| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Find stationary point coordinate |
| Difficulty | Standard +0.8 This question combines differentiation using quotient rule with exponentials, algebraic manipulation to derive an iterative formula, interval verification, and iterative solution. While each component is standard A-level material, the multi-step nature, the non-trivial algebraic rearrangement in part (i), and the requirement to apply Newton-Raphson/fixed-point iteration systematically elevate this above average difficulty. It's more demanding than routine calculus but doesn't require exceptional insight. |
| 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 |
|---|---|---|
| (i) Use quotient rule or equivalent | M1 | |
| Obtain \(\frac{2x(1 + e^{3x}) - 3x^2 e^{3x}}{(1 + e^{3x})^2}\) or equivalent | A1 | |
| Equate first derivative to zero and attempt rearrangement to \(x = \ldots\) | DM1 | |
| Obtain \(x = \frac{2}{3}\left(1 + e^{-3x}\right)\) with sufficient detail and no errors seen (AG) | A1 | [4] |
| (ii) Consider sign of \(x - \frac{2}{3}\left(1 + e^{-3x}\right)\) at 0.7 and 0.8 or equivalent | M1 | |
| Obtain correct values (–0.05 and 0.07 or equivalents) and conclude appropriately | A1 | [2] |
| (iii) 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] |
**(i)** Use quotient rule or equivalent | M1 |
Obtain $\frac{2x(1 + e^{3x}) - 3x^2 e^{3x}}{(1 + e^{3x})^2}$ or equivalent | A1 |
Equate first derivative to zero and attempt rearrangement to $x = \ldots$ | DM1 |
Obtain $x = \frac{2}{3}\left(1 + e^{-3x}\right)$ with sufficient detail and no errors seen (AG) | A1 | [4]
**(ii)** Consider sign of $x - \frac{2}{3}\left(1 + e^{-3x}\right)$ at 0.7 and 0.8 or equivalent | M1 |
Obtain correct values (–0.05 and 0.07 or equivalents) and conclude appropriately | A1 | [2]
**(iii)** 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]{293e1e27-77e9-4b19-a152-96d71b75346e-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]}}