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Is |x-1| < 1?

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mokap25
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Is |x-1| < 1? [#permalink] New post   17 Aug 2011, 05:36
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Is |x-1| < 1?

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

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Hi someone help solve this, shouldnt be very difficult.


I know how to evaluate statement 1 and statement 2, looking to just understand how do we test a combined statement i,e C when we have multiple inequality equations. Do we test 4 values...

Maybe somebody could explain while solving this question..

Once again how do we evaluate c- statement combined together..is my main concern

Thanks
Kaps


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E
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anordinaryguy
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Re: Is |x-1| < 1? [#permalink] New post   Updated on: 17 Aug 2011, 11:06
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First I did calculation partially and got the answer C, but than when I was trying to type solution for you, I get more insight into the question.

We need to prove

Is |x-1| < 1? => -1<(x-1)<1 => 0<x<2 ?

(1) (x-1)^2 <= 1

=> 0 <= X <= 2 (Correction in earlier calculation)
=> Doesn't prove 0<x<2, Not Sufficient

(2) x^2 - 1 > 0

=> (x > -1 and x > 1) OR (x < -1 and x < 1)
=> x > 1 or x < -1
=> Doesn't prove 0<x<2, Not Sufficient

Combining these two we get

1 < X <= 2

=> Doesn't prove 0<x<2, because x = 2 is a possible solution in combined equality.

So E is the answer

Originally posted by anordinaryguy on 17 Aug 2011, 10:24.
Last edited by anordinaryguy on 17 Aug 2011, 11:06, edited 1 time in total.
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Re: Is |x-1| < 1? [#permalink] New post   17 Aug 2011, 10:33
mokap25 wrote:
Is |x-1| < 1?

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


From statement 1: If (x-1)^2 <= 1, the value of x varies from 0 to 2 i.e. x is non-negative.

(x-1)^2 <= 1
(x-1) (x-1) <= 1

i) x-1 <= 1
x-1 <= 1
x <= 2

If x is 2, |x-1| = |2-1| = 1 ..........No.
If x is 1.5, |x-1| = |1.5-1| = 0.5.. Yes.

ii) x-1 => 1
x => 0

If x is 0, |x-1| = |0-1| = 1 ..........No.
If x is 0.5, |x-1| = |0.5-1| = 0.5.. Yes.

Not sufficient......

From statement 2: x^2 - 1 > 0
x^2 > 1

i) either, x > 1
ii) or x < -1

If x is 1.1, |x-1| = |1.1-1| = 0.1 .......... yes.
If x is -1.1, |x-1| = |-1.1-1| = -2.1. No.

From 1 and 2: the value of x varies from 0 to 2 but still insufficient.

E.
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jansam123
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Re: Is |x-1| < 1? [#permalink] New post   17 Aug 2011, 11:43
Both Statements 1 & 2 (Individually & Combined) yields range of numbers that do not satisfy the Question Stem.

We can simplyfy the question stem
Is |x-1| < 1? to
Is 0<x<2 to fit and check the range of numbers yielded by the statement 1 & 2.
Also X can be a fraction which further increases the possible value ranges of X .

So I agree with Answer E.
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Re: Is |x-1| < 1? [#permalink] New post   18 Aug 2011, 07:37
I struggled a lot with these types of questions and actually drawing a number line helped me a lot.

If we look at the question we know either (1), (2) or (1)+(2) must prove 0<x<2. Draw that on a line, compare it to (1), (2) and then for c, compare whether the range of (1) + (2)'s (where 1 & 2 overlap) falls purely in 0<x<2.
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Re: Is |x-1| < 1? [#permalink] New post   13 Jun 2017, 23:08
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Is |x-1| < 1?

Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we can not say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

Hope it's clear.

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Re: Is |x-1| < 1? [#permalink]
13 Jun 2017, 23:08
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Bunuel
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