Question Number 185052 by emmanuelson123 last updated on 16/Jan/23
Commented by Frix last updated on 16/Jan/23
$$\forall{x}\in\left[\sqrt{\mathrm{2}},\:+\infty\right):\:−\mathrm{1}\leqslant\frac{\mathrm{1}}{\mathrm{1}−\lfloor{x}^{\mathrm{2}} \rfloor}<\mathrm{0}\:\Rightarrow \\ $$$$\forall{x}\in\left[\sqrt{\mathrm{2}},\:+\infty\right):\:\lfloor\frac{\mathrm{1}}{\mathrm{1}−\lfloor{x}^{\mathrm{2}} \rfloor}\rfloor=−\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{1};\:\sqrt{\mathrm{2}}\leqslant{x}<+\infty \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=\pi\:\Rightarrow\:{x}=\sqrt{\pi−\mathrm{1}} \\ $$$${f}^{−\mathrm{1}} \left(\pi\right)=\sqrt{\pi−\mathrm{1}} \\ $$
Commented by mnjuly1970 last updated on 16/Jan/23
$$\:\:{excellent}\:{sir} \\ $$
Commented by Frix last updated on 16/Jan/23
$$\mathrm{Thank}\:\mathrm{you} \\ $$