Higher June 2018 Paper 2 Q1
1
(a) Make \(a\) the subject of the formula \(M = ac - bd\) (2)
(b) Solve the inequality \(5x - 4 \lt 39\) (2)
(c) Factorise fully \(18e^2f^3 - 12e^3f\) (2)
| Scheme | Marks |
|---|---|
\(ac = M + bd\) or \(-ac = -M - bd\) or \(\dfrac{M}{c} = a - \dfrac{bd}{c}\) | M1 |
| \(a = \dfrac{M + bd}{c}\) | A1 |
| (2) |
Notes
M1: For a correct first stage
A1: oe, eg \(a = \dfrac{M}{c} + \dfrac{bd}{c}\), \(a = \dfrac{-M - bd}{-c}\)
[must have been seen with \(a\) = to award accuracy mark]
| Scheme | Marks |
|---|---|
| \(5x \lt 39 + 4\) oe | M1 |
| \(x \lt 8\dfrac{3}{5}\) | A1 |
| (2) |
Notes
M1: Accept as equation or with the wrong inequality sign. Also award M1 for an answer of 8.6 or 8.6 with an = sign or the incorrect inequality sign.
A1: Accept \(x \lt \dfrac{43}{5}\) or \(x \lt 8.6\) or [−∞, 8.6)
| Scheme | Marks |
|---|---|
| eg \(6e^2(3f^3 - 2ef)\), eg \(2f(9e^2f^2 - 6e^3)\) eg \(ef(18ef^2 - 12e^2)\) | M1 |
| \(6e^2f(3f^2 - 2e)\) | A1 |
| (2) | |
| (6 marks) |
Notes
M1: Any correct partially factorised expression with at least 2 terms in the common factor or for the correct common factor and a 2 term expression inside the brackets with just one error