OCR MEI AS Paper 2 2019 June — Question 9 10 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeLinear transformation to find constants
DifficultyModerate -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.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines

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]
\(t\)01234
\(V\)85006970572046903840
\(\log _ { 10 } V\)3.933.843.763.673.58
\captionsetup{labelformat=empty} \caption{Fig. 9.1}
\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]
\includegraphics[alt={},max width=\textwidth]{11e5167f-9f95-4494-9b66-b59fdce8b1ef-6_776_682_1886_246} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
\end{figure}
  1. Find an expression for \(\log _ { 10 } V\) in terms of \(t\).
  2. 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 \(\pounds 3150\).
  3. 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 \(\pounds 500\).
  4. According to the model, after how many years will Adam sell his car to Webuyoldcars?

Question 9:
Part (a):
AnswerMarks 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
Part (b):
AnswerMarks 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
Part (c):
AnswerMarks Guidance
\(3151.29\ (\approx 3150)\) so the model is a (very) good fitE1 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):
AnswerMarks Guidance
\(8500 \times 0.82^t = 500\) soiM1 Ft from (b) for M1M1
take logs of both sidesM1 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 yearsA1
# 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]}}