June 2016 Paper 1 Q7
7 Work out the integer values of \(x\) for which \(\qquad x^2 - 20x + 96 \lt 0\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x \pm a)(x \pm b)\) | M1 | \(ab = 96\) or \(a + b = -20\) |
| \((x - 8)(x - 12)\) or (\(x\) =) 8 and (\(x\) =) 12 | A1 | |
| 9, 10 and 11 | A1 | A0 if extra values seen |
| Alternative method 2 | ||
| \((x - 10)^2 - 100\ (+ 96)\ (\lt 0)\) | M1 | oe eg \(\ (x - 10)^2 - 4\ (\lt 0)\) |
| \(8 \lt x \lt 12\) or (\(x\) =) 8 and (\(x\) =) 12 | A1 | |
| 9, 10 and 11 | A1 | A0 if extra values seen |
| Alternative method 3 | ||
| \(\dfrac{20 \pm \sqrt{\{(20)^2 - 4 \times 1 \times 96\}}}{2}\) or \(\dfrac{20 \pm \sqrt{\{(-20)^2 - 4 \times 1 \times 96\}}}{2}\) | M1 | accept \((20)^2\) or \((-20)^2\) for \(b^2\) in the discriminant |
| 8 and 12 | A1 | |
| 9, 10 and 11 | A1 | A0 if extra values seen |
Additional guidance
9, 10 and 11 using Trial and Improvement - all correct is 3 marks, otherwise 0 marks.
No working ... treat as Trial and Improvement
For alt 3 ... substitution in the formula must be correct