4400 Higher June 2007 Paper 4 Q17
17 The functions f and g are defined as follows.
\[\mathrm{f}(x) = \frac{1}{x + 2}\]\[\mathrm{g}(x) = \sqrt{x - 1}\](a)
(i) State which value of \(x\) cannot be included in the domain of f.
(ii) State which values of \(x\) cannot be included in the domain of g. (3)
(b) Calculate \(\mathrm{fg}(10)\) (3)
(c) Express the inverse function \(\mathrm{g}^{-1}\) in the form \(\mathrm{g}^{-1}(x) = \ldots\) (4)
| Scheme | Marks |
|---|---|
| (i) Answer: −2 | B1 |
| (ii) Answer: \(x \lt 1\) | B2 |
| (3) |
Notes
B1: or \(x \ne -2\) or \(x = -2\)
B2: B1 for \(x \leqslant 1\) or 0, −1, −2, −3 . . .
| Scheme | Marks |
|---|---|
| \(\sqrt{9}\) or \(\sqrt{10 - 1}\) | M1 |
| \(\dfrac{1}{\text{her}\sqrt{9} + 2}\) | M1 |
| \(\frac{1}{5}\) or 0.2 | A1 |
| (3) |
Notes
M1: or \(\dfrac{1}{\sqrt{x - 1} + 2}\)
A1: ignore ans = −1
| Scheme | Marks |
|---|---|
| \(y = \sqrt{x - 1}\) | M1 |
| \(y^2 = x - 1\) | M1 |
| \(x = y^2 + 1\) | M1 |
| (\(\mathrm{g}^{-1}(x) =\)) \(x^2 + 1\) oe | A1 |
| (4) | |
| (10 marks) |
Notes
Alternative
| −1, \(\sqrt{\ }\) | M1 |
| Reverse order | M1dep |
| squ, + 1 | M1 |
Alternative
| \(y = \sqrt{x - 1}\) | M1 |
| \(x = \sqrt{y - 1}\) | M1dep |
| \(x^2 = y - 1\) | M1 |
condone \(\sqrt{x} - 1\) if next step correct
\(y^2 + 1\) M3
\(y = x^2 + 1\) M3
\(x = x^2 + 1\) M3
SC (\(\mathrm{g}^{-1}(x) =\)) \((x + 1)^2\): B1