Higher January 2022 Paper 1 Q6
6
(a) Solve the inequality \(5x - 7 \leqslant 2\) (2)
(b)
(i) Factorise \(y^2 - 2y - 35\) (2)
(ii) Hence, solve \(y^2 - 2y - 35 = 0\) (1)
| Scheme | Marks |
|---|---|
| \(5x \leqslant 2 + 7\) or \(5x \leqslant 9\) or \(\dfrac{5x}{5} - \dfrac{7}{5} \leqslant \dfrac{2}{5}\) oe | M1 |
| \(x \leqslant 1.8\) | A1 |
| (2) |
Notes
M1: allow any sign instead of \(\leqslant\)
or for an answer of 1.8 oe
or \(x\) and 1.8 oe with the incorrect sign
A1: oe
| Scheme | Marks |
|---|---|
| (i) \((y \pm 7)(y \pm 5)\) | M1 |
| (i) \((y - 7)(y + 5)\) | A1 |
| (ii) 7, −5 | B1ft |
| (3) | |
| (5 marks) |
Notes
M1: for \((y \pm 7)(y \pm 5)\)
or \((y + a)(y + b)\) where \(ab = -35\)
or \(a + b = -2\)
A1: isw if student goes on to solve the equation in this part
B1ft: answer must ft from their \((y + a)(y + b)\) in (b)(i). Award B0 for 7, −5 if no marks scored in (i)