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Is |x-1| < 1?

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12 Feb 2012, 21:49
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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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Apr 2013, 05:55, edited 2 times in total.
Edited the question and added the OA
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12 Feb 2012, 22:09
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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, 22:16
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, 22:24
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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, 05:23
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, 05:27
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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, 05:56
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, 23:23
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, 01:35
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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 youll 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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09 Apr 2013, 02:12
Zarrolou wrote:
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 youll see it s right.
In the other two cases you pick the internal values, it would have been $$-1<x<1$$

Hope it helps

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

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10 May 2013, 17:55
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, 18:41
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!
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Re: Is |x-1| < 1? [#permalink]

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12 Jun 2013, 20:27
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, 20:56
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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, 22:32
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, 22:56
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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, 23:01
Expert's post
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, 16:10
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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