chetan2u wrote:
Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?
(1) |x| < 2
(2) x^3 = x
\(x^3 < x^2.......x^2-x^3>0.....x^2(1-x)>0\)
therefore 1-x>0 or x<1 but \(x\neq{0}\)
(1) |x| < 2
so -2<x<2...
possible values = -1,0,1
but x<1 and not equal to 1, so only value left -1
sufficient
(2) x^3 = x
\(x^3=x.....x^3-x=0......x(x^2-1)=0\)
so x=o or x^2=1 that is -1 and 1
but as seen above only -1 is possible out of -1,0 and 1
sufficient
D
hi there
chetan2u regarding STATEMENT ONE, i dont get how it can be sufficient.
This is how i understand it
|x| < 2
When X is positive x < 2 so x can be 1 , 0 , -1 etc
now plug in initital question
-1^3 < -1^2, ----> -1<1 TRUE
1^3 < 1^2 ---- > 1<1
NOT TRUE When x is negative -|x| < 2
-x < 2 (divide by -1)
x > -2 so X can be -1, 0, 1 etc
plug in into initial question
-1^3 < -1^2 ----- > -1 < 1 TRUE
1^3 < 1^2 ----- > 1 < 1
NOT TRUE So what`s wrong with my approach ?
thanks and have a great day