Aj85 wrote:
Siddhans ,
Regarding your second doubt . your wrote " same applies for 2 ... sincex^2>x x^3>x^2"
No, X^2 > X does not necessarily means x^3 will be greater than X^2 and so on...suppose x= -2 (minus 2) then x^2 = +4 which means x^2 >x but x^3 will be -8 which is not more than +4. so you have to consider +- and fraction value while solving such questions.
Can we solve this algebrically
Question says\(x^3 > x^2\) ?
Thus we can reduce this to \(x^3 - x^2\) > 0?
\(X^2 (x-1)> 0?\)------ (1)
Now this equation 1 of of the form xy >0 so either x and y are both positive or both negative ...
since x^2 is always +ve this x-1 is also positive
so equation 1 become x> 0 or x> 1
thus the question becomes is x >1 ??
Now lets come to the statements
1) x>0
This just tells us x> 0 but we dont know if x > 1 ---- Insufficient
2) \(x^2 > x\)
\(X^2 - x > 0\)
thus, x (x-1) >0
x and (x-1) must have same sign either both positive or both negative
Case 1 : x and x-1 both positive
x > 0 or x-1 >0
x> 0 or x > 1
thus from case 1 we can conclude x>1
Case 2: x and x-1 both negative
x<0 or x-1<0
x<0 or x<1
thus x < 0 from case 2
combining case 1 and 2 we get :
i dont know how to proceed ahead after this... can someone help???