Higher January 2023 Paper 2 Q25
25 The function f is such that \(\;\mathrm{f}(x) = 3x^2 - 12x + 7\;\) where \(\;x \leqslant 2\)
Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\;\mathrm{f}^{-1}(x) = \ldots\)
(4)
| Scheme | Marks |
|---|---|
\(y = 3(x^2 - 4x) + 7\) or \(y = 3\left(x^2 - 4x + \dfrac{7}{3}\right)\) or \(\dfrac{y - 7}{3} = x^2 - 4x\) or \(y = 3(x - 2)^2\ldots\) | M1 |
eg \(y = 3\left((x - 2)^2 - 2^2\right) + 7\) or \(y = 3\left((x - 2)^2 - 2^2 + \dfrac{7}{3}\right)\) or \(y = 3(x - 2)^2 - 5\) oe or | M1 |
\((x - 2)^2 = \dfrac{y + 5}{3}\) oe eg \((x - 2)^2 = \dfrac{y - 7}{3} + 4\) or \(x - 2 = (\pm)\sqrt{\dfrac{y + 5}{3}}\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(2 - \sqrt{\dfrac{x + 5}{3}}\) | A1 oe |
| (4) | |
| (4 marks) |
Notes
M1: for a correct equation for a first step to complete the square
A1 oe: NB: note only negative square root.
Must be in terms of \(x\)
any equivalent form
Note: Allow candidates to swap \(x\) and \(y\) when finding the inverse
| Scheme | Marks |
|---|---|
| \(3x^2 - 12x + (7 - y) = 0\) | M1 |
| \((x =)\;\dfrac{12 \pm \sqrt{144 - 12(7 - y)}}{6}\) | M1 |
| \((x =)\;2 \pm \sqrt{\dfrac{60 + 12y}{36}}\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(2 - \sqrt{\dfrac{x + 5}{3}}\) | A1 oe |
Notes
M1: for a correct first step
A1 oe: NB: note only negative square root.
Must be in terms of \(x\)
any equivalent form