Higher June 2018 Paper 1R Q20
20 Two functions, f and g are defined as
\(\mathrm{f} : x \mapsto 1 + \dfrac{1}{x}\) for \(x \gt 0\)
\(\mathrm{g} : x \mapsto \dfrac{x + 1}{2}\) for \(x \gt 0\)
Given that \(\mathrm{h} = \mathrm{fg}\)
express the inverse function \(\mathrm{h}^{-1}\) in the form \(\mathrm{h}^{-1} : x \mapsto \ldots\)
(4)
| Scheme | Marks |
|---|---|
| \(\mathrm{h} = \mathrm{f}\left(\dfrac{x + 1}{2}\right) = 1 + \dfrac{1}{\frac{x + 1}{2}} \left(= 1 + \dfrac{2}{x + 1}\right)\) | M1 |
\(\left(y = 1 + \dfrac{2}{x + 1}\right)\) \(y - 1 = \dfrac{2}{x + 1}\) or \(y(x + 1) = 1(x + 1) + 2\) | M1 |
| \(x + 1 = \dfrac{2}{y - 1}\) or \(xy - x = 3 - y\) | M1 |
\(x = \dfrac{2}{y - 1} - 1\) or \(x = \dfrac{3 - y}{y - 1}\) Answer: \(\dfrac{2}{x - 1} - 1\) or \(\dfrac{3 - x}{x - 1}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for \(1 + \dfrac{1}{\frac{x + 1}{2}}\)
M1: (dep on M1) for a correct first step to change the subject
M1: (dep on M1)
A1: oe
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse
| Scheme | Marks |
|---|---|
| \(\mathrm{h} = \mathrm{f}\left(\dfrac{x + 1}{2}\right) = 1 + \dfrac{1}{\frac{x + 1}{2}} \left(= 1 + \dfrac{2}{x + 1} = \dfrac{x + 3}{x + 1}\right)\) | M1 |
\(\left(y = \dfrac{x + 3}{x + 1}\right)\) \(y(x + 1) = (x + 3)\) | M1 |
| \(xy - x = 3 - y\) | M1 |
\(x = \dfrac{3 - y}{y - 1}\) Answer: \(\dfrac{3 - x}{x - 1}\) | A1 |
Notes
M1: for \(1 + \dfrac{1}{\frac{x + 1}{2}}\)
M1: (dep on M1) for a correct first step to change the subject
M1: (dep on M1)
A1: oe