Question Number 146619 by mnjuly1970 last updated on 14/Jul/21
$$ \\ $$$$\:\:\:\:{solve}\:::\:\:\:\left(\:{x}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{x}\:\right]\:=\:\left[\:{x}^{\:\mathrm{2}} −\:{x}\:−\mathrm{6}\:\right] \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{note}::\:\:\:\left[{x}\:\right]\::=\:{max}\:\left\{\:{q}\:\in\:\mathbb{Z}\:\mid\:{q}\:\leqslant\:{x}\:\right\} \\ $$
Answered by mindispower last updated on 18/Jul/21
$$\left[{x}\right]={n}=\left[{x}^{\mathrm{2}} −{x}−\mathrm{6}\right] \\ $$$$\Rightarrow{n}<{x}<{n}+\mathrm{1},{n}\leqslant{x}^{\mathrm{2}} −{x}−\mathrm{6}<{n}+\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{6}<{x}+\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{7}<\mathrm{0} \\ $$$$\Rightarrow{x}\in\left[\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}},\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\right] \\ $$$${if}\:{x}\in\left[\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}},−\mathrm{1}\left[\cup\left[−\mathrm{1},\mathrm{0}\left[\cup\left[\mathrm{0},\mathrm{1}\left[\cup\left[\mathrm{1},\mathrm{2}\left[\cup\left[\mathrm{2},\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\left[\right.\right.\right.\right.\right.\right.\right.\right.\right.\right. \\ $$$${case}\:{one}\:\left[{x}\right]=−\mathrm{2} \\ $$$$\Rightarrow−\mathrm{2}\leqslant{x}^{\mathrm{2}} −{x}−\mathrm{6}<−\mathrm{1}\cap\left[−\mathrm{2},−\mathrm{1}\left[\right.\right. \\ $$$$\left[{x}\right]=−\mathrm{1} \\ $$$$−\mathrm{1}\leqslant{x}^{\mathrm{2}} −{x}−\mathrm{6}<\mathrm{0}\:\:\cap\left[−\mathrm{1},\mathrm{0}\left[\right.\right. \\ $$$${find}\:\mathrm{5}\:{cases}\:{solve}\:{one}\:{by}\:{one} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$