| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard fixed-point iteration question requiring routine techniques: sketching graphs to show uniqueness, substituting values to verify a root's location, algebraic rearrangement (which is trivial here), and applying an iterative formula. All steps are procedural with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09c 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. \(y = 3e^x\) or \(y = 8 - 2x\) | B1 | |
| Sketch a second relevant graph and justify the given statement | B1 | [2] |
| (ii) Consider sign of \(3e^x - 8 + 2x\) at \(x = 0.7\) and \(x = 0.8\), or equivalent | M1 | |
| Complete the argument correctly with appropriate calculations (\(f(0.7) = -0.559\), \(f(0.8) = 0.277\) or equivalent) | A1 | [2] |
| (iii) Show that given equation is equivalent to \(x = \ln\left(\frac{8 - 2x}{3}\right)\), or vice versa | B1 | [1] |
| (iv) Use the iterative formula correctly at least once | M1 | |
| Obtain final answer 0.768 | A1 | |
| Show sufficient iterations to justify its accuracy to 3 d.p. | [3] | |
| Or show there is a sign change in the interval \((0.6675, 0.6685)\) | B1 | [3] |
**(i)** Make a recognisable sketch of a relevant graph, e.g. $y = 3e^x$ or $y = 8 - 2x$ | B1 |
Sketch a second relevant graph and justify the given statement | B1 | [2]
**(ii)** Consider sign of $3e^x - 8 + 2x$ at $x = 0.7$ and $x = 0.8$, or equivalent | M1 |
Complete the argument correctly with appropriate calculations ($f(0.7) = -0.559$, $f(0.8) = 0.277$ or equivalent) | A1 | [2]
**(iii)** Show that given equation is equivalent to $x = \ln\left(\frac{8 - 2x}{3}\right)$, or vice versa | B1 | [1]
**(iv)** Use the iterative formula correctly at least once | M1 |
Obtain final answer 0.768 | A1 |
Show sufficient iterations to justify its accuracy to 3 d.p. | | [3]
Or show there is a sign change in the interval $(0.6675, 0.6685)$ | B1 | [3]6 (i) By sketching a suitable pair of graphs, show that the equation
$$3 \mathrm { e } ^ { x } = 8 - 2 x$$
has only one root.\\
(ii) Verify by calculation that this root lies between $x = 0.7$ and $x = 0.8$.\\
(iii) Show that this root also satisfies the equation
$$x = \ln \left( \frac { 8 - 2 x } { 3 } \right)$$
(iv) Use the iterative formula $x _ { n + 1 } = \ln \left( \frac { 8 - 2 x _ { n } } { 3 } \right)$ to determine this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
\hfill \mbox{\textit{CAIE P2 2013 Q6 [8]}}