October 2020 Paper 2 Q15
15 Functions \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) are defined as follows.
\(\mathrm{f}(x) = \sqrt{x}\) for \(x > 0\) and \(\mathrm{g}(x) = x^3 - x - 6\) for \(x > 2\).
The function \(\mathrm{h}(x)\) is defined as
\(\mathrm{h}(x) = \mathrm{fg}(x)\).
(a) Find \(\mathrm{h}(x)\) in terms of \(x\) and state its domain. [2]
(b) Find \(\mathrm{h}(3)\). [1]
Fig. 15 shows \(\mathrm{h}(x)\) and \(\mathrm{h}^{-1}(x)\), together with the straight line \(y = x\).
(c) Determine the gradient of \(y = \mathrm{h}^{-1}(x)\) at the point where \(y = 3\). [4]
| Scheme | Marks | AO |
|---|---|---|
| [\(\mathrm{h}(x)\) or \(\mathrm{fg}(x) =\)] \(\sqrt{x^3 - x - 6}\) oe | B1 | 1.1 |
| \(x > 2\) | B1 | 1.1 |
| [2] |
Notes
B1: expression
mark the final answer
B1: domain
| Scheme | Marks | AO |
|---|---|---|
| \(\sqrt{18}\) oe isw FT their \(\mathrm{h}(x)\) | B1 | 1.1 |
| [1] |
Notes
B1: allow 4.2426406872… rounded to 2 or more sf
| Scheme | Marks | AO |
|---|---|---|
| \(\frac{1}{2} \times \dfrac{3x^2 - 1}{\sqrt{(x^3 - x - 6)}}\) or \(\dfrac{3x^2 - 1}{2\mathrm{h}(x)}\) oe | M1 A1 | 3.1a 1.1 |
| their \(\dfrac{\mathrm{d}h}{\mathrm{d}x}\) evaluated at \(x = 3\) | M1 | 1.1 |
| \(\dfrac{3\sqrt{2}}{13}\) or 0.326356975932 rounded to 2 sf or better | A1 | 3.2a |
| [4] |
Notes
M1: chain rule used
allow one slip in differentiation, eg sign error
A1: all correct
M1: \(\mathrm{h}(x)\) must be correct for first M1
Alternative
| Scheme | Marks |
|---|---|
| OR \(x^2 = y^3 - y - 6 \Rightarrow 2x\dfrac{\mathrm{d}x}{\mathrm{d}y} = 3y^2 - 1\) oe | M1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{3y^2 - 1}{2x}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x}{3y^2 - 1}\) | A1 |
| substitution of \(y = 3\) and \(x = \) their \(\sqrt{18}\) | M1 |
| \(\dfrac{3\sqrt{2}}{13}\) or 0.326356975932 rounded to 2 sf or better | A1 |
M1: allow one slip eg sign error
rearrangement to find \(\mathrm{h}^{-1}(x)\) explicitly in terms of \(x\) followed by differentiation does not score