| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Finding x from given y value |
| Difficulty | Standard +0.3 This is a standard exponential modelling question requiring substitution of given values to find constants, then solving a logarithmic equation. The algebra is straightforward (part b is scaffolded as 'show that'), and part c requires routine rearrangement and use of logarithms. Slightly above average due to the multi-step nature and exponential manipulation, but all techniques are standard C3 material with no novel insight required. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks |
|---|---|
| \(A = 1500\) | B1 |
| Answer | Marks |
|---|---|
| Sub \(t = 2, V = 13500 \Rightarrow 16000e^{-2k} = 12000\) | M1 |
| \(\Rightarrow e^{-2k} = \frac{3}{4}\) oe \(0.75\) | A1 |
| \(\Rightarrow k = -\frac{1}{2}\ln\frac{3}{4}, = \ln\sqrt{\frac{4}{3}} = \ln\left(\frac{2}{\sqrt{3}}\right)\) | dM1, A1* |
| Answer | Marks |
|---|---|
| Sub \(6000 = 16000e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} + 1500 \Rightarrow e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} = C\) | M1 |
| \(\Rightarrow e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} = \frac{45}{160} = 0.28125\) | A1 |
| \(\Rightarrow T = -\frac{\ln\left(\frac{45}{160}\right)}{\ln\left(\frac{2}{\sqrt{3}}\right)} = 8.82\) | M1 A1 |
| Answer | Marks |
|---|---|
| Sub \(t = 2, V = 13500 \Rightarrow 13500 - 16000e^{-2k} + 1500' \Rightarrow 16000e^{-2k} = 1200\) | M1 |
| \(\Rightarrow \ln 1600 - 2k = \ln 1200\) | A1 |
| \(\Rightarrow k = -\frac{1}{2}\ln\frac{1200}{1600}, = \ln\sqrt{\frac{4}{3}} = \ln\left(\frac{2}{\sqrt{3}}\right)\) | dM A1* |
**(a)**
| $A = 1500$ | B1 |
(1 mark)
**(b)**
| Sub $t = 2, V = 13500 \Rightarrow 16000e^{-2k} = 12000$ | M1 |
| $\Rightarrow e^{-2k} = \frac{3}{4}$ oe $0.75$ | A1 |
| $\Rightarrow k = -\frac{1}{2}\ln\frac{3}{4}, = \ln\sqrt{\frac{4}{3}} = \ln\left(\frac{2}{\sqrt{3}}\right)$ | dM1, A1* |
(4 marks)
**(c)**
| Sub $6000 = 16000e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} + 1500 \Rightarrow e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} = C$ | M1 |
| $\Rightarrow e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} = \frac{45}{160} = 0.28125$ | A1 |
| $\Rightarrow T = -\frac{\ln\left(\frac{45}{160}\right)}{\ln\left(\frac{2}{\sqrt{3}}\right)} = 8.82$ | M1 A1 |
(4 marks)
**Alt (b):**
| Sub $t = 2, V = 13500 \Rightarrow 13500 - 16000e^{-2k} + 1500' \Rightarrow 16000e^{-2k} = 1200$ | M1 |
| $\Rightarrow \ln 1600 - 2k = \ln 1200$ | A1 |
| $\Rightarrow k = -\frac{1}{2}\ln\frac{1200}{1600}, = \ln\sqrt{\frac{4}{3}} = \ln\left(\frac{2}{\sqrt{3}}\right)$ | dM A1* |
(4 marks)
(9 marks total)
**Guidance notes:**
- You may mark parts (a) and (b) together
- B1: Sight of $A = 1500$
- M1: Substitutes $t = 2, V = 13500 \Rightarrow 13500 = 16000e^{-2k} + 1500$ and proceeds to $Pe^{-2k} = ...$ or $Qe^{-2k} = ...$. Condone slips, e.g. $V$ may be 1350.
- A1: $e^{-2k} = \frac{3}{4}$ oe $0.75$ or $e^{-2k} = \frac{4}{3}$
- dM1: For taking ln's and proceeding to $k = ...$. For example $k = -\frac{1}{2}\ln\frac{3}{4}$ oe. May be implied by the correct decimal answer awrt 0.144.
- A1*: cso $k = \ln\left(\frac{2}{\sqrt{3}}\right)$ (brackets not required) with a correct intermediate line of either $\frac{1}{2}\ln\frac{4}{3}, \frac{1}{2}\ln 4 - \frac{1}{2}\ln 3, \ln\sqrt{\frac{4}{3}}$ or $\ln\left(\frac{3}{4}\right)^{\frac{1}{2}}$
- M1: Sub $V = 6000 \Rightarrow 6000 = 16000e^{-krt} + 1500$ and proceeds to $e^{-krt} = c, c > 0$
- A1: $e^{-\ln\left(\frac{2}{\sqrt{3}}\right)t} = \frac{45}{160} = 0.28125$
- M1: Correct order of operations using ln's and division leading to a value of $T$. It is implied by awrt 8.8
- A1: cso 8.82 only following correct work. This is not awrt. Allow a solution using an inequality as long as it arrives at the solution 8.82.
Trial and improvement solutions are also acceptable with appropriate scoring.
---\begin{enumerate}
\item The value of a car is modelled by the formula
\end{enumerate}
$$V = 16000 \mathrm { e } ^ { - k t } + A , \quad t \geqslant 0 , t \in \mathbb { R }$$
where $V$ is the value of the car in pounds, $t$ is the age of the car in years, and $k$ and $A$ are positive constants.
Given that the value of the car is $\pounds 17500$ when new and $\pounds 13500$ two years later,\\
(a) find the value of $A$,\\
(b) show that $k = \ln \left( \frac { 2 } { \sqrt { 3 } } \right)$\\
(c) Find the age of the car, in years, when the value of the car is $\pounds 6000$
Give your answer to 2 decimal places.\\
\hfill \mbox{\textit{Edexcel C3 2018 Q3 [9]}}