| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Optimization with sectors |
| Difficulty | Standard +0.8 This is a multi-part question combining geometry, trigonometry, numerical methods, and proof. Part (a) requires setting up area formulas and algebraic manipulation to reach a given result. Part (b) involves change of sign method with specific interval bounds. Parts (c)(i-iii) test Newton-Raphson application, iteration, and understanding of method failure. While individual components are standard A-level techniques, the combination of geometric setup, proof, and numerical analysis with explanation of failure cases elevates this above average difficulty. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09a Sign change methods: locate roots1.09d Newton-Raphson method |
| 10 |
| \(\theta = \sin 2 \theta\) |
| 10 (c) (i) Using \(\theta _ { 1 } = \frac { \pi } { 5 }\) as a first approximation for \(\theta\) apply the Newton-Raphson method twice |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area of sector \(= \frac{1}{2}r^2\theta\) | B1 | AO 1.2. \(r\) can be any letter or \(OA\) or \(OB\) |
| \(\frac{1}{2}ab\sin C = \left(\frac{1}{2}\right)\frac{1}{2}r^2\theta\) | M1 | AO 3.1a. Forms equation relating area of triangle \(OAC\) and sector using \(\frac{1}{2}bh = k\frac{1}{2}r^2\theta\) where \(k>0\) |
| Area of triangle \(= \frac{1}{2}r\cos\theta \times r\sin\theta\) | B1 | AO 2.2a. Deduces area of triangle is \(\frac{1}{2}r\cos\theta \times r\sin\theta\); must use trigonometry for height and base |
| \(2\sin\theta\cos\theta = \theta\), therefore \(\theta = \sin 2\theta\) | R1 | AO 2.1. Completes reasoned argument with clear use of double angle identity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(f(\theta)=\theta - \sin2\theta\); \(f\!\left(\frac{\pi}{5}\right)=-0.3227\ldots < 0\); \(f\!\left(\frac{2\pi}{5}\right)=0.6688\ldots > 0\) | M1 | AO 1.1a. Rearranges to \(\theta - \sin2\theta=0\); evaluates at \(\frac{\pi}{5}\) and \(\frac{2\pi}{5}\) |
| Hence solution lies between \(\frac{\pi}{5}\) and \(\frac{2\pi}{5}\) | R1 | AO 2.1. Reasoned argument with reference to change of sign; must refer to \(\frac{\pi}{5}\) and \(\frac{2\pi}{5}\) in conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(\theta)=1-2\cos2\theta\) | B1 | AO 1.1b. Differentiates \(\sin2\theta\) to obtain \(2\cos2\theta\) OE |
| \(\theta_{n+1} = \theta_n - \dfrac{\theta_n - \sin2\theta_n}{1-2\cos2\theta_n}\) | M1 | AO 1.1a. Correct Newton-Raphson expression; accept use of \(\frac{\pi}{5}\); condone missing/incorrect subscript |
| \(\theta_2 = 1.4732575\ldots \approx 1.473\) | A1 | AO 1.1b. AWRT \(\theta_2\ 1.473\) |
| \(\theta_3 = 1.0413241\ldots \approx 1.041\) | A1 | AO 1.1b. AWRT \(\theta_3\ 1.041\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use more iterations | E1 | AO 2.4. Explains that more iterations could be used; accept "keep on using Newton–Raphson, keep re-iterating" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'\left(\frac{\pi}{6}\right) = 0\) | E1 | States that the derivative equals zero at \(x = \frac{\pi}{6}\) |
| The value is on a stationary point, so Newton-Raphson does not converge to a particular root | E1 | Accept: too close to a stationary point; the value is on a stationary point; the tangent does not cross the \(x\)-axis; it converges to a different root; the formula is undefined |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of sector $= \frac{1}{2}r^2\theta$ | B1 | AO 1.2. $r$ can be any letter or $OA$ or $OB$ |
| $\frac{1}{2}ab\sin C = \left(\frac{1}{2}\right)\frac{1}{2}r^2\theta$ | M1 | AO 3.1a. Forms equation relating area of triangle $OAC$ and sector using $\frac{1}{2}bh = k\frac{1}{2}r^2\theta$ where $k>0$ |
| Area of triangle $= \frac{1}{2}r\cos\theta \times r\sin\theta$ | B1 | AO 2.2a. Deduces area of triangle is $\frac{1}{2}r\cos\theta \times r\sin\theta$; must use trigonometry for height and base |
| $2\sin\theta\cos\theta = \theta$, therefore $\theta = \sin 2\theta$ | R1 | AO 2.1. Completes reasoned argument with clear use of double angle identity |
---
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $f(\theta)=\theta - \sin2\theta$; $f\!\left(\frac{\pi}{5}\right)=-0.3227\ldots < 0$; $f\!\left(\frac{2\pi}{5}\right)=0.6688\ldots > 0$ | M1 | AO 1.1a. Rearranges to $\theta - \sin2\theta=0$; evaluates at $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ |
| Hence solution lies between $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ | R1 | AO 2.1. Reasoned argument with reference to change of sign; must refer to $\frac{\pi}{5}$ and $\frac{2\pi}{5}$ in conclusion |
---
## Question 10(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(\theta)=1-2\cos2\theta$ | B1 | AO 1.1b. Differentiates $\sin2\theta$ to obtain $2\cos2\theta$ OE |
| $\theta_{n+1} = \theta_n - \dfrac{\theta_n - \sin2\theta_n}{1-2\cos2\theta_n}$ | M1 | AO 1.1a. Correct Newton-Raphson expression; accept use of $\frac{\pi}{5}$; condone missing/incorrect subscript |
| $\theta_2 = 1.4732575\ldots \approx 1.473$ | A1 | AO 1.1b. AWRT $\theta_2\ 1.473$ |
| $\theta_3 = 1.0413241\ldots \approx 1.041$ | A1 | AO 1.1b. AWRT $\theta_3\ 1.041$ |
---
## Question 10(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use more iterations | E1 | AO 2.4. Explains that more iterations could be used; accept "keep on using Newton–Raphson, keep re-iterating" |
## Question 10(c)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'\left(\frac{\pi}{6}\right) = 0$ | E1 | States that the derivative equals zero at $x = \frac{\pi}{6}$ |
| The value is on a stationary point, so Newton-Raphson does not converge to a particular root | E1 | Accept: too close to a stationary point; the value is on a stationary point; the tangent does not cross the $x$-axis; it converges to a different root; the formula is undefined |
---10 The diagram shows a sector of a circle $O A B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-16_758_796_360_623}
The point $C$ lies on $O B$ such that $A C$ is perpendicular to $O B$.\\
Angle $A O B$ is $\theta$ radians.\\
10
\begin{enumerate}[label=(\alph*)]
\item Given the area of the triangle $O A C$ is half the area of the sector $O A B$, show that
$$\theta = \sin 2 \theta$$
10
\item Use a suitable change of sign to show that a solution to the equation
$$\theta = \sin 2 \theta$$
lies in the interval given by $\theta \in \left[ \frac { \pi } { 5 } , \frac { 2 \pi } { 5 } \right]$\\
\begin{center}
\begin{tabular}{ | l | }
\hline
10
\item The Newton-Raphson method is used to find an approximate solution to the equation \\
$\theta = \sin 2 \theta$ \\
10 (c) (i) Using $\theta _ { 1 } = \frac { \pi } { 5 }$ as a first approximation for $\theta$ apply the Newton-Raphson method twice \\
\end{tabular}
\end{center} to find the value of $\theta _ { 3 }$
Give your answer to three decimal places.\\
10 (c) (ii) Explain how a more accurate approximation for $\theta$ can be found using the Newton-Raphson method.\\
10 (c) (iii) Explain why using $\theta _ { 1 } = \frac { \pi } { 6 }$ as a first approximation in the Newton-Raphson method\\[0pt]
[2 marks] does not lead to a solution for $\theta$.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2022 Q10 [12]}}