Answer with Step-by-step explanation:
a.[tex]x^2=1[/tex]
[tex]x=\sqrt{1}=\pm 1[/tex]
x is a real number
Therefore, the set={-1,1}
b.x is a positive integer less than 12
Therefore, the value of x are
1,2,3,4,5,6,7,8,9,10,11
Set={1,2,3,4,5,6,7,8,9,10,11}
c.x is a square of an integer and x<100
Therefore, the value of x can be
1,4,9,16,25,36,49,64,81
Set={1,4,9,16,25,36,49,64,81}
d.x is an integer such that [tex]x^2=2[/tex]
[tex]x=\pm\sqrt{2}[/tex]
But x is an integer
[tex]\sqrt 2[/tex] and [tex]-\sqrt 2[/tex] are not integers.
Therefore, there is no integer element which satisfied [tex]x^2=2[/tex]
Set=Null set=[tex]\phi[/tex]
