Higher June 2022 Paper 1 Q24
24
(a) Simplify fully \(\quad \dfrac{6}{a} - \dfrac{11}{4a}\) [2 marks]
(b) Simplify fully \(\quad (y^2 - 3y) \times \dfrac{y^2 + 10y + 21}{y^2 - 9}\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\left(\dfrac{6}{a} =\right) \dfrac{24}{4a}\) or converts both fractions to a common denominator or correct unsimplified fraction eg \(\dfrac{26}{8a}\) or \(\dfrac{13a}{4a^2}\) or \(\dfrac{3.25}{a}\) | M1 | oe eg \(\dfrac{48}{8a}\) and \(\dfrac{22}{8a}\) or \(\dfrac{24a}{4a^2}\) and \(\dfrac{11a}{4a^2}\) |
| \(\dfrac{13}{4a}\) | A1 |
Additional guidance
| Do not ignore further work eg \(\dfrac{13}{4a}\) followed by answer \(\dfrac{3.25}{a}\) | M1A0 |
| Allow a division sign rather than a fraction line for M1 only eg \(26 \div 8a\) eg \(13 \div 4a\) | M1A0 M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| \(y(y - 3)\) | M1 | |
| \((y + 7)(y + 3)\) | M1 | |
| \((y + 3)(y - 3)\) | M1 | |
| \(y(y + 7)\) or \(y^2 + 7y\) | A1 | SC1 \(y^4 - 3y^3 + 10y^3 - 30y^2 + 21y^2 - 63y\) or \(y^4 + 7y^3 - 9y^2 - 63y\) |
Additional guidance
| \(y(y + 7)\) or \(y^2 + 7y\) with no other working | M1M1M1A1 |
| Answer \(\dfrac{y(y + 7)}{1}\) or \(\dfrac{y^2 + 7y}{1}\) | M1M1M1A0 |
| Ignore the consistent use of a different variable within a factorisation | |
| Award SC1 only if there are no correct factorisations eg correct factorisation to \((y + 7)(y + 3)\) and correct expansion to \(y^4 - 3y^3 + 10y^3 - 30y^2 + 21y^2 - 63y\) | M1 only |