| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| 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 iteration question with routine steps: sketching graphs to show one root, verifying the root location by substitution, showing convergence algebraically (substituting the fixed point), and performing iterations. The trigonometric parts are also standard bookwork derivatives and identities. All techniques are direct applications of syllabus content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.06a Exponential function: a^x and e^x graphs and properties1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Make a recognisable sketch of a relevant graph, e.g. \(y = 2 - x\) | B1 | |
| Sketch an appropriate second graph, e.g. \(y = e^{2x}\), and justify the given statement | B1 | [2] |
| (ii) Consider sign of \(e^{2x} - (2 - x)\) at \(x = 0\) and \(x = 0.5\), or equivalent | M1 | |
| Complete the argument correctly with correct calculations | A1 | [2] |
| (iii) Show that \(e^{2x} = 2 - x\) is equivalent to \(x = \frac{1}{2}\ln(2 - x)\), or vice versa | B1 | [1] |
| (iv) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(0.27\) | A1 | |
| Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval \((0.265, 0.275)\) | A1 | [3] |
**(i)** Make a recognisable sketch of a relevant graph, e.g. $y = 2 - x$ | B1 |
Sketch an appropriate second graph, e.g. $y = e^{2x}$, and justify the given statement | B1 | [2]
**(ii)** Consider sign of $e^{2x} - (2 - x)$ at $x = 0$ and $x = 0.5$, or equivalent | M1 |
Complete the argument correctly with correct calculations | A1 | [2]
**(iii)** Show that $e^{2x} = 2 - x$ is equivalent to $x = \frac{1}{2}\ln(2 - x)$, or vice versa | B1 | [1]
**(iv)** Use the iterative formula correctly at least once | M1 |
Obtain final answer $0.27$ | A1 |
Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval $(0.265, 0.275)$ | A1 | [3]7 (i) By sketching a suitable pair of graphs, show that the equation
$$\mathrm { e } ^ { 2 x } = 2 - x$$
has only one root.\\
(ii) Verify by calculation that this root lies between $x = 0$ and $x = 0.5$.\\
(iii) Show that, if a sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 2 - x _ { n } \right)$$
converges, then it converges to the root of the equation in part (i).\\
(iv) Use this iterative formula, with initial value $x _ { 1 } = 0.25$, to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.\\
(i) By differentiating $\frac { \cos x } { \sin x }$, show that if $y = \cot x$ then $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x$.\\
(ii) By expressing $\cot ^ { 2 } x$ in terms of $\operatorname { cosec } ^ { 2 } x$ and using the result of part (i), show that
$$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \cot ^ { 2 } x \mathrm {~d} x = 1 - \frac { 1 } { 4 } \pi$$
(iii) Express $\cos 2 x$ in terms of $\sin ^ { 2 } x$ and hence show that $\frac { 1 } { 1 - \cos 2 x }$ can be expressed as $\frac { 1 } { 2 } \operatorname { cosec } ^ { 2 } x$. Hence, using the result of part (i), find
$$\int \frac { 1 } { 1 - \cos 2 x } \mathrm {~d} x$$
\hfill \mbox{\textit{CAIE P2 2010 Q7 [8]}}