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devinawilliam83
Joined: 25 Aug 2011
Last visit: 01 Mar 2013
Posts: 113
Location: India
GMAT 1: 730 Q49 V40
WE:Operations (Insurance)
GMAT 1: 730 Q49 V40
Posts: 113
Updated on: 25 Oct 2023, 08:11
Originally posted by devinawilliam83 on 12 Feb 2012, 20:49.
Last edited by Bunuel on 25 Oct 2023, 08:11, edited 4 times in total.
Last edited by Bunuel on 25 Oct 2023, 08:11, edited 4 times in total.
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Is \(|x - 1| < 1\) ?
(1) \((x - 1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)
(1) \((x - 1)^2 \leq 1\)
(2) \(x^2 - 1 > 0\)
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Is \(|x - 1| < 1\) ?
Is \(|x - 1| < 1\) ?
Is \(-1 < x - 1 < 1\) ?
Is \(0 < x < 2\) ?
(1) \((x - 1)^2 \leq 1\).
Given that both sides of the inequality are non-negative, we can take the square root of both sides, resulting in \(|x-1| \le 1\). Consequently, \(|x-1|\) can be less than 1 (answer YES), as well as equal to 1 when \(x=2\) or \(x=0\) (answer NO). Hence, this statement is insufficient.
Notice here though that \(|x-1| \leq 1\) implies \(0 \leq x \leq 2\).
(2) \(x^2 - 1 > 0\).
Rearranging: \(x^2 > 1\). Again, since both sides of the inequality are non-negative, we can take the square root of both sides, leading to \(|x| > 1\). This means that \(x < -1\) or \(x > 1\). If \(x=1.5\), then the answer is YES, but if \(x=2\), then the answer is NO. This is also insufficient.
(1)+(2) From (1) we deduce that \(0 \leq x \leq 2\), and from (2) we infer that \(x < -1\) or \(x > 1\). As a result, \(1 < x \leq 2\). If \(x\) is any number between 1 and 2, not inclusive, the answer to the question is YES. However, if \(x=2\), the answer to the question is NO. Not sufficient.
Answer: E
Hope it's clear.
_________________
Is \(|x - 1| < 1\) ?
Is \(-1 < x - 1 < 1\) ?
Is \(0 < x < 2\) ?
(1) \((x - 1)^2 \leq 1\).
Given that both sides of the inequality are non-negative, we can take the square root of both sides, resulting in \(|x-1| \le 1\). Consequently, \(|x-1|\) can be less than 1 (answer YES), as well as equal to 1 when \(x=2\) or \(x=0\) (answer NO). Hence, this statement is insufficient.
Notice here though that \(|x-1| \leq 1\) implies \(0 \leq x \leq 2\).
(2) \(x^2 - 1 > 0\).
Rearranging: \(x^2 > 1\). Again, since both sides of the inequality are non-negative, we can take the square root of both sides, leading to \(|x| > 1\). This means that \(x < -1\) or \(x > 1\). If \(x=1.5\), then the answer is YES, but if \(x=2\), then the answer is NO. This is also insufficient.
(1)+(2) From (1) we deduce that \(0 \leq x \leq 2\), and from (2) we infer that \(x < -1\) or \(x > 1\). As a result, \(1 < x \leq 2\). If \(x\) is any number between 1 and 2, not inclusive, the answer to the question is YES. However, if \(x=2\), the answer to the question is NO. Not sufficient.
Answer: E
Hope it's clear.
_________________
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General Discussion
devinawilliam83
Joined: 25 Aug 2011
Last visit: 01 Mar 2013
Posts: 113
Location: India
GMAT 1: 730 Q49 V40
WE:Operations (Insurance)
GMAT 1: 730 Q49 V40
Posts: 113
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..
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..
devinawilliam83
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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Posts: 4,849
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.
Answer: E
Hope this helps!
CrackVerbal Academics Team
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.
Answer: E
Hope this helps!
CrackVerbal Academics Team
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