June 2024 Paper 1 Q7
7 Show that
\[\frac{3 + \sqrt{8n}}{1 + \sqrt{2n}}\]can be written as
\[\frac{4n - 3 + \sqrt{2n}}{2n - 1}\]where \(n\) is a positive integer. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Simplifies \(\sqrt{8n}\) to \(2\sqrt{2n}\) | B1 | 1.1b |
| Multiplies by \(\dfrac{1 - \sqrt{2n}}{1 - \sqrt{2n}}\) or \(\dfrac{\sqrt{2n} - 1}{\sqrt{2n} - 1}\) | M1 | 1.1a |
| Obtains correct single fraction with denominator of \(1 - 2n\) or \(2n - 1\) | A1 | 1.1b |
| Completes reasoned argument to obtain \(\dfrac{4n - 3 + \sqrt{2n}}{2n - 1}\) AG Condone eg \(\sqrt{8}n\) or \(\sqrt{2}n\) except if seen on their final line. | R1 | 2.1 |
| (4 marks) |