| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Topic | Fixed Point Iteration |
| Type | Derive equation from area/geometry |
| Difficulty | Standard +0.8 This question requires deriving a transcendental equation from geometric area relationships (sector minus triangle equals triangle), then applying fixed point iteration. Part (i) demands careful geometric reasoning with sector/triangle area formulas and trigonometry. Part (ii) is more routine iteration mechanics but requires understanding convergence and verification. The geometric setup and equation derivation elevate this above standard iterative problems, though the iteration itself is straightforward once the equation is established. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05o Trigonometric equations: solve in given intervals1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
**Question 6(i) and 6(ii)**
**(i)**
- $\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r\sin\theta \cdot r\cos\theta$ **B1** Obtain area of segment as $\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta$
- $\frac{1}{2}r^2(\theta - \sin\theta) = \frac{1}{2}r^2\sin\theta\cos\theta$; $\theta - \sin\theta = \sin\theta\cos\theta$ **B1** Obtain area of triangle as $\frac{1}{2}r\sin\theta \cdot r\cos\theta$
- $\theta = \sin\theta(1+\cos\theta)$ **A.G.** **M1** Dep on both B marks. Equate areas and simplify
- **A1** $\theta = \sin\theta(1+\cos\theta)$ **A.G.**
**(ii)**
- $1.296...$ **B1** Correct first iterate, from starting with 1. Allow 1.30
- $\ldots 1.224, 1.260, 1.243, 1.252, 1.247, 1.249\ldots$ **M1** Correct iteration process at least 3 times. Allow working to 3sf
- Hence root is $1.25$ **A1** Obtain 1.25 (must be 3sf)
OR (Newton-Raphson)
- $\theta_{n+1} = \theta_n - \frac{\sin\theta_n + \sin\theta_n\cos\theta_n - \theta_n}{\cos\theta_n + \cos^2\theta_n - \sin^2\theta_n - 1}$ **B1** Correct Newton-Raphson formula. Allow any equiv e.g. with double angles
- $1.338, 1.2535, 1.2488, 1.2488\ldots$ **M1** Attempt to use N-R formula at least 3 times. Must come from starting with $f(\theta)=0$
- Hence root is $1.25$ **A1** Obtain 1.25 (must be 3sf)
Sign change verification:
- $1.245 < 1.251$; $1.255 > 1.246$ or $f(1.245) = 5.623\times10^{-3}$; $f(1.255) = -9.235\times10^{-3}$ **M1** Substitute 1.245 and 1.255 (or better) into both sides of equation, or equiv
- Inequality sign change, hence root is 1.25 to 3sf **A1** Conclude correctly, referring to sign change. A0 if not confirming 3sf6\\
\includegraphics[max width=\textwidth, alt={}, center]{f4b66aaa-16b9-4b15-b3f5-b9657fe98274-3_545_557_269_794}
The diagram shows a sector $A O B$ of a circle, centre $O$ and radius $r$. Angle $A O B$ is $\theta$ radians. The point $C$ lies on $O B$, and $A C$ is perpendicular to $O B$. The area of the triangle $A O C$ is equal to the area of the segment bounded by the chord $A B$ and the $\operatorname { arc } A B$.\\
(i) Show that $\theta = \sin \theta ( 1 + \cos \theta )$.
The equation $\theta = \sin \theta ( 1 + \cos \theta )$ has only one positive root.\\
(ii) Use an iterative process based on this equation to find the value of the root correct to 3 significant figures. Use a starting value of 1 and show the result of each iteration. Use a change of sign to verify that the value you have found is correct to 3 significant figures.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2018 Q6 [12]}}