Higher June 2019 Paper 1 Q21
21 The functions f and g are such that
\(\mathrm{f}(x) = 3x - 1 \quad\) and \(\quad \mathrm{g}(x) = x^2 + 4\)
(a) Find \(\mathrm{f}^{-1}(x)\) (2)
Given that \(\mathrm{fg}(x) = 2\mathrm{gf}(x)\),
(b) show that \(15x^2 - 12x - 1 = 0\) (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{x + 1}{3}\) | M1 | first step to change the subject of \(y = 3x - 1\) or \(x = 3y - 1\), eg \(y + 1 = 3x\) |
| A1 | oe |
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | for method to find \(\mathrm{fg}(x)\), eg \(\mathrm{fg}(x) = 3(x^2 + 4) - 1\) |
| M1 | for method to find \(\mathrm{gf}(x)\), eg \(\mathrm{gf}(x) = (3x - 1)^2 + 4\) | |
| M1 | (dep on previous two M marks) for setting up equation, eg \(3(x^2 + 4) - 1 = 2[(3x - 1)^2 + 4]\) | |
| M1 | (dep 2nd M1) for correct expansion of \((3x - 1)^2\) eg \(9x^2 - 3x - 3x + 1\) | |
| C1 | for \(15x^2 - 12x - 1 = 0\) from correct working |