| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.8 This is a standard logarithmic transformation question requiring students to linearize an exponential model and use graphical methods. Part (i) is routine algebraic manipulation of logarithms, parts (ii)-(iii) involve straightforward plotting and reading gradient/intercept from a graph, and part (iv) is simple substitution. The question is easier than average as it follows a well-practiced template with no conceptual surprises or problem-solving required. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines |
| Year | 2001 | 2002 | 2003 | 2004 | 2005 |
| \(t\) | 1 | 2 | 3 | 4 | 5 |
| \(P\) | 7900 | 8800 | 10000 | 11300 | 12800 |
| Answer | Marks | Guidance |
|---|---|---|
| i \(\log_{10} P = \log_{10} a + \log_{10} b^{Pt}\) | B1 | condone omission of base |
| \(\log_{10} 10^6 = bt\) | B1 | |
| intercept indicated as \(\log_{10}a\) | B1 | |
| ii \(3.9(0), 3.94, 4(00), 4.05, 4.11\) | T1 | to 3 sf or more; condone one error |
| plots ft | P1 | 1 mm ruled and reasonable |
| line of best fit ft | L1 | |
| iii (gradient \(=\)) \(0.04\) to \(0.06\) seen | M1 | |
| (intercept \(=\)) \(3.83\) to \(3.86\) seen | M1 | |
| \((a =) 6760\) to \(7245\) seen | A1 | |
| \(P = 7000 \times 10^{0.05t}\) oe | A1 | \(7000 \times 1.12^t\) left A2; SC \(P = 10^{0.05t + 3.85}\) left A2 |
| iv \(17 000\) to \(18 500\) | B2 | \(14 000\) to \(22 000\) B1 |
i $\log_{10} P = \log_{10} a + \log_{10} b^{Pt}$ | B1 | condone omission of base |
$\log_{10} 10^6 = bt$ | B1 | |
intercept indicated as $\log_{10}a$ | B1 | | $3$ |
ii $3.9(0), 3.94, 4(00), 4.05, 4.11$ | T1 | to 3 sf or more; condone one error |
plots ft | P1 | 1 mm ruled and reasonable |
line of best fit ft | L1 | | $3$ |
iii (gradient $=$) $0.04$ to $0.06$ seen | M1 | |
(intercept $=$) $3.83$ to $3.86$ seen | M1 | |
$(a =) 6760$ to $7245$ seen | A1 | |
$P = 7000 \times 10^{0.05t}$ oe | A1 | $7000 \times 1.12^t$ left A2; SC $P = 10^{0.05t + 3.85}$ left A2 | $4$ |
iv $17 000$ to $18 500$ | B2 | $14 000$ to $22 000$ B1 | $2$ | $12$ |Answer the whole of this question on the insert provided.
A colony of bats is increasing. The population, $P$, is modelled by $P = a \times 10^{bt}$, where $t$ is the time in years after 2000.
\begin{enumerate}[label=(\roman*)]
\item Show that, according to this model, the graph of $\log_{10} P$ against $t$ should be a straight line of gradient $b$. State, in terms of $a$, the intercept on the vertical axis. [3]
\item The table gives the data for the population from 2001 to 2005.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Year & 2001 & 2002 & 2003 & 2004 & 2005 \\
\hline
$t$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$P$ & 7900 & 8800 & 10000 & 11300 & 12800 \\
\hline
\end{tabular}
Complete the table of values on the insert, and plot $\log_{10} P$ against $t$. Draw a line of best fit for the data. [3]
\item Use your graph to find the equation for $P$ in terms of $t$. [4]
\item Predict the population in 2008 according to this model. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2006 Q12 [12]}}