4MA0 Higher January 2018 Paper 3R Q23
23 f is the function such that \(\mathrm{f}(x) = (x + 1)^2\) for \(x > 0\)
(a) Find the value of \(a\) for which \(\mathrm{f}(a) = \dfrac{25}{9}\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
g is the function such that \(\mathrm{g}(x) = \dfrac{1}{x}\) for \(x > 0\)
(b) Show that \(x^2\mathrm{fg}(x) = \mathrm{f}(x)\) for all \(x > 0\) (2)
h is the function such that \(\mathrm{hf}(x) = x\) for all \(x > 1\)
(c) Find the function h in the form \(\mathrm{h}(x) = \ldots\) (2)
| Scheme | Marks |
|---|---|
| \((a + 1)^2 = \dfrac{25}{9}\) | M1 |
| \(a + 1 = (\pm)\dfrac{5}{3}\) | M1 |
| \(\dfrac{2}{3}\) | A1 |
| (3) |
Notes
M1: or \(a + 1 = \dfrac{5}{3}\) or solving a correct quadratic equation of the form \(ax^2 + bx + c = 0\) e.g. \(a^2 + 2a - \dfrac{16}{9} = 0\) followed by \(\left(a - \dfrac{2}{3}\right)\left(a + \dfrac{8}{3}\right)(= 0)\)
or \(a = \dfrac{-2 \pm \sqrt{2^2 - 4 \times 1 \times -\dfrac{16}{9}}}{2}\) (allow 1 sign error)
A1: (DEP on at least M1) for \(\dfrac{2}{3}\) as the only value
| Scheme | Marks |
|---|---|
| \(\mathrm{fg}(x) = \mathrm{f}\left(\dfrac{1}{x}\right) = \left(\dfrac{1}{x} + 1\right)^2\) | M1 |
\(x^2\left(\dfrac{1}{x} + 1\right)^2 = \) etc. Answer: Shown | A1 |
| (2) |
| Scheme | Marks |
|---|---|
\(y = (x + 1)^2 \therefore \sqrt{y} = x + 1\) or \(x = (y + 1)^2 \therefore \sqrt{x} = y + 1\) | M1 |
| \(\sqrt{x} - 1\) | A1 |
| (2) |
Notes
M1: (accept \(\pm\sqrt{y} = x + 1\) or \(\pm\sqrt{x} = y + 1\))