| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a straightforward multi-part calculus question requiring quotient rule differentiation, setting derivative to zero, algebraic manipulation to rearrange the stationary point equation, interval verification by substitution, and applying a given iterative formula. All steps are standard A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07q Product and quotient rules: differentiation1.09a 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use quotient rule or equivalent | M1 | |
| Obtain \(\dfrac{2x(1+e^{3x}) - 3x^2e^{3x}}{(1+e^{3x})^2}\) or equivalent | A1 | |
| Equate first derivative to zero and attempt rearrangement to \(x = \ldots\) | DM1 | |
| Obtain \(x = \dfrac{2}{3}(1 + e^{-3x})\) with sufficient detail and no errors seen | A1 | [4] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Consider sign of \(x - \dfrac{2}{3}(1+e^{-3x})\) at \(0.7\) and \(0.8\) or equivalent | M1 | |
| Obtain correct values (\(-0.05\) and \(0.07\) or equivalents) and conclude appropriately | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer \(0.739\) | A1 | |
| Show sufficient iterations to 5 decimal places to justify result or show a sign change in the interval \((0.7385, 0.7395)\) | A1 | [3] |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use quotient rule or equivalent | M1 | |
| Obtain $\dfrac{2x(1+e^{3x}) - 3x^2e^{3x}}{(1+e^{3x})^2}$ or equivalent | A1 | |
| Equate first derivative to zero and attempt rearrangement to $x = \ldots$ | DM1 | |
| Obtain $x = \dfrac{2}{3}(1 + e^{-3x})$ with sufficient detail and no errors seen | A1 | [4] AG |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Consider sign of $x - \dfrac{2}{3}(1+e^{-3x})$ at $0.7$ and $0.8$ or equivalent | M1 | |
| Obtain correct values ($-0.05$ and $0.07$ or equivalents) and conclude appropriately | A1 | [2] |
## Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the iterative formula correctly at least once | M1 | |
| Obtain final answer $0.739$ | A1 | |
| Show sufficient iterations to 5 decimal places to justify result or show a sign change in the interval $(0.7385, 0.7395)$ | A1 | [3] |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{72d50061-ead5-466a-96fc-2203438d1407-3_296_675_945_735}
The diagram shows part of the curve $y = \frac { x ^ { 2 } } { 1 + \mathrm { e } ^ { 3 x } }$ and its maximum point $M$. The $x$-coordinate of $M$ is denoted by $m$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and hence show that $m$ satisfies the equation $x = \frac { 2 } { 3 } \left( 1 + \mathrm { e } ^ { - 3 x } \right)$.\\
(ii) Show by calculation that $m$ lies between 0.7 and 0.8 .\\
(iii) Use an iterative formula based on the equation in part (i) to find $m$ correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
\hfill \mbox{\textit{CAIE P2 2014 Q6 [9]}}