Higher June 2024 Paper 3 Q19
19 \(\mathrm{f}(x) = x - 3 \qquad \mathrm{g}(x) = 4x - 7\)
(a) Work out the value of \(\ \mathrm{fg}(6)\) [2 marks]
(b) Solve \(\quad (\mathrm{f}(x))^2 = \mathrm{g}(x)\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(4 \times 6 - 7\) or \(24 - 7\) or 17 | M1 | |
| 14 | A1 | |
| Alternative method 2 | ||
| \(4x - 7 - 3\) or \(4x - 10\) | M1 | |
| 14 | A1 | |
| Answer | Mark | Comments |
|---|---|---|
| \((x - 3)^2 = 4x - 7\) or \(x^2 - 6x + 9 = 4x - 7\) | M1 | oe equation |
| \(x^2 - 10x + 16\ (= 0)\) | M1dep | oe their 3-term quadratic equation with terms collected correctly |
| \((x - 2)(x - 8)\) or \(\dfrac{--10 \pm \sqrt{(-10)^2 - 4 \times 1 \times 16}}{2 \times 1}\) or \(5 \pm \sqrt{9}\) | M1 | oe correct for their 3-term quadratic |
| \(x = 2\) and \(x = 8\) | A1 |
Additional guidance
| \((x - 3)^2 = 4x - 7 \quad x^2 + 9 = 4x - 7 \quad x^2 - 4x + 16\ (= 0)\) | M1M1 |
| \((x - 3)^2 = 4x - 7 \quad x^2 + 9 = 4x - 7 \quad x^2 - 4x + 2\ (= 0) \quad x = 2 \pm \sqrt{2}\) correct answers imply 3rd M | M1M0M1A0 |
| \((x - 3)^2 = 4x - 7 \quad x^2 + 9 = 4x - 7 \quad x^2 - 4x + 2\ (= 0)\) | M1M0 |