Higher November 2021 Paper 1 Q23
23 Rearrange \(\quad y = \dfrac{1}{\sqrt{x + 1}} \quad\) to make \(x\) the subject. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(y\sqrt{x + 1} = 1\) or \(\sqrt{x + 1} = \dfrac{1}{y}\) or \(y^2 = \dfrac{1}{x + 1}\) | M1 | |
| \(y^2(x + 1) = 1\) or \(y^2x + y^2 = 1\) or \(y^2x = 1 - y^2\) or \(x + 1 = \dfrac{1}{y^2}\) or \(\dfrac{1}{y^2} - 1\) or \(\dfrac{1 - y^2}{y^2}\) | M1dep | |
| \(x = \dfrac{1}{y^2} - 1\) or \(x = \dfrac{1 - y^2}{y^2}\) | A1 | oe in the form \(x =\) |
Additional guidance
| Correct answer in working repeated on answer line without \(x =\) | |
| eg \(x = \dfrac{1}{y^2} - 1\) seen in working with answer \(\dfrac{1}{y^2} - 1\) | M1M1A1 |
| Allow \(\left(\dfrac{1}{y}\right)^2\) for \(\dfrac{1}{y^2}\) throughout | |
| Allow \(1^2\) for 1 throughout |