Higher November 2017 Paper 1 Q6
6 Work out the value of \(\quad \left(\sqrt{3}\right)^2 \times \left(\sqrt{2}\right)^2\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\left(\left(\sqrt{3}\right)^2 =\right)\) 3 and \(\left(\left(\sqrt{2}\right)^2 =\right)\) 2 or \(\left(\sqrt{6}\right)^2\) or \(\sqrt{6^2}\) or \(\sqrt{36}\) or \(\sqrt{9} \times \sqrt{4}\) or \(\sqrt{9 \times 4}\) | M1 | |
| 6 | A1 |
Additional guidance
| \(3 \times 2 = 6\) with answer eg \(\sqrt{6}\) or \(6^4\) | M0A0 |
| Condone \(\sqrt{3} = 1.7\), \(1.7^2 = 3\) and \(\sqrt{2} = 1.4\), \(1.4^2 = 2\), otherwise \(\sqrt{3}\) or \(\sqrt{2}\) or \(3^2\) or \(2^2\) incorrectly evaluated does not score, even if answer is 6 | |
| eg \(\sqrt{3} = 1.5\), \(1.5^2 = 3\) | M0A0 |
| \(\sqrt{2} = 1\), \(1^2 = 2\), answer 6 | M0A0 |
| \(3^2 = 6\), \(\sqrt{6} = 3\) | M0 |
| \(\left(\sqrt{6}\right)^4\) | M0A0 |
| \(\sqrt{2} = 1\) | M0 |