Higher June 2022 Paper 1 Q16
16 Without using a calculator, show that \(\dfrac{12}{\sqrt{2} - 1} - \left(\sqrt{2}\right)^5 = 2\sqrt{32} + 12\)
Show your working clearly.
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{12}{\sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1}\) or \(\dfrac{12}{\sqrt{2} - 1} \times \dfrac{-\sqrt{2} - 1}{-\sqrt{2} - 1}\) and \(4\sqrt{2}\) or \(2\sqrt{8}\) or \(\sqrt{32}\) oe | M1 |
E.g. \(12\sqrt{2} + 12 - 4\sqrt{2}\) or \(8\sqrt{2} + 12\) \(12\sqrt{2} + 12 - 2\sqrt{8}\) or \(12\sqrt{2} + 12 - \sqrt{32}\) oe | M1 |
E.g. \(12\sqrt{2}(+12) - 4\sqrt{2} = 8\sqrt{2}(+12) = 2\sqrt{4^2 \times 2}(+12) = 2\sqrt{32}(+12)\) or \(12\sqrt{2}(+12) - 2\sqrt{8} = 6\sqrt{8}(+12) - 2\sqrt{8} = 4\sqrt{8}(+12) = 2\sqrt{4 \times 8}(+12) = 2\sqrt{32}(+12)\) or \(12\sqrt{2}(+12) - \sqrt{32} = 3\sqrt{4^2 \times 2}(+12) - \sqrt{32} = 2\sqrt{32}(+12)\) oe Note \(8\sqrt{2} = 2\sqrt{4^2 \times 2}\) or \(2\sqrt{16 \times 2}\) or \(\sqrt{32 \times 4}\) or \(\sqrt{64 \times 2}\) \(12\sqrt{2} = 3\sqrt{4^2 \times 2}\) or \(3\sqrt{16 \times 2}\) or \(\sqrt{32 \times 9}\) Answer: Shown | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for showing a correct method for rationalising the denominator and dealing with \(\left(\sqrt{2}\right)^5\)
M1: dep expression must be in surd form
A1: dep on M2 for showing working to given answer (they may dismiss the +12 and just deal with the surd part for this stage)