Higher January 2021 Paper 2R Q17
17 The functions f and g are defined as
\(\mathrm{f}(x) = x^2 + 6\)
\(\mathrm{g}(x) = x - 10\)
(a) Find \(\mathrm{fg}(3)\) (2)
(b) Solve the equation \(\mathrm{fg}(x) = \mathrm{f}(x)\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
The function h is defined as \(\mathrm{h}(x) = \dfrac{2x - 4}{x}\)
(c) State the value of \(x\) that cannot be included in the domain of h (1)
(d) Express the inverse function \(\mathrm{h}^{-1}\) in the form \(\mathrm{h}^{-1}(x) = \ldots\) (3)
| Scheme | Marks |
|---|---|
| \(\mathrm{g}(3) = -7\) or \(\mathrm{f}(3 - 10) = (3 - 10)^2 + 6\) or \(3^2 - 20 \times 3 + 106\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 55 | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \((x - 10)^2 + 6 = x^2 + 6\) | M1 |
| \(x^2 - 10x - 10x + 100\) oe | M1 |
| Working required Answer: 5 | A1 |
| (3) |
Notes
M1: Using \(\mathrm{f}(x - 10)\) and setting equal to \(x^2 + 6\)
M1: for \((x - 10)^2\) expanded correctly.
A1: dep 1st M1
| Scheme | Marks |
|---|---|
| 0 | B1 |
| (1) |
Notes
B1: accept \(x \neq 0\) or \(x = 0\)
| Scheme | Marks |
|---|---|
| eg \(yx = 2x - 4\) oe or \(4 = 2x - yx\) or \(xy = 2y - 4\) oe \(4 = 2y - yx\) | M1 |
eg \(4 = x(2 - y)\) oe or \(\dfrac{4}{x} = 2 - y\) or \(\dfrac{4}{2 - y} = x\) or \(4 = y(2 - x)\) oe \(\dfrac{4}{y} = 2 - x\) \(\dfrac{4}{2 - x} = y\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{4}{2 - x}\) or \(\dfrac{-4}{x - 2}\) | A1 |
| (3) | |
| (9 marks) |
Notes
M1: Removing denominator
equation may be rearranged
M1: for correct factorisation or implied factorisation
A1: oe