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Sagot :
Step-by-step explanation:
Middle option.
(x + 1 ≤ 1) ∩ (x + 1 ≥ 1)
If you work both sides separately you get
(x ≤ 0) ∩ (x ≥ 0)
which reduces nicely to
{x | x = 0}
Let's check one by one
#1
[tex]\\ \sf\longmapsto x+1<-1\implies x<-2[/tex]
[tex]\\ \sf\longmapsto x+1<1\implies x<0[/tex]
- Rejected
#2
[tex]\\ \sf\longmapsto x+1\leqslant 1\implies x\leqslant 0[/tex]
[tex]\\ \sf\longmapsto x+1\geqslant 1\implies x\geqslant 0[/tex]
Option B is correct
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