| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | March |
| Marks | 8 |
| 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 on stationary points and iteration. Part (a) requires differentiating xe^(2x) - 5x and rearranging (standard product rule and logarithms), part (b) is simple substitution, and part (c) is routine iteration with a given formula. All techniques are standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.09a 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 |
| Use correct product rule | M1 | |
| Obtain correct derivative in any form | A1 | e.g. \(\frac{dy}{dx} = e^{2x} + 2xe^{2x} - 5\) |
| Equate derivative to zero and obtain \(\alpha = \frac{1}{2}\ln\left(\frac{5}{1+2\alpha}\right)\) | A1 | Given answer – need to see \(e^{2x} = 5/(1+2x)\) or \(\ln e^{2x} = \ln(5/(1+2x))\) in working. Must be in terms of \(\alpha\) not \(x\). Allow \(\alpha\) to be used before equating to 0. |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Calculate the value of a relevant expression or values of a pair of expressions at \(x = 0.4\) and \(x = 0.5\) | M1 | Need to attempt BOTH values and have one correct. |
| Complete the argument correctly with correct calculated values | A1 | e.g. \(0.4 < 0.51[08]\) and \(0.5 > 0.458\) or \(0.46\) or \(0.45\); or \(-0.11[08] < 0\) and \(0.042 > 0\). If use original derivative \(-0.994\) \((0.4)\) and \(0.437\) \((0.5)\). |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the iterative process \(\alpha_{n+1} = \frac{1}{2}\ln\left(\frac{5}{1+2\alpha_n}\right)\) correctly at least twice anywhere in iteration process | M1 | Obtain one value and then substitute it into the formula to obtain a second value. |
| Obtain final answer \(0.47\) | A1 | |
| Show sufficient iterations to 4 d.p. to justify \(0.47\) to 2 d.p. or show there is a sign change in the interval \((0.465, 0.475)\) | A1 | \(0.4, 0.5108, 0.4528, 0.4823, 0.4670, 0.4749\), \(0.45, 0.4838, 0.4663, 0.4753, 0.4707, 0.4730\), \(0.5, 0.4581, 0.4795, 0.4685, 0.4742\). Allow self correction. |
| Total | 3 | SC B1 No working \(0.47\) |
## Question 7:
### Part 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct product rule | M1 | |
| Obtain correct derivative in any form | A1 | e.g. $\frac{dy}{dx} = e^{2x} + 2xe^{2x} - 5$ |
| Equate derivative to zero and obtain $\alpha = \frac{1}{2}\ln\left(\frac{5}{1+2\alpha}\right)$ | A1 | Given answer – need to see $e^{2x} = 5/(1+2x)$ or $\ln e^{2x} = \ln(5/(1+2x))$ in working. Must be in terms of $\alpha$ not $x$. Allow $\alpha$ to be used before equating to 0. |
| **Total** | **3** | |
### Part 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculate the value of a relevant expression or values of a pair of expressions at $x = 0.4$ and $x = 0.5$ | M1 | Need to attempt BOTH values and have one correct. |
| Complete the argument correctly with correct calculated values | A1 | e.g. $0.4 < 0.51[08]$ and $0.5 > 0.458$ or $0.46$ or $0.45$; or $-0.11[08] < 0$ and $0.042 > 0$. If use original derivative $-0.994$ $(0.4)$ and $0.437$ $(0.5)$. |
| **Total** | **2** | |
### Part 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative process $\alpha_{n+1} = \frac{1}{2}\ln\left(\frac{5}{1+2\alpha_n}\right)$ correctly at least twice anywhere in iteration process | M1 | Obtain one value and then substitute it into the formula to obtain a second value. |
| Obtain final answer $0.47$ | A1 | |
| Show sufficient iterations to 4 d.p. to justify $0.47$ to 2 d.p. or show there is a sign change in the interval $(0.465, 0.475)$ | A1 | $0.4, 0.5108, 0.4528, 0.4823, 0.4670, 0.4749$, $0.45, 0.4838, 0.4663, 0.4753, 0.4707, 0.4730$, $0.5, 0.4581, 0.4795, 0.4685, 0.4742$. Allow self correction. |
| **Total** | **3** | SC B1 No working $0.47$ |
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\includegraphics[max width=\textwidth, alt={}, center]{446573d3-73b1-482a-a3f6-1abddfdd90d0-10_620_517_260_774}
The diagram shows the curve $\mathrm { y } = \mathrm { xe } ^ { 2 \mathrm { x } } - 5 \mathrm { x }$ and its minimum point $M$, where $x = \alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ satisfies the equation $\alpha = \frac { 1 } { 2 } \ln \left( \frac { 5 } { 1 + 2 \alpha } \right)$.
\item Verify by calculation that $\alpha$ lies between 0.4 and 0.5.
\item Use an iterative formula based on the equation in part (a) to determine $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q7 [8]}}