Higher June 2022 Paper 2 Q25
25 \(\text{f}(x) = 2x + 5\)
Show that \(\quad 3\text{f}(x) - 12\text{f}^{-1}(x) \quad\) simplifies to an integer. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x = 2y + 5\) or \(x - 5 = 2y\) or \(y - 5 = 2x\) or \(\dfrac{y - 5}{2}\) | M1 | oe eg \(x = 2\text{f}^{-1} + 5\) or \(\text{f}(x) - 5 = 2x\) |
| \(\dfrac{x - 5}{2}\) | A1 | oe eg \(\dfrac{x}{2} - \dfrac{5}{2}\) may be implied eg by \(12\text{f}^{-1}(x) = 6(x - 5)\) implied by \(\dfrac{y - 5}{2}\) if \(\dfrac{x - 5}{2}\) used in subsequent working |
| Correctly expands \(3(2x + 5) - 12 \times\) their \(\dfrac{x - 5}{2}\) to a linear expression | M1 | \(6x + 15 - 6x + 30\) if M1A1 their \(\dfrac{x - 5}{2}\) must be a function of \(x\) their \(\dfrac{x - 5}{2}\) cannot be \(2x + 5\) implied by a correct linear expression or value for \(3(2x + 5) - 12 \times\) their \(\dfrac{x - 5}{2}\) |
| \(\dfrac{x - 5}{2}\) and 45 | A1 |
Additional guidance
| 45 with no working | Zero |
| 45 from wrong working does not score 4 marks – mark the working seen | |
| First A1 Condone \(y = \dfrac{x - 5}{2}\) or \(\text{f} = \dfrac{x - 5}{2}\) or \(\text{f}(x) = \dfrac{x - 5}{2}\) or \(x = \dfrac{x - 5}{2}\) | |
| For \(6x + 15 - 6x + 30\) allow \(\dfrac{12x + 30 - 12x + 60}{2}\) but not \(6x + 15 - \dfrac{12x - 60}{2}\) | |
| \(x = 2y + 5 \qquad \dfrac{x + 5}{2}\) | M1A0 |
| \(6x + 15 - \dfrac{12x}{2} - \dfrac{60}{2}\) (implied by \(-15\)) | M1A0 |
| \(-2x - 5\) | M0A0 |
| \(6x + 15 + 24x + 60\) (implied by \(30x + 75\)) | M1A0 |