Higher January 2022 Paper 2 Q17
17 Show that \(\dfrac{\sqrt{12}}{\sqrt{3} + 2}\)
can be written in the form \(a + \sqrt{b}\) where \(a\) and \(b\) are integers.
(3)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{\sqrt{12}}{\sqrt{3} + 2} \times \dfrac{\sqrt{3} - 2}{\sqrt{3} - 2}\) | M1 |
eg \(\dfrac{\left(\sqrt{36} - 2\sqrt{12}\right)}{3 - 4}\) or \(\dfrac{6 - 2\sqrt{12}}{-1}\) or \(-6 + 2\sqrt{12}\) or \(\dfrac{6 - 4\sqrt{3}}{-1}\) or \(-6 + 4\sqrt{3}\) | M1 |
Working required Answer: \(-6 + \sqrt{48}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: rationalise denominator – award for seeing multiplication by \(\dfrac{\sqrt{3} - 2}{\sqrt{3} - 2}\) or \(\dfrac{-\sqrt{3} + 2}{-\sqrt{3} + 2}\)
M1: dep M1 correctly simplifying numerator and denominator.
(denominator could be 3 – 4 or –1)
A1: dep M2 must be in correct form (including \(\sqrt{48}\))
allow \(a = -6\) and \(b = 48\)