Higher June 2025 Paper 1 Q24
24 Prove that \(\quad \dfrac{60x^4 - 15x^2}{-2x - 1} \times \dfrac{1}{6x - 3} \quad\) can never be positive. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Partially or fully factorises numerator | M1 | eg \(15x(4x^3 - x)\) or \(x^2(60x^2 - 15)\) or \(15x^2(2x - 1)(2x + 1)\) |
| Factorises at least one denominator or correct multiplication of the denominators | M1 | eg \(-(2x + 1)\) and/or \(3(2x - 1)\) eg \(-12x^2 + (6x - 6x +)\ 3\) may be in a grid |
| Converts numerators and denominators into terms which can be fully cancelled | M1dep | dep on M1M1 eg \(\dfrac{15x^2(2x - 1)(2x + 1)}{-(2x + 1)3(2x - 1)}\) or \(\dfrac{15x^2(2x - 1)(2x + 1)}{-(2x + 1)} \times \dfrac{1}{3(2x - 1)}\) or \(\dfrac{5x^2(12x^2 - 3)}{-(12x^2 - 3)}\) factorisation and cancelling may be done in stages |
| \(\dfrac{15x^2}{-3}\) or \(-5x^2\) with M3 awarded and explanation that \(x^2\) cannot be negative | A1 | oe with full cancelling of algebraic terms condone explanation that \(x^2\) must be positive |
Additional guidance
| \(\dfrac{20x^4 - 5x^2}{(-2x - 1)(2x - 1)}\) or \(\dfrac{20x^4 - 5x^2}{-4x^2 + 1}\) implies M1M1 as 3 has been cancelled from both | M1M1 |
| \(\dfrac{5x^2(2x + 1)}{-(2x + 1)}\) implies complete factorisation of numerator and denominator as 3 and \((2x - 1)\) have been cancelled from both | M1M1M1 |