Higher June 2022 Paper 3 Q19
19 Show that \(\dfrac{3x}{x + 2} - \dfrac{2x + 1}{x - 2} - 1\) can be written in the form \(\dfrac{ax + b}{x^2 - 4}\)
where \(a\) and \(b\) are integers. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{-11x + 2}{x^2 - 4}\) | M1 | for writing at least one of the 3 terms with a denominator of \((x^2 - 4)\) or \((x - 2)(x + 2)\) eg \(\dfrac{3x(x - 2)}{x^2 - 4}\) oe or \(\dfrac{(x + 2)(2x + 1)}{x^2 - 4}\) oe or \(\dfrac{x^2 - 4}{x^2 - 4}\) |
| M1 | for \(\dfrac{3x(x - 2)}{x^2 - 4} - \dfrac{(x + 2)(2x + 1)}{x^2 - 4} - \dfrac{x^2 - 4}{x^2 - 4}\) oe or for \(\dfrac{x^2 - 11x - 2}{x^2 - 4}\ (-\ 1)\) or for \(\dfrac{[x^2 - 11x - 2]}{x^2 - 4} - \dfrac{x^2 - 4}{x^2 - 4}\) | |
| M1 | for a numerator of \(3x^2 - 6x - 2x^2 - 5x - 2 - x^2 + 4\) | |
| A1 | for \(\dfrac{-11x + 2}{x^2 - 4}\) |
Additional guidance
Students may work with a denominator of \((x - 2)(x + 2)\) for the first 3 marks
\([x^2 - 11x - 2]\) denotes their expansion of \(3x(x - 2) - (x + 2)(2x + 1)\)
May be simplified
Accept \(a = -11\) and \(b = 2\)