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Question Number 184938 by mnjuly1970 last updated on 14/Jan/23

$$$$$$\mathrm{If},\{\begin{array}{l}f:[\sqrt{2},+\mathrm{\infty })\to \mathbb{R}\\ f\left(x\right)=x^2+\lfloor \frac{1}{1-\lfloor x^2\rfloor }\rfloor \end{array}$$$$$$$$\Rightarrow \lfloor f^{-1}\left(\pi \right)\rfloor =?$$$$$$$$$$

Answered by Frix last updated on 14/Jan/23

$$\sqrt{\pi +1}$$

Commented by mnjuly1970 last updated on 14/Jan/23

$$thanksalot...$$

Commented by Frix last updated on 14/Jan/23

$$\mathrm{Sorry}\mathrm{for}\mathrm{the}\mathrm{missing}\mathrm{solution},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{the}$$$$\mathrm{result}\mathrm{of}\mathrm{my}\mathrm{playing}\mathrm{around}\mathrm{with}\mathrm{it}$$

Answered by mnjuly1970 last updated on 14/Jan/23

$$mysolution:$$$$x⩾\sqrt{2}\Rightarrow x^2⩾2\Rightarrow \lfloor x{}^2\rfloor ⩾2$$$$\Rightarrow 1-\lfloor x^2\rfloor ⩽-1<0$$$$\Rightarrow 0>\frac{1}{1-\lfloor x^2\rfloor }⩾-1$$$$f\left(x\right)=x^2-1$$$$f^{-1}\left(x\right)=\sqrt{x+1}$$$$\lfloor f^{-1}\left(\pi \right)\rfloor =\lfloor \sqrt{\pi +1}\rfloor =2$$$$$$

Commented by aba last updated on 14/Jan/23

$$\mathrm{good}$$

Commented by Mastermind last updated on 15/Jan/23

$$\mathrm{Is}\mathrm{this}\mathrm{a}\mathrm{solution}\mathrm{to}\mathrm{my}\mathrm{question}$$$$\mathrm{pls},\mathrm{could}\mathrm{you}\mathrm{please}\mathrm{make}\mathrm{it}\mathrm{easier}$$$$\mathrm{than}\mathrm{this}?$$

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