4400 Higher June 2010 Paper 4 Q20
20 \(\mathrm{f}(x) = (x - 1)^2\)
(a) Find \(\mathrm{f}(8)\) (1)
(b) The domain of \(\mathrm{f}\) is all values of \(x\) where \(x \geqslant 7\)
Find the range of \(\mathrm{f}\). (2)
Find the range of \(\mathrm{f}\). (2)
\(\mathrm{g}(x) = \dfrac{x}{x - 1}\)
(c) Solve the equation \(\mathrm{g}(x) = 1.2\) (2)
(d)
(i) Find \(\mathrm{g}^{-1}(x)\).
(ii) Hence write down \(\mathrm{gg}(x)\) in terms of \(x\).
(6)| Scheme | Marks |
|---|---|
| 49 | B1 |
| (1) |
Notes
B1: cao
| Scheme | Marks |
|---|---|
| \((7 - 1)^2\) or 36 seen | M1 |
| \(\mathrm{f}(x) \geqslant 36\) or \(y \geqslant 36\) | A1 |
| (2) |
Notes
A1: allow \(\mathrm{f} \geqslant 36\) \(\quad x \geqslant 36\): M1A0 (don’t accept >)
| Scheme | Marks |
|---|---|
\(\dfrac{x}{x - 1} = 1.2\) \(x = 1.2(x - 1)\) | M1 |
| 6 | A1 |
| (2) |
Notes
M1: Do not accept \(\mathrm{g}(1.2) = 6\) method
A1: cao Answer only = M0 A0 Algebra method reqd.
| Scheme | Marks |
|---|---|
| (i) \(y = \dfrac{x}{x - 1}\) | M1 |
| \(y(x - 1) = x\) \(xy - y = x\) | M1 |
| \(xy - x = y\) \(x(y - 1) = y\) | M1 |
| \(x = \dfrac{y}{y - 1}\) | M1 |
| \(\dfrac{x}{x - 1}\) | A1 |
| (ii) Answer: \(x\) | B1 |
| (6) | |
| (11 marks) |
Notes
M1: \(x = \dfrac{y}{y - 1}\)
M1: \(x(y - 1) = y\)
\(xy - x = y\)
M1: \(xy - y = x\)
\(y(x - 1) = x\)
B1: accept \([x/(x - 1)]/[(x/(x - 1) - 1]\) do not isw