Higher January 2020 Paper 2 Q16
16 Make \(x\) the subject of \(\;y = \sqrt{\dfrac{x + 1}{x - 4}}\)
(4)
| Scheme | Marks |
|---|---|
| \(y^2 = \dfrac{x + 1}{x - 4}\) | M1 |
| \(y^2(x - 4) = x + 1\) or \(y^2x - 4y^2 = x + 1\) | M1 |
| \(y^2x - x = 4y^2 + 1\) or \(-4y^2 - 1 = x - y^2x\) or \(x(y^2 - 1) = 4y^2 + 1\) or \(-4y^2 - 1 = x(1 - y^2)\) | M1 |
| \(x = \dfrac{4y^2 + 1}{y^2 - 1}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for squaring
M1: for removing the fraction
M1: for expanding the bracket and rearranging for \(x\) so that the terms in \(x\) are on one side of the correct equation
A1: for \(x = \dfrac{4y^2 + 1}{y^2 - 1}\) or \(x = \dfrac{-4y^2 - 1}{1 - y^2}\)
(need to see \(x\) = somewhere)