| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| 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 multi-part question on fixed point iteration requiring standard techniques: sketching graphs to show existence/uniqueness of a root, verifying bounds by substitution, algebraic rearrangement using basic trigonometric identities (cot x = 1/tan x), and applying an iterative formula with a calculator. All steps are routine for P3 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.06a 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 |
|---|---|---|
| Make recognisable sketch of a relevant graph, e.g. \(y = 2\cot x\) | B1 | |
| Sketch an appropriate second graph, e.g. \(y = 1 + e^x\) correctly and justify the given statement | B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Consider graph \(2\cot x - 1 - e^x\) at \(x = 0.5\) and \(x = 1\), or equivalent | M1 | |
| Complete the argument with appropriate calculations | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Show that the given equation is equivalent to \(x = \tan^{-1}\left(\frac{2}{1+e^x}\right)\), or vice versa. | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(0.61\) | A1 | |
| Show sufficient iterations to justify its accuracy to 2d.p., or show there is a sign change in the interval \((0.605, 0.615)\) | A1 | 3 |
**(i)**
| Make recognisable sketch of a relevant graph, e.g. $y = 2\cot x$ | B1 |
| Sketch an appropriate second graph, e.g. $y = 1 + e^x$ correctly and justify the given statement | B1 | 2 |
**(ii)**
| Consider graph $2\cot x - 1 - e^x$ at $x = 0.5$ and $x = 1$, or equivalent | M1 |
| Complete the argument with appropriate calculations | A1 | 2 |
**(iii)**
| Show that the given equation is equivalent to $x = \tan^{-1}\left(\frac{2}{1+e^x}\right)$, or vice versa. | B1 | 1 |
**(iv)**
| Use the iterative formula correctly at least once | M1 |
| Obtain final answer $0.61$ | A1 |
| Show sufficient iterations to justify its accuracy to 2d.p., or show there is a sign change in the interval $(0.605, 0.615)$ | A1 | 3 |\begin{enumerate}[label=(\roman*)]
\item By sketching a suitable pair of graphs, show that the equation
$$2\cot x = 1 + e^x,$$
where $x$ is in radians, has only one root in the interval $0 < x < \frac{1}{2}\pi$. [2]
\item Verify by calculation that this root lies between 0.5 and 1.0. [2]
\item Show that this root also satisfies the equation
$$x = \tan^{-1}\left(\frac{2}{1 + e^x}\right).$$ [1]
\item Use the iterative formula
$$x_{n+1} = \tan^{-1}\left(\frac{2}{1 + e^{x_n}}\right),$$
with initial value $x_1 = 0.7$, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2006 Q6 [8]}}