Higher June 2025 Paper 1 Q15
15 \(\left(\sqrt{3}\right)^5 = k\sqrt{3}\) where \(k\) is an integer.
(a) Find the value of \(k\) (1)
(b) Show that \(\dfrac{21}{3 - \sqrt{2}}\) can be written in the form \(c + \sqrt{d}\)
where \(c\) and \(d\) are integers.
Show each stage of your working clearly. (3)
where \(c\) and \(d\) are integers.
Show each stage of your working clearly. (3)
| Scheme | Marks |
|---|---|
| 9 | B1 |
| (1) |
Notes
B1: allow \(9\sqrt{3}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{21}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}}\) or \(\dfrac{21}{3 - \sqrt{2}} \times \dfrac{-3 - \sqrt{2}}{-3 - \sqrt{2}}\) | M1 |
eg \(\dfrac{21\left(3 + \sqrt{2}\right)}{9 - 3\sqrt{2} + 3\sqrt{2} - 2}\) or \(\dfrac{21\left(3 + \sqrt{2}\right)}{3^2 - 2}\) or \(\dfrac{21\left(3 + \sqrt{2}\right)}{9 - 2}\) or \(\dfrac{21\left(3 + \sqrt{2}\right)}{7}\) or \(\dfrac{63 + 21\sqrt{2}}{9 - 2}\) or \(\dfrac{63 + 21\sqrt{2}}{7}\) | M1 |
Working required Answer: \(9 + \sqrt{18}\) | A1 |
| (3) | |
| (4 marks) |
Notes
M1: for explicitly multiplying the numerator and the denominator by \(3 + \sqrt{2}\) or \(-3 - \sqrt{2}\)
M1: dep on M1 (denominator may be 4 terms which all need to be correct)
\(\dfrac{21}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}} = 9 + 3\sqrt{2}\) scores M1M0
A1: dep on M2
SCB1 for \(9 + \sqrt{18}\) gained with no method marks awarded
SCB2 for \(9 + \sqrt{18}\) gained if you would award 1st M1 but not 2nd M1 (total 2 marks)