Higher January 2023 Paper 2 Q17
17 Given that \(\;8\sqrt{m} + \sqrt{49m} - \sqrt{9m} = k\sqrt{m}\)
where \(k\) is an integer and \(m\) is a prime number,
(a) work out the value of \(k\) (1)
(b) Show that \(\;\dfrac{5 - \sqrt{18}}{1 - \sqrt{2}}\;\) can be written in the form \(\;a + b\sqrt{2}\)
where \(a\) and \(b\) are integers.
Show each stage of your working clearly. (3)
where \(a\) and \(b\) are integers.
Show each stage of your working clearly. (3)
| Scheme | Marks |
|---|---|
| 12 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\dfrac{5 - \sqrt{18}}{1 - \sqrt{2}} \times \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}}\) or \(\dfrac{5 - \sqrt{18}}{1 - \sqrt{2}} \times \dfrac{-1 - \sqrt{2}}{-1 - \sqrt{2}}\) oe | M1 |
\(\dfrac{5 - \sqrt{36} + 5\sqrt{2} - \sqrt{18}}{1 + \sqrt{2} - \sqrt{2} - 2}\) or \(\dfrac{5 - 6 - 3\sqrt{2} + 5\sqrt{2}}{-1}\) or \(\dfrac{-5 + 6 + 3\sqrt{2} - 5\sqrt{2}}{1}\) oe NB: allow \(\sqrt{18}\) or \(3\sqrt{2}\); \(\;\sqrt{36}\) or 6 or \(\sqrt{6}\sqrt{6}\) | M1 |
working required Answer: \(1 - 2\sqrt{2}\) | A1 |
| (3) | |
| (4 marks) |
Notes
M1: Multiplying numerator and denominator by \(1 + \sqrt{2}\)
M1: Showing correct expansions (not necessarily as a fraction)
A1: dep on M2 (ie all stages of working must be shown convincingly)
or for stating \(a = 1\) and \(b = -2\)