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Is |x| < 1 ?

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Is |x| < 1 ?

by lalmanistl » Mon Jan 25, 2010 9:10 pm
Is |x| < 1 ?

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

(2) |x minus 3| > 0


please help me to solve this question.

thanks in advance.
Last edited by lalmanistl on Mon Jan 25, 2010 9:39 pm, edited 2 times in total.
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by fibbonnaci » Mon Jan 25, 2010 9:34 pm
please post the question properly. the mathematical symbols are not clear.
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by papgust » Mon Jan 25, 2010 10:54 pm
lalmanistl wrote:Is |x| < 1 ?

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

(2) |x minus 3| > 0
Is -1 < x < 1?

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

Square both sides,
(x+1)^2 = 4 (x-1)^2
x^2 + 2x + 1 = 4x^2 - 8x + 4
3x^2 - 10x + 3 = 0
x=3,1/3
1/3 lies b/w -1 and 1 But 3 does not lie b/w -1 and 1.
Insufficient.

(2) |x - 3| > 0
x - 3 > 0 ; -x + 3 > 0
x > 3; x < 3
x = 3.
x does not lie b/w -1 and 1. Sufficient.

Hence B
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by VikingWarrior » Tue Jan 26, 2010 12:02 am
Papgust, I differ in my solution.
My answer would be C.
As you have correctky pointed out that from 1. we can derive x=3 or x=1/3
from 2 we can derive that x is not=3 i.e. for absolute value of (x-3) to be more than 0 x can be any no. but 3.
Thus combining 1 and 2 we get -1<x<1

Of course the assumption is that x is an integer
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by tnguyen » Tue Jan 26, 2010 1:00 am
can someone explain why squaring both sides is the first and correct step?
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by ajith » Tue Jan 26, 2010 1:28 am
papgust wrote:
(2) |x - 3| > 0
x - 3 > 0 ; -x + 3 > 0
x > 3; x < 3
x = 3.
x does not lie b/w -1 and 1. Sufficient.

Hence B
|x-3| is always positive no matter what the value of x is so, I do not think the conclusion that x do not lie between 1 and -1 is true


Hence E - Both statement together is not sufficient
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by ajith » Tue Jan 26, 2010 1:33 am
tnguyen wrote:can someone explain why squaring both sides is the first and correct step?
Because of the mode operation.

say if x is -ve and y is +ve

|x| = |y| means -x = y

if x is -ve and y is -ve

it means

-x = -y

So we cannot be sure whether x = y or x = -y (if we have no idea about the signs of x and y)

but squaring solves that problem and makes both side positive

x^2 = y^2 (no matter what the signs of x and y are)
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by VikingWarrior » Tue Jan 26, 2010 2:02 am
Ajith,

|x-y|>0 for all integers x and y provided x is not equal to y
Absolute value cannot have a negative sign.

|x-3|>0 is only suggesting that x-3 cannot be equal to 0 therefore x is not equal to 3. Add this information to statement 1 and you have the answer.
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by ajith » Tue Jan 26, 2010 2:08 am
VikingWarrior wrote: |x-3|>0 is only suggesting that x-3 cannot be equal to 0 therefore x is not equal to 3. Add this information to statement 1 and you have the answer.
Agree there!
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