| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| 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 question requiring standard differentiation of ln and product rule, algebraic rearrangement to show a given result, interval verification by substitution, and routine fixed-point iteration. All techniques are standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt use of product rule to differentiate \(x\ln(4x+1)\) | \*M1 | |
| Obtain \(\ln(4x+1) + \frac{4x}{4x+1} - 3\) | A1 | |
| Equate first derivative to zero and attempt correct rearrangement to obtain the form \(4x+1 = \frac{ax+b}{\ln(4x+1)}\) | DM1 | OE |
| Confirm \(x = \frac{2x+0.75}{\ln(4x+1)} - 0.25\) | A1 | AG – necessary detail needed. Allow verification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider sign of \(x - \frac{2x+0.75}{\ln(4x+1)} + 0.25\) or equivalent for 1.8 and 1.9 | M1 | |
| Obtain \(-0.017...\) and \(0.035...\) or equivalents and justify conclusion | A1 | AG – necessary detail needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use iteration process correctly at least once | M1 | |
| Obtain final answer 1.83 | A1 | answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in the interval \([1.825, 1.835]\) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt use of product rule to differentiate $x\ln(4x+1)$ | \*M1 | |
| Obtain $\ln(4x+1) + \frac{4x}{4x+1} - 3$ | A1 | |
| Equate first derivative to zero and attempt correct rearrangement to obtain the form $4x+1 = \frac{ax+b}{\ln(4x+1)}$ | DM1 | OE |
| Confirm $x = \frac{2x+0.75}{\ln(4x+1)} - 0.25$ | A1 | AG – necessary detail needed. Allow verification |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider sign of $x - \frac{2x+0.75}{\ln(4x+1)} + 0.25$ or equivalent for 1.8 and 1.9 | M1 | |
| Obtain $-0.017...$ and $0.035...$ or equivalents and justify conclusion | A1 | AG – necessary detail needed |
---
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use iteration process correctly at least once | M1 | |
| Obtain final answer 1.83 | A1 | answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in the interval $[1.825, 1.835]$ | A1 | |
---5 The curve with equation $y = x \ln ( 4 x + 1 ) - 3 x$ has one stationary point $P$.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of $P$ satisfies the equation
$$x = \frac { 2 x + 0.75 } { \ln ( 4 x + 1 ) } - 0.25$$
\item Show by calculation that the $x$-coordinate of $P$ lies between 1.8 and 1.9.
\item Use an iterative formula, based on the equation in part (a), to find the $x$-coordinate of $P$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2022 Q5 [9]}}