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# Is |x| < 1?

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Joined: 07 Feb 2010
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Updated on: 27 Jul 2015, 15:11
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Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

My soln

we have to find -1<x<1 ....

From 1 , we have |x + 1| = 2|x - 1|
Assume -1 < x < 1 so the expression on RHS will be -ve while LHS will be +ve
=> x+1 =2(1-x)

solving we get x=1/3 so we can easily say that |x| is <1

From 2 we cannot find whether |x| <1

PLease let me know where I am wrong

Originally posted by boros2203 on 12 Feb 2010, 04:20.
Last edited by Bunuel on 27 Jul 2015, 15:11, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Is |x| < 1?  [#permalink]

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12 Feb 2010, 15:52
2
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Is $$|x| < 1$$, means is $$x$$ in the range (-1,1)? or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
There will be two checkpoints (-1 and 1), thus three ranges to test:
A. $$x<-1$$ --> $$-x-1=2(-x+1)$$ --> $$x=3$$ not good, as $$x<-1$$;

B. $$-1\leq{x}\leq{1}$$ --> $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$ is in the range $$-1<x<1$$. OK.

C. $$x>1$$ --> $$x+1=2(x-1)$$ --> $$x=3$$ is in the range $$x>1$$. OK

So we got TWO values of $$x$$, $$\frac{1}{3}$$ and $$3$$. One ($$\frac{1}{3}$$), is in the range (-1,1) and another ($$3$$), is out of this range. Not sufficient.

(2) $$|x - 3|\neq0$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) From (1): $$x=\frac{1}{3}$$ or $$x=3$$ and from (2) $$x\neq{3}$$ --> means $$x$$ can have only one value $$\frac{1}{3}$$, which IS in the range (-1,1). Sufficient.

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Re: Is |x| < 1?  [#permalink]

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12 Feb 2010, 04:52
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IMO C

you have done right, but do consider the cases x>1 and x<-1 also
and they will give x=3 and using 2nd equation this isnt possible.

Thus both taken together are sufficient.
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Re: Is |x| < 1?  [#permalink]

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12 Feb 2010, 10:26
1
Thanks Gurpreet , i was missing this second part
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Re: Is |x| < 1?  [#permalink]

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12 Feb 2010, 15:23
IMO A

(1) |x + 1| = 2|x - 1|
when x + 1>=0, you get
x+1=2(x-1)
x+1=2x-2
x=3

when x + 1<=0, you get
x+1=-2(x-1)
x+1=-2x+2
x=1/3 which does not satisfy the inequality

(2)|x - 3| ≠ 0
=> x is all but 3
This is not sufficient to answer

So, A -- Statement 1 alone is sufficient
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Updated on: 06 Mar 2016, 11:49
testprep2010 wrote:
IMO A

(1) |x + 1| = 2|x - 1|
when x + 1>=0, you get
x+1=2(x-1)
x+1=2x-2
x=3

when x + 1<=0, you get
x+1=-2(x-1)
x+1=-2x+2
x=1/3 which does not satisfy the inequality

(2)|x - 3| ≠ 0
=> x is all but 3
This is not sufficient to answer

So, A -- Statement 1 alone is sufficient

I originally thought A was the correct answer.
But after redrawing the graphs, I realized I misinterpreted the graphs.
Condition A gives two options, one of which falls between -1 and 1 and the other, 3, doesn't.
Since B is basically eliminates 3 as a candidate, we are left with one option. Hence, the answer is C.

Originally posted by jjsverbal on 06 Mar 2016, 00:39.
Last edited by jjsverbal on 06 Mar 2016, 11:49, edited 1 time in total.
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Re: Is |x| < 1?  [#permalink]

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06 Mar 2016, 03:24
1
Square both sides of statement (1) and we get x=3 or x=1/3.
So statement (1) is insufficient.

statement (2) tells us that X is not equal to 3. Clearly, statement (2) alone is insufficient.

But if we combine statement (1) and statement (2), we get x= 1/3 which is definitely less than 1.

Hence C .
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Re: Is |x| < 1?  [#permalink]

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09 Mar 2016, 09:15
1
Here is my approach
here we need to see whether x lies in the zone => (-1,1)
statement 1 is a double modulus equality => Remember whenever you encounter a double modulus equality (not inequality ) we can have only two cases => one where both of the signs will be +ve and the other when one is positive and the other is negative(take any one positive and the other negative)

as per this logic => x=3 and x=1/3 => 3 does not satisfy our bound but 1/3 does
hence insufficient
statement 2 is insufficient too as x can be 0 or 50
combining them we get x≠3 so x=1/3 => which satisfies our bound => C is sufficient
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Re: Is |x| < 1?  [#permalink]

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09 Jun 2018, 02:07
Bunuel wrote:
Is $$|x| < 1$$, means is $$x$$ in the range (-1,1)? or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
There will be two checkpoints (-1 and 1), thus three ranges to test:
A. $$x<-1$$ --> $$-x-1=2(-x+1)$$ --> $$x=3$$ not good, as $$x<-1$$;

B. $$-1\leq{x}\leq{1}$$ --> $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$ is in the range $$-1<x<1$$. OK.

C. $$x>1$$ --> $$x+1=2(x-1)$$ --> $$x=3$$ is in the range $$x>1$$. OK

So we got TWO values of $$x$$, $$\frac{1}{3}$$ and $$3$$. One ($$\frac{1}{3}$$), is in the range (-1,1) and another ($$3$$), is out of this range. Not sufficient.

(2) $$|x - 3|\neq0$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) From (1): $$x=\frac{1}{3}$$ or $$x=3$$ and from (2) $$x\neq{3}$$ --> means $$x$$ can have only one value $$\frac{1}{3}$$, which IS in the range (-1,1). Sufficient.

Hi Bunuel,

For the 2nd statement, since it is modulus, don't we need to consider this as well: -(x-3)!=0 so x!=-3 ?
So there will two not equals,
x!=3 or x!=-3.
Please let me know whether I am right in this logic.
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Re: Is |x| < 1?  [#permalink]

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09 Jun 2018, 05:02
For the 2nd statement, since it is modulus, don't we need to consider this as well: -(x-3)!=0 so x!=-3?
So there will two not equals,
x!=3 or x!=-3.
Please let me know whether I am right in this logic.

The error is in the highlighted statement above.

-(x-3)!=0

=> - x + 3 !=0

=> -x != - 3

=> x != 3

Does that make sense?
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