Higher November 2019 Paper 3 Q28
28 \(\mathrm{f}(x) = 2x - 3 \quad\) and \(\quad \mathrm{g}(x) = x^2\)
Show that \(\quad \mathrm{f}^{-1}(55) = \mathrm{fg}(4)\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{y + 3}{2} = x\) or \(x = 2y - 3\) and \(x + 3 = 2y\) or \(2x - 3 = 55\) | M1 | |
| \(\dfrac{x + 3}{2}\) or \(\dfrac{55 + 3}{2}\) | A1 | |
| \(2x^2 - 3\) or \(2 \times 4^2 - 3\) or \(2 \times 16 - 3\) | M1 | |
| \(\dfrac{55 + 3}{2} = 29\) and \(2 \times 4^2 - 3 = 29\) or \(2 \times 16 - 3 = 29\) | A1 |
Additional guidance
| 29 with no working or only from incorrect working | M0A0M0A0 |