Is x^3 > x^2?
1. x > 0
2. x^2 > x
Can some one explain how to solve this?
I hope yo know the basic structure of DS questions? You need to first check the suffecieny of statement 1...then for statement 2....and if neither is insuffecient on its own...then test for suffeciency together etc...
his means x could be a number like 1/2 (which between 0 and 1) or could be a number like 2 (>1)...
for x=1/2, \(x^3<x^2\) as 1/8<1/4
for x=2, \(X^3>X^2\) as 8>4.
As both cases are possible, statement 1 alone is not suffecient to say if \(x^3>x^2\) or not
St2: this says x^2>x => x(x-1)>0 =>x>1 or x<0 =>x could be numbers like -1/2,-2, 2
for x =-1/2, x^3<x^2
for x= -2, x^3<x^2
for x= 2, X^3>x^2....
hence not suffecient to say if\(x^3\) is greater than\(X^2\) or not
St1 and St2 together: X>0 AND (X>1 or X<0)....only numbers satisfying both cases (or in other words only common area for both cases on number line) is for x>1....for all x greater than 1. x^3>x^2...so suffecient to say YES to the question asked.