| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard A-level techniques: finding a minimum using differentiation (product rule), algebraic rearrangement to show an iterative form, and applying a given iteration formula. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use product rule | M1* | |
| Obtain derivative in any correct form | A1 | |
| Equate derivative to zero and solve for \(x\) | M1(dep*) | |
| Obtain answer \(x = -\frac{1}{2}\) correctly | A1 | |
| Obtain \(y = -\frac{1}{2e}\) or exact equivalent | A1 | [5] |
| (ii) Show that \(20 = xe^{2x}\) is equivalent to \(x = \frac{1}{2}\ln(20/x)\) or vice versa | B1 | [1] |
| (iii) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer 1.35 | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p. | A1 | [3] |
**(i)** Use product rule | M1* |
Obtain derivative in any correct form | A1 |
Equate derivative to zero and solve for $x$ | M1(dep*) |
Obtain answer $x = -\frac{1}{2}$ correctly | A1 |
Obtain $y = -\frac{1}{2e}$ or exact equivalent | A1 | [5]
**(ii)** Show that $20 = xe^{2x}$ is equivalent to $x = \frac{1}{2}\ln(20/x)$ or vice versa | B1 | [1]
**(iii)** Use the iterative formula correctly at least once | M1 |
Obtain final answer 1.35 | A1 |
Show sufficient iterations to justify its accuracy to 2 d.p. | A1 | [3]7\\
\includegraphics[max width=\textwidth, alt={}, center]{b9556031-871d-4dd3-9523-e3438a41339f-3_655_685_262_730}
The diagram shows the curve $y = x \mathrm { e } ^ { 2 x }$ and its minimum point $M$.\\
(i) Find the exact coordinates of $M$.\\
(ii) Show that the curve intersects the line $y = 20$ at the point whose $x$-coordinate is the root of the equation
$$x = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x } \right)$$
(iii) Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x _ { n } } \right)$$
with initial value $x _ { 1 } = 1.3$, to calculate the root correct to 2 decimal places, giving the result of each iteration to 4 decimal places.
\hfill \mbox{\textit{CAIE P2 2009 Q7 [9]}}