| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Exponential relation to line equation |
| Difficulty | Moderate -0.3 This is a straightforward logarithmic manipulation question requiring students to take ln of both sides, rearrange to y = mx + c form, then use two points to find constants. The algebraic steps are routine for P2 level with no conceptual surprises, making it slightly easier than average but still requiring competent execution of standard techniques. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(2y\ln a = 3x + k\) and conclude that gradient is \(\dfrac{3}{2\ln a}\) | B1 | AG – necessary detail needed |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate \(\dfrac{3}{2\ln a}\) to gradient of line | M1 | |
| Obtain \(\dfrac{3}{2\ln a} = \dfrac{2.85}{2.9}\) or equivalent and hence obtain \(a = 4.6\) or \(a = e^{\frac{29}{19}}\) | A1 | Allow greater accuracy |
| Substitute appropriate values to find value of \(k\) | M1 | |
| Obtain \(k = 1.7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(0.95(2\ln a) = 3(0.4) + k\) or \(a^{1.9} = e^{1.2+k}\) | M1 | OE |
| Obtain \(3.80(2\ln a) = 3(3.3) + k\) or \(a^{7.6} = e^{9.9+k}\) | M1 | OE |
| Obtain \(a = 4.6\) or \(a = e^{\frac{29}{19}}\) | A1 | Allow greater accuracy |
| Obtain \(k = 1.7\) | A1 | |
| 4 |
## Question 1:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $2y\ln a = 3x + k$ and conclude that gradient is $\dfrac{3}{2\ln a}$ | **B1** | AG – necessary detail needed |
| | **1** | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate $\dfrac{3}{2\ln a}$ to gradient of line | **M1** | |
| Obtain $\dfrac{3}{2\ln a} = \dfrac{2.85}{2.9}$ or equivalent and hence obtain $a = 4.6$ or $a = e^{\frac{29}{19}}$ | **A1** | Allow greater accuracy |
| Substitute appropriate values to find value of $k$ | **M1** | |
| Obtain $k = 1.7$ | **A1** | |
**Alternative Method for Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $0.95(2\ln a) = 3(0.4) + k$ or $a^{1.9} = e^{1.2+k}$ | **M1** | OE |
| Obtain $3.80(2\ln a) = 3(3.3) + k$ or $a^{7.6} = e^{9.9+k}$ | **M1** | OE |
| Obtain $a = 4.6$ or $a = e^{\frac{29}{19}}$ | **A1** | Allow greater accuracy |
| Obtain $k = 1.7$ | **A1** | |
| | **4** | |1 The variables $x$ and $y$ satisfy the equation $a ^ { 2 y } = \mathrm { e } ^ { 3 x + k }$, where $a$ and $k$ are constants.\\
The graph of $y$ against $x$ is a straight line.
\begin{enumerate}[label=(\alph*)]
\item Use logarithms to show that the gradient of the straight line is $\frac { 3 } { 2 \ln a }$.
\item Given that the straight line passes through the points $( 0.4,0.95 )$ and $( 3.3,3.80 )$, find the values of $a$ and $k$.\\
\includegraphics[max width=\textwidth, alt={}, center]{dcc483e9-630e-4f02-ad8c-4a27c0720fc6-03_2723_33_99_21}
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2024 Q1 [5]}}