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# x| < 1? 1) |x+1| = 2|x-1| 2) |x-3| does not equal 0

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x| < 1? 1) |x+1| = 2|x-1| 2) |x-3| does not equal 0 [#permalink]

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01 May 2011, 20:59
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67% (03:33) correct 33% (02:19) wrong based on 9 sessions

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dc123 wrote:
|x| < 1?

1) |x+1| = 2|x-1|

2) |x-3| does not equal 0

Therefore we are looking to see if -1<x<1????

Can someone plz help. I dont know why its C

In absolute equations, one should validate the equation for all possible cases w.r.t the signs of expression embedded in the modulus.

Q:
|X|<1
-1<X<1

(1) |x+1| = 2|x-1|

Case 1:
(x+1)>=0 AND (x-1)>=0
x+1=2(x-1)
-x=-3
x=3

Case 2:
(x+1)>=0 AND (x-1)<0
x+1=-2(x-1))
x+1=-2x+2
3x=1
x=1/3

Case 3:
(x+1)<0 AND (x-1)>=0
-(x+1)=2(x-1)
-x-1=2x-2
-3x=-1
x=1/3

Case 4:
(x+1)<0 AND (x-1)<0
-(x+1)=-2(x-1)
-x-1=-2x+2
x=3

Thus, x can be 3 or 1/3.
Not Sufficient.

2. x != 3
Not Sufficient.

Combining both;
x=1/3, which is between -1 and 1.
Sufficient.

Ans: "C"
[Reveal] Spoiler: OA

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02 May 2011, 04:43
1. 3 regions to be checked x<-1, -1<x< 1 and x>1

for x>1 , (x+1) = 2(x-1)
x= 3
for x<-1, (x+1) = -2(x-1)
x= 1/3 which is not possible.Hence no solution in this region.

for -1<xx<1, (x+1) = -2(x-1)
x= 1/3

No sufficient.

2. x !=3 not sufficient

1+2 gives x = 1/3 Hence C
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02 May 2011, 10:22
i did it a bit different. plz tell me if u think im wrong doing so:

|x| < 1?

1) |x+1| = 2|x-1|

2) |x-3| does not equal 0

the question is - x^2<1
so
x^2+2x+1=4(x^2-2x+1)
====>
3x^2-10x+3=0
so x = 3 or 1/3

st. 2:
x<>3
not helping me.

combination
x=1/3

is that approach wrong?
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02 May 2011, 18:08
(1)

Square both sides :

x^+1+2x = 4(x^2 -2x + 1)

3x^2 -10x +3 = 0

3x^2 - 9x -x + 3 = 0

3x(x-3) -1(x-3) = 0

(x-3)(3x-1) = 0

x = 3, x = 1/3

Not sufficient

(2) is not sufficient as well

(1) + (2)

x = 1/3

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02 May 2011, 18:57
dc123 wrote:
|x| < 1?

1) |x+1| = 2|x-1|

2) |x-3| does not equal 0

Therefore we are looking to see if -1<x<1????

Can someone plz help. I dont know why its C

You can do these questions in less than a minute if you understand the approach I discuss here:
http://www.veritasprep.com/blog/2011/01 ... edore-did/

On to your question.
Is |x| < 1?
Yes, essentially this just asks you whether x lies between -1 and 1. |x| represents the distance from 0 which will be less than 1 if x lies within -1 < x < 1.

Statement 1: |x+1| = 2|x-1|
Draw the number line. This equation says that distance from -1 is twice the distance from 1. At which point is the distance from -1, twice the distance from 1? If you split the distance (2 units) between -1 and 1 into 3 parts, 1 part away from 1 and 2 parts away from -1 will be the point 1/3. The diagram below will show you this situation. Similarly the distance between -1 and 1 is 2 so if you go 2 units further to the right of 1, you get the point 3 which will be 4 units away from -1 i.e. twice the distance from 1.
Attachment:

Ques6.jpg [ 4.74 KiB | Viewed 1142 times ]

x could be 1/3 or 3. Hence x may or may not lie between -1 and 1. Not sufficient.

Statement 2: |x-3| does not equal 0
This statement just says that x is not 3. It is not sufficient alone.

Both together, we get that x must be either 3 or 1/3 and it is not 3. Then x must be 1/3 and must lie between -1 and 1. Sufficient.
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Re: DS.....   [#permalink] 02 May 2011, 18:57
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# x| < 1? 1) |x+1| = 2|x-1| 2) |x-3| does not equal 0

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