Give your answer in the form \(\qquad \dfrac{\sqrt{a} - \sqrt{b}}{c} \qquad\) where \(a\), \(b\) and \(c\) are integers. [4 marks]
Mark scheme
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 \(\;\dfrac{2 \times \dfrac{1}{\sqrt{2}} - 1}{4 \times \dfrac{\sqrt{3}}{1}}\)
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