Higher November 2019 Paper 1 Q22
22. The function \(f\) is defined, for \(x \ne 1\), by \(f(x) = \dfrac{7}{x-1}\).
(a)
(i) Explain why \(x \ne 1\) for this function. [1]
(ii) Show that \(f^{-1}(x) = \dfrac{a}{x} + b\), where \(a\) and \(b\) are integers. [2]
(b) The functions \(g\) and \(h\) are defined for all real \(x\) by\[\begin{aligned} g(x) &= \sqrt[3]{x+1}\,, \\ h(x) &= 9x^3\,. \end{aligned}\]Solve \(hg(x) = f(x)\). [5]
| Scheme | Marks | Comments |
|---|---|---|
| (i) Valid explanation e.g. ‘7 divided by 0 is not defined.’ | E1 | Accept e.g. ‘You cannot divide by 0.’ ‘The denominator is 0.’ is not sufficient |
| (ii) Swop variables and change subject or vice versa | M1 | Allow one sign or arithmetic slip |
| \(y = \dfrac{7}{x} + 1 \quad \left(f^{-1}(x) = \dfrac{7}{x} + 1\right)\) | A1 | CAO |
| Scheme | Marks | Comments |
|---|---|---|
| \((hg(x) =)\ 9(x + 1)\) | B2 | B1 for \(9\left(\sqrt[3]{x+1}\right)^3\) |
| \(9(x + 1)(x - 1) = 7\) | M1 | FT ‘their hg’ if of equivalent difficulty |
| \(x^2 = \dfrac{7}{9} + 1\) oe | M1 | |
| \(x = \pm\dfrac{4}{3}\) oe | A1 | CAO |
| (8) |