Functions: Existence of inverse — JPJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
Functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by \[\begin{aligned} \mathrm{f}\!: x &\mapsto [\ln(x-1)]^{2}+2,\quad x\geq a,\\ \mathrm{g}\!: x &\mapsto 4+3x-x^{2},\quad x\leq \tfrac{3}{2}. \end{aligned}\]
- It is given that the function \(\mathrm{f}^{-1}\) exists. State the smallest value of \(a\).
- Find an expression for \(\mathrm{g}^{-1}(x)\), stating its domain.
Using the value of \(a\) found in part (i),
- determine whether the composite function \(\mathrm{g}^{-1}\mathrm{f}^{-1}\) exists.
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(i) [1 mark]
Smallest value of \(a = 2\).
(For \(\mathrm{f}^{-1}\) to exist, \(\mathrm{f}\) must be a one-one function.)
(ii) [3 marks]
Let \(y = 4+3x-x^{2}\). \[\begin{aligned} y &= -\left[x^{2}-3x-4\right]\\ &= -\left[\left(x-\tfrac{3}{2}\right)^{2}-\left(\tfrac{3}{2}\right)^{2}-4\right]\\ &= -\left[\left(x-\tfrac{3}{2}\right)^{2}-\tfrac{25}{4}\right]\\ y &= -\left(x-\tfrac{3}{2}\right)^{2}+\tfrac{25}{4} \end{aligned}\]
\[\begin{aligned} \left(x-\tfrac{3}{2}\right)^{2} &= \tfrac{25}{4}-y\\ x-\tfrac{3}{2} &= \pm\sqrt{\tfrac{25}{4}-y}\\ x &= \tfrac{3}{2}\pm\sqrt{\tfrac{25}{4}-y} \end{aligned}\]Since \(x \leq \tfrac{3}{2}\), \(x = \tfrac{3}{2}-\sqrt{\tfrac{25}{4}-y}\).
Thus, \(\mathrm{g}^{-1}: x \mapsto \dfrac{3}{2}-\sqrt{\dfrac{25}{4}-x},\ x \leq \dfrac{25}{4}\). \[ D_{\mathrm{g}^{-1}} = R_{\mathrm{g}} = \left(-\infty,\,\tfrac{25}{4}\right]. \]
(iii) [1 mark] \[ R_{\mathrm{f}^{-1}} = D_{\mathrm{f}} = [2,\infty) \] \[ D_{\mathrm{g}^{-1}} = R_{\mathrm{g}} = \left(-\infty,\,\tfrac{25}{4}\right] \] Since \(R_{\mathrm{f}^{-1}} \not\subseteq D_{\mathrm{g}^{-1}}\), \(\mathrm{g}^{-1}\mathrm{f}^{-1}\) does not exist.