Higher June 2023 Unit 1 Q11
11.
(a) Given that \(y\) is inversely proportional to \(x\) and that \(y = 0{\cdot}2\) when \(x = 160\), find an expression for \(y\) in terms of \(x\). [3]
(b) Use the expression you found in part (a) to complete the following table. [2]
| \(x\) | 160 | 128 | |
|---|---|---|---|
| \(y\) | 0·2 | 0·8 |
| Scheme | Marks | Comments |
|---|---|---|
| \(y\ \alpha\ \dfrac{1}{x}\) OR \(y = \dfrac{k}{x}\) | B1 | Allow \(y\ \alpha\ \dfrac{k}{x}\) . |
| \(0{\cdot}2 = \dfrac{k}{160}\) OR \(k = 32\) | M1 | M1 implies B1. FT from \(y\ \alpha\ \dfrac{1}{x^n}\) with \(n \gt 1\) and \(n\) = integer. (Use of \(n = 2\) leads to \(k = 5120\).) |
| \(y = \dfrac{32}{x}\) | A1 | May be seen in part (b). |
| Scheme | Marks | Comments | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Check working space if table is empty. Table takes precedence over working space. | |||||||||
| B1 | For \(y\)-value, accept equivalents to 0·25 including \(\dfrac{32}{128}\). Mark final answer. | |||||||||
| B1 | For \(x\)-value, do not accept \(\dfrac{32}{0{\cdot}8}\) or \(\dfrac{320}{8}\). Mark final answer. FT from ‘their \(k\)’ (using \(y = \dfrac{k}{x}\) ) or FT from \(y\ \alpha\ \dfrac{1}{x^n}\), with \(n \gt 1\) and \(n\) = integer. Use of \(n = 2\) leads to answers of 0·3125 and 80. No FT from continued use of direct proportion from part (a). If no marks in part (a), allow B1 B1 for answers of \(y = 0{\cdot}25\), \(x = 40\) in part (b). |