Higher June 2022 Paper 2 Q20
20. \(R\) is proportional to \(P\left(1 - \dfrac{P}{100}\right)\).
When \(P\) is 50, \(R\) is 0·02.
(a) Find a formula for \(R\) in terms of \(P\). [3]
(b) Find the positive value of \(P\) for which \(R\) is 0. [2]
| Scheme | Marks | Comments |
|---|---|---|
| \(R \propto P\left(1 - \dfrac{P}{100}\right)\) or \(R = kP\left(1 - \dfrac{P}{100}\right)\) | M1 | Allow for \(R \propto kP\left(1 - \dfrac{P}{100}\right)\) |
| \(0.02 = 50k\left(1 - \dfrac{50}{100}\right)\) oe | M1 | Award of this M1 implies the previous M1 |
| \(R = 0.0008P\left(1 - \frac{P}{100}\right)\) or \(R = \frac{P}{1250}\left(1 - \frac{P}{100}\right)\) oe | A1 | \(k = 0.0008\) or \(\dfrac{1}{1250}\) only Do not allow e.g. \(R \propto 0.0008P\left(1 - \frac{P}{100}\right)\) Mark final answer but allow the mark for sight of the correct equation in (b) |
| Scheme | Marks | Comments |
|---|---|---|
| \((0.0008 \times)\ P\left(1 - \frac{P}{100}\right) = 0\) or better | M1 | Must see an equation used here FT ‘their equation’ of the form \(R = kP\left(1 - \frac{P}{100}\right)\) but allow \(P\left(1 - \frac{P}{100}\right) = 0\) |
| \(P = 100\) | A1 | Ignore \(P = 0\) if \(P = 100\) given. |
| (5) |