| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard linearisation question requiring students to find the equation of a line from two points, then convert back to exponential form. The steps are routine: calculate gradient, write linear equation, use antilog to find exponential model. Part (c) requires simple substitution and comparison, while part (d) involves solving a logarithmic inequality. All techniques are standard AS-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines |
| \(t\) | 0 | 1 | 2 | 3 | 4 |
| \(V\) | 8500 | 6970 | 5720 | 4690 | 3840 |
| \(\log _ { 10 } V\) | 3.93 | 3.84 | 3.76 | 3.67 | 3.58 |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = \frac{3.58 - 3.93}{4 - 0}\) | M1 | \((\pm)\ 0.0875\) oe |
| \(\log_{10} V = -0.0875t + 3.93\) | M1 | Allow any value for \(-0.0875\); condone omission of base |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = 10^{-0.0875t + 3.93}\) | M1 | Or \(\log V = \log A + \log b^t\) |
| \(A = 10^{3.93} = 8500\) | A1 | \(8511.38\); allow \(8500\) taken from table (M1A1) |
| \(b = 10^{-0.0875} = 0.82\) | A1 | Accept \(V = 8500 \times 0.82^t\); \(0.81752\); sc B1 if both answers not to 2 sf |
| Answer | Marks | Guidance |
|---|---|---|
| \(3151.29\ (\approx 3150)\) so the model is a (very) good fit | E1 | Must see value from model OR find value of \(t\) (approx. 4.98) when model gives 3150; e.g. 3107.93 from more accurate values |
| Answer | Marks | Guidance |
|---|---|---|
| \(8500 \times 0.82^t = 500\) soi | M1 | Ft from (b) for M1M1 |
| take logs of both sides | M1 | e.g. \(\log 8500 + \log 0.82^t = \log 500\); can be implied by 15 or \(14.(\ldots)\); or \(\log 0.82^t = \log\!\left(\frac{500}{8500}\right)\) |
| \((t = 14.27\) so) after 15 years | A1 |
# Question 9:
## Part (a):
$m = \frac{3.58 - 3.93}{4 - 0}$ | M1 | $(\pm)\ 0.0875$ oe
$\log_{10} V = -0.0875t + 3.93$ | M1 | Allow any value for $-0.0875$; condone omission of base
| A1 |
## Part (b):
$V = 10^{-0.0875t + 3.93}$ | M1 | Or $\log V = \log A + \log b^t$
$A = 10^{3.93} = 8500$ | A1 | $8511.38$; allow $8500$ taken from table (M1A1)
$b = 10^{-0.0875} = 0.82$ | A1 | Accept $V = 8500 \times 0.82^t$; $0.81752$; sc B1 if both answers not to 2 sf
## Part (c):
$3151.29\ (\approx 3150)$ so the model is a (very) good fit | E1 | Must see value from model OR find value of $t$ (approx. 4.98) when model gives 3150; e.g. 3107.93 from more accurate values
## Part (d):
$8500 \times 0.82^t = 500$ soi | M1 | Ft from (b) for M1M1
take logs of both sides | M1 | e.g. $\log 8500 + \log 0.82^t = \log 500$; can be implied by 15 or $14.(\ldots)$; or $\log 0.82^t = \log\!\left(\frac{500}{8500}\right)$
$(t = 14.27$ so) after 15 years | A1 |
9 In 2012 Adam bought a second hand car for $\pounds 8500$. Each year Adam has his car valued. He believes that there is a non-linear relationship between $t$, the time in years since he bought the car, and $V$, the value of the car in pounds. Fig. 9.1 shows successive values of $V$ and $\log _ { 10 } V$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$t$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$V$ & 8500 & 6970 & 5720 & 4690 & 3840 \\
\hline
$\log _ { 10 } V$ & 3.93 & 3.84 & 3.76 & 3.67 & 3.58 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 9.1}
\end{center}
\end{table}
Adam uses a spreadsheet to plot the points ( $t , \log _ { 10 } V$ ) shown in Fig. 9.1, and then generates a line of best fit for these points. The line passes through the points $( 0,3.93 )$ and $( 4,3.58 )$. A copy of his graph is shown in Fig. 9.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{11e5167f-9f95-4494-9b66-b59fdce8b1ef-6_776_682_1886_246}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $\log _ { 10 } V$ in terms of $t$.
\item Find a model for $V$ in the form $V = A \times b ^ { t }$, where $A$ and $b$ are constants to be determined. Give the values of $A$ and $b$ correct to 2 significant figures.
In 2017 Adam's car was valued at $\pounds 3150$.
\item Determine whether the model is a good fit for this data.
A company called Webuyoldcars pays $\pounds 500$ for any second hand car. Adam decides that he will sell his car to this company when the annual valuation of his car is less than $\pounds 500$.
\item According to the model, after how many years will Adam sell his car to Webuyoldcars?
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2019 Q9 [10]}}