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Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x

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Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 14 Jul 2010, 05:36
2
5
00:00
A
B
C
D
E

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  65% (hard)

Question Stats:

53% (01:48) correct 47% (02:07) wrong based on 235 sessions

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Is 1 > |x - 1| ?

(1) (x - 1)^2 > 1
(2) 0 > x

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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 14 Jul 2010, 05:47
2
Is \(|x-1|<1\)?

Is \(|x-1|<1\)? --> is \(-1<x-1<1\)? --> add 1 to all three parts: is \(0<x<2\)?

(1) \((x-1)^2>1\) --> since both sides are non-negative, we can take the square root from both parts (recall that \(\sqrt{x^2}=|x|\)): \(|x-1|>1\) --> directly gives a NO answer to the question. Sufficient.

(2) \(x<0\) --> \(x\) is not in the range \(0<x<2\). So, again we have a NO answer to the question. Sufficient.

Answer: D.

Hope it's clear.
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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 15 Jul 2010, 03:51
Can i also just square the given term?

yielding:
(x-1)^2<1 ?

--> Then I basically get the oppositie of whats stated in ST1 (hence answer is NO)
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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 15 Jul 2010, 06:58
AndreG wrote:
Can i also just square the given term?

yielding:
(x-1)^2<1 ?

--> Then I basically get the oppositie of whats stated in ST1 (hence answer is NO)


Since both parts of inequality \(1>|x-1|\) are non-negative we can safely do that --> \(1>(x-1)^2\).
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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 24 Jun 2013, 23:33
2
Is 1 > |x-1| ?
Rewrite the question as: is \(0<x<2\)?

I. (x-1)^2 > 1
so x<0 or x>2, in both cases x is outside the interval 0-2.
Sufficient

II. 0 > x
x is less then 0, so it's outside the 0-2 interval.
Sufficient
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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 20 May 2015, 04:46
For those who have difficulty dealing with modulus in inequality, you can use the definition of modulus to interpret the range of the variable inside the modulus.

In this question we are asked if |x -1 | < 1. We know that modulus denotes the magnitude of the distance of a number from a particular point. So |x - 1| means the distance of x from 1. Since | x - 1| < 1 it means that x lies at a distance of less than 1 unit on either side of 1.

Hence 1 - 1 < x < 1 + 1 i.e. 0 < x < 2 which is what the question is asking us.

Hope this helps :)

Regards
Harsh
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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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New post 20 May 2015, 18:38
Hi All,

When absolute values 'interact' with inequalities, it can sometimes be difficult to automatically spot all of the possible values that "fit." In these situations, you can usually TEST VALUES to figure out which values "fit" and which do not.

Here, we're asked if 1 > |X-1|. This is a YES/NO question.

Before dealing with the two Facts, we can think about what would gives us a "YES" answer and what would give us a "NO" answer...

X = 1 is an obvious YES answer, since 1 > 0.
X = 2 gives us a NO answer, since 1 is NOT > 1
X = 0 gives us a NO answer too.

So 0 and 2 are the "borders"; with a bit more 'playing around', you'll see that any value BETWEEN 0 and 2 will give us a YES answer.
All other values, including 0 and 2, will give us a NO answer.

Fact 1: (X-1)^2 > 1

Neither 0 nor 2 will fit this inequality (since 1 is NOT > 1). Any value BETWEEN 0 and 2 will turn the parentheses into a positive fraction, which also doesn't fit (since a positive fraction is NOT > 1). By eliminating all of the values that would lead to a NO answer, the only values that fit will create a YES answer. The answer will ALWAYS be YES.
Fact 1 is SUFFICIENT.

Fact 2: 0 > X

This Fact also eliminates all of the values that would lead to a NO answer, so all that is left are YES answers.
Fact 2 is SUFFICIENT

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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x  [#permalink]

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Re: Is 1 > |x - 1| ? (1) (x - 1)^2 > 1 (2) 0 > x   [#permalink] 04 Feb 2019, 08:14
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