Higher January 2022 Paper 2 Q25
25 The function g is defined as
\(\mathrm{g} : x \mapsto 5 + 6x - x^2\) with domain \(\{x : x \geqslant 3\}\)
(a) Express the inverse function \(\mathrm{g}^{-1}\) in the form \(\mathrm{g}^{-1} : x \mapsto \ldots\) (4)
(b) State the domain of \(\mathrm{g}^{-1}\) (1)
| Scheme | Marks |
|---|---|
| \((x - 3)^2\) or \((3 - x)^2\) or \((y - 3)^2\) or \((3 - y)^2\) | M1 |
| 14 or – 14 | M1 |
| \(3 \pm \sqrt{14 - x}\) or \(3 \pm \sqrt{14 - y}\) | M1 |
| \(3 + \sqrt{14 - x}\) | A1 |
| (4) |
Notes
M1: As part of an expression in \(x\) or \(y\) or an equation in \(x\) and \(y\)
M1: Can be \(\pm\) or – or +
A1: oe must be in \(x\)
ALTERNATIVE
| Scheme | Marks |
|---|---|
| Alternative method: \(x^2 - 6x + (y - 5) = 0\) oe or \(y^2 - 6y + (x - 5) = 0\) oe | M1 |
| \(y = \dfrac{6 \pm \sqrt{36 - 4(x - 5)}}{2}\) or \(x = \dfrac{6 \pm \sqrt{36 - 4(y - 5)}}{2}\) | M1 |
| \(3 \pm \sqrt{14 - x}\) or \(3 \pm \sqrt{14 - y}\) | M1 |
| \(3 + \sqrt{14 - x}\) | A1 |
Notes
M1: rearrange to form a quadratic in \(x\) or \(y\)
terms can be in any order but must be in an equation equal to zero
M1: correct substitution into quadratic formula
M1: Can be \(\pm\) or – or +
A1: oe must be in \(x\)
| Scheme | Marks |
|---|---|
| \(x \leqslant 14\) | B1 |
| (1) | |
| (5 marks) |
Notes
B1: oe must ft from part (a) dep on an answer in correct form