| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Definite integral with logarithmic form |
| Difficulty | Moderate -0.8 Both parts are direct applications of standard integral formulas with minimal manipulation. Part (a) requires recognizing the logarithmic form f'(x)/f(x) and adjusting the constant, while part (b) is a straightforward exponential integral with linear substitution. These are textbook exercises testing recall of standard results rather than problem-solving ability. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(2\ln(2x-5)\) | B1 | |
| Apply limits correctly | M1 | For integral of form \(k\ln(2x-5)\) |
| Use one relevant logarithm property correctly | M1 | For integral of form \(k\ln(2x-5)\) |
| Apply second logarithm property correctly and obtain \(\ln 25\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate to obtain \(\frac{1}{2}e^{2x-5}\) | B1 | |
| Obtain final answer \(\frac{1}{2}e^{15} - \frac{1}{2}e^{3}\) | B1FT | or exact equivalent, FT on *their* \(ke^{2x-5}\) |
| Total: 2 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $2\ln(2x-5)$ | B1 | |
| Apply limits correctly | M1 | For integral of form $k\ln(2x-5)$ |
| Use one relevant logarithm property correctly | M1 | For integral of form $k\ln(2x-5)$ |
| Apply second logarithm property correctly and obtain $\ln 25$ | A1 | |
| **Total: 4** | | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate to obtain $\frac{1}{2}e^{2x-5}$ | B1 | |
| Obtain final answer $\frac{1}{2}e^{15} - \frac{1}{2}e^{3}$ | B1FT | or exact equivalent, FT on *their* $ke^{2x-5}$ |
| **Total: 2** | | |3
\begin{enumerate}[label=(\alph*)]
\item Find $\int _ { 4 } ^ { 10 } \frac { 4 } { 2 x - 5 } \mathrm {~d} x$, giving your answer in the form $\ln a$, where $a$ is an integer.
\item Find the exact value of $\int _ { 4 } ^ { 10 } \mathrm { e } ^ { 2 x - 5 } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q3 [6]}}