| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.3 This is a straightforward exponential model question requiring substitution to find constants and solving a logarithmic equation. Part (a) involves direct substitution with no conceptual challenge, and part (b) is a standard logarithm application. Slightly easier than average due to the guided structure and routine techniques, though the multi-step nature prevents it from being trivial. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 12499\) | B1 (1 mark) | Stated in (i) or (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k^{36} = \frac{7000}{\text{their } A}\) | M1 | \(p = \frac{7000}{12499} = 0.560044803\) |
| \(k = \sqrt[36]{0.56(00448...)} = 0.9840251(26)\) or \((0.56(00448...))^{\frac{1}{36}}\) or \(k = \sqrt[36]{\frac{7000}{12499}}\), \(k = 0.984025\) | A1 (2 marks) | Correct expression for \(k\) or \(7\)th dp seen. \(k = 10^{\frac{1}{36}\log p}\) or \(k = 10^{-0.00699...}\); \(k = e^{\frac{1}{36}\ln p}\) or \(k = e^{-0.016103...}\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(k^t = \frac{5000}{\text{their } A}\) | M1 | \(\frac{5000}{12499} = 0.400032...\); condone 4999 |
| \(t\log(k) = \log\left(\frac{5000}{A}\right)\), \((t = 56.89)\) | m1 | Correct use of logs |
| \(n = 57\) | A1 (3 marks) | \(n\) integer; \(n = 57\); CAO |
| Alternative: trial and improvement on \(5000 = 12499 \times 0.984025^t\); 2 values of \(t \geq 40\); 1 value of \(t\), \(50 < t < 60\); \(n = 57\) | (M1)(m1)(A1) |
# Question 4(a)(i):
$A = 12499$ | B1 (1 mark) | Stated in (i) or (ii)
# Question 4(a)(ii):
$k^{36} = \frac{7000}{\text{their } A}$ | M1 | $p = \frac{7000}{12499} = 0.560044803$
$k = \sqrt[36]{0.56(00448...)} = 0.9840251(26)$ or $(0.56(00448...))^{\frac{1}{36}}$ or $k = \sqrt[36]{\frac{7000}{12499}}$, $k = 0.984025$ | A1 (2 marks) | Correct expression for $k$ or $7$th dp seen. $k = 10^{\frac{1}{36}\log p}$ or $k = 10^{-0.00699...}$; $k = e^{\frac{1}{36}\ln p}$ or $k = e^{-0.016103...}$; AG
# Question 4(b):
$k^t = \frac{5000}{\text{their } A}$ | M1 | $\frac{5000}{12499} = 0.400032...$; condone 4999
$t\log(k) = \log\left(\frac{5000}{A}\right)$, $(t = 56.89)$ | m1 | Correct use of logs
$n = 57$ | A1 (3 marks) | $n$ integer; $n = 57$; CAO
**Alternative:** trial and improvement on $5000 = 12499 \times 0.984025^t$; 2 values of $t \geq 40$; 1 value of $t$, $50 < t < 60$; $n = 57$ | (M1)(m1)(A1) |
**Special case:** answer only: $n=57$ (3/3); $n=56$ (0/3); $n=56.9$ (2/3)
---4 A car depreciates in value according to the model
$$V = A k ^ { t }$$
where $\pounds V$ is the value of the car $t$ months from when it was new, and $A$ and $k$ are constants. Its value when new was $\pounds 12499$ and 36 months later its value was $\pounds 7000$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the value of $A$.
\item Show that the value of $k$ is 0.984025 , correct to six decimal places.
\end{enumerate}\item The value of this car first dropped below $\pounds 5000$ during the $n$th month from new. Find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q4 [6]}}