Higher January 2021 Paper 1 Q22
22 The function f is such that \(\mathrm{f}(x) = x^2 - 8x + 5\) where \(x \leqslant 4\)
Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\)
(3)
| Scheme | Marks |
|---|---|
| \(y = (x - 4)^2 - 4^2\;(+5)\) oe or \(x = (y - 4)^2 - 4^2\;(+5)\) | M1 |
| \(y = 4 \pm \sqrt{11 + x}\) or \(x = 4 \pm \sqrt{11 + y}\) | A1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(4 - \sqrt{x + 11}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a correct first step in order to complete the square
A1: allow \(y = 4 + \sqrt{11 + x}\)
or \(x = 4 + \sqrt{11 + y}\)
A1: oe
| Scheme | Marks |
|---|---|
\(x^2 - 8x + (5 - y) = 0\) \((x =)\;\dfrac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times (5 - y)}}{2 \times 1}\) or \(y^2 - 8y + (5 - x) = 0\) \((y =)\;\dfrac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times (5 - x)}}{2 \times 1}\) | M1 |
| \(y = 4 \pm \sqrt{11 + x}\) or \(x = 4 \pm \sqrt{11 + y}\) | A1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(4 - \sqrt{x + 11}\) | A1 |
Notes
M1: for a correct first step in preparation for use of quadratic formula and substitution into the quadratic formula (allow one sign error)
A1: allow \(y = 4 + \sqrt{11 + x}\)
or \(x = 4 + \sqrt{11 + y}\)
A1: oe
| Scheme | Marks |
|---|---|
Using \(ax^2 + bx + c = a(x + p)^2 + q\) \(\left(p = \dfrac{b}{2a}\right) = \dfrac{-8}{2}\;(= -4)\) and \(q = (4)^2 - 8(4) + 5\;(= -11)\) | M1 |
| \(y = 4 \pm \sqrt{11 + x}\) or \(x = 4 \pm \sqrt{11 + y}\) | A1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(4 - \sqrt{x + 11}\) | A1 |
Notes
M1: for finding \(p\) and \(q\)
A1: allow \(y = 4 + \sqrt{11 + x}\)
or \(x = 4 + \sqrt{11 + y}\)
A1: oe