4400 Higher November 2010 Paper 3 Q21
21
(a) \(\mathrm{f}(x) = 2x + 1\)
Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\) (2)
Express the inverse function \(\mathrm{f}^{-1}\) in the form \(\mathrm{f}^{-1}(x) = \ldots\) (2)
(b) \(\mathrm{g}(x) = 2 + x\)
\(\mathrm{h}(x) = x^2\)
Solve the equation \(\mathrm{hg}(x) = \mathrm{h}(x)\). (3)
\(\mathrm{h}(x) = x^2\)
Solve the equation \(\mathrm{hg}(x) = \mathrm{h}(x)\). (3)
| Scheme | Marks |
|---|---|
\(y = 2x + 1\) \(x = \dfrac{y - 1}{2}\) | M1 |
| \(\mathrm{f}^{-1}(x) = (x - 1)/2\) oe | A1 |
| (2) |
Notes
M1: \(x = 2y + 1\)
\(y = \dfrac{x - 1}{2}\)
A1: answer only = M1A1
| Scheme | Marks |
|---|---|
| \((2 + x)^2 = x^2\) | M1 |
| \(4 + 4x + x^2 = x^2\) | M1 |
| \(x = -1\) | A1 |
| (3) | |
| (5 marks) |
Notes
M1: M1 for \((2 + x)^2\)
M1: or \(2 + x = -x\) (from rooting both sides)
A1: Answer only = M0A0A0