Higher June 2017 Paper 1 Q29
29 Simplify \(\qquad \dfrac{2\sin 45^\circ - \tan 45^\circ}{4\tan 60^\circ}\)
Give your answer in the form \(\qquad \dfrac{\sqrt{a} - \sqrt{b}}{c} \qquad\) where \(a\), \(b\) and \(c\) are integers. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sin 45 = \dfrac{\sqrt{2}}{2}\) or \(\dfrac{1}{\sqrt{2}}\) or \(\tan 45 = 1\) or \(\dfrac{1}{1}\) or \(\tan 60 = \sqrt{3}\) or \(\dfrac{\sqrt{3}}{1}\) | B1 | oe stated or in correct place in expression or implied by multiplier of 2 or 4 |
| \(\sin 45 = \dfrac{\sqrt{2}}{2}\) or \(\dfrac{1}{\sqrt{2}}\) and \(\tan 45 = 1\) or \(\dfrac{1}{1}\) and \(\tan 60 = \sqrt{3}\) or \(\dfrac{\sqrt{3}}{1}\) | B1 | oe stated or in correct place in expression or implied by multiplier of 2 or 4 eg \(\;\dfrac{2 \times \dfrac{1}{\sqrt{2}} - 1}{4 \times \dfrac{\sqrt{3}}{1}}\) |
| \(\dfrac{\sqrt{2} - 1}{4\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}\) | M1 | oe rationalisation of their denominator eg \(\;\dfrac{\dfrac{2}{\sqrt{2}} - 1}{4\sqrt{3}} \times \dfrac{4\sqrt{3}}{4\sqrt{3}}\) |
| \(\dfrac{\sqrt{6} - \sqrt{3}}{12}\) | A1 | oe in the form \(\dfrac{\sqrt{6a^2} - \sqrt{3a^2}}{12a}\) where \(a\) is a positive integer eg \(\;\dfrac{\sqrt{24} - \sqrt{12}}{24}\) (when \(a = 2\)) |
Additional guidance
| \(\dfrac{2 \times \dfrac{1}{\sqrt{2}} - 1}{4\sqrt{3}}\) or \(\dfrac{\sqrt{2} - 1}{4\sqrt{3}}\) or \(\dfrac{\sqrt{2} - 1}{\sqrt{48}}\) | B1B1 |
| \(\dfrac{\sqrt{48}(\sqrt{2} - 1)}{\sqrt{48}\sqrt{48}}\) or \(\dfrac{\sqrt{48}(\sqrt{2} - 1)}{48}\) | B1B1M1 |
| \(\dfrac{\sqrt{96} - \sqrt{48}}{48}\) | B1B1M1A1 |
| B1B1 awarded, incorrect simplification, then correct method to rationalise | B1B1M1 |