C3 January 2009 Q2
2.\[\mathrm{f}(x) = \frac{2x + 2}{x^2 - 2x - 3} - \frac{x + 1}{x - 3}\]
(a) Express \(\mathrm{f}(x)\) as a single fraction in its simplest form. (4)
(b) Hence show that \(\mathrm{f}^{\prime}(x) = \dfrac{2}{(x - 3)^2}\) (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{2x + 2}{x^2 - 2x - 3} - \dfrac{x + 1}{x - 3} = \dfrac{2x + 2}{(x - 3)(x + 1)} - \dfrac{x + 1}{x - 3}\) \(= \dfrac{2x + 2 - (x + 1)(x + 1)}{(x - 3)(x + 1)}\) | M1 A1 |
| \(= \dfrac{(x + 1)(1 - x)}{(x - 3)(x + 1)}\) | M1 |
| \(= \dfrac{1 - x}{x - 3}\) Accept \(-\dfrac{x - 1}{x - 3}\), \(\dfrac{x - 1}{3 - x}\) | A1 |
| (4) |
Alternative to (a)
| \(\dfrac{2x + 2}{x^2 - 2x - 3} = \dfrac{2(x + 1)}{(x - 3)(x + 1)} = \dfrac{2}{x - 3}\) | M1 A1 |
| \(\dfrac{2}{x - 3} - \dfrac{x + 1}{x - 3} = \dfrac{2 - (x + 1)}{x - 3}\) | M1 |
| \(= \dfrac{1 - x}{x - 3}\) | A1 (4) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1 - x}{x - 3}\right) = \dfrac{(x - 3)(-1) - (1 - x)1}{(x - 3)^2}\) | M1 A1 |
| \(= \dfrac{-x + 3 - 1 + x}{(x - 3)^2} = \dfrac{2}{(x - 3)^2}\ \ \ast\) cso | A1 |
| (3) | |
| (7 marks) |
Alternatives to (b) ①
| \(\mathrm{f}(x) = \dfrac{1 - x}{x - 3} = -1 - \dfrac{2}{x - 3} = -1 - 2(x - 3)^{-1}\) | |
| \(\mathrm{f}^{\prime}(x) = (-1)(-2)(x - 3)^{-2}\) | M1 A1 |
| \(= \dfrac{2}{(x - 3)^2}\ \ \ast\) cso | A1 (3) |
Alternatives to (b) ②
| \(\mathrm{f}(x) = (1 - x)(x - 3)^{-1}\) | |
| \(\mathrm{f}^{\prime}(x) = (-1)(x - 3)^{-1} + (1 - x)(-1)(x - 3)^{-2}\) | M1 |
| \(= -\dfrac{1}{x - 3} - \dfrac{1 - x}{(x - 3)^2} = \dfrac{-(x - 3) - (1 - x)}{(x - 3)^2}\) | A1 |
| \(= \dfrac{2}{(x - 3)^2}\ \ \ast\) | A1 (3) |
Notes
(corrected from the printed mark scheme: in ② the first term of \(\mathrm{f}^{\prime}(x)\) is printed as \((-1)(x - 3)\); it should be \((-1)(x - 3)^{-1}\), as the next line shows)