C3 January 2013 Q8
8. The value of Bob’s car can be calculated from the formula
\[V = 17000\mathrm{e}^{-0.25t} + 2000\mathrm{e}^{-0.5t} + 500\]
where \(V\) is the value of the car in pounds (£) and \(t\) is the age in years.
Give your answer in pounds per year to the nearest pound. (4)
| Scheme | Marks |
|---|---|
| (£) 19500 | B1 |
| (1) |
Notes
B1 19500. The £ sign is not important for this mark
| Scheme | Marks |
|---|---|
| \(9500 = 17000e^{-0.25t} + 2000e^{-0.5t} + 500\) \(17e^{-0.25t} + 2e^{-0.5t} = 9\) \((\times e^{0.5t}) \Rightarrow 17e^{0.25t} + 2 = 9e^{0.5t}\) | |
| \(0 = 9e^{0.5t} - 17e^{0.25t} - 2\) | M1 |
| \(0 = (9e^{0.25t} + 1)(e^{0.25t} - 2)\) | M1 |
| \(e^{0.25t} = 2\) | A1 |
| \(t = 4\ln(2)\) oe | A1 |
| (4) |
Notes
M1 Substitute V=9500, collect terms and set on 1 side of an equation =0. Indices must be correct
Accept \(17000e^{-0.25t} + 2000e^{-0.5t} - 9000 = 0\) and \(17000x + 2000x^2 - 9000 = 0\) where \(x = e^{-0.25t}\)
M1 Factorise the quadratic in \(e^{0.25t}\) or \(e^{-0.25t}\)
For your information the factorised quadratic in \(e^{-0.25t}\) is \((2e^{-0.25t} - 1)(e^{-0.25t} + 9) = 0\)
Alternatively let \('x' = e^{0.25t}\) or otherwise and factorise a quadratic equation in x
A1 Correct solution of the quadratic. Either \(e^{0.25t} = 2\) or \(e^{-0.25t} = \dfrac{1}{2}\) oe.
A1 Correct exact value of t. Accept variations of \(4\ln(2)\), such as \(\ln(16)\), \(\dfrac{\ln\left(\frac{1}{2}\right)}{-0.25}\), \(\dfrac{\ln(2)}{0.25}\), \(-4\ln\left(\dfrac{1}{2}\right)\)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}V}{\mathrm{d}t}\right) = -4250e^{-0.25t} - 1000e^{-0.5t}\) | M1A1 |
| When t=8 Decrease = 593 (£/year) | M1A1 |
| (4) | |
| (9 marks) |
Notes
M1 Differentiates \(V = 17000e^{-0.25t} + 2000e^{-0.5t} + 500\) by the chain rule.
Accept answers of the form \(\left(\dfrac{\mathrm{d}V}{\mathrm{d}t}\right) = \pm Ae^{-0.25t} \pm Be^{-0.5t} \quad A, B \text{ are constants} \neq 0\)
A1 Correct derivative \(\left(\dfrac{\mathrm{d}V}{\mathrm{d}t}\right) = -4250e^{-0.25t} - 1000e^{-0.5t}\).
There is no need for it to be simplified so accept
\(\left(\dfrac{\mathrm{d}V}{\mathrm{d}t}\right) = 17000 \times -0.25e^{-0.25t} + 2000 \times -0.5e^{-0.5t}\) oe
M1 Substitute t=8 into their \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\).
This is not dependent upon the first M1 but there must have been some attempt to differentiate.
Do not accept t=8 in V
A1 \(\pm 593\). Ignore the sign and the units. If the candidate then divides by 8, withhold this mark. This would not be isw. Be aware that sub t=8 into V first and then differentiating can achieve 593. This is M0A0M0A0.