| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs ln(x) linear graph |
| Difficulty | Moderate -0.8 This is a straightforward application of linear regression with logarithmic data. Part (a) requires calculating gradient and intercept from two points (basic coordinate geometry), and part (b) requires converting back from logarithmic form using index laws. Both are routine AS-level techniques with no problem-solving or insight required—simpler than average A-level questions. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| \(\log _ { 10 } l\) | - 0.097 | 0.146 |
| \(\log _ { 10 } T\) | 0.254 | 0.376 |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient \(n = \frac{0.376 - 0.254}{0.146 - (-0.097)} = 0.50\) | M1, A1 | Allow www. Do not allow for reciprocal of gradient. Allow for 0.5 or 0.502...seen |
| So \(k = 10^{0.303} = 2.0\) (2sf) | M1, A1 | Finding the intercept and attempting to find \(k\). Must be 2 significant figures. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.254 = \log k - 0.097n\); \(0.376 = \log k + 0.146n\) | M1 | Setting up a pair of simultaneous equations |
| \(n = 0.50\) | A1 | Allow for 0.5 or 0.502...seen |
| So \(k = 10^{0.303} = 2.0\) (2sf) | M1, A1 | Finding the intercept and attempting to find \(k\). Must be 2 significant figures. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_{10} T = \log_{10} k + \log_{10} l^n = \log_{10} kl^n\); So \(T = kl^n = 2.0\, l^{0.50}\) | M1, A1 | Uses laws of logs for powers and products. FT their \(n\) and positive value for \(k\) |
## Question 11:
### Part (a):
Gradient $n = \frac{0.376 - 0.254}{0.146 - (-0.097)} = 0.50$ | M1, A1 | Allow www. Do not allow for reciprocal of gradient. Allow for 0.5 or 0.502...seen
So $k = 10^{0.303} = 2.0$ (2sf) | M1, A1 | Finding the intercept and attempting to find $k$. Must be 2 significant figures.
**Alternative method:**
$0.254 = \log k - 0.097n$; $0.376 = \log k + 0.146n$ | M1 | Setting up a pair of simultaneous equations
$n = 0.50$ | A1 | Allow for 0.5 or 0.502...seen
So $k = 10^{0.303} = 2.0$ (2sf) | M1, A1 | Finding the intercept and attempting to find $k$. Must be 2 significant figures.
**[4]**
### Part (b):
$\log_{10} T = \log_{10} k + \log_{10} l^n = \log_{10} kl^n$; So $T = kl^n = 2.0\, l^{0.50}$ | M1, A1 | Uses laws of logs for powers and products. FT their $n$ and positive value for $k$
**[2]**11 A student records the time a pendulum takes to swing for different lengths of pendulum.
The student decides to plot a graph of $\log _ { 10 } T$ against $\log _ { 10 } l$ where $T$ is the time in seconds that the pendulum takes to return to its start position and $l$ is the length in metres of the pendulum. They use a model for $\log _ { 10 } T$ in terms of $\log _ { 10 } l$ of the form $\log _ { 10 } T = \log _ { 10 } \mathrm { k } + \mathrm { n } \log _ { 10 } \mathrm { l }$.
The student records the following data points.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
$\log _ { 10 } l$ & - 0.097 & 0.146 \\
\hline
$\log _ { 10 } T$ & 0.254 & 0.376 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Determine the values of $k$ and $n$ that best model the data. Give your values correct to 2 significant figures.
\item Using these values of $k$ and $n$, write the student's model as an equation expressing $T$ in terms of $l$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q11 [6]}}