| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2003 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show convergence to specific root |
| Difficulty | Standard +0.3 This is a standard A-level fixed point iteration question requiring graph sketching to show existence of a root, algebraic manipulation to verify the iteration formula converges to the correct root, and mechanical application of the iteration. All techniques are routine for P3/C3 level with no novel insights required, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Make recognisable sketch of \(y = \sec x\) or \(y = 3 - x^2\), for \(0 < x < \frac{1}{2}\pi\) | B1 | Award B1 for sketch with positive \(y\)-intercept and correct concavity. Correct sketch of \(y = \cos x\) can only earn B1 in presence of \(1/(3-x^2)\). Allow a correct single graph and its intersection with \(y=0\) to earn full marks. |
| Sketch the other graph correctly and justify the given statement | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply equation \(\alpha = \cos^{-1}(1/(3-\alpha^2))\) or \(\cos\alpha = 1/(3-\alpha^2)\) | B1 | |
| Rearrange this in the form given in part (i) i.e. \(\sec\alpha = 3 - \alpha^2\) | B1 | Or work vice versa |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use the iterative formula with \(0 \leq x_1 \leq \sqrt{2}\) | M1 | |
| Obtain final answer 1.03 | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval \((1.025, 1.035)\) | A1 |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Make recognisable sketch of $y = \sec x$ or $y = 3 - x^2$, for $0 < x < \frac{1}{2}\pi$ | B1 | Award B1 for sketch with positive $y$-intercept and correct concavity. Correct sketch of $y = \cos x$ can only earn B1 in presence of $1/(3-x^2)$. Allow a correct single graph and its intersection with $y=0$ to earn full marks. |
| Sketch the other graph correctly and justify the given statement | B1 | |
**Total: [2]**
---
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply equation $\alpha = \cos^{-1}(1/(3-\alpha^2))$ or $\cos\alpha = 1/(3-\alpha^2)$ | B1 | |
| Rearrange this in the form given in part (i) i.e. $\sec\alpha = 3 - \alpha^2$ | B1 | Or work vice versa |
**Total: [2]**
---
## Question 5(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the iterative formula with $0 \leq x_1 \leq \sqrt{2}$ | M1 | |
| Obtain final answer 1.03 | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval $(1.025, 1.035)$ | A1 | |
**Total: [3]**
---5 (i) By sketching suitable graphs, show that the equation
$$\sec x = 3 - x ^ { 2 }$$
has exactly one root in the interval $0 < x < \frac { 1 } { 2 } \pi$.\\
(ii) Show that, if a sequence of values given by the iterative formula
$$x _ { n + 1 } = \cos ^ { - 1 } \left( \frac { 1 } { 3 - x _ { n } ^ { 2 } } \right)$$
converges, then it converges to a root of the equation given in part (i).\\
(iii) Use this iterative formula, with initial value $x _ { 1 } = 1$, to determine the root in the interval $0 < x < \frac { 1 } { 2 } \pi$ correct to 2 decimal places, showing the result of each iteration.
\hfill \mbox{\textit{CAIE P3 2003 Q5 [7]}}