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Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1

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gsingh0711
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Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   Updated on: 30 Jul 2018, 20:44
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Given that x is a real number, is −1 < x < 3?

(1) ||x − 1| − 1| < 1

(2) (x + 1)(x − 3) < 0
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D

Originally posted by gsingh0711 on 30 Jul 2018, 14:51.
Last edited by Bunuel on 30 Jul 2018, 20:44, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   30 Jul 2018, 14:53
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Explanation:

As we know | x | = x if x > 0, and | x | = −x if x < 0.

Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)

Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2

Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)

Considering the RHS of equation,
| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3

Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.

Considering statement 2:
(x + 1)(x − 3) < 0

LHS of the inequality (x + 1)(x − 3) is equal to zero if x = −1 or 3. Hence, for any value of x between −1 and 3, the LHS is negative. This can be verified by picking any value of x between −1 and 3, say x = 0. Then x + 1 is a positive value and x − 3 is a negative value. Hence, the product of x + 1 and x − 3 is negative.

Hence, statement 2 ⇒ −1 < x < 3

Since we are getting a definite answer from above statement , statement 2 itself is sufficient to provide the answer.

Answer: D.

I think, I deserve a Kudus:D Lol
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Re: Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   02 Aug 2018, 03:07
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[quote="gsingh0711"]Explanation:

As we know | x | = x if x > 0, and | x | = −x if x < 0.

Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)

Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2

Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)


| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3

Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.

If X is not equal to 1,how come x lies between -1 and 3?
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Re: Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   06 Dec 2018, 03:42
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prashant6923 wrote:
gsingh0711 wrote:
Explanation:

As we know | x | = x if x > 0, and | x | = −x if x < 0.

Considering statement 1:
| | x − 1 | − 1 | < 1
⇒ −1 < | x − 1 | − 1 < 1
(If | N | < 1, then N lies between −1 and 1.)

Adding 1 to every side of the inequality ⇒ 0 < | x − 1 | < 2

Considering the LHS of equation,
| x − 1 | > 0 ⇒ x ≠ 1
(Absolute value of any number is always positive as long as the value within modulus is not zero.)

| x − 1 | < 2
⇒ −2 < x − 1 < 2
(If | N | < 2, then N lies between −2 and 2.)
⇒ −1 < x < 3

Since we are getting a definite answer from above statement , statement 1 itself is sufficient to provide the answer.

If X is not equal to 1,how come x lies between -1 and 3?



I agree on that.
gsingh0711

The statement1, "1" should be excluded to say it's sufficient. Statement 2 is sufficient, thus correct answer should be B.
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Re: Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   06 Dec 2018, 07:20
Bunuel chetan2u VeritasKarishma could you comment here?
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Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   19 Jun 2019, 06:04
gsingh0711 wrote:
Given that x is a real number, is −1 < x < 3?

(1) ||x − 1| − 1| < 1

(2) (x + 1)(x − 3) < 0


Dear IanStewart
Bunuel

In statement 1, when x = 1, then the inequality is incorrect. Hence, the solution will be -1<x<1 & 1<x<3. Should statement 1 be sufficient? I doubt because the question includes x=1 in the range −1 < x < 3, and so should be insufficient. What do you think?

Can you help please?

Thanks in advance
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Re: Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 [#permalink] New post   19 Jun 2019, 07:54
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Mo2men wrote:
[
In statement 1, when x = 1, then the inequality is incorrect. Hence, the solution will be -1<x<1 & 1<x<3. Should statement 1 be sufficient? I doubt because the question includes x=1 in the range −1 < x < 3, and so should be insufficient. What do you think?


Yes, it should definitely be sufficient. It might be easier to see why with a 250-level example question:

Is x > 0?
1. x = 5
2. x > 3

If you see a question like this, the answer is D. The question is not asking "can x be equal to any positive number at all?" The question is asking "can you be certain that x is a positive number?" Using Statement 1, x clearly cannot be equal to *any* positive number at all -- x is exactly equal to 5. But that's certainly sufficient, because you can be absolutely sure that x is positive. Similarly, using Statement 2, there are positive values, like 1.5 or 2, that x cannot equal. But Statement 2 is also sufficient, because you can be completely certain that x > 0 is true.

Back to the original question - if you can be completely certain that x is somewhere between -1 and 3, you have sufficient information. It's irrelevant if x cannot equal 1, say. If a statement told you "x = 2", or "0 < x < 0.5", that would be sufficient: you would know with absolute certainty that x is somewhere between -1 and 3.
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Re: Given that x is a real number, is 1 < x < 3? (1) | |x 1| 1 | < 1 [#permalink] New post   10 Jan 2024, 04:01
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Re: Given that x is a real number, is 1 < x < 3? (1) | |x 1| 1 | < 1 [#permalink]
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