Higher June 2021 Paper 1 Q24
24 The functions f and g are defined as
\(\mathrm{f}(x) = 5x^2 - 10x + 7 \qquad\) where \(x \geqslant 1\)
\(\mathrm{g}(x) = 7x - 6\)
(a) Find \(\mathrm{fg}(2)\) (2)
(b) Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\) (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{g}(2) = 7 \times 2 - 6\ (= 8)\) or \(5(7 \times 2 - 6)^2 - 10(7 \times 2 - 6) + 7\) | M1 |
| Working not required, so correct answer scores full marks Answer: 247 | A1 |
| (2) |
| Scheme | Marks |
|---|---|
eg \(y = 5(x^2 - 2x) + 7\) or \(y = 5\left(x^2 - 2x + \dfrac{7}{5}\right)\) oe or eg \(x = 5(y^2 - 2y) + 7\) or \(x = 5\left(y^2 - 2y + \dfrac{7}{5}\right)\) | M1 |
eg \(y = 5[(x - 1)^2 - 1^2] + 7\) or \(y = 5\left((x - 1)^2 - 1^2 + \dfrac{7}{5}\right)\) oe or eg \(x = 5\left((y - 1)^2 - 1^2\right) + 7\) or \(x = 5\left((y - 1)^2 - 1^2 + \dfrac{7}{5}\right)\) oe | M1 |
\((x - 1)^2 = \dfrac{y - 2}{5}\) oe or \((y - 1)^2 = \dfrac{x - 2}{5}\) oe | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(1 + \sqrt{\dfrac{x - 2}{5}}\) | A1 |
| (4) | |
| (6 marks) |
Notes
M1: or eg \(\dfrac{y - 7}{5} = x^2 - 2x\)
M1: or eg \(\dfrac{y - 7}{5} = (x - 1)^2 - 1^2\)
M1: or eg \((x - 1)^2 = \dfrac{y - 7}{5} + 1\)
A1: Must be in terms of \(x\), oe eg \(1 + \sqrt{\dfrac{x - 7}{5} + 1}\)
(NB: \(\mathrm{f}^{-1}(x) = 1 \pm \sqrt{\dfrac{x - 2}{5}}\) is 3 marks)
| Scheme | Marks |
|---|---|
| Let \(x = 5y^2 - 10y + 7\ [\Leftrightarrow]\ 5y^2 - 10y + (7 - x) = 0\) oe | M1 |
| \([y =]\ \dfrac{10 \pm \sqrt{100 - 20(7 - x)}}{10}\) | M1 |
| \(1 \pm \sqrt{\dfrac{x - 2}{5}}\) | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(1 + \sqrt{\dfrac{x - 2}{5}}\) | A1 |
Notes
A1: Must be in terms of \(x\)