Higher November 2021 Paper 1 Q25
25 Show that \(\quad \dfrac{\sqrt{150} - \sqrt{6}}{\sqrt{2} \times \sqrt{3}} \quad\) simplifies to an integer. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| (\(\sqrt{150}\) =) \(\sqrt{25}\sqrt{6}\) or \(5\sqrt{6}\) or (\(\sqrt{2} \times \sqrt{3}\) =) \(\sqrt{6}\) | M1 | numerator allow \(\sqrt{2}\sqrt{3}\) for \(\sqrt{6}\) denominator |
| \(\dfrac{\sqrt{25}\sqrt{6} - \sqrt{6}}{\sqrt{6}}\) or \(\dfrac{5\sqrt{6} - \sqrt{6}}{\sqrt{6}}\) or \(\dfrac{4\sqrt{6}}{\sqrt{6}}\) | M1dep | allow consistent use of \(\sqrt{2}\sqrt{3}\) for \(\sqrt{6}\) |
| 4 with M1M1 awarded | A1 | |
| Alternative method 2 | ||
| \(\sqrt{6}(\sqrt{25} - 1)\) or \(\sqrt{6}(5 - 1)\) or \(4\sqrt{6}\) or (\(\sqrt{2} \times \sqrt{3}\) =) \(\sqrt{6}\) | M1 | numerator allow \(\sqrt{2}\sqrt{3}\) for \(\sqrt{6}\) denominator |
| \(\dfrac{\sqrt{6}(\sqrt{25} - 1)}{\sqrt{6}}\) or \(\dfrac{\sqrt{6}(5 - 1)}{\sqrt{6}}\) | M1dep | allow consistent use of \(\sqrt{2}\sqrt{3}\) for \(\sqrt{6}\) |
| 4 with M1M1 awarded | A1 | |
| Alternative method 3 | ||
| \(\dfrac{\sqrt{150} - \sqrt{6}}{\sqrt{2} \times \sqrt{3}} \times \dfrac{\sqrt{6}}{\sqrt{6}}\) | M1 | allow \(\dfrac{\sqrt{2}\sqrt{3}}{\sqrt{2}\sqrt{3}}\) for \(\dfrac{\sqrt{6}}{\sqrt{6}}\) |
| \(\dfrac{\sqrt{900} - 6}{6}\) | M1dep | oe rationalised |
| 4 with M1M1 awarded | A1 | |
Additional guidance
| Condone answer 4 and \(-6\) from use of \(\sqrt{25} = \pm 5\) | M1M1A1 |