Question Number 169569 by mnjuly1970 last updated on 03/May/22
$$ \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:=\:{x}\:−\lfloor\frac{{x}}{\mathrm{2}}\rfloor−\lfloor\frac{{x}}{\mathrm{3}}\rfloor−\lfloor\frac{{x}}{\mathrm{6}}\rfloor \\ $$$$\:\:\:\:\:\:\:\:\:\:{R}_{\:{f}} \:=\:? \\ $$$$\:\:\:\:\: \\ $$
Answered by mahdipoor last updated on 16/Jul/22
![x=6k+j 0≤j∈R<6 k∈Z ⇒f(x)=6k+j−[3k+(j/2)]−[2k+(j/3)]−[k+(j/6)]= j−[(j/2)]−[(j/3)]−[(j/6)]=j−[(j/2)]−[(j/3)] { ((0≤j<2⇒ 0≤f(x)=j−0−0<2)),((2≤j<3⇒ 1≤f(x)=j−1−0<2)),((3≤j<4⇒ 1≤f(x)=j−1−1<2)),((4≤j<6⇒ 1≤f(x)=j−2−1<3)),((⇒⇒ R_f =[0,3))) :}](https://www.tinkutara.com/question/Q169574.png)
$${x}=\mathrm{6}{k}+{j}\:\:\:\:\:\:\mathrm{0}\leqslant{j}\in{R}<\mathrm{6}\:\:\:\:\:{k}\in\mathrm{Z} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{6}{k}+{j}−\left[\mathrm{3}{k}+\frac{{j}}{\mathrm{2}}\right]−\left[\mathrm{2}{k}+\frac{{j}}{\mathrm{3}}\right]−\left[{k}+\frac{{j}}{\mathrm{6}}\right]= \\ $$$${j}−\left[\frac{{j}}{\mathrm{2}}\right]−\left[\frac{{j}}{\mathrm{3}}\right]−\left[\frac{{j}}{\mathrm{6}}\right]={j}−\left[\frac{{j}}{\mathrm{2}}\right]−\left[\frac{{j}}{\mathrm{3}}\right] \\ $$$$\begin{cases}{\mathrm{0}\leqslant{j}<\mathrm{2}\Rightarrow\:\mathrm{0}\leqslant{f}\left({x}\right)={j}−\mathrm{0}−\mathrm{0}<\mathrm{2}}\\{\mathrm{2}\leqslant{j}<\mathrm{3}\Rightarrow\:\mathrm{1}\leqslant{f}\left({x}\right)={j}−\mathrm{1}−\mathrm{0}<\mathrm{2}}\\{\mathrm{3}\leqslant{j}<\mathrm{4}\Rightarrow\:\mathrm{1}\leqslant{f}\left({x}\right)={j}−\mathrm{1}−\mathrm{1}<\mathrm{2}}\\{\mathrm{4}\leqslant{j}<\mathrm{6}\Rightarrow\:\mathrm{1}\leqslant{f}\left({x}\right)={j}−\mathrm{2}−\mathrm{1}<\mathrm{3}}\\{\Rightarrow\Rightarrow\:{R}_{{f}} =\left[\mathrm{0},\mathrm{3}\right)}\end{cases} \\ $$
Commented by mnjuly1970 last updated on 03/May/22
$${please}\:\:{recheck}\:{Sir}\:{mahdipour} \\ $$