| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Inverse function with exponentials |
| Difficulty | Easy -1.2 This is a straightforward two-part question requiring basic manipulation of exponential equations using logarithms. Part (i) involves routine rearrangement (divide by A, take ln of both sides, divide by 0.02), and part (ii) is direct substitution into the formula. Both parts are standard textbook exercises with no problem-solving or insight required, making this easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ln\frac{y}{A} = \ln e^{0.02t}\) or \(\ln y = \ln A + \ln e^{0.02t}\) | M1 | First stage of taking logs correct |
| \(t = 50\ln\frac{y}{A}\) oe | A1 | \(t=(\ln y - \ln A)/0.02\); \(t=(\log y - \log A)/0.02\log e\) |
| \(t = 21\) or \(20.6 - 20.625\) | B1 | awrt 20.6 |
## Question 3:
$\ln\frac{y}{A} = \ln e^{0.02t}$ or $\ln y = \ln A + \ln e^{0.02t}$ | **M1** | First stage of taking logs correct
$t = 50\ln\frac{y}{A}$ oe | **A1** | $t=(\ln y - \ln A)/0.02$; $t=(\log y - \log A)/0.02\log e$
$t = 21$ or $20.6 - 20.625$ | **B1** | awrt 20.6
---3 You are given that $y = A e ^ { 0.02 t }$.
\begin{itemize}
\item Make $t$ the subject of the formula.
\item Find the value of $t$ when $y = 10 ^ { 8 }$ and $A = 6.62 \times 10 ^ { 7 }$.
\end{itemize}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2022 Q3 [3]}}