Higher November 2020 Paper 1R Q16
16 The functions f and g are defined as
\[\begin{aligned} \mathrm{f} &: x \mapsto 5x - 7\\ \mathrm{g} &: x \mapsto \dfrac{5x}{x + 4} \end{aligned}\](a) Write down the value of \(x\) that must be excluded from any domain of g (1)
(b) Find gf(2.6) (2)
(c) Solve fg(\(x\)) = 2 (3)
(d) Express the inverse function \(\mathrm{g}^{-1}\) in the form \(\mathrm{g}^{-1} : x \mapsto \ldots\) (3)
| Scheme | Marks |
|---|---|
| −4 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| (f(2.6) =) 5 × 2.6 − 7 (= 6) or \(\mathrm{gf}(x) = \dfrac{5(5x - 7)}{5x - 7 + 4}\) oe | M1 |
| 3 | A1 |
| (2) |
Notes
M1: for finding f(2.6) or gf(\(x\))
| Scheme | Marks |
|---|---|
| \(5\left(\dfrac{5x}{x + 4}\right) - 7 = 2\) or \(\dfrac{5x}{x + 4} = \dfrac{2 + 7}{5}\) oe | M1 |
| \(25x = 9(x + 4)\) oe | M1 |
| 2.25 | A1 |
| (3) |
Notes
M1: for removing the denominator (\(x\) + 4) in a correct equation
A1: oe
ALT (c)
| Scheme | Marks |
|---|---|
| fg(\(x\)) = 2 \(\Rightarrow\) g(\(x\)) = \(\mathrm{f}^{-1}\)(2) (=9/5) and attempt at \(\mathrm{f}^{-1}\) or \(\mathrm{f}^{-1}\)(2) | M1 |
| \(x = \mathrm{g}^{-1}\)(“9/5”) | M1 |
| 2.25 | A1 |
Notes
A1: oe
| Scheme | Marks |
|---|---|
\(y = \dfrac{5x}{x + 4}\) or \(x = \dfrac{5y}{y + 4}\) \(y(x + 4) = 5x\) \(x(y + 4) = 5y\) | M1 |
| e.g. \(4y = x(5 - y)\) or e.g. \(4x = y(5 - x)\) | M1 |
| \(\dfrac{4x}{5 - x}\) | A1 |
| (3) | |
| (9 marks) |
Notes
M1: for a correct rearrangement and factorising
A1: oe e.g. \(\dfrac{-4x}{x - 5}\)