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Updated on: 25 Oct 2023, 09:11
Originally posted by devinawilliam83 on 12 Feb 2012, 21:49.
Last edited by Bunuel on 25 Oct 2023, 09:11, edited 4 times in total.
Last edited by Bunuel on 25 Oct 2023, 09:11, edited 4 times in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
47% (02:24) correct
53%
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based on 543
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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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12 Feb 2012, 22:09
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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.
General Discussion
12 Feb 2012, 22:16
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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..
