| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| 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 standard multi-part question covering routine A-level techniques: sketching graphs to show root existence, verifying roots by substitution, rearranging equations for iteration, and performing iterative calculations. The second part on trigonometric form and solving is also textbook standard. While it requires multiple steps and careful execution, no novel insight or problem-solving is needed—all techniques are direct applications of syllabus methods. |
| 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) Draw correct sketch of \(y = e^{2x}\) | B1 | |
| Draw correct sketch of \(y = 14 - x^2\) | B1 | |
| Indicate two real roots only from correct sketches | B1 | [3] |
| (ii) Consider sign of \(e^{2x} + x^2 - 14\) for \(1.2\) and \(1.3\) or equivalent | M1 | |
| Justify conclusion with correct calculations (\(f(1.2) = -1.54, f(1.3) = 1.15\)) | A1 | [2] |
| (iii) Confirm given answer \(x = \frac{1}{2}\ln(14 - x^2)\) | B1 | [1] |
| (iv) Use the iteration process correctly at least once | M1 | |
| Obtain final answer \(1.26\) | A1 | |
| Show sufficient iterations to \(4\) decimal places to justify answer or show a sign change in the interval \((1.255, 1.256)\) [1.2 → 1.2653 → 1.2588 → 1.2595; 1.25 → 1.2604 → 1.2593 → 1.2594; 1.3 → 1.2522 → 1.2598 → 1.2594] | A1 | [3] |
**(i)** Draw correct sketch of $y = e^{2x}$ | B1 |
Draw correct sketch of $y = 14 - x^2$ | B1 |
Indicate two real roots only from correct sketches | B1 | [3]
**(ii)** Consider sign of $e^{2x} + x^2 - 14$ for $1.2$ and $1.3$ or equivalent | M1 |
Justify conclusion with correct calculations ($f(1.2) = -1.54, f(1.3) = 1.15$) | A1 | [2]
**(iii)** Confirm given answer $x = \frac{1}{2}\ln(14 - x^2)$ | B1 | [1]
**(iv)** Use the iteration process correctly at least once | M1 |
Obtain final answer $1.26$ | A1 |
Show sufficient iterations to $4$ decimal places to justify answer or show a sign change in the interval $(1.255, 1.256)$ [1.2 → 1.2653 → 1.2588 → 1.2595; 1.25 → 1.2604 → 1.2593 → 1.2594; 1.3 → 1.2522 → 1.2598 → 1.2594] | A1 | [3]7 (i) By sketching a suitable pair of graphs, show that the equation
$$\mathrm { e } ^ { 2 x } = 14 - x ^ { 2 }$$
has exactly two real roots.\\
(ii) Show by calculation that the positive root lies between 1.2 and 1.3.\\
(iii) Show that this root also satisfies the equation
$$x = \frac { 1 } { 2 } \ln \left( 14 - x ^ { 2 } \right) .$$
(iv) Use an iteration process based on the equation in part (iii), with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places.\\
(i) Express $4 \sin \theta - 6 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. Give the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.\\
(ii) Solve the equation $4 \sin \theta - 6 \cos \theta = 3$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
(iii) Find the greatest and least possible values of $( 4 \sin \theta - 6 \cos \theta ) ^ { 2 } + 8$ as $\theta$ varies.
\hfill \mbox{\textit{CAIE P2 2011 Q7 [9]}}