| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from integral condition |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining basic integration of e^(-x), simple algebraic manipulation to show a given result, and mechanical application of an iterative formula. All steps are routine for P2 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) EITHER: Integrate \(1 - e^{-x}\) obtaining \(x + e^{-x}\) | M1 | |
| Obtain indefinite integral \(x - e^{-x}\) | A1 | |
| Substitute limits \(x = 0, x = p\) correctly | M1 | |
| Obtain answer \(p + e^{-p} - 1\), or equivalent | A1 | |
| OR: Integrate \(e^{-x}\) obtaining \(\pm e^{-x}\) | M1 | |
| Substitute limits \(x = 0, x = p\) correctly | M1 | |
| Obtain area below curve is \(1 - e^{-p}\) | A1 | |
| Obtain answer \(p + e^{-p} - 1\), or equivalent | A1 | [4 marks] |
| (ii) Show that \(p + e^{-p} - 1 = 1\) is equivalent to \(p = 2 - e^{-p}\) or vice versa | B1 | [1 mark] |
| (iii) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(1.84\) | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p. | A1 | [3 marks] |
**(i)** **EITHER:** Integrate $1 - e^{-x}$ obtaining $x + e^{-x}$ | M1 |
Obtain indefinite integral $x - e^{-x}$ | A1 |
Substitute limits $x = 0, x = p$ correctly | M1 |
Obtain answer $p + e^{-p} - 1$, or equivalent | A1 |
**OR:** Integrate $e^{-x}$ obtaining $\pm e^{-x}$ | M1 |
Substitute limits $x = 0, x = p$ correctly | M1 |
Obtain area below curve is $1 - e^{-p}$ | A1 |
Obtain answer $p + e^{-p} - 1$, or equivalent | A1 | [4 marks]
**(ii)** Show that $p + e^{-p} - 1 = 1$ is equivalent to $p = 2 - e^{-p}$ or vice versa | B1 | [1 mark]
**(iii)** Use the iterative formula correctly at least once | M1 |
Obtain final answer $1.84$ | A1 |
Show sufficient iterations to justify its accuracy to 2 d.p. | A1 | [3 marks]7\\
\includegraphics[max width=\textwidth, alt={}, center]{67a12825-d7ce-4853-ada4-b8d3009331b5-3_531_759_262_694}
The diagram shows the curve $y = \mathrm { e } ^ { - x }$. The shaded region $R$ is bounded by the curve and the lines $y = 1$ and $x = p$, where $p$ is a constant.\\
(i) Find the area of $R$ in terms of $p$.\\
(ii) Show that if the area of $R$ is equal to 1 then
$$p = 2 - \mathrm { e } ^ { - p }$$
(iii) Use the iterative formula
$$p _ { n + 1 } = 2 - \mathrm { e } ^ { - p _ { n } }$$
with initial value $p _ { 1 } = 2$, to calculate the value of $p$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\hfill \mbox{\textit{CAIE P2 2009 Q7 [8]}}