| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Iterative formula from integral equation |
| Difficulty | Standard +0.8 This question requires integration by parts to derive the equation, then uses iterative methods to solve a transcendental equation. While integration by parts of x²ln(x) is standard P3 content, the algebraic manipulation to reach the given form and the iterative solution method elevate this above average difficulty. The multi-stage nature (integrate, rearrange, verify bounds, iterate) and the need to handle logarithmic equations make this moderately challenging but still within expected P3 scope. |
| 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 | Mark | Guidance |
| Commence integration and reach \(ax^3 \ln x + b\int x^3 \cdot \frac{1}{x}\,dx\) | \*M1 | OE. Allow omission of \(dx\) |
| Obtain \(\frac{1}{3}x^3 \ln x - \frac{1}{3}\int x^3 \cdot \frac{1}{x}\,dx\) | A1 | OE. Allow omission of \(dx\) |
| Complete integration and obtain \(\frac{1}{3}x^3 \ln x - \frac{1}{9}x^3\) | A1 | Allow \(-\frac{1}{3}\left(\frac{1}{3}x^3\right)\) |
| Use limits correctly and equate to 4, having integrated twice | DM1 | \(\frac{1}{3}a^3 \ln a - \frac{1}{9}a^3 - (0 - \frac{1}{9}) = 4\); allow one sign error OR one numerical error, but 0 may be absent or expressed as \(\frac{a^3}{3}\ln 1\). Allow \(-\frac{1}{3}\left(\frac{1}{3}ax^3\right)\) and \(-\frac{1}{3}\left(\frac{1}{3}\right)\) |
| Obtain given result correctly | A1 | \(a = \left(\dfrac{35}{3\ln a - 1}\right)^{\frac{1}{3}}\) AG. After substitution, any errors even if corrected A0. Need to see at least one line of working between substitution and the given answer. |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Calculate the values of a relevant expression or pair of expressions at \(a = 2.4\) and \(a = 2.8\). All values must be correct for M1 (numerical question) | M1 | |
| Justify the given statement with correct calculated values | A1 | \(2.4 < 2.7(8)\) and \(2.8 > 2.5(6)\) sign change here insufficient. OR \(-0.3(8)\) and \(0.2(4)\): \(< 0,\, > 0\) or change of sign |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the iterative process \(a_{n+1} = \left(\dfrac{35}{3\ln a_n - 1}\right)^{\frac{1}{3}}\) correctly at least twice | M1 | |
| Obtain final answer \(a = 2.64\) | A1 | Must be 2 dp |
| Show sufficient iterations to 4 dp to justify 2.64 to 2 dp, or show there is a sign change in \((2.635,\, 2.645)\) | A1 | \(2.635\): \((35/(3\ln a -1))^{1/3} - a = 0.0029(4) > 0\); \(\quad 2.645\): \((35/(3\ln a -1))^{1/3} - a = -0.012 < 0\) |
| Total | 3 |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Commence integration and reach $ax^3 \ln x + b\int x^3 \cdot \frac{1}{x}\,dx$ | \*M1 | OE. Allow omission of $dx$ |
| Obtain $\frac{1}{3}x^3 \ln x - \frac{1}{3}\int x^3 \cdot \frac{1}{x}\,dx$ | A1 | OE. Allow omission of $dx$ |
| Complete integration and obtain $\frac{1}{3}x^3 \ln x - \frac{1}{9}x^3$ | A1 | Allow $-\frac{1}{3}\left(\frac{1}{3}x^3\right)$ |
| Use limits correctly and equate to 4, having integrated twice | DM1 | $\frac{1}{3}a^3 \ln a - \frac{1}{9}a^3 - (0 - \frac{1}{9}) = 4$; allow one sign error OR one numerical error, but 0 may be absent or expressed as $\frac{a^3}{3}\ln 1$. Allow $-\frac{1}{3}\left(\frac{1}{3}ax^3\right)$ and $-\frac{1}{3}\left(\frac{1}{3}\right)$ |
| Obtain given result correctly | A1 | $a = \left(\dfrac{35}{3\ln a - 1}\right)^{\frac{1}{3}}$ AG. After substitution, any errors even if corrected A0. Need to see at least one line of working between substitution and the given answer. |
| **Total** | **5** | |
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Calculate the values of a relevant expression or pair of expressions at $a = 2.4$ and $a = 2.8$. All values must be correct for M1 (numerical question) | M1 | |
| Justify the given statement with correct calculated values | A1 | $2.4 < 2.7(8)$ and $2.8 > 2.5(6)$ sign change here insufficient. OR $-0.3(8)$ and $0.2(4)$: $< 0,\, > 0$ or change of sign |
| **Total** | **2** | |
---
## Question 10(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the iterative process $a_{n+1} = \left(\dfrac{35}{3\ln a_n - 1}\right)^{\frac{1}{3}}$ correctly at least twice | M1 | |
| Obtain final answer $a = 2.64$ | A1 | Must be 2 dp |
| Show sufficient iterations to 4 dp to justify 2.64 to 2 dp, or show there is a sign change in $(2.635,\, 2.645)$ | A1 | $2.635$: $(35/(3\ln a -1))^{1/3} - a = 0.0029(4) > 0$; $\quad 2.645$: $(35/(3\ln a -1))^{1/3} - a = -0.012 < 0$ |
| **Total** | **3** | |10 The constant $a$ is such that $\int _ { 1 } ^ { a } x ^ { 2 } \ln x \mathrm {~d} x = 4$.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = \left( \frac { 35 } { 3 \ln a - 1 } \right) ^ { \frac { 1 } { 3 } }$.
\item Verify by calculation that $a$ lies between 2.4 and 2.8.
\item Use an iterative formula based on the equation in part (a) to determine $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q10 [10]}}