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Is |x|<1 ? [#permalink]

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Is |x|<1 ?

(1)|x + 1|= 2|x - 1|
(2) |x - 3| ≠ 0

Can anybody suggest a general approach to solve this kind of problems ? I am facing problems to solve these type of questions where equations are containing absolute values together with inequalities... please help.
[Reveal] Spoiler: OA

Last edited by msand on 31 Dec 2009, 10:30, edited 1 time in total.

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Re: DS question [#permalink]

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Is \(|x| < 1\)?

Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?

(1) \(|x + 1| = 2|x - 1|\)
Two key points: \(x=-1\) and \(x=1\) (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).

So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) \(|x - 3|\neq{0}\)
Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.

(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.

Answer: C.

Check Walker's post: Absolute value and also to practice in absolute value questions check the link of Inequalities in my signature.

Hope it helps.
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Re: Tough Absolute value problem [#permalink]

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New post 07 Jun 2010, 14:31
|X|<1 means -1<x<1

a. |x+2| = 2 |X-1|

Case 1. X+2 = 2(X-1) => X=3
Case 2. X+2 = -2(X-1) => X=1/3

Not Sufficient

b. | X-3| not equals 0 means X not equals 3 => Not sufficient

From both a & B,

X = 3, 1/3, and X not equals 3 => Hence X can have only one value X=1/3 which lies between -1 and 1

So, Ans. C

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Re: DS question [#permalink]

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New post 08 Jun 2010, 06:16
Thank you Bunuel for your great explanation!

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Re: DS question [#permalink]

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New post 26 Jun 2010, 03:14
Bunuel wrote:
Is \(|x| < 1\)?


A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).




I am not clear about how you derived this.
\(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\)

Same with other two ranges (green and red)

Appreciate your time.

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Re: DS question [#permalink]

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JoyLibs wrote:
Bunuel wrote:
Is \(|x| < 1\)?


A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).




I am not clear about how you derived this.
\(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\)

Same with other two ranges (green and red)

Appreciate your time.


We have: \(|x + 1| = 2|x - 1|\).

Absolute value properties:
If \(x\geq{0}\), then \(|x|=x\) and if \(x\leq{0}\), then \(|x|=-x\).

For the range \(x<-1\) (blue range) --> \(x+1<0\) so \(|x+1|=-(x+1)\) and \((x - 1)<0\) so \(|x-1|=-(x-1)\) --> \(|x + 1| = 2|x - 1|\) becomes: \(-(x+1)=2(-x+1)\)

The same for other ranges.

Check this for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: DS question [#permalink]

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New post 26 Jun 2010, 04:18
Thanks a lot

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Re: absolute value [#permalink]

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New post 23 Oct 2010, 09:09
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tatane90 wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

thanks in advance


(1) |x+1| = 2|x-1|
Lets solve this thing
+/-(x+1) = +/-2 (x-1)

Case 1 : x+1 = 2(x-1) .. x=3
Case 2 : x+1 = -2(x-1) .. 3x=1 .. x=1/3
Case 3 : -x-1 = 2(x-1) .. 3x=1 .. x=1/3
Case 4 : -x-1 = -2x+2 .. x=3

So x is either 3 or 1/3. Not enough to answer question

(2) This only implies x cannot be 3. Clearly insufficient

(1+2) x can only be 1/3. Sufficient to answer YES

Answer : (c)
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Re: how the answer C why not B [#permalink]

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Re: Is |x|<1 ? [#permalink]

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Re: Is |x|<1 ? [#permalink]

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New post 27 Nov 2013, 06:40
msand wrote:
Is |x|<1 ?

(1)|x + 1|= 2|x - 1|
(2) |x - 3| ≠ 0

Can anybody suggest a general approach to solve this kind of problems ? I am facing problems to solve these type of questions where equations are containing absolute values together with inequalities... please help.


Question --> Is |x|<1 ?
Is -1<x<1?

Statement 1
Square both sides
You end up with quadratic 3x^2-10x+30
Factorize (3x-1)(x-3)
x= 3 or x=1/3.
Two answers, therefore Insuff

Statement 2
x cannot be 3

Statement 1 and 2 together
Then it has to be x 1/3 and it is within the range so answer is YES

Suff

Hence C the correct answer
Cheers!
J :)

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Re: Is |x|<1 ? [#permalink]

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New post 16 Dec 2014, 02:51
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Re: Is |x|<1 ?   [#permalink] 16 Dec 2014, 02:51
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