| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Standard +0.3 This is a straightforward differentiation question using the quotient rule on an exponential function, followed by finding a stationary point by setting the derivative to zero. Both parts are standard textbook exercises requiring routine application of techniques with no novel insight, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | Differentiate using quotient rule | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ex +2)2 | A1 | OE |
| Substitute x=0 in first derivative and attempt evaluation | M1 |
| Answer | Marks |
|---|---|
| 9 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | Equate first derivative to zero and attempt factorisation or | |
| equivalent | M1 | |
| Solve a three-term quadratic equation in ex to obtain ex =... | M1 | (2ex +9)(2ex −1)=0. |
| Answer | Marks |
|---|---|
| 2 | A1 |
| Obtain y-coordinate 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/22 Cambridge International AS Level – Mark Scheme February/March 2023
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2023 Page 5 of 11
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | Differentiate using quotient rule | M1 | OE
(ex +2)8e2x −(4e2x +9)ex
Obtain
(ex +2)2 | A1 | OE
Substitute x=0 in first derivative and attempt evaluation | M1
Obtain 11
9 | A1
4
--- 6(b) ---
6(b) | Equate first derivative to zero and attempt factorisation or
equivalent | M1
Solve a three-term quadratic equation in ex to obtain ex =... | M1 | (2ex +9)(2ex −1)=0.
Obtain x-coordinate ln1 or −ln2
2 | A1
Obtain y-coordinate 4 | A1
4
Question | Answer | Marks | Guidance\includegraphics{figure_6}
The diagram shows the curve with equation $y = \frac{4e^{2x} + 9}{e^x + 2}$. The curve has a minimum point $M$ and crosses the $y$-axis at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of the gradient of the curve at $P$. [4]
\item Find the exact coordinates of $M$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q6 [8]}}