IMO D
-1<x<1
1. x>x^2
This is possible only if x is between 0 and 1 . Between 0 and -1, x<x^2
Although we dont have to consider x>1, x<-1 , even in those situations x^2 >x
0<x<1
Sufficient
2. x>x^3
Again this is possible only if x is between 0 and 1 considering range of x of -1<x<1
We also know that x >x^3 when x is less than -1 but we dont have to consider that possibilities as x is only between 1 and -1
Between 0 and -1, x^3>x
Sufficient
D
-1<x<1
1. x>x^2
This is possible only if x is between 0 and 1 . Between 0 and -1, x<x^2
Although we dont have to consider x>1, x<-1 , even in those situations x^2 >x
0<x<1
Sufficient
2. x>x^3
Again this is possible only if x is between 0 and 1 considering range of x of -1<x<1
We also know that x >x^3 when x is less than -1 but we dont have to consider that possibilities as x is only between 1 and -1
Between 0 and -1, x^3>x
Sufficient
D
















