Higher November 2020 Paper 1 Q13
13
(a) \(s\) and \(t\) are positive integers.
\((x + s)(x - t) \quad\) is expanded and simplified.
The answer is \(\quad x^2 + kx - 40 \quad\) where \(k\) is a positive integer.
Work out the smallest possible value of \(k\). [2 marks]
(b) Faisal tries to solve \(\quad (x + 2)(x - 7) = 0\)
Here is his working.
| \((x + 2) = 0\) | or | \((x - 7) = 0\) | |
| Answer | \(x = 2\) | or | \(x = 7\) |
Give a reason why his answer is wrong. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \((x + 8)(x - 5)\) or \((k =)\ 3\) or \((x + 5)(x - 8)\) or \((k =)\ {-}3\) or \((x + 10)(x - 4)\) or \((k =)\ 6\) or \((x + 4)(x - 10)\) or \((k =)\ {-}6\) or \((x + 20)(x - 2)\) or \((k =)\ 18\) or \((x + 2)(x - 20)\) or \((k =)\ {-}18\) or \((x + 40)(x - 1)\) or \((k =)\ 39\) or \((x + 1)(x - 40)\) or \((k =)\ {-}39\) or \(s = 8\) and \(t = 5\) or \(8 - 5\) | M1 | oe correct factorisation |
| 3 | A1 | condone embedded answer \(x^2 + 3x - 40\) |
Additional guidance
| \(x^2 + sx - tx - st\) with no further working | M0A0 |
| Ignore incorrect factorisations in working |
| Answer | Mark | Comments |
|---|---|---|
| Valid reason | B1 | eg it should be \(-2\) or \(4 \times -5\) isn’t 0 or \((2 + 2)(2 - 7) = -20\) or \(2 + 2 = 4\) or \(2 + 2 \neq 0\) |
Additional guidance
| ‘He didn’t change the sign on the left’ | B1 |
| ‘If you substitute 2 it does not give 0’ | B1 |
| \(x = 2\) is wrong | B1 |
| \(x = -2\) (and \(x = 7\)) | B1 |
| \(x = -2\) and \(x = -7\) | B0 |
| ‘One solution is wrong’ or ‘Only one answer is correct’ | B0 |
| \(x = 2\) | B0 |
| Ignore statements which do not contradict a correct answer |