| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| 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 decay application requiring substitution of given values to find constants A and k, then solving a logarithmic equation. The steps are routine and well-practiced in C3: finding A from initial conditions is immediate (A=10000), finding k requires one logarithm calculation, and part (ii) is a standard 'solve for t' exercise. No conceptual difficulty or novel problem-solving required. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
The value £V of a car is modelled by the equation $V = Ae^{-kt}$, where $t$ is the age of the car in years and $A$ and $k$ are constants. Its value when new is £10000, and after 3 years its value is £6000.
(i) Find the values of $A$ and $k$. [5]
M1: Use initial condition $V(0) = 10000$
A1: $A = 10000$
M1: Use condition at $t = 3$: $6000 = 10000 e^{-3k}$
M1: Solve $e^{-3k} = 0.6$ or $e^{-3k} = \frac{3}{5}$
A1: $k = -\frac{1}{3} \ln(0.6) = \frac{1}{3} \ln\left(\frac{5}{3}\right)$ (exact form required)
(ii) Find the age of the car when its value is £2000. [2]
M1: Set up equation $2000 = 10000 e^{-kt}$ or $e^{-kt} = 0.2$
A1: $t = \frac{\ln 5}{k}$ or $t = 3\ln 5 / \ln(5/3)$ (or equivalent exact form)
---3 The value $\pounds V$ of a car is modelled by the equation $V = A \mathrm { e } ^ { - k t }$, where $t$ is the age of the car in years and $A$ and $k$ are constants. Its value when new is $\pounds 10000$, and after 3 years its value is $\pounds 6000$.\\
(i) Find the values of $A$ and $k$.\\
(ii) Find the age of the car when its value is $\pounds 2000$.
\hfill \mbox{\textit{OCR MEI C3 2007 Q3 [7]}}