A2 June 2023 Paper 1 Q6
6 In this question you must show detailed reasoning.
The power output, \(p\) watts, of a machine at time \(t\) hours after it is switched on can be modelled by the equation \(p = 20 - 20\tanh(1.44t)\) for \(t \geqslant 0\).
Determine, according to the model, the mean power output of the machine over the first half hour after it is switched on. Give your answer correct to 2 decimal places. [4]
| Scheme | Marks |
|---|---|
| DR \(\text{Mean value} = \dfrac{1}{0.5}\displaystyle\int_0^{0.5}\left(20 - 20\tanh(1.44t)\right)\mathrm{d}t\) | B1 |
| \(= \left[40t - \dfrac{40}{1.44}\ln\left(\left|\mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\right|\right)\right]_0^{0.5}\) | M1 |
| \(= \left(20 - \dfrac{40}{1.44}\left(\ln\left(\mathrm{e}^{0.72} + \mathrm{e}^{-0.72}\right) - \ln 2\right)\right)\) | A1 |
| \(= \left(20 - \dfrac{40}{1.44}\ln\dfrac{2.5412}{2}\right) = 13.35\ (\mathrm{W})\) | A1 |
| [4] |
Notes
B1: For using the definition of the mean value of \(p\) wrt \(t\), correct limits and 1/0.5.
M1: For \(\int \tanh 1.44t\,\mathrm{d}t = k\ln\left|\mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\right| (+c)\) \(k\) can \(= 1\)
Or \(\int \tanh 1.44t\,\mathrm{d}t = k\ln|\cosh 1.44t| (+c)\)
May see integration by substitution, eg. \(u = \mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\) or \(u = \cosh 1.44t\). If so, award this mark for \(k\ln|u|\) seen
A1: For fully correct integration of \(\tanh 1.44t\).
Either for \(\int \tanh 1.44t\,\mathrm{d}t = \frac{1}{1.44}\ln\left|\mathrm{e}^{1.44t} + \mathrm{e}^{-1.44t}\right| (+c)\)
or \(\int \tanh 1.44t\,\mathrm{d}t = \frac{1}{1.44}\ln|\cosh 1.44t| (+c)\)
or \(\int \tanh 1.44t\,\mathrm{d}t = \frac{1}{1.44}\ln|u| (+c)\) with \(u\) as above.
Condone missing modulus.
A1: cao, with clear working.