Higher June 2019 Paper 2 Q13
13 Show that \(6 + \left[(x + 5) \div \dfrac{x^2 + 3x - 10}{x - 1}\right]\) simplifies to \(\dfrac{ax - b}{cx - d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{7x - 13}{x - 2}\) | B1 | for factorising eg \((x + 5)(x - 2)\) |
| M1 | for a method to divide \((x + 5)\) by the algebraic fraction eg \((x + 5) \times \dfrac{(x - 1)}{x^2 + 3x - 10}\) | |
| M1 | for finding 2 fractions with a common denominator or a single fraction eg \(\dfrac{6(x - 2)}{x - 2} + \dfrac{(x - 1)}{x - 2}\) or \(\dfrac{6(x - 2) + (x - 1)}{x - 2}\) or \(\dfrac{6(x^2 + 3x - 10)}{x^2 + 3x - 10} + \dfrac{(x + 5)(x - 1)}{x^2 + 3x - 10}\) or \(\dfrac{6(x^2 + 3x - 10) + (x + 5)(x - 1)}{x^2 + 3x - 10}\) | |
| A1 | \(\dfrac{7x - 13}{x - 2}\) |
Additional guidance
Condone incorrect factorising (both M marks)