Higher June 2018 Paper 5 Q20
20 In the following equation, \(n\) is an integer greater than 1.
\[\left(\sqrt{2}\right)^n = k\sqrt{2}\](a)
(i) Find \(k\) when \(n = 7\). [2]
(ii) Find \(n\) when \(k = 64\). [2]
(b) Show that \(\dfrac{14}{3 - \sqrt{2}}\) can be written in the form \(a + b\sqrt{2}\). [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) 8 | 2 | M1 for \(\left[\left(\sqrt{2}\right)^7 =\right]\ 2^3 \times \sqrt{2}\) | For M1 accept 2 × 2 × 2 for \(2^3\) Final answer \(8\sqrt{2}\) scores M1 |
| (ii) 13 | 2 | B1 for 2 correct trials with \(n \gt 3\) correctly evaluated or M1 for \(\left(\sqrt{2}\right)^{12} = 2^6\) oe or for \(\dfrac{n - 1}{2} = 6\) oe | e.g. \(\left(\sqrt{2}\right)^6 = 8\) and \(\left(\sqrt{2}\right)^9 = 16\sqrt{2}\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{14}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}}\) or better | M1 | If written in a single fraction, must have brackets | |
| \(\dfrac{14\left(3 + \sqrt{2}\right)}{7}\) | B3 | or M2 for \(\dfrac{14\left(3 + \sqrt{2}\right)}{9 + 3\sqrt{2} - 3\sqrt{2} - \left(\sqrt{2}\right)^2}\) or better or M1 for numerator or denominator correct | For B marks or method marks, allow numerator brackets expanded For M1, allow denominator unsimplified but not 9 – 2 or 7 if from wrong working Allow M1 for either numerator or denominator even if not in fraction |
| \(2\left(3 + \sqrt{2}\right)\) or \(6 + 2\sqrt{2}\) | A1 | Dep on M1B3 earned | |