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Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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04 Nov 2010, 18:49

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Is X between 0 and 1 ?

(1) x^2 is less than x (2) x^3 is positive

I am curious how to rephrase Statement 1 using inequalities. I rewrote it as \(x^2 - x < 0\) , which then gives me \(x(x-1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly?

I am curious how to rephrase Statement 1 using inequalities. I rewrote it as \(x^2 - x < 0\) , which then gives me \(x(x-1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly?

Hmm i think i am more confused after reading that the first time through....

I think i am missing the reason why the inequality sign for \(x < 0\) should actually be\(x > 0\). I determined x\(< 1\) because i set the inequality of \(x - 1 < 0\) and after subtracting from both sides give me \(x < 1\) . What is different about \(x < 0\) becoming \(x > 0\) ?

Hmm i think i am more confused after reading that the first time through....

I think i am missing the reason why the inequality sign for \(x < 0\) should actually be\(x > 0\). I determined x\(< 1\) because i set the inequality of \(x - 1 < 0\) and after subtracting from both sides give me \(x < 1\) . What is different about \(x < 0\) becoming \(x > 0\) ?

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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I am curious how to rephrase Statement 1 using inequalities. I rewrote it as \(x^2 - x < 0\) , which then gives me \(x(x-1) < 0\). If x < 0 and x < 1 then \(0>x<1\). Wouldnt this statement be insufficient? or am i writing that dual inequality incorrectly?

Is x between 0 and 1?

Is \(0<x<1\)?

(1) x^2 is less than x --> \(x^2<x\) --> \(x(x-1)<0\):

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

So \(x(x-1)<0\) holds true when \(0<x<1\). Sufficient.

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!

Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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07 Jun 2012, 17:44

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.

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

These are just two different approaches.
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Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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08 Jun 2012, 21:14

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.

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.

Is X between 0 and 1? 1. x^2 is less than x. 2. x^3 is positive

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.

Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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05 Nov 2014, 14:17

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...

Re: Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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26 Dec 2015, 15:27

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Is X between 0 and 1 ? (1) x^2 is less than x (2) x^3 is [#permalink]

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30 Jul 2016, 09:55

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nancy77 wrote:

Hi, when I first tackled this problem, I took the square root of both sides so that gave me the equation of x < sqrt(x).

Is it wrong to approach it this way? I now understand this is a positives and negatives problem based on the solutions above...

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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