4MA0 Higher June 2014 Paper 3R Q11
11 The functions f and g are defined as \[\mathrm{f}(x) = \dfrac{1}{2}x + 4\] \[\mathrm{g}(x) = \dfrac{2x}{x + 1}\]
(a) Work out \(\mathrm{f}(6)\) (1)
(b) Work out \(\mathrm{fg}(-3)\) (2)
(c) \(\mathrm{g}(a) = -2\)
Work out the value of \(a\). (2)
Work out the value of \(a\). (2)
(d) Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\) (3)
| Scheme | Marks |
|---|---|
| 7 | B1 |
| (1) |
Notes
B1: cao
| Scheme | Marks |
|---|---|
| \(\mathrm{g}(-3) = \dfrac{2 \times -3}{-3 + 1}\left(= \dfrac{-6}{-2} = 3\right)\) | M1 |
| \(5\frac{1}{2}\) | A1 |
| (2) |
Notes
M1: for correct expression for \(\mathrm{g}(-3)\)
A1: or 5.5 or 11/2
| Scheme | Marks |
|---|---|
| \(-2(a + 1) = 2a\) or \(-2a - 2 = 2a\) or \(4a = -2\) | M1 |
| \(-\frac{1}{2}\) | A1 |
| (2) |
Notes
M1: for correct equation with fraction cleared (allow \(x\))
A1: Accept \(-0.5\)
| Scheme | Marks |
|---|---|
| \(y = \frac{1}{2}x + 4\) or \(x = \frac{1}{2}y + 4\) (any two different variables) or flow diagram \(\div 2 \to +4\) | M1 |
\(x = 2(y - 4)\) or \(x = 2y - 8\) Or \(y = 2x - 8\) or \(y = 2(x - 4)\) or \(x = \dfrac{y - 4}{0.5}\) or \(y = \dfrac{x - 4}{0.5}\) | M1 |
| \(2(x - 4)\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: For full method to correctly change the subject.
A1: or \(2x - 8\) or \(\dfrac{x - 4}{0.5}\) oe
Do not allow other variables.
SC: Award B2 for \(f^{-1}(x) = 2y - 8\) if no working shown.