Problem: The set of all real solutions of the inequality β£xβ1β£+β£x+2β£<5 is
Answer Choices:
A. {x:β3<x<2}
A. {x:β3<x<2}
B. {x:β1<x<2}
C. {x:β2<x<1}
D. {x:β23β<x<21β}
E. Ο (empty)
Solution:
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If β2<x<1, then xβ1<0 and x+2>0 so that the given inequality requires βx+1+x+2<5 or 3<5 which is alvays true. The inequality is true also when x=1 and x=β2.
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If x>1, then xβ1>0 and x+2>0 en that the given inequality becomes (xβ1)+(x+2)=2x+1<5.x<2.β΄1<x<2.
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If x<β2, then x+2<0 and xβ1<0 and the given inequality becomes β(xβ1)β(x+2)=2xβ1<5.β΄β2x<6,x>β3. We have accordingly β3<x<2 as the set of all real solutions.