Question Number 113154 by I want to learn more last updated on 11/Sep/20
$$\mathrm{If}\:\:\:\:\:\:\lfloor\mathrm{x}\:\:+\:\:\sqrt{\mathrm{5}}\rfloor\:\:\:=\:\:\:\lfloor\mathrm{x}\rfloor\:\:+\:\:\lfloor\mathrm{5}\rfloor \\ $$$$\mathrm{then}\:\:\:\:\:\lfloor\mathrm{x}\rfloor\:\:−\:\:\mathrm{x}\:\:\:\:\mathrm{would}\:\mathrm{be}\:\mathrm{greater}\:\mathrm{than} \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\sqrt{\mathrm{5}}\:\:−\:\:\mathrm{2}\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\:\:\:\sqrt{\mathrm{5}}\:\:−\:\:\mathrm{3}\:\:\:\:\:\:\left(\mathrm{c}\right)\:\:\:\:\sqrt{\mathrm{5}}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\:\:\:\sqrt{\mathrm{5}}\:\:+\:\:\mathrm{1}\:\:\:\:\:\:\:\left(\mathrm{e}\right)\:\:\:\sqrt{\mathrm{5}}\:\:−\:\:\mathrm{1} \\ $$
Answered by 1549442205PVT last updated on 11/Sep/20
$$\left[\mathrm{x}\right]−\mathrm{x}<\mathrm{0}\:,\sqrt{\mathrm{5}}−\mathrm{2}>\mathrm{0},\sqrt{\mathrm{5}}>\mathrm{0},\sqrt{\mathrm{5}}+\mathrm{1}>\mathrm{0},\sqrt{\mathrm{5}}−\mathrm{1}>\mathrm{0} \\ $$$$\sqrt{\mathrm{5}}−\mathrm{3}<\mathrm{0}\:,\mathrm{so}\:\left[\mathrm{x}\right]−\mathrm{x}>\sqrt{\mathrm{5}}−\mathrm{3} \\ $$$$\left.\boldsymbol{\mathrm{Choose}}\:\boldsymbol{\mathrm{b}}\right) \\ $$
Commented by I want to learn more last updated on 11/Sep/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$