Foundation January 2021 Paper 1 Q12
12
(a) Expand \(\;x(4 - x)\) (1)
\(t = ab - c\)
\(a = 1.5 \qquad b = 2.4 \qquad c = -5.6\)
(b) Work out the value of \(t\). (2)
(c) Make \(d\) the subject of \(\;y = dx - e\) (2)
| Scheme | Marks |
|---|---|
| \(4x - x^2\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| e.g. 1.5 × 2.4 – (−5.6) or 1.5 × 2.4 + 5.6 or 3.6 + 5.6 oe | M1 |
| 9.2 | A1 |
| (2) |
Notes
M1: for a correct substitution
A1: accept \(\dfrac{46}{5}\) or \(9\dfrac{1}{5}\)
| Scheme | Marks |
|---|---|
| \(y + e = dx\) oe or \(\dfrac{y}{x} = d - \dfrac{e}{x}\) | M1 |
| \(d = \dfrac{y + e}{x}\) | A1 |
| (2) | |
| (5 marks) |
Notes
M1: for a correct first step
A1: oe e.g. \(d = \dfrac{y}{x} + \dfrac{e}{x}\)