Problem: The relation x2(x2β1)β§0 is true for:
Where xβ§ a means that x can take on all values greater than a and the value equal to a, while xβ¦a has a corresponding meaning with "less than."
Answer Choices:
A. xβ§1 only
B. β1β¦xβ¦1
C. x=0,x=1,x=β1
D. x=0,xβ¦β1,xβ§1
E. xβ§0 only
Solution:
First note that the equality x2(x2β1)=0 is satisfied by 0,0,+1,β1. Since x2 is non-negative, then x2(x2β1)>0 implies x2β1>0. β΄x2>1; β΄β£xβ£>1, that is, x>1 or x<β1. Combining these, we have x=0,xβ€β1,xβ₯1.