June 2016 Paper 2 Q24
24 \(\mathrm{f}(x) = \dfrac{x}{2x + 1} \quad\) for positive values of \(x\).
Work out \(\quad \mathrm{f}(x + 1) - \mathrm{f}(x)\)
Give your answer as a fraction in its simplest form.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{x + 1}{2(x + 1) + 1}\) or \(\dfrac{x + 1}{2x + 3}\) | M1 | oe |
| \((2x + 1)(2x + 3)\) | M1 | oe Correct common denominator for their two fractions Their two fractions must have different algebraic denominators |
| \((2x + 1)(x + 1)\) and \(x(2x + 3)\) | M1dep | oe Correct expressions for their numerators dep on 2nd M1 |
| \(2x^2 + 2x + x + 1 - 2x^2 - 3x\) | M1dep | Subtracts their numerators Must expand all brackets Allow one sign error dep on 2nd M1 |
| \(\dfrac{1}{(2x + 3)(2x + 1)}\) or \(\dfrac{1}{4x^2 + 8x + 3}\) | A1 | oe fraction in its simplest form |
Additional guidance
| Correct answer followed by incorrect further work However, \(\dfrac{1}{(2x + 3)(2x + 1)}\) with an attempt to expand the denominator should not be penalised even if their expansion is not correct | M4 A0 |
| 2nd and 3rd M1 \((2x + 1)\) must be used | |
| \(\dfrac{2x^2 + 3x + 1}{(2x + 3)(2x + 1)} - \dfrac{2x^2 + 3x}{(2x + 3)(2x + 1)}\) (no further correct work) | M1 M1 M1 M0 A0 |
| Allow = 0 throughout, including in the answer |