Higher November 2021 Paper 1 Q19
19 Show that \(\dfrac{8 + \sqrt{12}}{5 + \sqrt{3}}\) can be written in the form \(\dfrac{a + \sqrt{3}}{b}\), where \(a\) and \(b\) are integers. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Result shown | M1 | (indep) for writing \(\sqrt{12}\) as \(2\sqrt{3}\) |
| M1 | for method to rationalise the denominator eg \(\dfrac{8 + \sqrt{12}}{5 + \sqrt{3}} \times \dfrac{5 - \sqrt{3}}{5 - \sqrt{3}}\) or \(\dfrac{8 + 2\sqrt{3}}{5 + \sqrt{3}} \times \dfrac{5 - \sqrt{3}}{5 - \sqrt{3}}\) oe | |
| M1 | (dep on previous M1) for expanding terms, condone one error in numerator or denominator eg \(\dfrac{40 - 8\sqrt{3} + 5\sqrt{12} - \sqrt{12}\sqrt{3}}{25 - 5\sqrt{3} + 5\sqrt{3} - \sqrt{3}\sqrt{3}}\) or \(\dfrac{40 - 8\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\sqrt{3}}{25 - 5\sqrt{3} + 5\sqrt{3} - \sqrt{3}\sqrt{3}}\) or \(\dfrac{34 + 2\sqrt{3}}{22}\) oe | |
| A1 | for a complete chain of reasoning leading to \(\dfrac{17 + \sqrt{3}}{11}\) |
Additional guidance
This mark can be awarded whenever this is seen, which might be later in the process.