| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive equation from integral condition |
| Difficulty | Standard +0.8 This question requires integration by parts to evaluate the integral, algebraic manipulation to derive the fixed point equation, and then iterative calculation. The integration by parts with exponential is standard A-level technique, but deriving the specific form and applying iteration adds moderate complexity beyond routine exercises. |
| Spec | 1.08i Integration by parts1.09a 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 |
| Commence integration and reach \(pxe^{-2x} + q\int e^{-2x}\,dx\) | *M1 | OE |
| Obtain \(-\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x}\,dx\) | A1 | OE |
| Complete integration and obtain \(-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\) | A1 | |
| Use limits correctly and equate to \(\frac{1}{8}\), having integrated twice | DM1 | \(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4} = \frac{1}{8}\) |
| Obtain \(a = \frac{1}{2}\ln(4a+2)\) correctly | A1 | AG |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Calculate the values of a relevant expression or pair of expressions at \(a=0.5\) and \(a=1\) | M1 | |
| Justify the given statement with correct calculated values | A1 | e.g. \(0.5 < 0.69\ldots\), \(1 > 0.89\ldots\); \(0.193 > 0\), \(-1.105 < 0\); \(0.066 < 0.125\), \(0.148 > 0.125\) if put limits in the integral. Condone if they use calculator for the definite integral. |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the iterative process \(a_{n+1} = \frac{1}{2}\ln(4a_n + 2)\) correctly at least once | M1 | |
| Obtain final answer \(0.84\) | A1 | |
| Show sufficient iterations to at least 4 d.p. to justify \(0.84\) to 2 d.p. or show that there is a sign change in \((0.835, 0.845)\) | A1 | e.g. \(0.75, 0.8047, 0.8261, 0.8343, 0.8373, 0.8385, 1, 0.8959, 0.8599, 0.8469, 0.8420, 0.8402\) |
| Total | 3 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Commence integration and reach $pxe^{-2x} + q\int e^{-2x}\,dx$ | *M1 | OE |
| Obtain $-\frac{1}{2}xe^{-2x} + \frac{1}{2}\int e^{-2x}\,dx$ | A1 | OE |
| Complete integration and obtain $-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}$ | A1 | |
| Use limits correctly and equate to $\frac{1}{8}$, having integrated twice | DM1 | $-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4} = \frac{1}{8}$ |
| Obtain $a = \frac{1}{2}\ln(4a+2)$ correctly | A1 | AG |
| **Total** | **5** | |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Calculate the values of a relevant expression or pair of expressions at $a=0.5$ and $a=1$ | M1 | |
| Justify the given statement with correct calculated values | A1 | e.g. $0.5 < 0.69\ldots$, $1 > 0.89\ldots$; $0.193 > 0$, $-1.105 < 0$; $0.066 < 0.125$, $0.148 > 0.125$ if put limits in the integral. Condone if they use calculator for the definite integral. |
| **Total** | **2** | |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the iterative process $a_{n+1} = \frac{1}{2}\ln(4a_n + 2)$ correctly at least once | M1 | |
| Obtain final answer $0.84$ | A1 | |
| Show sufficient iterations to at least 4 d.p. to justify $0.84$ to 2 d.p. or show that there is a sign change in $(0.835, 0.845)$ | A1 | e.g. $0.75, 0.8047, 0.8261, 0.8343, 0.8373, 0.8385, 1, 0.8959, 0.8599, 0.8469, 0.8420, 0.8402$ |
| **Total** | **3** | |
---9 The constant $a$ is such that $\int _ { 0 } ^ { a } x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = \frac { 1 } { 8 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = \frac { 1 } { 2 } \ln ( 4 a + 2 )$.
\item Verify by calculation that $a$ lies between 0.5 and 1 .
\item Use an iterative formula based on the equation in (a) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q9 [10]}}