\(x(x-1) < 0\) gives you the solution 0 < x < 1.

check out the link below for the explanation:
http://gmatclub.com/forum/inequalities-trick-91482.html#p804990
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x(x - 1) < 0 is not the same as x <0 and (x - 1)< 0

When I multiply two terms, the result is negative if and only if one of them is negative and the other is positive. When I multiply x with (x - 1), the result x(x - 1) will be negative (less than 0) in two cases:

Case I: x < 0 (x is negative) but (x - 1) > 0 (x - 1 is positive)
(x - 1) > 0 implies x > 1
But this is not possible. x cannot be less than 0 and greater than 1 at the same time.

Case II: x > 0 (x is positive) but (x - 1) < 0 (x - 1 is negative)
(x - 1) < 0 implies x < 1
This will happen when x lies between 0 and 1. i.e. when 0 < x < 1.

The link gives you the shortcut of solving inequalities of this type.
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Ok I see it now! I was not taking into consideration the 2 cases that you've just made clear for me. Now I see how x(x-1) < 0 must become 0<x<1 . Thanks guys!

Merging similar topics.



In (1) as x^2<x then x can not be negative, because if it is then we would have x^2>0>x.
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I have a question is this. Why have we considered both the options.

x(x-1)<0:

Multiples must have opposite signs:
1. x<0 and x-1>0, or x>1 --> no solution (x can not be simultaneously less than zero and more than 1);
2. x>0 and x-1<0, or x<1 --> 0<x<1;

In the link to the post when we find the roots of the quadratic equation and if we know the sign is "<" we can directly right the roots as "root 1" < x < "root 2". The same way in this case also there are 2 roots 0 and 1 so we can directly write it this way. 0<x<1. Why do we have to consider case 1 also.

Rahul

Hi Bunuel, Am I right in construing when I say that x(x-1)<0, which means the roots are 0, 1 and since it is "<" the solution must lie between 0 and 1 and hence, 0<x<1. Please confirm.

From F.S 1, we have \(x^2<x\)

or \(x*(x-1)<0\) . This is possible only if they have different signs. Thus, either x<0 AND (x-1)>0[ This is not possible as x can't be more than 1 and yet be negative] or x>0 AND (x-1)<0. This gives us that 0<x<1. Sufficient.

From F.S 2, we know that \(x^3\) >0. Thus, cancelling out x^2 from both sides, we have x>0. Insufficient.

A.
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Could someone please explain why it's not possible:

x^2 < x
try x=1/2 => 1/4 < 1/2 Yes, 0 < x < 1
try x=-1 => 1 > -1 No, x < 0 < 1

Why can x be only positive in this case since it can be negative and squared? It is not implied in"0 < x <1" that x must be a positive number? The question asks whether x is between 0 and 1, in case 1 x can be -1 and still satisfy the equation...

EDIT: Sorry, I realized that statement 1 must be correct in itself...

Even if you take x < sqrt(x), you know that :

1) x has to be positive because sqrt of -ve number is always imaginary.
2) for x to be less than its square root , it has to be less than 1 and greater than 0. because any number greater than 1 would have its square root less than itself.

Thus, your approach is also fine.
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Target question: Is x between 0 and 1 ?

Statement 1:x² is less than x
In other words, x² < x
We can apply some inequality rules here.
Since x² must be POSITIVE here, we can take x² < x and divide both sides by x²
We get: 1 < 1/x
Since 1/x is greater than 1, we can conclude that 1/x is positive, which means x is POSITIVE (i.e., x > 0)
Since x is POSITIVE, we can take 1 < 1/x and multiply both sides by x to get: x < 1
When we combine our two inequalities, we get 0 < x < 1
In other words, x IS between 0 and 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³ is positive
There are several values of x that satisfy statement 2. Here are two:
Case a: x = 1/2 (so, x³ = (1/2)³ = 1/8). In this case, x IS between 0 and 1
Case b: x = 2 (so, x³ = 2³ = 8). In this case, x is NOT between 0 and 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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