| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Standard +0.3 This is a standard log-linear modelling question requiring conversion between logarithmic and exponential forms, finding equation parameters from two points, and a straightforward model evaluation. The techniques are routine for A-level (reading from graph, using log laws, substituting values) 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 form1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\log_{10} V = 3 \Rightarrow V = 10^3\) | M1 | Sets \(\log_{10} V = 3\) and attempts to find value of \(a\) or expression for \(V\) when \(t = 0\) |
| \((V =)\) £1000 | A1 | £1000 cao (including units), do not accept £\(10^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. \((\log_{10} b =) \dfrac{2.79 - 3}{10 - 0} = -0.021\) or \(\log_{10} V = 3 - 0.021t\) or \(10^{2.79} =\) "\(1000\)"\(b^{10}\) | M1 | Finds gradient between two points, score for expression \(\frac{2.79-3}{10-0}\); or finds equation for \(\log_{10} V\) in terms of \(t\); or forms equation \(10^{2.79} =\) "\(1000\)"\(b^{10}\). Do not condone sign slips. |
| e.g. \(b = 10^{-0.021}\ (= 0.952796\ldots)\) or \(V = 10^3 \times 10^{-0.021t}\) or \(b = \sqrt[10]{\text{"0.61659\ldots"}}\) | M1 | Attempts to find value or expression for \(b\) using their gradient or equation |
| \(V = 1000 \times 0.953^t\) | A1ft | Complete correct equation, follow through on "1000". Accept \(V =\) "\(10^3\)"\(\times (\text{awrt } 0.953)^t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| e.g. \(V = 1000 \times\) "\(0.953\)"\(^{24}\ (= £315)\) or e.g. \(\log_{10} V = 3 -\) "\(0.021\)"\(\times 24 \Rightarrow V = \ldots\ (= £313)\) | M1 | Full and valid attempt to substitute \(t = 24\) into model of form \(V = ab^t\) or \(\log_{10} V = p + qt\); or substitute \(V = 320\) into model to find \(t\) |
| which is close (to £320) so it is a suitable model | A1 | Compares awrt £313–£315 with £320, or awrt \(t = 23.5 - 23.7\) with \(t = 24\), and makes valid conclusion with reason. Must have correct calculations, a reason such as "values are close/similar/approximately equal", and a statement that it is a "good" or "accurate" model. Do not allow comments suggesting model is not reliable. |
## Question 11:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_{10} V = 3 \Rightarrow V = 10^3$ | M1 | Sets $\log_{10} V = 3$ and attempts to find value of $a$ or expression for $V$ when $t = 0$ |
| $(V =)$ £1000 | A1 | £1000 cao (including units), do not accept £$10^3$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $(\log_{10} b =) \dfrac{2.79 - 3}{10 - 0} = -0.021$ **or** $\log_{10} V = 3 - 0.021t$ **or** $10^{2.79} =$ "$1000$"$b^{10}$ | M1 | Finds gradient between two points, score for expression $\frac{2.79-3}{10-0}$; or finds equation for $\log_{10} V$ in terms of $t$; or forms equation $10^{2.79} =$ "$1000$"$b^{10}$. Do not condone sign slips. |
| e.g. $b = 10^{-0.021}\ (= 0.952796\ldots)$ **or** $V = 10^3 \times 10^{-0.021t}$ **or** $b = \sqrt[10]{\text{"0.61659\ldots"}}$ | M1 | Attempts to find value or expression for $b$ using their gradient or equation |
| $V = 1000 \times 0.953^t$ | A1ft | Complete correct equation, follow through on "1000". Accept $V =$ "$10^3$"$\times (\text{awrt } 0.953)^t$ |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. $V = 1000 \times$ "$0.953$"$^{24}\ (= £315)$ **or** e.g. $\log_{10} V = 3 -$ "$0.021$"$\times 24 \Rightarrow V = \ldots\ (= £313)$ | M1 | Full and valid attempt to substitute $t = 24$ into model of form $V = ab^t$ or $\log_{10} V = p + qt$; or substitute $V = 320$ into model to find $t$ |
| which is close (to £320) so it is a suitable model | A1 | Compares awrt £313–£315 with £320, or awrt $t = 23.5 - 23.7$ with $t = 24$, and makes valid conclusion with reason. Must have correct calculations, a reason such as "values are close/similar/approximately equal", and a statement that it is a "good" or "accurate" model. Do not allow comments suggesting model is not reliable. |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-28_590_739_219_671}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The value, $V$ pounds, of a mobile phone, $t$ months after it was bought, is modelled by
$$V = a b ^ { t }$$
where $a$ and $b$ are constants.\\
Figure 2 shows the linear relationship between $\log _ { 10 } V$ and $t$.\\
The line passes through the points $( 0,3 )$ and $( 10,2.79 )$ as shown.\\
Using these points,
\begin{enumerate}[label=(\alph*)]
\item find the initial value of the phone,
\item find a complete equation for $V$ in terms of $t$, giving the exact value of $a$ and giving the value of $b$ to 3 significant figures.
Exactly 2 years after it was bought, the value of the phone was $\pounds 320$
\item Use this information to evaluate the reliability of the model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q11 [7]}}