| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| 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, interval verification by substitution, algebraic rearrangement to show convergence to the root, and iterative calculation. All steps follow routine procedures taught in P3 with no novel insight required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05i 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 recognizable sketch of a relevant graph | B1 | |
| Sketch the other relevant graph and justify the given statement | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use calculations to consider the value of a relevant expression at \(x = 1.4\) and \(x = 1.6\), or the values of relevant expressions at \(x = 1.4\) and \(x = 1.6\) | M1 | |
| Complete the argument correctly with correct calculated values | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State \(x = 2\sin^{-1}\left(\frac{3}{x+3}\right)\) | B1 | |
| Rearrange this in the form \(\text{cosec}\frac{1}{2}x = \frac{1}{3}x + 1\) | B1 | If working in reverse, need \(\sin\frac{x}{2} = \left(\frac{3}{x+3}\right)\) for first B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(1.471\) | A1 | |
| Show sufficient iterations to 5 d.p. to justify \(1.471\) to 3 d.p., or show there is a sign change in the interval \((1.4705, 1.4715)\) | A1 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Make recognizable sketch of a relevant graph | B1 | |
| Sketch the other relevant graph and justify the given statement | B1 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use calculations to consider the value of a relevant expression at $x = 1.4$ and $x = 1.6$, or the values of relevant expressions at $x = 1.4$ and $x = 1.6$ | M1 | |
| Complete the argument correctly with correct calculated values | A1 | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State $x = 2\sin^{-1}\left(\frac{3}{x+3}\right)$ | B1 | |
| Rearrange this in the form $\text{cosec}\frac{1}{2}x = \frac{1}{3}x + 1$ | B1 | If working in reverse, need $\sin\frac{x}{2} = \left(\frac{3}{x+3}\right)$ for first B1 |
### Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer $1.471$ | A1 | |
| Show sufficient iterations to 5 d.p. to justify $1.471$ to 3 d.p., or show there is a sign change in the interval $(1.4705, 1.4715)$ | A1 | |
---6 (i) By sketching a suitable pair of graphs, show that the equation
$$\operatorname { cosec } \frac { 1 } { 2 } x = \frac { 1 } { 3 } x + 1$$
has one root in the interval $0 < x \leqslant \pi$.\\
(ii) Show by calculation that this root lies between 1.4 and 1.6.\\
(iii) Show that, if a sequence of values in the interval $0 < x \leqslant \pi$ given by the iterative formula
$$x _ { n + 1 } = 2 \sin ^ { - 1 } \left( \frac { 3 } { x _ { n } + 3 } \right)$$
converges, then it converges to the root of the equation in part (i).\\
(iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
\hfill \mbox{\textit{CAIE P3 2016 Q6 [9]}}