C3 June 2008 Q4
4. The function f is defined by\[\mathrm{f} : x \mapsto \frac{2(x - 1)}{x^2 - 2x - 3} - \frac{1}{x - 3}, \qquad x > 3.\]
(a) Show that \(\mathrm{f}(x) = \dfrac{1}{x + 1}\), \(x > 3\). (4)
(b) Find the range of f. (2)
(c) Find \(\mathrm{f}^{-1}(x)\). State the domain of this inverse function. (3)
The function g is defined by\[\mathrm{g} : x \mapsto 2x^2 - 3, \qquad x \in \mathbb{R}.\]
(d) Solve \(\mathrm{fg}(x) = \dfrac{1}{8}\). (3)
| Scheme | Marks |
|---|---|
| \(x^2 - 2x - 3 = (x - 3)(x + 1)\) | B1 |
| \(\mathrm{f}(x) = \dfrac{2(x - 1) - (x + 1)}{(x - 3)(x + 1)}\) \(\left(\text{or } \dfrac{2(x - 1)}{(x - 3)(x + 1)} - \dfrac{x + 1}{(x - 3)(x + 1)}\right)\) | M1 A1 |
| \(= \dfrac{x - 3}{(x - 3)(x + 1)} = \dfrac{1}{x + 1}\ \ \ast\) cso | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\left(0, \dfrac{1}{4}\right)\) Accept \(0 < y < \dfrac{1}{4},\ 0 < \mathrm{f}(x) < \dfrac{1}{4}\) etc. | B1 B1 |
| (2) |
| Scheme | Marks |
|---|---|
| Let \(y = \mathrm{f}(x)\) \(y = \dfrac{1}{x + 1}\) \(x = \dfrac{1}{y + 1}\) \(yx + x = 1\) | |
| \(y = \dfrac{1 - x}{x}\) or \(\dfrac{1}{x} - 1\) \(\mathrm{f}^{-1}(x) = \dfrac{1 - x}{x}\) | M1 A1 |
| Domain of \(\mathrm{f}^{-1}\) is \(\left(0, \dfrac{1}{4}\right)\) ft their part (b) | B1 ft |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{fg}(x) = \dfrac{1}{2x^2 - 3 + 1}\) | |
| \(\dfrac{1}{2x^2 - 2} = \dfrac{1}{8}\) | M1 |
| \(x^2 = 5\) | A1 |
| \(x = \pm\sqrt{5}\) both | A1 |
| (3) | |
| (12 marks) |