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# Is |x-1| < 1?

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Manager
Joined: 25 Aug 2011
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Updated on: 12 Apr 2013, 04:55
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Difficulty:

95% (hard)

Question Stats:

43% (02:20) correct 57% (02:23) wrong based on 423 sessions

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Is |x-1| < 1?

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

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Originally posted by devinawilliam83 on 12 Feb 2012, 20:49.
Last edited by Bunuel on 12 Apr 2013, 04:55, edited 2 times in total.
Edited the question and added the OA
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12 Feb 2012, 21:09
4
2
devinawilliam83 wrote:
. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

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.

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

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12 Feb 2012, 21:16
Thanks, Just 1 question.
can I solve the equation in 1. (x-1)^2<=1 by taking sq root on both sides? so the equaton becomes |x-1|<=1 or
-1<=x-1<=1..
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Re: Is |x-1| < 1? (1) (x-1)^2 <= 1 (2) x^2 - 1 > 0  [#permalink]

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12 Feb 2012, 21:24
1
3
devinawilliam83 wrote:
Thanks, Just 1 question.
can I solve the equation in 1. (x-1)^2<=1 by taking sq root on both sides? so the equaton becomes |x-1|<=1 or
-1<=x-1<=1..

Since both parts of the inequality are nonnegative we can take square root and write as you did.

GENERAL RULES FOR THAT:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.
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28 Mar 2013, 04:23
Bunuel wrote:
devinawilliam83 wrote:
. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

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.

Hope it's clear.

Hi Bunuel,

Is it not B?

x^2 - 1 > 0 --> x<-1 or x>1

x can be 2,3,4.... or -2,-3,-4...

So every time, |x-1| > 1.

Isn't this sufficient to answer that |x-1| will never be less than 1?
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28 Mar 2013, 04:27
1
Spaniard wrote:
Bunuel wrote:
devinawilliam83 wrote:
. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

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.

Hope it's clear.

Hi Bunuel,

Is it not B?

x^2 - 1 > 0 --> x<-1 or x>1

x can be 2,3,4.... or -2,-3,-4...

So every time, |x-1| > 1.

Isn't this sufficient to answer that |x-1| will never be less than 1?

We are not told that x is an integer. So, check for x=1.5.
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28 Mar 2013, 04:56
Bunuel wrote:
Spaniard wrote:
. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

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.

Hope it's clear.

Hi Bunuel,

Is it not B?

x^2 - 1 > 0 --> x<-1 or x>1

x can be 2,3,4.... or -2,-3,-4...

So every time, |x-1| > 1.

Isn't this sufficient to answer that |x-1| will never be less than 1?[/quote]

We are not told that x is an integer. So, check for x=1.5.[/quote]

Rookie mistake. Thanks
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08 Apr 2013, 22:23
Bunuel wrote:
devinawilliam83 wrote:
. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

Correct answer is neither A, nor C, it's E.

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.

Hope it's clear.

Hi,
Could someone please explain how you arrived at (2) in the above statement. How is X^2 - 1>0 ---> X<-1 or X>1?? I am not understanding how you arrived at X<-1
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09 Apr 2013, 00:35
1
SauravD wrote:
(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient

Hi,
Could someone please explain how you arrived at (2) in the above statement. How is X^2 - 1>0 ---> X<-1 or X>1?? I am not understanding how you arrived at X<-1

$$x^2-1>0, x^2>1$$

I have an old trick to solve those, basically you solve $$x^2=1, x=+-1$$, now you look at the sign of x^2 and at the operator. If they are (+,>) or (-,<) you pick the external values : $$x>1 and x<-1$$
You can insert numbers and you`ll see it s right.
In the other two cases you pick the internal values, it would have been $$-1<x<1$$

Hope it helps
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Re: Is |x-1| < 1?  [#permalink]

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10 May 2013, 16:55
devinawilliam83 wrote:
Is |x-1| < 1?

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

according to me only 1 is sufficient to answer this.the OA is C. any thoughts?

A very good question. The question is asking you : Does x lie between 0 and 1.

From 1 ) Open the given condition and we have x(x-2) <=0. This says x lies between 0 and 2 including both end points. Not sufficient to answer the question.
2) this says x > 1 or x < -1 . Again Insufficient.

Combining 1 and 2 , we can't find the answer as x=2 , will prove the condition false and x = 1.5 will prove it to be true.

Hence E.
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Re: Is |x-1| < 1?  [#permalink]

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12 Jun 2013, 19:27
Can someone explain to me how you arrive at 0<x≤2? I simplified (x-1)^2≤1 down to x(x-2)≤0 and I see that x≤2 but how do you get x<0?

Also, for #1, why couldn't x be one in addition to 0 and 2?

Thanks!
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Re: Is |x-1| < 1?  [#permalink]

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12 Jun 2013, 19:56
1
WholeLottaLove wrote:
Can someone explain to me how you arrive at 0<x≤2? I simplified (x-1)^2≤1 down to x(x-2)≤0 and I see that x≤2 but how do you get x<0?

Also, for #1, why couldn't x be one in addition to 0 and 2?

Thanks!

What you have is x(x-2)≤0--> Thus, the two factors x and (x-2) are of opposite signs-->

Either x>0 AND x<2 --> 0<x<2
OR
x<0 AND x>2--> Invalid Solution.

Thus, taking the equality in consideration, we have 0≤x≤2.

x can be ANY value between 0 and 2, inclusive.
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Re: Is |x-1| < 1?  [#permalink]

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12 Jun 2013, 21:32
WholeLottaLove wrote:
Can someone explain to me how you arrive at 0<x≤2? I simplified (x-1)^2≤1 down to x(x-2)≤0 and I see that x≤2 but how do you get x<0?

Also, for #1, why couldn't x be one in addition to 0 and 2?

Thanks!

$$x(x-2)\leq{0}$$ holds true for $$0\leq{x}\leq{2}$$.tips-and-tricks-inequalities-150873.html

x can have any value in the interval [0-2], incluse 1
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Re: Is |x-1| < 1?  [#permalink]

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12 Jun 2013, 21:56
WholeLottaLove wrote:
Can someone explain to me how you arrive at 0<x≤2? I simplified (x-1)^2≤1 down to x(x-2)≤0 and I see that x≤2 but how do you get x<0?

Thanks!

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: Is |x-1| < 1?  [#permalink]

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12 Jun 2013, 22:01
WholeLottaLove wrote:
Can someone explain to me how you arrive at 0<x≤2? I simplified (x-1)^2≤1 down to x(x-2)≤0 and I see that x≤2 but how do you get x<0?

Also, for #1, why couldn't x be one in addition to 0 and 2?

Thanks!

Solution says: x is in the range (0,2) inclusive. So, x could be 1 too. But if x is 0 or 2 |x-1| < 1 won't hold true.
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Re: Is |x-1| < 1?  [#permalink]

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27 Jun 2013, 15:10
Is |x-1| < 1?

Two cases: positive and negative

Positive: x≥1:
(x - 1) < 1
x - 1 < 1
x < 2

Negative: x<1
-(x - 1) < 1
-x + 1 < 1
-x < 0
x > 0

SO

0 < x < 2

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

(1) (x-1)^2 ≤ 1
(x-1)^2 ≤ 1
√(x-1)^2 ≤ √1
|x - 1| ≤ 1

Two cases:
Positive: x≥1:
(x - 1) ≤ 1
x ≤ 2

Negative: x<1:
-(x - 1) ≤ 1
-x + 1 ≤ 1
-x ≤ 0
x ≥ 0

SO

0 ≤ x ≤ 2

The stem tells us that 0 < x < 2 however #1 says that 0 ≤ x ≤ 2. In other words, the stem says x must be BETWEEN 0 and 2 while #1 says that x could be 0 or 2.
INSUFFICIENT.

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

x>1 OR x<-1

The stem tells us that 0 < x< 2 but #2 says that x > 1 or x < -1 which means x could fall in the correct range but it might not also.
INSUFFICIENT

1+2)
is 0 < x < 2?

0 ≤ x ≤ 2
AND
x>1 OR x<-1

Tells us that x is between zero and two inclusive and also that x > 1 which means x could = 2 which isn't allowed in the stem (0 < x < 2)
INSUFFICIENT

(E)
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30 Mar 2017, 04:28
Is |x-1| < 1?

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

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30 Mar 2017, 04:36
Mo2men wrote:
Is |x-1| < 1?

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

Merging topics. Please refer to the previous discussion.
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31 Mar 2017, 04:39
1
Top Contributor
2
Hi,

When dealing with inequalities and absolute values, it helps to know these standard forms.

1. If x^2 < a^2 then the range for x will always be -a < x < a.

2. If x^2 > a^2 then the range for x will always be x > a and x < -a.

3. If |x| < a then the range for x will always be -a < x < a

4. If |x| > a then the range for x will always be x > a and x < -a

Note : The above standard forms hold good even if we have <= or > =. Also the ranges for x^2 < a^2 and x^2 > a^2 are the same for |x| < a and |x| > a. The reason for this is because |x| = $$\sqrt{x^2}$$

So the question here asks us 'Is |x-1| < 1'?

We can break down |x-1|< 1 to -1 < x - 1 < 1. Adding 1 throughout we can rephrase the question to

'Is 0 < x < 2'

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

(x - 1)^2 <=1 -----> -1 <= x - 1 <= 1. Adding 1 throughout we get 0 <= x <= 2. This clearly is not sufficient. As x can be equal to 0 and 2.

Statement 2 : x^2 - 1 > 0

x^2 > 1 -----> x > 1 and x < -1. This again is clearly insufficient.

Combining 1 and 2 : From Statement 1 we have 0 <= x <= 2 and from Statement 2 we have x > 1 and x < -1. So to satisfy both the statements the only possible range for x will be 1 < x <= 2, which again is insufficient since the target question is 'Is 0 < x < 2'. x can be 1,5 which gives us a YES and x can also be 2 which gives us a NO.

Hope this helps!

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Re: Is |x-1| < 1?  [#permalink]

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14 Jun 2017, 17:50
Hi,

I got the right answers as my final equations however, I chose D instead of E. For S1, I got x= 0 or 2. If you plug these two in the original equation, you can "no" as an answer which I thought was sufficient. For S2, I got x= -1 or 1. Again, I plugged both of these numbers in and got "no" as an answer as well. Could anyone please explain how both of these equations are insufficient even when we are getting "no" as an answer? Please help!
Re: Is |x-1| < 1? &nbs [#permalink] 14 Jun 2017, 17:50

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