June 2023 Paper 3 Q3
3 In this question you must show detailed reasoning.
Find the value of \(k\) such that \(\dfrac{1}{\sqrt{5}+\sqrt{6}} + \dfrac{1}{\sqrt{6}+\sqrt{7}} = \dfrac{k}{\sqrt{5}+\sqrt{7}}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\sqrt{6}-\sqrt{5}}{6-5} + \dfrac{\sqrt{7}-\sqrt{6}}{7-6}\) oe | M1 | 3.1a |
| \(\sqrt{7}-\sqrt{5}\) or \(\dfrac{\sqrt{5}-\sqrt{7}}{-1}\) | A1 | 1.1 |
| \(\dfrac{k}{\sqrt{5}+\sqrt{7}} = \dfrac{k\left(\sqrt{7}-\sqrt{5}\right)}{7-5} = \sqrt{7}-\sqrt{5}\) so \(k = 2\) | A1 | 2.2a |
| [3] |
Notes
M1: Rationalising denominators.
This is the minimum working needed for M1
Accept 1 for “6 – 5” and –1 for “5 – 6” etc
A1: Finding \(k\) convincingly after M1
Alternative
| Scheme | Marks |
|---|---|
| \(\left(\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right) + \left(\sqrt{5}+\sqrt{6}\right)\left(\sqrt{5}+\sqrt{7}\right)\) \(= k\left(\sqrt{5}+\sqrt{6}\right)\left(\sqrt{6}+\sqrt{7}\right)\) | M1 |
| \(\left(2\sqrt{5}\sqrt{6} + 2\sqrt{5}\sqrt{7} + 2\sqrt{6}\sqrt{7} + 12\right)\) \(= k\left(\sqrt{5}\sqrt{6} + \sqrt{5}\sqrt{7} + \sqrt{6}\sqrt{7} + 6\right)\) | A1 |
| \(2\left(\sqrt{5}\sqrt{6} + \sqrt{5}\sqrt{7} + \sqrt{6}\sqrt{7} + 6\right)\) \(= k\left(\sqrt{5}\sqrt{6} + \sqrt{5}\sqrt{7} + \sqrt{6}\sqrt{7} + 6\right)\) so \(k = 2\) | A1 |
M1: For dealing appropriately with fractions (working with both sides) e.g. clearing the fractions or making k the subject with the RHS as a single fraction.
\(k = \dfrac{\left(\sqrt{5}+\sqrt{7}+2\sqrt{6}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{6}\right)\left(\sqrt{6}+\sqrt{7}\right)}\)
A1: Expanding brackets and collecting like surds
\(k = \dfrac{12 + 2\sqrt{5}\sqrt{7} + 2\sqrt{5}\sqrt{6} + 2\sqrt{6}\sqrt{7}}{6 + \sqrt{5}\sqrt{7} + \sqrt{5}\sqrt{6} + \sqrt{6}\sqrt{7}}\)
A1: Finding \(k\) convincingly after M1