Higher June 2018 Paper 1 Q3
3
(a) Simplify \(y^5 \times y^9\) (1)
(b) Simplify \((2m^3)^4\) (2)
(c) Solve \(5(x + 3) = 3x - 4\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
(d)
(i) Factorise \(x^2 + 2x - 24\) (2)
(ii) Hence, solve \(x^2 + 2x - 24 = 0\) (1)
| Scheme | Marks |
|---|---|
| \(y^{14}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(16m^{12}\) | B2 |
| (2) |
Notes
B2: if not B2 then
B1 for \(am^{12}\) or \(16m^b\) or \(2^4m^{12}\) \(b \ne 0, 12\) \(a \ne 1, 16\)
| Scheme | Marks |
|---|---|
\(5x + 15 = 3x - 4\) or \(x + 3 = \dfrac{3x}{5} - \dfrac{4}{5}\) | M1 |
| e.g. \(5x - 3x = -4 - 15\) | M1 |
Working required Answer: \(-\dfrac{19}{2}\) oe | A1 |
| (3) |
Notes
M1: for removing bracket in a correct equation or dividing all terms by 5 in a correct equation
M1: ft from \(ax + b = cx + d\) for correctly isolating terms in \(x\) on one side of equation and constant terms on the other side
A1: dep on at least M1
| Scheme | Marks |
|---|---|
| M1 | |
| \((x - 4)(x + 6)\) | A1 |
| (2) |
Notes
M1: for \((x + a)(x + b)\) where either \(ab = -24\) or \(a + b = +2\)
e.g. \((x - 6)(x + 4)\)
| Scheme | Marks |
|---|---|
| 4, −6 | B1 |
| (1) | |
| (9 marks) |
Notes
B1: cao or ft from any \((x + p)(x + q)\)