Higher January 2023 Paper 1R Q16
16 The function f is such that
\(\mathrm{f}(x) = \dfrac{2}{3x - 5}\;\) where \(\;x \ne \dfrac{5}{3}\)
(a) Find \(\;\mathrm{f}\left(\dfrac{1}{3}\right)\) (1)
(b) Find \(\;\mathrm{f}^{-1}(x)\) (2)
The function g is such that
\(\mathrm{g}(x) = 5x^2 - 20x + 23\)
(c) Express \(\mathrm{g}(x)\) in the form \(\;a(x - b)^2 + c\) (3)
| Scheme | Marks |
|---|---|
| −0.5 | B1 |
| (1) |
Notes
B1: oe eg \(-\dfrac{1}{2}, \dfrac{-1}{2}, \dfrac{1}{-2}\), −1/2
| Scheme | Marks |
|---|---|
\((3x - 5)y = 2\) or \((3y - 5)x = 2\) or \(3xy - 5y = 2\) or \(3xy - 5x = 2\) oe or \(3y - 5 = \dfrac{2}{x}\) or \(3x - 5 = \dfrac{2}{y}\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{2 + 5x}{3x}\) | A1oe |
| (2) |
Notes
M1: remove denominator or get to the stage \(3y - 5 = \dfrac{2}{x}\) or \(3x - 5 = \dfrac{2}{y}\)
A1oe: eg \(\dfrac{2}{3x} + \dfrac{5}{3}\) or \(\dfrac{\frac{2}{x} + 5}{3}\) must be in terms of \(x\)
| Scheme | Marks |
|---|---|
| \(5(x^2 - 4x)\ldots\ldots\) or \(5(x^2 - 4x\ldots\ldots)\) or \(5(x - 2)^2\ldots\) | M1 |
\(5\left[(x - 2)^2 - (-2)^2\right]\ldots\ldots\) or \(5\left[(x - 2)^2 - (-2)^2\ldots\ldots\right]\) or \(5(x - 2)^2 - 20\ldots\ldots\) or \(5\left[(x - 2)^2 + \dfrac{3}{5}\right]\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(5(x - 2)^2 + 3\) | A1 |
| (3) | |
| (6 marks) |
Notes
M1: \((-2)^2\) can be \(2^2\) or 4 or \(\left(\pm\dfrac{4}{2}\right)^2\)
| Scheme | Marks |
|---|---|
| \(ax^2 - 2abx + ab^2 + c\) | M1 |
| 2 of: \(a = 5 \qquad 2ab = 20\) oe \(\qquad ab^2 + c = 23\) oe | M1 |
| \(5(x - 2)^2 + 3\) | A1 |
Notes
M1: for multiplying out \(a(x - b)^2 + c\) to obtain \(ax^2 - 2abx + ab^2 + c\) oe
M1: for equating coefficients and making 2 correct statements