2021 RI P2 Q1

2021 RI P2 Q1

8 marks
Prelims

Functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by

\(\mathrm{f}:x\mapsto {{e}^{{{(x-1)}^{2}}}}\), \(x\in \mathbb{R}\),

\(\mathrm{g}:x\mapsto \frac{1}{2-x}\), \(x\in \mathbb{R}\), \(1\le x<2\).

  1. Sketch the graph of \(y=\mathrm{f}(x)\).[1]
  2. If the domain of \(\mathrm{f}\) is restricted to \(x\ge k\), state with a reason the least value of \(k\) for which the function \({{\mathrm{f}}^{-1}}\) exists.[2]

In the rest of the question, the domain of \(\mathrm{f}\) is \(x\ge k\), using the value of \(k\) found in part (ii).

  1. Find \({{\mathrm{g}}^{-1}}(x)\) and show that the composite function \({{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}\) exists.[4]
  2. Find the range of \({{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}\).[1]

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Answer:(i) Minimum \((1,1)\), \(y\)-intercept \((0,\mathrm e)\) (ii) \(k=1\) (iii) \(\mathrm g^{-1}(x)=2-\dfrac1x\), \(x\geq1\), and \(\mathrm g^{-1}\mathrm f^{-1}\) exists (iv) \(R_{\mathrm g^{-1}\mathrm f^{-1}}=[1,2)\)

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