| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2020 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find coordinate from gradient condition |
| Difficulty | Standard +0.3 This is a straightforward application of differentiation (product rule) followed by standard fixed-point iteration. Part (a) requires routine algebraic manipulation to rearrange dy/dx=15, part (b) is simple substitution to verify a bracket, and part (c) is mechanical iteration with no convergence issues. The question is slightly above average due to the product rule with exponential and the multi-step nature, but requires no novel insight or problem-solving beyond textbook methods. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate using the product rule | \*M1 | |
| Obtain \(3x^2e^{0.2x} + 0.2x^3e^{0.2x}\) | A1 | OE |
| Equate first derivative to 15 and rearrange to \(x = \ldots\) | DM1 | |
| Confirm \(x = \sqrt{\dfrac{75e^{-0.2x}}{15+x}}\) | A1 | AG – necessary detail needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Consider sign of \(x - \sqrt{\dfrac{75e^{-0.2x}}{15+x}}\) or equivalent for 1.7 and 1.8 | M1 | |
| Obtain \(-0.08\ldots\) and \(0.03\ldots\) or equivalents and justify conclusion | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use iterative process correctly at least once | M1 | Answer required to exactly 4 sf |
| Obtain final answer 1.771 | A1 | |
| Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval \([1.7705,\ 1.7715]\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate using the product rule | \*M1 | |
| Obtain $3x^2e^{0.2x} + 0.2x^3e^{0.2x}$ | A1 | OE |
| Equate first derivative to 15 and rearrange to $x = \ldots$ | DM1 | |
| Confirm $x = \sqrt{\dfrac{75e^{-0.2x}}{15+x}}$ | A1 | AG – necessary detail needed |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider sign of $x - \sqrt{\dfrac{75e^{-0.2x}}{15+x}}$ or equivalent for 1.7 and 1.8 | M1 | |
| Obtain $-0.08\ldots$ and $0.03\ldots$ or equivalents and justify conclusion | A1 | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use iterative process correctly at least once | M1 | Answer required to exactly 4 sf |
| Obtain final answer 1.771 | A1 | |
| Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval $[1.7705,\ 1.7715]$ | A1 | |
---6 A curve has equation $y = x ^ { 3 } \mathrm { e } ^ { 0.2 x }$ where $x \geqslant 0$. At the point $P$ on the curve, the gradient of the curve is 15 .
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of $P$ satisfies the equation $x = \sqrt { \frac { 75 \mathrm { e } ^ { - 0.2 x } } { 15 + x } }$.
\item Use the equation in part (a) to show by calculation that the $x$-coordinate of $P$ lies between 1.7 and 1.8.
\item Use an iterative formula, based on the equation in part (a), to find the $x$-coordinate of $P$ correct to 4 significant figures. Give the result of each iteration to 6 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2020 Q6 [9]}}