| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs non-x variable: find constants |
| Difficulty | Moderate -0.8 This is a straightforward logarithmic transformation question requiring students to take ln of both sides and rearrange to y = mx + c form. Part (i) involves simple algebraic manipulation (gradient = 2/3), and part (ii) requires substituting x=0 to find A = e^1.5 ≈ 4.48. Both parts are routine applications of standard techniques with no problem-solving insight needed, making this easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply \(3 \ln y = \ln A + 2x\) at any stage | B1 | — |
| State gradient is \(\frac{2}{3}\), or equivalent | B1 | [2] |
| (ii) Substitute \(x = 0\), \(\ln y = 0.5\) and solve for \(A\) | M1 | — |
| Obtain \(A = 4.48\) | A1 | [2] |
**(i)** State or imply $3 \ln y = \ln A + 2x$ at any stage | B1 | —
State gradient is $\frac{2}{3}$, or equivalent | B1 | [2]
**(ii)** Substitute $x = 0$, $\ln y = 0.5$ and solve for $A$ | M1 | —
Obtain $A = 4.48$ | A1 | [2]
---The variables $x$ and $y$ satisfy the equation $y^3 = Ae^{2x}$, where $A$ is a constant. The graph of $\ln y$ against $x$ is a straight line.
\begin{enumerate}[label=(\roman*)]
\item Find the gradient of this line. [2]
\item Given that the line intersects the axis of $\ln y$ at the point where $\ln y = 0.5$, find the value of $A$ correct to 2 decimal places. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2010 Q2 [4]}}