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Is |x - 1| < 1 ? 1. (x - 1)^2 <= 1 2. x^2 - 1

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Joined: 05 Oct 2008
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Is |x - 1| < 1 ? 1. (x - 1)^2 <= 1 2. x^2 - 1 [#permalink]

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01 Nov 2008, 21:55
Is $$|x - 1| 0$$

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01 Nov 2008, 22:27
expanding #1, x-1 <= 1 OR x <= 2, which doesn't help us solve is Mod(x-1) < 1. So A and D are out.

expanding #2, x^2 > 1 or x > 1, again not sufficient enough to answer Mod(x-1) < 1. So B is out.

But the 2 statements together, we get 1 < x <= 2, again not sufficient to say whether Mod(x-1) < 1. So my choice is E
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01 Nov 2008, 23:55
1
study wrote:
Is $$|x - 1| < 1$$ ?

1. $$(x - 1)^2 <= 1$$
2. $$x^2 - 1 > 0$$

|x - 1| < 1 has two cases

case 1) x - 1 is positive, then x - 1 < 1 ===> x < 2
case 2) x - 1 is negative, then -(x - 1)<1 ===> x-1 > -1 ===> x>0

so the question is asking if x>0 and x<2, in other words does x lie in between 0 and 2 ( 0 and 2 NOT included)

when we have x²<1 --- we write it as |x|<1

so stmt1) $$(x - 1)^2 <= 1$$ ===> |(x - 1)| <= 1

case 1) (x - 1) is positive, then x-1<=1 or x <=2
case 2) (x-1) is negative, then -(x-1)<=1 or x-1 >= -1 ====> x>=0

this means x>=0 and x<=2, in other words x lies in between 0 and 2 ( here 0 and 2 are INCLUDED, because we have "=" sign). This is NOT sufficient to say whether x lies between 0 and 2, with 0 and 2 not included.

Stmt2) $$x^2 - 1 > 0$$ ===> $$x^2 > 1$$ ===> |x|>1

case 1) x is positive , then x >1
case 2) x is negative, then -x>1 ===> x < -1

Again, not sufficient. X can be between 0 and 2 and can be out of the range.

(1) and (2) COMBINED

When we have such problems, we combine the range we get from the 2 statements and check if the overlapping range lies within the range which the question is asking for.

statement (1) says x lies between 0 and 2 (both inclusive)
statement (2) says x lies outside -1 and 1.

the overlapping range would be x>1 and x<=2.

still this info is not sufficient to say whether x lies in the range with 2 not included. Since we have <=2, x could be = 2 and thus not satisfy. Other values of x >1 and <2 satisfy.

Not sufficient. so E
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02 Nov 2008, 03:06
E.

From stmt1: |x-1| <= 1 and is not sufficient as |x-1| can also equal 1.
From stmt2, |x-1| > 0 and is not sufficient as |x-1| can also be > 1.

Combining two, also, insufficient.
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02 Nov 2008, 06:13
scthakur wrote:
E.

From stmt1: |x-1| <= 1 and is not sufficient as |x-1| can also equal 1.
From stmt2, |x-1| > 0 and is not sufficient as |x-1| can also be > 1.

Combining two, also, insufficient.

scthakur .. we can take x²-1 > 0 as |x-1|>0 ??
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02 Nov 2008, 11:47
fresinha12 wrote:
amitdgr wrote:
scthakur wrote:
E.

From stmt1: |x-1| <= 1 and is not sufficient as |x-1| can also equal 1.
From stmt2, |x-1| > 0 and is not sufficient as |x-1| can also be > 1.

Combining two, also, insufficient.

scthakur .. we can take x²-1 > 0 as |x-1|>0 ??

NO..

i guess scthakur got it wrong then .....
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02 Nov 2008, 23:58
amitdgr wrote:
scthakur wrote:
E.

From stmt1: |x-1| 0 and is not sufficient as |x-1| can also be > 1.

Combining two, also, insufficient.

scthakur .. we can take x²-1 > 0 as |x-1|>0 ??

(x^2-1) > 0 means (x-1)(x+1) > 0. This means, either (x-1) is >0 or (x-1) 0.

I hope, I am clear.

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Re: Absolute values   [#permalink] 02 Nov 2008, 23:58
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