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for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

Well for each values of x, use Â½, -1/2, 2, -2, 3, -3 for trials and you will see how it is E. A little exhaustive but gives the right answer. .

OA can't be "E"...in ur values of X all ur +ve numbers do not satisfy both statements...1/2 doesn't satisfy state 2 and 2,3 don't satisfy state 1....only x < 0 satisfies both.

from i, x = more than any negative value that is less than 1= -ve and less than 1>.
from ii, x=either -ve or +ve but more than 1.
to satisfy both, x must less than 0 but more than -1 = -1<x<0
suppose x = -0.5,
from i, X^3<X^2
(-0.5)^3<(-0.5)^2
-0.125<0.25

from ii, X^3<X^4
(-0.5)^3<(-0.5)^4
-0.125<0.0625. therefore it is C.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

X<0? 1) X^3<X^2 2) X^3<X^4

From 1)

x^3 - x^2 <0 x^2(x-1)<0

==>> x^2<0 and (x-1)>0 (Since both of them cannot be negative)

or x^2>0 and (x-1)<0

So you cannot tell anything about x

from 2)

x^3(1-x)<0

Again the same logic

x^3<0 and (1-x)>0

or

x^3>0 and (1-x)<0

which again isnt sufficient.

Since(1) and (2) are giving multiple situations we cannot tell if x<0.

So I believe the answer is E.

Right approach, but x^2 can never be negative. We are talking about real numbers.

Assume x>0, then you can multiple the two equations, and you get:
x^6<x^6 which is impossible. In other words it is impossible that x>0. You know that x can't be equal to zero either. So x must be less than zero. (C).
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
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Last edited by HongHu on 14 Apr 2005, 20:08, edited 1 time in total.

for these who choose E, please elaborate why E is right.

otherwise, the answer given might be wrong.

Thanks

X<0? 1) X^3<X^2 2) X^3<X^4

From 1)

x^3 - x^2 <0 x^2(x-1)<0

==>> x^2<0 and (x-1)>0 (Since both of them cannot be negative)

or x^2>0 and (x-1)<0

So you cannot tell anything about x

from 2)

x^3(1-x)<0

Again the same logic

x^3<0 and (1-x)>0

or

x^3>0 and (1-x)<0

which again isnt sufficient.

Since(1) and (2) are giving multiple situations we cannot tell if x<0.

So I believe the answer is E.

Right approach, but x^2 can never be negative. We are talking about real numbers.

Assume x>0, then you can multiple the two equations, and you get: x^6<x^6 which is impossible. In other words it is impossible that x>0. You know that x can be equal to zero either. So x must be less than zero. (C).

I would have cancelled x on both sides had this question came before I read your cameo on inqualities in this forum.Atleast the approach is right now.....Thanks HH