Higher November 2021 Paper 2 Q15
15 The functions f and g are such that
\(\mathrm{f}(x) = 2x - 3\)
\(\mathrm{g}(x) = \dfrac{x}{3x + 1}\)
(a) State the value of \(x\) that cannot be included in any domain of g (1)
(b) Find \(\mathrm{gf}(x)\)
Simplify your answer. (2)
Simplify your answer. (2)
(c) Express the inverse function \(\mathrm{g}^{-1}\) in the form \(\mathrm{g}^{-1}(x) = \ldots\) (3)
| Scheme | Marks |
|---|---|
| \(-\dfrac{1}{3}\) | B1 |
| (1) |
Notes
B1: oe allow \(-0.\dot{3}\) or −0.33 or better
allow \(x = -\dfrac{1}{3}\) or \(x \neq -\dfrac{1}{3}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{2x - 3}{3(2x - 3) + 1}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{2x - 3}{6x - 8}\) | A1 |
| (2) |
Notes
M1: for substituting f(\(x\)) into g(\(x\))
Allow \(\dfrac{\mathrm{f}}{3\mathrm{f} + 1}\)
A1: oe (do not isw incorrect cancelling)
| Scheme | Marks |
|---|---|
| \(y(3x + 1) = x\) and \(3xy + y = x\) or \(x(3y + 1) = y\) and \(3xy + x = y\) | M1 |
| \(x(1 - 3y) = y\) or \(x(3y - 1) = -y\) or \(y(1 - 3x) = x\) or \(y(3x - 1) = -x\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{x}{1 - 3x}\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for moving the denominator to the other side of the equation and expanding correctly
M1: for collecting and factorising the variable on one side in a correct equation
A1: oe eg \(-\dfrac{x}{3x - 1}\) or \(\dfrac{-x}{-1 + 3x}\) oe