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???