June 2019 Paper 1 Q4
4 \(p(x - 1) + 2(3x + k) \equiv 4(x + 2) \qquad\) where \(p\) and \(k\) are integers.
Work out the values of \(p\) and \(k\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(px - p + 6x + 2k = 4x + 8\) or \(px + 6x = 4x\) or \(p + 6 = 4\) | M1 | oe |
| \(p = -2\) | A1 | This could imply first M mark if not seen |
| \(2k - \text{their } p = 8\) or \(2k = \text{their } p + 8\) | M1 | oe could be awarded by substituting a value of \(x\) with \(p = -2\) |
| \(k = 3\) | A1ft | need to check back for ft mark |
| Alternative method 2 | ||
| A correct equation obtained by substituting a value for \(x\) in the identity | M1 | eg \(\ x = 0 \quad 2k - p = 8\) \(\phantom{\text{eg }} x = 1 \quad p - p + 6 + 2k = 12\) \(\phantom{\text{eg }} x = 2 \quad 2p - p + 12 + 2k = 16\) |
| A second correct equation obtained by substituting a value for \(x\) in the identity | M1 | oe could go back to equating coefficients at this stage |
| \(p = -2\) | A1 | |
| \(k = 3\) | A1 | may come from one equation by substituting \(x = 1\) |
Additional guidance
| Correct expansion, then \(p + 6 = 4\) followed by \(p = 2\) (incorrect) would give \(k = 5\) on ft … allow ft mark for \(k\) | M1, A0 M1, A1ft |
| In Alt 2 substituting \(x = 1\) leads to \(k = 3\) (a second equation would be needed to gain further marks) | M1, A1 |