2023 MI PRU1 P1 Q8
2023 MI PRU1 P1 Q8
12 marks
Promo
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by
\(\mathrm{f}:x\mapsto {{x}^{2}}-2x+5\), for \(x\in \mathbb{R}\),
\(\mathrm{g}:x\mapsto \frac{1}{x-1}\), for \(x\in \mathbb{R}\), \(x\ne 1\).
- Show that the composite function \(\mathrm{fg}\) exists.[2]
- Find an expression for \(\mathrm{fg}\left( x \right)\) and state its domain. (There is no need to express \(\mathrm{fg}\left( x \right)\) as a single algebraic fraction.)[2]
- Find the range of \(\mathrm{fg}\).[2]
For the rest of the question, the domain of \(\mathrm{f}\) is now restricted to \(\left( -\infty ,1 \right]\).
- Find \({{\mathrm{f}}^{-1}}\left( x \right)\) and state the domain of \({{\mathrm{f}}^{-1}}\).[3]
- Find the range of values of \(x\) that satisfies the equation \(\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)=x\).[1]
- It is given that \({{\mathrm{g}}^{-1}}\left( a \right)=b\), where \(a\) and \(b\) are constants and \(a\ne 0\), \(b\ne 1\). Without finding an expression for \({{\mathrm{g}}^{-1}}\left( x \right)\), find \(a\) in terms of \(b\).[2]
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Answer:(i) \({R_{\mathrm{g}} = (-\infty, 0) \cup (0, \infty)}\). \({D_{\mathrm{f}} = (-\infty, \infty)}\). Since \({R_{\mathrm{g}} \subseteq D_{\mathrm{f}}}\), \({\mathrm{fg}}\) exists. (ii) \({D_{\mathrm{fg}} = D_{\mathrm{g}} = (-\infty, 1) \cup (1, \infty)}\)
(iii) \({R_{\mathrm{fg}} = [4, \infty)}\) (iv) \({D_{\mathrm{f}^{-1}} = R_{\mathrm{f}} = [4, \infty)}\) (v) \({x \geq 4}\) (vi) \({a = \frac{1}{b - 1}}\)
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