| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Logarithmic equation solving |
| Difficulty | Moderate -0.8 This is a straightforward application question requiring substitution into a linear equation in ln X, solving simultaneous equations for two unknowns, then using the model for direct and inverse calculations. All steps are routine with no conceptual challenges beyond basic logarithm manipulation—easier than a typical A-level question which would require more problem-solving or proof elements. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks |
|---|---|
| \(B = 31\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(71 = A\ln 12 + 31\) | M1 | |
| \(A = 16\) | A1 | CAO. May see \(16.097\ldots\) A0 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 79 or 78.9 or better | B1 | from \(16\ln 20 + 31\) and \(78.9317\). Condone \(79.2\ldots\) from unrounded value of \(A\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(120 = 16\ln X + 31\) | M1 | FT their \(A\) and \(B\); allow sign error |
| \(\exp(\text{their } \frac{120-31}{16})\) | M1 | Allow slip in rearrangement |
| awrt \(X = 260\) | A1 | CAO. \(A = 16.097\ldots\) leads to 250, A0 |
| [3] |
**Question 5(a):**
$B = 31$ | B1 |
**[1]**
## Question 5:
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $71 = A\ln 12 + 31$ | M1 | |
| $A = 16$ | A1 | CAO. May see $16.097\ldots$ A0 |
| **[2]** | | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| 79 or 78.9 or better | B1 | from $16\ln 20 + 31$ and $78.9317$. Condone $79.2\ldots$ from unrounded value of $A$ |
| **[1]** | | |
### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $120 = 16\ln X + 31$ | M1 | FT their $A$ and $B$; allow sign error |
| $\exp(\text{their } \frac{120-31}{16})$ | M1 | Allow slip in rearrangement |
| awrt $X = 260$ | A1 | CAO. $A = 16.097\ldots$ leads to 250, A0 |
| **[3]** | | |
---5 In 2019 scientists developed a model for comparing the ages of humans and dogs.\\
According to the model,\\
$Y = A \ln X + B$\\
where $X =$ dog age in years and $Y =$ human age in years.\\
For the model, it is known that when $X = 1 , Y = 31$ and when $X = 12 , Y = 71$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $B$.
\item Determine the value of $A$, correct to the nearest whole number.
Use the model, with the exact value of $B$ and the value of $A$ correct to the nearest whole number, to answer parts (c) and (d).
\item Find the human age corresponding to a dog age of 20 years.
\item Determine the dog age corresponding to a human age of 120 years.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2021 Q5 [7]}}