Higher June 2018 Paper 2 Q28
28 \(\mathrm{f}(x) = 5 - x \quad\) and \(\quad \mathrm{g}(x) = 3x + 7\)
(a) Simplify \(\quad \mathrm{f}(2x) + \mathrm{g}(x - 1)\) [3 marks]
(b) Solve \(\quad \mathrm{g}^{-1}(x) = 2x\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(5 - 2x\) | B1 | may be implied |
| \(3(x - 1) + 7\) or \(3x + 4\) | M1 | oe ignore incorrect expansion if \(3(x - 1) + 7\) seen |
| \(9 + x\) | A1 |
Additional guidance
| Working out \(2\mathrm{f}(x)\) | B0 |
| Working out \(\mathrm{g}(x + 1)\) | M0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(x - 7 = 3y\) or \(y - 7 = 3x\) | M1 | allow \(x - 7 = 3g\) or \(g - 7 = 3x\) |
| \(\dfrac{x - 7}{3}\) or \(\dfrac{y - 7}{3}\) | A1 | oe allow \(\dfrac{g - 7}{3}\) |
| \(-1.4\) or \(-\dfrac{7}{5}\) | A1 | oe |
| Alternative method 2 | ||
| \(3(2x) + 7\) | M1 | oe |
| \(x = 3(2x) + 7\) or \(x = 6x + 7\) | A1 | oe equation |
| \(-1.4\) or \(-\dfrac{7}{5}\) | A1 | oe |
Additional guidance
| Beware \(-3x - 7 = 2x\) leading to \(-1.4\) | M0A0A0 |