4MA0 Higher June 2017 Paper 3 Q19
19 \(\mathrm{f}(x) = \dfrac{4}{x - 3}\)
\(\mathrm{g}(x) = \dfrac{x - 2}{x}\)
(a) Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\) (3)
(b) Solve \(\mathrm{fg}(a) = 1\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
\(x = \dfrac{4}{y - 3}\) \(x(y - 3) = 4\) or \(y = \dfrac{4}{x - 3}\) \(y(x - 3) = 4\) | M1 |
\(xy = 4 + 3x\) or \(y - 3 = \dfrac{4}{x}\) or \(xy = 4 + 3y\) or \(x - 3 = \dfrac{4}{y}\) | M1 |
| \(\dfrac{4 + 3x}{x}\) oe | A1 |
| (3) |
Notes
M1: for \(x(y - 3) = 4\) or \(y(x - 3) = 4\)
M1: (implies the first M1)
A1: for \(\dfrac{4 + 3x}{x}\) oe e.g. \(\dfrac{4}{x} + 3\)
| Scheme | Marks |
|---|---|
E.g. \((\mathrm{fg}(a) =)\;\dfrac{4}{\dfrac{a - 2}{a} - 3}\) or \(4 = \dfrac{a - 2}{a} - 3\) or \(\dfrac{4a}{a - 2 - 3a}\;(= 1)\) | M1 |
| E.g. \(4a = a - 2 - 3a\) or \(7a = a - 2\) | M1 |
| \(a = -\dfrac{1}{3}\) oe | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for a correct expression for \(\mathrm{fg}(a)\)
M1: for a correct equation where the fraction has been removed.
A1: dep on M1
Accept −0.333(333...) rounded or truncated to at least 3SF
Condone the use of \(x\) rather than \(a\)
Alternative scheme
| Scheme | Marks |
|---|---|
E.g. \(\mathrm{g}(a) = \mathrm{f}^{-1}(1)\) or \(\mathrm{g}(a) = \dfrac{4 + 3 \times 1}{1}\) oe or \(\dfrac{4 + 3 \times 1}{1} = \dfrac{a - 2}{a}\) or \(7 = \dfrac{a - 2}{a}\) | M1 |
| E.g. \(7a = a - 2\) | M1 |
| \(a = -\dfrac{1}{3}\) oe | A1 |
Notes
M1: for use of \(\mathrm{f}^{-1}\mathrm{fg}(a) = \mathrm{f}^{-1}(1)\)
NB. ft for “\(\mathrm{f}^{-1}\)”
M1: for a correct equation where the fraction has been removed.
NB. ft for “\(\mathrm{f}^{-1}\)”
A1: dep on M1
Accept −0.333(333...) rounded or truncated to at least 3SF