| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| 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 quotient rule differentiation, algebraic manipulation to rearrange the stationary point condition, interval testing by substitution, and applying a given iterative formula. All techniques are standard A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate using quotient rule | *M1 | OE |
| Obtain \(\dfrac{(1+3x)\,2e^{2x} - (1+e^{2x})\,3}{(1+3x)^2}\) | A1 | OE |
| Equate first derivative to zero and arrange as far as \(x = \ldots\) | DM1 | |
| Confirm \(x = \frac{1}{6} + \frac{1}{2}e^{-2x}\) | A1 | Answer given – necessary detail needed. Must have exact terms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Consider sign of \(x - \frac{1}{6} - \frac{1}{2}e^{-2x}\) | M1 | OE |
| Obtain \(-0.06\ldots\,(-0.064959\ldots)\) and \(0.08\ldots\,(0.0800\ldots)\) or equivalents and justify conclusion | A1 | Answer given – necessary detail needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Consider \(f(x) = \frac{1}{6} - \frac{1}{2}e^{-2x}\) and obtain \(f(0.35) = 0.42\,(0.4149\ldots)\) and \(f(0.45) = 0.37\,(0.36995\ldots)\) | (M1) | |
| Conclude \(f(0.35) < 0.45\) and \(f(0.45) > 0.35\) so root lies in given interval | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Consider the sign of *their* \(\dfrac{dy}{dx}\) from part (a) | (M1) | |
| Obtain \(-0.19\ldots\,(-0.187\ldots)\) and \(0.21\ldots\,(0.2139\ldots)\) or equivalents and justify conclusion | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use iterative process correctly at least once | M1 | |
| Obtain final answer \(0.394\) | A1 | Answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval \([0.3935,\ 0.3945]\) | A1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate using quotient rule | *M1 | OE |
| Obtain $\dfrac{(1+3x)\,2e^{2x} - (1+e^{2x})\,3}{(1+3x)^2}$ | A1 | OE |
| Equate first derivative to zero and arrange as far as $x = \ldots$ | DM1 | |
| Confirm $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ | A1 | Answer given – necessary detail needed. Must have exact terms |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider sign of $x - \frac{1}{6} - \frac{1}{2}e^{-2x}$ | M1 | OE |
| Obtain $-0.06\ldots\,(-0.064959\ldots)$ and $0.08\ldots\,(0.0800\ldots)$ or equivalents and justify conclusion | A1 | Answer given – necessary detail needed |
**Alternative Method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider $f(x) = \frac{1}{6} - \frac{1}{2}e^{-2x}$ and obtain $f(0.35) = 0.42\,(0.4149\ldots)$ and $f(0.45) = 0.37\,(0.36995\ldots)$ | (M1) | |
| Conclude $f(0.35) < 0.45$ **and** $f(0.45) > 0.35$ so root lies in given interval | (A1) | |
**Alternative Method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Consider the sign of *their* $\dfrac{dy}{dx}$ from part (a) | (M1) | |
| Obtain $-0.19\ldots\,(-0.187\ldots)$ and $0.21\ldots\,(0.2139\ldots)$ or equivalents and justify conclusion | (A1) | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use iterative process correctly at least once | M1 | |
| Obtain final answer $0.394$ | A1 | Answer required to exactly 3 s.f. |
| Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval $[0.3935,\ 0.3945]$ | A1 | |
---5 A curve has equation $\mathrm { y } = \frac { 1 + \mathrm { e } ^ { 2 \mathrm { x } } } { 1 + 3 \mathrm { x } }$. The curve has exactly one stationary point $P$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { dy } } { \mathrm { dx } }$ and hence show that the $x$-coordinate of $P$ satisfies the equation $x = \frac { 1 } { 6 } + \frac { 1 } { 2 } \mathrm { e } ^ { - 2 x }$.
\item Show by calculation that the $x$-coordinate of $P$ lies between 0.35 and 0.45 .
\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.\\
\includegraphics[max width=\textwidth, alt={}, center]{971a1d8d-a82e-4a3a-b72d-3c147e4f30bb-10_451_647_258_699}
The diagram shows the curve with equation $\mathrm { y } = \sqrt { \sin 2 \mathrm { x } + \sin ^ { 2 } 2 \mathrm { x } }$ for $0 \leqslant x \leqslant \frac { 1 } { 6 } \pi$. The shaded region is bounded by the curve and the straight lines $x = \frac { 1 } { 6 } \pi$ and $y = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2024 Q5 [9]}}