Higher June 2019 Paper 1R Q18
18
(a) Expand and simplify \((x + 2)(2x + 3)(x - 7)\)
Show your working clearly. (3)
Show your working clearly. (3)
(b) Make \(m\) the subject of \(p^2 = \dfrac{x + m}{2m - y}\) (3)
| Scheme | Marks |
|---|---|
| \((x + 2)(2x + 3) = 2x^2 + 3x + 4x + 6\) \((2x + 3)(x - 7) = 2x^2 - 14x + 3x - 21\) \((x + 2)(x - 7) = x^2 - 7x + 2x - 14\) | M1 |
| \((2x^2 + 7x + 6)(x - 7) = 2x^3 - 14x^2 + 7x^2 - 49x + 6x - 42\) \((2x^2 - 11x - 21)(x + 2) = 2x^3 + 4x^2 - 11x^2 - 22x - 21x - 42\) \((x^2 - 5x - 14)(2x + 3) = 2x^3 + 3x^2 - 10x^2 - 15x - 28x - 42\) | M1dep |
| Working required Answer: \(2x^3 - 7x^2 - 43x - 42\) | A1 |
| (3) |
Notes
M1: For multiplying a pair of brackets and getting 3 out of 4 terms correct.
M1dep: For multiplying the product of the first 2 brackets (ft from the 1st stage) by the 3rd bracket, and getting at least 3 out of 6 or 4 out of 8 terms correct
A1: Fully correct. isw extra work as long as correct e.g. \(x(2x^2 - 7x - 43) - 42\)
| Scheme | Marks |
|---|---|
| Alternative (all in one method) \((x + 2)(2x + 3)(x - 7) =\) \(2x^3 - 14x^2 + 3x^2 - 21x + 4x^2 - 28x + 6x - 42\) | M2 |
| Working required Answer: \(2x^3 - 7x^2 - 43x - 42\) | A1 |
Notes
M2: For at least 6 out of 8 correct terms
(M1 for 4 or 5 out of 8 correct terms)
| Scheme | Marks |
|---|---|
| \(p^2(2m - y) = x + m\) \(2p^2m - p^2y = x + m\) | M1 |
| e.g. \(2p^2m - m = x + p^2y\) \(m(2p^2 - 1) = x + p^2y\) | M1 |
| \(m = \dfrac{x + p^2y}{2p^2 - 1}\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: Multiplying by denominator and expanding bracket
M1: Collect terms in \(m\) and factorise in a correct equation
A1: oe eg \(m = \dfrac{-x - p^2y}{1 - 2p^2}\)
must have \(m =\)