Higher June 2022 Paper 2 Q15
15
Show clear algebraic working. (4)
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
eg \(\dfrac{2(4x + 5) - 3(3 - 2x)}{6}\;(= 13)\) oe or \(\dfrac{2(4x + 5)}{6} - \dfrac{3(3 - 2x)}{6}\;(= 13)\) \(2(4x + 5) - 3(3 - 2x) = 13 \times 3 \times 2\) oe | M1 |
| eg \(8x + 10 - 9 + 6x = 78\) oe eg \(14x + 1 = 78\) | M1ft |
| eg \(8x + 6x = 78 - 10 + 9\) oe eg \(14x = 77\) | M1ft |
| Working required Answer: 5.5 | A1 |
| (4) |
Notes
M1: Writing fractions over a common denominator or removing denominator
If the student has removed the denominator at this stage then a correct method must be shown or implied
M1ft: ft dep on previous M1 removing brackets and fractions correctly in an equation
M1ft: ft dep on previous M1 terms in \(x\) on one side and number terms the other
A1: oe eg \(\dfrac{11}{2}\) dep on M2
Allow one error in removal of brackets
| Scheme | Marks |
|---|---|
\((2y + 5)(y - 6)\) or \(\dfrac{--7 \pm \sqrt{(-7)^2 - 4 \times 2 \times -30}}{2 \times 2}\) \(2\left[\left(y - \dfrac{7}{4}\right)^2 - \dfrac{49}{16}\right] - 30\;(= 0)\) oe | M1 |
| \((y =)\;6\), \((y =)\;-2.5\) | A1 |
| Working required Answer: \(-2.5 \leqslant y \leqslant 6\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1: A correct method to solve the quadratic - allow factorisation that gives 2 out of 3 terms correct when expanded or use of quadratic formula – if using formula, allow one sign error and allow if simplified as far as \(\dfrac{7 \pm \sqrt{49 + 240}}{4}\) or use of completing the square with one sign error as far as shown
A1: Correct critical values dep on M1
A1: oe eg \(y \geqslant -2.5\) (and) \(y \leqslant 6\) or \([-2.5, 6]\)
(do not penalise change of variable eg \(y\) to \(x\))
dep on M1