Higher June 2019 Paper 2 Q24
24 The function f is such that \(\mathrm{f}(x) = 3x - 2\)
(a) Find \(\mathrm{f}(5)\) (1)
The function g is such that \(\mathrm{g}(x) = 2x^2 - 20x + 9\) where \(x \geqslant 5\)
(b) Express the inverse function \(\mathrm{g}^{-1}\) in the form \(\mathrm{g}^{-1}(x) = \ldots\) (4)
| Scheme | Marks |
|---|---|
| 13 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
\(y = 2(x^2 - 10x) + 9\) or \(y = 2\left(x^2 - 10x + \dfrac{9}{2}\right)\) | M1 |
e.g. \(y = 2\left((x - 5)^2 - 5^2\right) + 9\) or \(y = 2\left((x - 5)^2 - 5^2 + \dfrac{9}{2}\right)\) or \(y = 2(x - 5)^2 - 41\) oe | M1 |
| \((x - 5)^2 = \dfrac{y + 41}{2}\) oe | M1 |
| \(5 + \sqrt{\dfrac{x + 41}{2}}\) | A1 |
| (4) | |
| (5 marks) |
Notes
M1: for a correct equation for a first step in order to complete the square
M1: dep
A1: oe
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse
| Scheme | Marks |
|---|---|
| \(2x^2 - 20x + (9 - y) = 0\) | M1 |
\(x = \dfrac{20 \pm \sqrt{400 - 8(9 - y)}}{4}\) or \(x = \dfrac{20 + \sqrt{400 - 8(9 - y)}}{4}\) | M1 |
| \(x = 5 \pm \sqrt{\dfrac{41 + y}{2}}\) oe | M1 |
| \(5 + \sqrt{\dfrac{x + 41}{2}}\) | A1 |
Notes
M1: for a correct first step
M1: dep
A1: oe
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse
| Scheme | Marks |
|---|---|
| \(2x^2 - 20x + (9 - y)\;(= 0)\) | M1 |
e.g. \(2\left((x - 5)^2 - 5^2\right) + 9 - y\) (= 0) or \(2\left((x - 5)^2 - 5^2 + \dfrac{9}{2}\right) - y\) (= 0) or \(2(x - 5)^2 - 41 - y\) (= 0) | M1 |
| \((x - 5)^2 = \dfrac{y + 41}{2}\) oe | M1 |
| \(5 + \sqrt{\dfrac{x + 41}{2}}\) | A1 |
Notes
M1: for a correct first step
M1: dep
A1: oe
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse