| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Standard +0.3 This is a standard A-level iteration question with routine steps: showing an equation holds (part a requires basic sector/triangle area formulas), applying a given iterative formula with calculator work (part b), and demonstrating convergence graphically using cobweb/staircase diagrams (parts c and d). All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| area \(OMB = \frac{1}{2}\left(\frac{1}{2}r\right)r\sin\theta\) | B1 (1.1) | Correct (possibly unsimplified) area of \(OMB\); could use other than \(r\) for radius |
| \(2\left(\frac{1}{2}r^2\theta - \frac{1}{4}r^2\sin\theta\right) = 3\left(\frac{1}{4}r^2\sin\theta\right)\) OR \(2\left(\frac{1}{2}r^2\theta\right) = 5\left(\frac{1}{4}r^2\sin\theta\right)\) OR \(3\left(\frac{1}{2}r^2\theta\right) = 5\left(\frac{1}{2}r^2\theta - \frac{1}{4}r^2\sin\theta\right)\) | M1 (3.1a) | Attempt to use ratio on two correct areas; must be two correct areas; must use correct ratio 2:3 if using \(OMB\), 2:5 if using \(OMB\) and \(OAB\), or 3:5 if using \(MAB\) and \(OAB\) |
| Correct equation in two variables (\(\theta\) and \(r\)) | A1 (2.1) | Any correct statement linking the two areas; could use \(2x^2\theta - x^2\sin\theta\) |
| \(\theta - \frac{1}{2}\sin\theta = \frac{3}{4}\sin\theta\) | A1 (2.1) | At least one line of working once ratio used |
| \(\theta = 1.25\sin\theta\) A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.599\) | B1 (1.1a) | Obtain correct first iterate; 3sf or better (more accurate: \(0.599281923...\)); condone truncating |
| \(0.705,\ 0.810\) | M1 (1.1a) | Attempt correct iterative process to find at least 2 more values; M0 if working in degrees |
| \(\text{root} = 1.13\) | A1 (1.1) | Obtain \(1.13\); must be 3sf; once M1 awarded, allow A1 for 1.13 even if incorrect iterate seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Draw \(y = \theta\) on diagram | B1* (3.1a) | Draw straight line starting at origin which intersects the graph; allow point of intersection to be greater than \(\theta = \frac{1}{2}\pi\); ignore incorrect labels such as \(y = x\) |
| Draw correct iterative process on diagram | B1 dep* (2.1) | Vertically into curve, then horizontally into straight line, as far as root; initial value should be before root; needs point of intersection to be before \(\theta = \frac{1}{2}\pi\) |
| State 'staircase' convergence | B1 (1.2) | Mark independently; staircase can be deduced from iterates in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Draw graph of \(y = \sin^{-1}0.8\theta\) for \(\theta \geq 0\) | B1* (3.1a) | Correct shape for \(y = \sin^{-1}k\theta\); one-to-one function starting at origin; increasing gradient for all \(\theta\); ignore \(\theta < 0\) |
| Draw \(y = \theta\) and show staircase divergence from root found in (b), on at least one side of root | B1 dep* (3.2a) | Straight line from origin to intersect their graph; diagram sufficient for B1 — no comment required |
# Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| area $OMB = \frac{1}{2}\left(\frac{1}{2}r\right)r\sin\theta$ | B1 (1.1) | Correct (possibly unsimplified) area of $OMB$; could use other than $r$ for radius |
| $2\left(\frac{1}{2}r^2\theta - \frac{1}{4}r^2\sin\theta\right) = 3\left(\frac{1}{4}r^2\sin\theta\right)$ OR $2\left(\frac{1}{2}r^2\theta\right) = 5\left(\frac{1}{4}r^2\sin\theta\right)$ OR $3\left(\frac{1}{2}r^2\theta\right) = 5\left(\frac{1}{2}r^2\theta - \frac{1}{4}r^2\sin\theta\right)$ | M1 (3.1a) | Attempt to use ratio on two correct areas; must be two correct areas; must use correct ratio 2:3 if using $OMB$, 2:5 if using $OMB$ and $OAB$, or 3:5 if using $MAB$ and $OAB$ |
| Correct equation in two variables ($\theta$ and $r$) | A1 (2.1) | Any correct statement linking the two areas; could use $2x^2\theta - x^2\sin\theta$ |
| $\theta - \frac{1}{2}\sin\theta = \frac{3}{4}\sin\theta$ | A1 (2.1) | At least one line of working once ratio used |
| $\theta = 1.25\sin\theta$ **A.G.** | | |
---
# Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.599$ | B1 (1.1a) | Obtain correct first iterate; 3sf or better (more accurate: $0.599281923...$); condone truncating |
| $0.705,\ 0.810$ | M1 (1.1a) | Attempt correct iterative process to find at least 2 more values; M0 if working in degrees |
| $\text{root} = 1.13$ | A1 (1.1) | Obtain $1.13$; must be 3sf; once M1 awarded, allow A1 for 1.13 even if incorrect iterate seen |
---
# Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Draw $y = \theta$ on diagram | B1* (3.1a) | Draw straight line starting at origin which intersects the graph; allow point of intersection to be greater than $\theta = \frac{1}{2}\pi$; ignore incorrect labels such as $y = x$ |
| Draw correct iterative process on diagram | B1 dep* (2.1) | Vertically into curve, then horizontally into straight line, as far as root; initial value should be before root; needs point of intersection to be before $\theta = \frac{1}{2}\pi$ |
| State 'staircase' convergence | B1 (1.2) | Mark independently; staircase can be deduced from iterates in (b) |
---
# Question 10(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Draw graph of $y = \sin^{-1}0.8\theta$ for $\theta \geq 0$ | B1* (3.1a) | Correct shape for $y = \sin^{-1}k\theta$; one-to-one function starting at origin; increasing gradient for all $\theta$; ignore $\theta < 0$ |
| Draw $y = \theta$ and show staircase divergence from root found in (b), on at least one side of root | B1 dep* (3.2a) | Straight line from origin to intersect their graph; diagram sufficient for B1 — no comment required |
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{38b515c2-4764-4b51-a1f5-9b48d46610f0-7_545_659_255_244}
The diagram shows a sector $O A B$ of a circle with centre $O$ and radius $O A$. The angle $A O B$ is $\theta$ radians. $M$ is the mid-point of $O A$. The ratio of areas $O M B : M A B$ is 2:3.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 1.25 \sin \theta$.
The equation $\theta = 1.25 \sin \theta$ has only one root for $\theta > 0$.
\item This root can be found by using the iterative formula $\theta _ { n + 1 } = 1.25 \sin \theta _ { n }$ with a starting value of $\theta _ { 1 } = 0.5$.
\begin{itemize}
\item Write down the values of $\theta _ { 2 } , \theta _ { 3 }$ and $\theta _ { 4 }$.
\item Hence find the value of this root correct to $\mathbf { 3 }$ significant figures.
\item The diagram in the Printed Answer Booklet shows the graph of $y = 1.25 \sin \theta$, for $0 \leqslant \theta \leqslant \pi$.
\item Use this diagram to show how the iterative process used in (b) converges to this root.
\item State the type of convergence.
\item Draw a suitable diagram to show why using an iterative process with the formula $\theta _ { n + 1 } = \sin ^ { - 1 } \left( 0.8 \theta _ { n } \right)$ does not converge to the root found in (b).
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2022 Q10 [12]}}