Higher November 2019 Unit 1 Q13
13.
(a) Show that \[6x(8x + 1) - 3(4x + 1)(4x - 1) \equiv 6x + 3.\] [3]
(b) Hence, write down the value of \(x\) for which \(6x(8x + 1) - 3(4x + 1)(4x - 1) = 0\). [1]
| Scheme | Marks | Comments |
|---|---|---|
| \(48x^2 + 6x - 48x^2 + 12x - 12x + 3\) OR \(48x^2 + 6x - 48x^2 + 3\). | B2 | Accept \(48x^2 + 6x - (48x^2 - 12x + 12x - 3)\) or \(48x^2 + 6x - (48x^2 - 3)\) B1 for \(16x^2\; [- 4x + 4x] - 1\) or \(48x^2\; [- 12x + 12x] - 3\) or \(-48x^2\; [+ 12x - 12x] + 3\). OR B1 if one error or incorrect (or extra or missing) term within entire expression. (An incorrect term may be implied e.g. \(-24x\) implies \(-12x - 12x\)). |
| \(6x + 3\) | B1 | Must be convincing. For last B1, do not accept \(48x^2 + 6x - (48x^2 - 12x + 12x - 3)\) or \(48x^2 + 6x - (48x^2 - 3)\) without further correct work seen before final \(6x + 3\). If no work seen in (a), allow marks in (a) for work shown in (b) |
| Scheme | Marks | Comments |
|---|---|---|
| \(-\dfrac{1}{2}\) or \(-\dfrac{3}{6}\) or – 0.5 or equivalent | B1 | Mark final answer. |