Higher January 2022 Paper 1 Q13
13
| Scheme | Marks |
|---|---|
| \(5x(x + 2) = 5x^2 + 10x\) or \((x + 2)(3x - 4) = 3x^2 - 4x + 6x - 8\;(= 3x^2 + 2x - 8)\) or \(5x(3x - 4) = 15x^2 - 20x\) | M1 |
| eg \([(5x^2 + 10x)(3x - 4) =]\;15x^3 - 20x^2 + 30x^2 - 40x\) or \([5x(3x^2 + 2x - 8) =]\;15x^3 + 10x^2 - 40x\) or \([(x + 2)(15x^2 - 20x) =]\;15x^3 - 20x^2 + 30x^2 - 40x\) | M1 |
| \(15x^3 + 10x^2 - 40x\) | A1 |
| (3) |
Notes
M1: for a correct intention to multiply all 3 factors by starting to multiply 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 \(15x^3 - 20x^2 + 30x^2 - 40x\) without firstly expanding two factors, allow two errors)
A1: isw correct factorisation eg \(5(3x^3 + 2x^2 - 8x)\)
do not isw incorrect factorisation eg \(15x^3 + 10x^2 - 40x = 3x^3 + 2x^2 - 8x\)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{2w^2}{y^5}\right)^{-3}\) or \(\left(\dfrac{y^{20}}{16w^8}\right)^{\frac{3}{4}}\) or \(\left(\dfrac{4096w^{24}}{y^{60}}\right)^{-\frac{1}{4}}\) | M1 |
\(\left(\dfrac{8w^6}{y^{15}}\right)^{-1}\) or \(\dfrac{2^{-3}w^{-6}}{y^{-15}}\) or \(\dfrac{\frac{1}{8}w^{-6}}{y^{-15}}\) or \(\left(\dfrac{y^5}{2w^2}\right)^3\) or \(\left(\dfrac{y^{60}}{4096w^{24}}\right)^{\frac{1}{4}}\) or \(\dfrac{0.125y^{15}}{w^6}\) or \(\dfrac{0.125w^{-6}}{y^{-15}}\) or \(\dfrac{0.125}{y^{-15}w^6}\) oe | M1 |
| \(\dfrac{y^{15}}{8w^6}\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for one of fourth rooting or reciprocating or cubing
M1: for two of fourth rooting or reciprocating or cubing
A1: allow \(\dfrac{y^{15}}{8w^6}\) or \(\dfrac{y^{15}w^{-6}}{8}\) or \(0.125y^{15}w^{-6}\) or \(\dfrac{1}{8}y^{15}w^{-6}\) or \(\dfrac{w^{-6}}{8y^{-15}}\) or \(\dfrac{1}{8y^{-15}w^6}\)
ALTERNATIVE
| Scheme | Marks |
|---|---|
| M2 | |
| \(\dfrac{y^{15}}{8w^6}\) | A1 |
Notes
M2: for 2 correct terms
(M1 for 1 correct term)
A1: allow \(\dfrac{y^{15}}{8w^6}\) or \(\dfrac{y^{15}w^{-6}}{8}\) or \(0.125y^{15}w^{-6}\) or \(\dfrac{1}{8}y^{15}w^{-6}\) or \(\dfrac{w^{-6}}{8y^{-15}}\) or \(\dfrac{1}{8y^{-15}w^6}\)