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Is |x|<1?

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Is |x|<1?

by apoorva.srivastva » Wed May 20, 2009 10:16 pm
Is |x|<1?

1.) |x+1| = 2 |x-1|
2.) |x-3| not equal to 3

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I woul like to know that if there is mod sign on both the sides of the linear equation ....can we get rid of the mod and slove the equation as a normal linear equation as in statement 1

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

so by my logic can we get rid of the mod sign

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

and slove further.....

ALSO PLEASE SUGGEST ME ANY PREP MATERIAL FOR MODULUS AND INEQUALITIES
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Re: Is |x|<1?

by Vemuri » Wed May 20, 2009 10:54 pm
Is the OA A?
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by apoorva.srivastva » Wed May 20, 2009 11:31 pm
NO the OA is not A
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Re: Is |x|<1?

by Vemuri » Thu May 21, 2009 2:31 am
Ok, let me put down my logic. I am not sure I understand why A is not the right answer.

The question is basically asking if -1<x<1, i.e. x is in between -1 & 1, i.e. -0.99....0.....0.99

Stmt1: |x+1| = 2 |x-1| can be reduced to either:
x+1 = 2x-2 ==> x=3
or
-x-1=-2x+2 ==> x=3
So, basically this statement is saying that x=3. So, we can answer the question with a definite NO. Sufficient.

Stmt2: |x-3| not equal to 3. Lets take an example. If |x-3|=6, then x=9 or x=-3. If |x-3|=2.9, then x=5.9 or 0.1. So, this statement is not sufficient to answer the question.

Am I missing something here?
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by mike22629 » Thu May 21, 2009 9:30 am
Use number line, the IMO is C.

l = absolute value sign

lx-(-1)l = 2lx-1l

Hence the distance between x and -1 is twice the distance between x and 1. Use number line to see how we can satisfy this.

1st possibility:

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

2nd possibility

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

In both situations, the distance between x and -1, and x and 1 is equal

Statement 2)

x does not equal 3, hence x = 1/3

IMO C
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by mike22629 » Thu May 21, 2009 9:34 am
Actually, misread question. IMO E.

2nd statement says lx-3l not equal to 3, which means its not equal to 6 or 0,

so x could be 1/3 or 3

I go with E.
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by mike22629 » Fri May 22, 2009 5:04 am
What is OA?
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by jack5047 » Fri May 22, 2009 11:07 am
A is the right answer..
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by sanjay_dce » Fri May 22, 2009 11:17 am
E is the ans
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by DeepakR » Fri May 22, 2009 7:38 pm
#1. |x+1| = 2 |x-1|
The ideal way would be to identify the critical points of the 2 absolute values and formulate the boundaries.

In this case x=-1 and x=1 are the boundaries. We have to consider the following cases:

When x<-1 then -(x+1)=-2(x-1) hence x=3.
When -1<x<1 then (x+1)=-2(x-1) hence x=1/3
When x>1 then (x+1)=2(x-1) hence x=3

So we have x=3 and x=1/3 as solutions. Hence we cannot conclusively say if x lies between 1 and -1.

#2. I think the question has to be lx - 3l &#8800; 0 in that case x<>3 so B insufficient.

Combining #1 and #2 we can conclusively say that x=1/3 and hence C.) is the answer.

If the question is correct |x-3| not equal to 3 then x<>6 or x<>0. Could be anything and hence E.)

- Deepak
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by apoorva.srivastva » Wed Jun 24, 2009 2:16 pm
OA is E
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by Domnu » Thu Jun 25, 2009 1:30 pm
The answer is definitely E. Here's why:

1) Solving, we get x = 1/3 or 3. This isn't decisive.
2) Solving, we get x not equal to 0 and x not equal to 6. Not decisive either.

Combining, we get x = 1/3 or 3 (and x not equal to 0 and x not equal to 6), but this second part is redundant. Still indecisive. So, we are at E.
Have you wondered how you could have found such a treasure? -T
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