4MA0 Higher June 2011 Paper 4 Q20
20 \[\begin{aligned} \mathrm{f}(x) &= \dfrac{2}{x} \\ \mathrm{g}(x) &= \dfrac{x + 1}{x} \end{aligned}\]
(a) State which value of \(x\) cannot be included in the domain of f or g. (1)
(b) Solve \(\mathrm{gf}(a) = 3\) (3)
(c) Express the inverse function \(\mathrm{g}^{-1}\) in the form \(\mathrm{g}^{-1}(x)\) (3)
| Scheme | Marks |
|---|---|
| (\(x\)=)0 | B1 |
| (1) |
Notes
B1: Accept (\(x\))≠0
Apart from questions 5b, 8, 15d, 20b, 21b, 23, 24b (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be taken to imply correct working.
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{2}{a} + 1\right) / \dfrac{2}{a} = 3\) | M1 |
| \(\dfrac{2}{a} + 1 = \dfrac{6}{a}\) or \(1 + \dfrac{a}{2} = 3\) oe | M1 |
| 4 | A1 |
| (3) |
Notes
M1: (Any letter in place of \(a\) acceptable) Solve g(\(x\))=3 (\(x\)=0.5)
M1: Solve f(a)=0.5
A1: dep on M2
| Scheme | Marks |
|---|---|
| \(y = \dfrac{x + 1}{x}\) | M1 |
\(x(y - 1) = 1\) \(x = \dfrac{1}{y - 1}\) | M1 |
| \(\dfrac{1}{x - 1}\) | A1 |
| (3) | |
| (Total 7 marks) |
Notes
M1: one occurrence of \(x\)
A1: reverse labels \(x\) and \(y\)
Alternative: \(x = \dfrac{y + 1}{y}\) reverse labels \(x\) and \(y\); \(y(x - 1) = 1\) one occurrence of \(y\)