Higher January 2021 Paper 1R Q15
15
(a) Use algebra to show that \(4.\dot{5}\dot{7} = 4\dfrac{19}{33}\) (2)
(b) Show that \(\dfrac{2}{6 - 3\sqrt{2}}\) can be written in the form \(\dfrac{a + \sqrt{a}}{b}\)
where \(a\) and \(b\) are integers.
Show your working clearly. (3)
where \(a\) and \(b\) are integers.
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
| E.g. \(x = 4.57\ldots\) and \(100x = 457.57\ldots\) or \(10x = 45.757\ldots\) and \(1000x = 4575.7\ldots\) or \(x = 0.57\ldots\) and \(100x = 57.57\ldots\) or \(10x = 5.757\ldots\) and \(1000x = 575.7\ldots\) | M1 |
E.g. \(100x - x = 457.57\ldots - 4.57\ldots = 453\) and \(\dfrac{453}{99} = \dfrac{151}{33}\) or \(4\dfrac{19}{33}\) or \(1000x - 10x = 4575.7\ldots - 45.757\ldots = 4530\) and \(\dfrac{4530}{990} = \dfrac{151}{33}\) or \(4\dfrac{19}{33}\) or \(100x - x = 57.57\ldots - 0.57\ldots = 57\) and \(\dfrac{57}{99}\) or \(\dfrac{19}{33}\) (so) \(4.\dot{5}\dot{7} = 4\dfrac{19}{33}\) or \(1000x - 10x = 575.7\ldots - 5.757\ldots = 570\) and \(\dfrac{570}{990}\) or \(\dfrac{57}{99}\) or \(\dfrac{19}{33}\) (so) \(4.\dot{5}\dot{7} = 4\dfrac{19}{33}\) Working required Answer: Shown | A1 |
| (2) |
Notes
M1: for selecting 2 recurring decimals that when subtracted give a whole number or terminating decimal eg 453 or 4530 etc
eg \(100x = 457.57\ldots\) and \(x = 4.57\ldots\) or \(1000x = 4575.7\ldots\) and \(10x = 45.757\ldots\) with intention to subtract. (If recurring dots not shown then allow \(10x = 45.757\), \(100x = 457.57\), and \(1000x = 4575.7\) to at least 5sf)
or
\(4 + 0.5757\) and eg \(x = 0.57\ldots\), \(100x = 57.57\ldots\) with intention to subtract.
A1: for completion to \(\dfrac{151}{33}\) or \(4\dfrac{19}{33}\)
| Scheme | Marks |
|---|---|
E.g. \(\dfrac{2}{6 - 3\sqrt{2}} \times \dfrac{6 + 3\sqrt{2}}{6 + 3\sqrt{2}}\) or \(\dfrac{2}{6 - 3\sqrt{2}} \times \dfrac{-6 - 3\sqrt{2}}{-6 - 3\sqrt{2}}\) | M1 |
\(\dfrac{12 + 6\sqrt{2}}{36 - 18\sqrt{2} + 18\sqrt{2} - 18}\) or \(\dfrac{12 + 6\sqrt{2}}{18}\) or \(\dfrac{12 + 6\sqrt{2}}{6^2 - \left(3\sqrt{2}\right)^2}\) or \(\dfrac{12 + 6\sqrt{2}}{6^2 - 9 \times 2}\) | M1 |
Working required Answer: \(\dfrac{2 + \sqrt{2}}{3}\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: for rationalising the denominator by multiplying numerator and denominator by \(6 + 3\sqrt{2}\) (or \(-6 - 3\sqrt{2}\))
M1: (numerator may be expanded or denominator may be 4 terms which need to be all correct)
A1: or for stating \(a = 2\) and \(b = 3\)
for \(\dfrac{2 + \sqrt{2}}{3}\) from correct working
dep on M2