Higher January 2023 Paper 2R Q12
12
(a) Simplify \(\;\dfrac{2}{y^0}\) (1)
(b) Simplify fully \(\;\left(16a^4\right)^{\frac{3}{4}}\) (2)
(c) Expand and simplify \(\;5x(3x + 4)(2x - 1)\) (3)
| Scheme | Marks |
|---|---|
| 2 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(8a^3\) | B2 |
| (2) |
Notes
B2: for \(8a^3\)
If not B2 then B1 for \(8a^k\) where \(k \ne 3\) or \(ka^3\) where \(k \ne 8\)
| Scheme | Marks |
|---|---|
| \(5x(3x + 4) = 15x^2 + 20x\) or \(5x(2x - 1) = 10x^2 - 5x\) or \((3x + 4)(2x - 1) = 6x^2 - 3x + 8x - 4\) \((= 6x^2 + 5x - 4)\) | M1 |
| \((15x^2 + 20x)(2x - 1) = 30x^3 - 15x^2 + 40x^2 - 20x\) oe \((10x^2 - 5x)(3x + 4) = 30x^3 + 40x^2 - 15x^2 - 20x\) oe \(5x(6x^2 + 5x - 4) = 30x^3 + 25x^2 - 20x\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(30x^3 + 25x^2 - 20x\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for a correct intention to multiply all 3 factors by multiplying 2 factors only, allow one error
M1: (dep)ft for expanding by the third factor, allow one error
(some may do the expansion in one stage and will get to \(30x^3 - 15x^2 + 40x^2 - 20x\) without firstly expanding two factors – this gains M2, allow one error)
A1: isw correct factorisation (\(30x^3 + 25x^2 - 20x\) must be seen previously to award 3 marks)
eg
\(5(6x^3 + 5x^2 - 4x)\)
\(x(30x^2 + 25x - 20)\)
\(5x(6x^2 + 5x - 4)\)
do not isw incorrect simplification
eg \(30x^3 + 25x^2 - 20x = 6x^3 + 5x^2 - 4x\) gets M2A0