| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This is a straightforward exponential model question requiring only direct substitution and basic logarithm manipulation. Part (a) is immediate recognition, part (b) involves solving k^56 = 62.5 using logarithms (a standard technique), and parts (c)(i)-(ii) are routine applications of the model. No problem-solving insight or novel approach is needed—purely mechanical application of exponential growth formulas taught in C4. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = 80\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5000 = 80 \times k^{56}\) | M1 | SC1 Verification. Need 62.51 or better |
| \(k = \sqrt[56]{\frac{5000}{80}} \approx 1.07664\) | M1A1 | Or using logs: M1 \(\ln\left(\frac{5000}{80}\right) = 56\ln k\); A1 \(k = e^{\ln\left(\frac{62.5}{56}\right)}\); Or 3/3 for \(k = 1.076636\); Or \(1.076637\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = 80 \times k^{106} = 200707\) | M1A1 | 200648 using full register \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln 10000 = \ln k^t\) | M1 | |
| \(t = \frac{\ln 10000}{\ln k} = 124.7 \Rightarrow 2024\) | M1A1 | M1 \(t\ln k = \ln 10000\); A1 CAO; Or trial and improvement M1 expression; M1 125, 124, A1 2024 |
## Question 4:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = 80$ | B1 | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5000 = 80 \times k^{56}$ | M1 | SC1 Verification. Need 62.51 or better |
| $k = \sqrt[56]{\frac{5000}{80}} \approx 1.07664$ | M1A1 | Or using logs: M1 $\ln\left(\frac{5000}{80}\right) = 56\ln k$; A1 $k = e^{\ln\left(\frac{62.5}{56}\right)}$; Or 3/3 for $k = 1.076636$; Or $1.076637$ seen |
### Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = 80 \times k^{106} = 200707$ | M1A1 | 200648 using full register $k$ |
### Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln 10000 = \ln k^t$ | M1 | |
| $t = \frac{\ln 10000}{\ln k} = 124.7 \Rightarrow 2024$ | M1A1 | M1 $t\ln k = \ln 10000$; A1 CAO; Or trial and improvement M1 expression; M1 125, 124, A1 2024 |
---4 On 1 January 1900, a sculpture was valued at $\pounds 80$.\\
When the sculpture was sold on 1 January 1956, its value was $\pounds 5000$.\\
The value, $\pounds V$, of the sculpture is modelled by the formula $V = A k ^ { t }$, where $t$ is the time in years since 1 January 1900 and $A$ and $k$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $A$.
\item Show that $k \approx 1.07664$.
\item Use this model to:
\begin{enumerate}[label=(\roman*)]
\item show that the value of the sculpture on 1 January 2006 will be greater than £200 000;
\item find the year in which the value of the sculpture will first exceed $\pounds 800000$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2006 Q4 [9]}}