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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.

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.

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.

Answer: E.

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?

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.

Answer: E.

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. _________________

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.

Answer: E.

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]

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.

Answer: E.

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

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 _________________

It is beyond a doubt that all our knowledge that begins with experience.

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 \(-1<x<1\)

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.

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

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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