Higher June 2023 Paper 6 Q19
19
(a) Show that \(\sqrt{11} \times \sqrt{22} = 11\sqrt{2}\). [1]
(b) Show that \(\frac{\sqrt{11}}{13 + \sqrt{22}}\) can be written in the form \(\frac{a\sqrt{11} - 11\sqrt{2}}{b}\) where \(a\) and \(b\) are integers. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(\sqrt{11}\sqrt{22}\) =] \(\sqrt{242} = \sqrt{121 \times 2}\) or \(\sqrt{121} \times \sqrt{2}\) [= \(11\sqrt{2}\)] or [\(\sqrt{11}\sqrt{22}\) =] \(\sqrt{11} \times \sqrt{11}\sqrt{2}\) or \(\sqrt{11} \times \sqrt{11 \times 2}\) or \(\sqrt{11 \times 11 \times 2}\) [= \(11\sqrt{2}\)] | 1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\frac{\sqrt{11}(13 - \sqrt{22})}{(13 + \sqrt{22})(13 - \sqrt{22})}\) | M1 | Condone missing bracket for this M1 if recovered later in numerator or denominator | Multiplying by \(\sqrt{22} - 13\) is eligible for M1 and then FT but A1 must be correct form Multiplying by \(13 + \sqrt{22}\) scores 0 |
| \(13\sqrt{11} - \sqrt{11}\sqrt{22}\) oe or better | M1 | May be in a grid | Equivalents likely to be seen for \(\sqrt{11}\sqrt{22}\) include \(\sqrt{242}\) and \(11\sqrt{2}\) |
| \(169\ [+13\sqrt{22} - 13\sqrt{22}] - 22\) | M1 | May be in a grid | |
| \(\frac{13\sqrt{11} - 11\sqrt{2}}{147}\) | A1 | Dep on M1M1M1 and no errors seen | An error is eg missing bracket in first M1 |