Question Number 86779 by M±th+et£s last updated on 31/Mar/20
![ssolve 1)x−[x]≥0 2)x−[x]≤0 3)x+[x]≥0 4)x+[x]≤0](Q86779.png)
$${ssolve} \\ $$$$\left.\mathrm{1}\right){x}−\left[{x}\right]\geqslant\mathrm{0} \\ $$$$\left.\mathrm{2}\right){x}−\left[{x}\right]\leqslant\mathrm{0} \\ $$$$\left.\mathrm{3}\right){x}+\left[{x}\right]\geqslant\mathrm{0} \\ $$$$\left.\mathrm{4}\right){x}+\left[{x}\right]\leqslant\mathrm{0}\: \\ $$
Answered by mr W last updated on 31/Mar/20
![(1) x≥[x] ⇒x≥0 (2) x≤[x] ⇒x≤0 (3) if x≥0, [x]=x−δ x+[x]=2x−δ≥0 x≥(δ/2) ⇒x≥0 if x<0, [x]=x+1−δ x+[x]=2x+1−δ≥0 x≥−(1/2)+(δ/2)≥0 ⇒x≥0 is solution (4) if x≥0, [x]=x−δ x+[x]=2x−δ≤0 x≤(δ/2)≤0 if x<0, [x]=x+1−δ x+[x]=2x+1−0 x≤−(1/2)+(δ/2)≤0 ⇒x≤0 is solution](Q86787.png)
$$\left(\mathrm{1}\right) \\ $$$${x}\geqslant\left[{x}\right] \\ $$$$\Rightarrow{x}\geqslant\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${x}\leqslant\left[{x}\right] \\ $$$$\Rightarrow{x}\leqslant\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${if}\:{x}\geqslant\mathrm{0}, \\ $$$$\left[{x}\right]={x}−\delta \\ $$$${x}+\left[{x}\right]=\mathrm{2}{x}−\delta\geqslant\mathrm{0} \\ $$$${x}\geqslant\frac{\delta}{\mathrm{2}} \\ $$$$\Rightarrow{x}\geqslant\mathrm{0} \\ $$$${if}\:\:{x}<\mathrm{0}, \\ $$$$\left[{x}\right]={x}+\mathrm{1}−\delta \\ $$$${x}+\left[{x}\right]=\mathrm{2}{x}+\mathrm{1}−\delta\geqslant\mathrm{0} \\ $$$${x}\geqslant−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\delta}{\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\Rightarrow{x}\geqslant\mathrm{0}\:{is}\:{solution} \\ $$$$ \\ $$$$\left(\mathrm{4}\right) \\ $$$${if}\:{x}\geqslant\mathrm{0}, \\ $$$$\left[{x}\right]={x}−\delta \\ $$$${x}+\left[{x}\right]=\mathrm{2}{x}−\delta\leqslant\mathrm{0} \\ $$$${x}\leqslant\frac{\delta}{\mathrm{2}}\leqslant\mathrm{0} \\ $$$${if}\:\:{x}<\mathrm{0}, \\ $$$$\left[{x}\right]={x}+\mathrm{1}−\delta \\ $$$${x}+\left[{x}\right]=\mathrm{2}{x}+\mathrm{1}−\mathrm{0} \\ $$$${x}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\delta}{\mathrm{2}}\leqslant\mathrm{0} \\ $$$$\Rightarrow{x}\leqslant\mathrm{0}\:{is}\:{solution} \\ $$
Commented by M±th+et£s last updated on 31/Mar/20
$${thank}\:{you}\:{sir} \\ $$
Commented by TawaTawa1 last updated on 31/Mar/20
$$\mathrm{Sir},\:\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\:\mathrm{Q86789} \\ $$
Commented by mr W last updated on 31/Mar/20
$${try}\:{to}\:{solve}\:{by}\:{yourself}\:{at}\:{first}... \\ $$