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

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

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New post Updated on: 30 Jul 2018, 19:44
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A
B
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D
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Question Stats:

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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

Originally posted by gsingh0711 on 30 Jul 2018, 13:51.
Last edited by Bunuel on 30 Jul 2018, 19:44, 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]

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New post 30 Jul 2018, 13:53
7
2
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]

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New post 02 Aug 2018, 02:07
1
[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]

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New post 06 Dec 2018, 02:42
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]

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New post 06 Dec 2018, 06:20
Bunuel chetan2u VeritasKarishma could you comment here?
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Re: Given that x is a real number, is −1 < x < 3? (1) | |x − 1| − 1 | < 1 &nbs [#permalink] 06 Dec 2018, 06:20
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