2013 SRJC J1 MYE Q7
2013 SRJC J1 MYE Q7
Mid Year Exam
The function \(\mathrm{f}\) is defined as
\(\mathrm{f}:x\mapsto -{{x}^{2}}-x+6\), \(x\in \mathbb{R}\).
- Explain why \({{\mathrm{f}}^{-1}}\) does not exist.
The function \(\mathrm{f}\) has an inverse if its domain is restricted to \(x < k\).
- Find the largest value of \(k\) and hence find \({{\mathrm{f}}^{-1}}\) in a similar form.
- Sketch, on the same diagram, the graphs of \(\mathrm{f}\), \({{\mathrm{f}}^{-1}}\) and \(\mathrm{f}{{\mathrm{f}}^{-1}}\).
- Hence, solve the equation \(\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)\), leaving your answer in exact form.
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Answer:(i) Since \(y=1\) cuts curve \(\mathrm{f}\) twice, \(\mathrm{f}\) is not one-one. Thus \({{\mathrm{f}}^{-1}}\) doesn’t exist. (ii) \({{\mathrm{f}}^{-1}}:x\mapsto -\frac{1}{2}-\sqrt{\frac{25}{4}-x}\), \(x\in \mathbb{R}\), \(x<\frac{25}{4}\)
(iii) \(\mathrm{f}{{\mathrm{f}}^{-1}}:x\mapsto x\); \(x<\frac{25}{4}\)
(iv) \(x=-1-\sqrt{7}\)
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