Higher November 2021 Paper 1 Q21
21 The functions f and g are such that
\(\mathrm{f}(x) = 3x^2 + 1\) for \(x \gt 0\) and \(\mathrm{g}(x) = \dfrac{4}{x^2}\) for \(x \gt 0\)
(a) Work out \(\mathrm{gf}(1)\) (2)
The function h is such that \(\mathrm{h} = (\mathrm{fg})^{-1}\)
(b) Find \(\mathrm{h}(x)\) (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{1}{4}\) | M1 | for \(\mathrm{f}(1) = 3 \times 1^2 + 1\ (= 4)\) and a clear intention to find \(\mathrm{g}(\text{``}4\text{''})\) or for \(\dfrac{4}{(3 \times 1^2 + 1)^2}\) or for stating \(\mathrm{gf}(x)\), eg \(\dfrac{4}{(3x^2 + 1)^2}\) oe |
| A1 | oe |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\sqrt[4]{\dfrac{48}{x - 1}}\) | M1 | for finding \(\mathrm{fg}(x)\), eg \(3 \times \left(\dfrac{4}{x^2}\right)^2 + 1\) or \(\dfrac{48}{x^4} + 1\) |
| M1 | for start of method to find the inverse of \(\mathrm{fg}(x)\), eg \(y - 1 = 3 \times \left(\dfrac{4}{x^2}\right)^2\) or \(y - 1 = \dfrac{48}{x^4}\) or \(x - 1 = \dfrac{48}{y^4}\) or \(x - 1 = 3 \times \left(\dfrac{4}{y^2}\right)^2\) | |
| M1 | for \(y^4 = \dfrac{48}{x - 1}\) or \(x^4 = \dfrac{48}{y - 1}\) or for a final answer of \(\sqrt[4]{\dfrac{48}{y - 1}}\) | |
| A1 | oe |