| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.3 This is a standard linear transformation question requiring students to convert between logarithmic and exponential forms using two given points. The steps are routine: find gradient, write log₁₀P equation, then convert to exponential form using properties of logarithms. While it requires multiple steps and careful algebraic manipulation, it follows a well-practiced technique with no novel insight needed, 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(\log_{10} a = 3.3\) or \(\log_{10} b = \frac{2.1-3.3}{6}\ (=-0.2)\) | M1 | Attempts equation in \(a\) or \(b\). Condone incorrectly evaluated gradient provided \(\frac{2.1-3.3}{6}\) was attempted. Do not penalise if base 10 missing. |
| \(a = 10^{3.3}\) or \(b = 10^{-0.2}\) | A1 | Correct unsimplified value for \(a\) or \(b\). |
| \(a = 10^{3.3}\) and \(b = 10^{"-0.2"}\), giving \(P = 10^{3.3-0.2x} = 10^{3.3} \times 10^{-0.2x}\) | dM1 | Correct method for both \(a\) and \(b\); uses laws of indices to achieve form \(P = 10^{3.3} \times 10^{"-0.2"x}\). Dependent on previous M. |
| \(P = 1995 \times 0.6310^x\) | A1 | \(a =\) awrt 1995 and \(b =\) awrt 0.6310 (condone 0.631) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| The concentration (in parts per million) 1 km from the chimney | B1 | Must refer to concentration (or e.g. parts per million) and 1km. Condone "concentration of smoke particles emitted 1km from the chimney". Do not accept "amount of smoke particles" or referring to when \(x=1\) (not in context). |
## Question 7:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\log_{10} a = 3.3$ or $\log_{10} b = \frac{2.1-3.3}{6}\ (=-0.2)$ | M1 | Attempts equation in $a$ or $b$. Condone incorrectly evaluated gradient provided $\frac{2.1-3.3}{6}$ was attempted. Do not penalise if base 10 missing. |
| $a = 10^{3.3}$ or $b = 10^{-0.2}$ | A1 | Correct unsimplified value for $a$ or $b$. |
| $a = 10^{3.3}$ and $b = 10^{"-0.2"}$, giving $P = 10^{3.3-0.2x} = 10^{3.3} \times 10^{-0.2x}$ | dM1 | Correct method for both $a$ and $b$; uses laws of indices to achieve form $P = 10^{3.3} \times 10^{"-0.2"x}$. Dependent on previous M. |
| $P = 1995 \times 0.6310^x$ | A1 | $a =$ awrt 1995 and $b =$ awrt 0.6310 (condone 0.631) |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| The **concentration** (in parts per million) **1 km** from the chimney | B1 | Must refer to **concentration** (or e.g. **parts per million**) and **1km**. Condone "concentration of smoke particles emitted 1km from the chimney". Do not accept "**amount** of smoke particles" or referring to when $x=1$ (not in context). |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23689deb-7eed-4022-848f-1278231a4056-18_614_878_296_555}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A chimney emits smoke particles.\\
On a particular day, the concentration of smoke particles in the air emitted by this chimney, $P$ parts per million, is measured at various distances, $x \mathrm {~km}$, from the chimney.
Figure 2 shows a sketch of the linear relationship between $\log _ { 10 } P$ and $x$ that is used to model this situation.
The line passes through the point ( $0,3.3$ ) and the point ( $6,2.1$ )
\begin{enumerate}[label=(\alph*)]
\item Find a complete equation for the model in the form
$$P = a b ^ { x }$$
where $a$ and $b$ are constants. Give the value of $a$ and the value of $b$ each to 4 significant figures.
\item With reference to the model, interpret the value of $a b$
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q7 [5]}}