| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2003 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Sketch graphs to show root existence |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question with standard components: sketching graphs to show root existence (routine), verifying the interval by substitution (trivial calculation), and applying a given iterative formula (mechanical process requiring no derivation or convergence analysis). All steps are procedural with no novel insight required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| Make recognisable sketch of \(y = 2^x\) or \(y = x^2\), for \(x < 0\) | B1 |
| Sketch the other graph correctly | B1 |
| Answer | Marks |
|---|---|
| Consider sign of \(2^x - x^2\) at \(x = -1\) and \(x = -0.5\), or equivalent | M1 |
| Complete the argument correctly with appropriate calculations | A1 |
| Answer | Marks |
|---|---|
| Use the iterative form correctly | M1 |
| Obtain final answer \(-0.77\) | A1 |
| Show sufficient iterations to justify its accuracy to 2 s.f., or show there is a sign change in the interval \((-0.775, -0.765)\) | A1 |
### (i)
Make recognisable sketch of $y = 2^x$ or $y = x^2$, for $x < 0$ | B1 |
Sketch the other graph correctly | B1 |
**Total: [2]**
### (ii)
Consider sign of $2^x - x^2$ at $x = -1$ and $x = -0.5$, or equivalent | M1 |
Complete the argument correctly with appropriate calculations | A1 |
**Total: [2]**
### (iii)
Use the iterative form correctly | M1 |
Obtain final answer $-0.77$ | A1 |
Show sufficient iterations to justify its accuracy to 2 s.f., or show there is a sign change in the interval $(-0.775, -0.765)$ | A1 |
**Total: [3]**\begin{enumerate}[label=(\roman*)]
\item By sketching a suitable pair of graphs, for $x < 0$, show that exactly one root of the equation $x^2 = 2^x$ is negative. [2]
\item Verify by calculation that this root lies between $-1.0$ and $-0.5$. [2]
\item Use the iterative formula
$$x_{n+1} = -\sqrt{(2^{x_n})}$$
to determine this root correct to 2 significant figures, showing the result of each iteration. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2003 Q5 [7]}}