Higher June 2022 Paper 1R Q21
21 Express \(\dfrac{3 + \sqrt{8}}{\left(\sqrt{2} - 1\right)^2}\) in the form \(p + \sqrt{q}\) where \(p\) and \(q\) are integers.
Show each stage of your working clearly.
(4)
| Scheme | Marks |
|---|---|
\((\sqrt{2} - 1)^2 = 2 - \sqrt{2} - \sqrt{2} + 1\;(= 3 - 2\sqrt{2})\) or \(\dfrac{(3 + \sqrt{8})}{(\sqrt{2} - 1)^2} \times \dfrac{(\sqrt{2} + 1)^2}{(\sqrt{2} + 1)^2}\) | M1 |
\(\dfrac{(3 + \sqrt{8})}{\text{``}{(3 - 2\sqrt{2})}\text{''}} \times \dfrac{(3 + 2\sqrt{2})}{(3 + 2\sqrt{2})}\) or \((\sqrt{2} - 1)^2 = 2 - \sqrt{2} - \sqrt{2} + 1\;(= 3 - 2\sqrt{2})\) or \((\sqrt{2} + 1)^2 = 2 + \sqrt{2} + \sqrt{2} + 1\;(= 3 + 2\sqrt{2})\) or \((\sqrt{2} - 1)(\sqrt{2} + 1) = 2 - \sqrt{2} + \sqrt{2} - 1\;(= 1)\) | M1 |
| eg \(\dfrac{9 + 6\sqrt{2} + 3\sqrt{8} + 8}{9 - 6\sqrt{2} + 6\sqrt{2} - 8}\) or \(\dfrac{9 + 12\sqrt{2} + 8}{9 - 8}\) or \(\dfrac{9 + 6\sqrt{2} + 3\sqrt{8} + 8}{1}\) or \(\dfrac{9 + 12\sqrt{2} + 8}{1}\) | M1 |
Working required Answer: \(17 + \sqrt{288}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: expand the denominator (accept \(2 - 2\sqrt{2} + 1\) - must see expansion) OR method to rationalise using \((\sqrt{2} + 1)^2\)
M1: oe ft \(3 - 2\sqrt{2}\) method to rationalise OR expansion of \((\sqrt{2} - 1)^2\) (accept \(2 - 2\sqrt{2} + 1\)) or \((\sqrt{2} + 1)^2\) (accept \(2 + 2\sqrt{2} + 1\)) or \((\sqrt{2} - 1)(\sqrt{2} + 1)\)
M1: dep on 2nd M1
correct expansion of brackets
A1: or \(p = 17\), \(q = 288\)
answer from fully correct working with intermediate steps of working seen