(\frac{1}{2}r^2\theta\right)$ | B1 | Uses $A = \frac{1}{2}ab\sin C$ for triangle $OAC$ or $OAB$; PI by equation |
| $\Rightarrow \frac{r^2}{4}\sin\theta = \frac{1}{8}r^2\theta$ | M1 | Forms equation relating area of $OAC$ and $ABC$ in form $Ar^2\sin\theta = Br^2\theta$ |
| $\Rightarrow 2r^2\sin\theta = r^2\theta$ | A1 | Obtains fully correct equation ACF |
| $\Rightarrow 2\sin\theta = \theta$ AG | R1 | Simplifies to obtain required equation; only award if all working correct with rigorous argument |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(\theta) = \theta - 2\sin\theta = 0$ | M1 | Rearranges to form $f(\theta) = 0$; PI by correct $\theta_2$ or $\theta_3$ |
| $\theta_{n+1} = \theta_n - \frac{\theta_n - 2\sin\theta_n}{1 - 2\cos\theta_n}$ | A1 | Differentiates their $f(\theta)$ or uses calculator; PI correct $\theta_2$ or $\theta_3$ |
| $\theta_2 = 2.094395\ldots$ | A1 | |
| $\theta_3 = 1.913222\ldots$ | | |
| $\theta_3 = 1.91322$ (5 d.p.) | A1 | Obtains correct $\theta_3$ |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.935\%$ | B1 | Obtains percentage error for $\theta_3$; AWRT $0.94\%$ |
---8 The diagram shows a sector of a circle $O A B$.\\
$C$ is the midpoint of $O B$.\\
Angle $A O B$ is $\theta$ radians.\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-10_700_963_536_534}
8
\begin{enumerate}[label=(\alph*)]
\item Given that the area of the triangle $O A C$ is equal to one quarter of the area of the sector $O A B$, show that $\theta = 2 \sin \theta$\\
8
\item Use the Newton-Raphson method with $\theta _ { 1 } = \pi$, to find $\theta _ { 3 }$ as an approximation for $\theta$. Give your answer correct to five decimal places.\\
8
\item Given that $\theta = 1.89549$ to five decimal places, find an estimate for the percentage error in the approximation found in part (b).\\
Turn over for the next question
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2018 Q8 [8]}}