4MA0 Higher June 2015 Paper 4 Q19
19
(a) Show that \(\left(5 - \sqrt{8}\right)\left(7 + \sqrt{2}\right) = 31 - 9\sqrt{2}\)
Show each stage of your working. (3)
Show each stage of your working. (3)
Given that \(c\) is a prime number,
(b) rationalise the denominator of \(\dfrac{3c - \sqrt{c}}{\sqrt{c}}\)
Simplify your answer. (2)
Simplify your answer. (2)
| Scheme | Marks |
|---|---|
\(5 \times 7 + 5\sqrt{2} - 7\sqrt{8} - \sqrt{8}\sqrt{2}\) Or \(35 + 5\sqrt{2} - 7\sqrt{8} - \sqrt{8}\sqrt{2}\) Or \(35 + 5\sqrt{2} - 7\sqrt{8} - \sqrt{16}\) Or \(35 + 5\sqrt{2} - 7(2\sqrt{2}) - \sqrt{16}\) | M1 |
| \(35 + 5\sqrt{2} - 14\sqrt{2} - 4\) or \(31 + 5\sqrt{2} - 14\sqrt{2}\) | M1dep |
| \(31 - 9\sqrt{2}\) | A1 |
| (3) |
Notes
M1: 4 terms, allow one sign error
Note 5×7 may be 35
\(-7\sqrt{8}\) may be \(-7 \times 2\sqrt{2}\) but not \(-14\sqrt{2}\)
\(\sqrt{8}\sqrt{2}\) may be \(\sqrt{16}\) or 4
or for \(\sqrt{8} = 2\sqrt{2}\)
A1: show from correct working
| Scheme | Marks |
|---|---|
| \(\dfrac{3c - \sqrt{c}}{\sqrt{c}} \times \dfrac{\sqrt{c}}{\sqrt{c}}\) or \(\dfrac{3c\sqrt{c} - c}{c}\) or \(\dfrac{\sqrt{c}(3\sqrt{c} - 1)}{\sqrt{c}}\) | M1 |
| \(3\sqrt{c} - 1\) | A1 |
| (2) |