C3 June 2017 Q1
1. Express \(\dfrac{4x}{x^2 - 9} - \dfrac{2}{x + 3}\) as a single fraction in its simplest form. (4)
| Scheme | Marks |
|---|---|
| \(x^2 - 9 = (x + 3)(x - 3)\) | B1 |
| \(\dfrac{4x}{x^2 - 9} - \dfrac{2}{(x + 3)} = \dfrac{4x - 2(x - 3)}{(x + 3)(x - 3)}\) | M1 |
| \(= \dfrac{2x + 6}{(x + 3)(x - 3)}\) | A1 |
| \(= \dfrac{2\cancel{(x + 3)}}{\cancel{(x + 3)}(x - 3)}\) | |
| \(= \dfrac{2}{(x - 3)}\) | A1 |
| (4) |
Notes
B1: \(x^2 - 9 = (x + 3)(x - 3)\) This can occur anywhere.
M1: For combining the two fractions with a common denominator. The denominator must be correct and at least one numerator must have been adapted. Accept as separate fractions. Condone missing brackets.
For example accept \(\dfrac{4x}{x^2 - 9} - \dfrac{2}{x + 3} = \dfrac{4x(x + 3) - 2(x^2 - 9)}{(x + 3)(x^2 - 9)}\)
accept separately \(\dfrac{4x}{(x + 3)(x - 3)} - \dfrac{2}{(x + 3)} = \dfrac{4x}{(x + 3)(x - 3)} - \dfrac{2x - 3}{(x + 3)(x - 3)}\) condoning missing bracket
condone \(\dfrac{4x}{x^2 - 9} - \dfrac{2}{x + 3} = \dfrac{4x(x + 3) - 2}{(x + 3)(x^2 - 9)}\) ..........as only one numerator has been adapted
A1: A correct intermediate form of \(\dfrac{\text{simplified linear}}{\text{simplified quadratic}}\)
Accept \(\dfrac{2x + 6}{(x + 3)(x - 3)}, \dfrac{2x + 6}{x^2 - 9}\), and even \(\dfrac{(2x + 6)\cancel{(x + 3)}}{(x^2 - 9)\cancel{(x + 3)}}\),
A1: Further factorises and cancels (which may be implied) to reach the answer \(\dfrac{2}{x - 3}\)
Do not penalise correct solutions that include incomplete lines Eg \(\dfrac{4x - 2(x - 3)}{(x + 3)(x - 3)} = \dfrac{4x - 2x + 6}{\ldots} = \dfrac{2x + 6}{(x + 3)(x - 3)} = \dfrac{2}{x - 3}\)
This is not a “show that” question.
Note: Watch out for an answer of \(\dfrac{2}{x + 3}\) probably scored from \(\dfrac{4x - 2(x - 3)}{(x + 3)(x - 3)} = \dfrac{2x - 6}{(x + 3)(x - 3)} = \dfrac{2(x - 3)}{(x + 3)(x - 3)}\)
This would score B1 M1 A0 A0