(|x| - 1)/(x - 1) =?

Question

(1) x < 0
(2) |x| = -x

chetan2u · Accepted Answer

Hi,
if we look at the eq..
(|x| - 1)/(x - 1) ..

the value depends on :-
1) if x is 0 or +ive, ans is 1, irrespective of value of x..
2) if x is -ive, the value of denominator and numerator changes, and the equations value will depend on value of x..

lets see the statements now..
(1) x < 0
here x is negative , so the value depends on value of x.. insuff

(2) |x| = -x
again value of x is 0 or -ive.. insuff..

combined .nothing new.. insuff
E..

just for knowledge...
Why x is -ive in statement 2..
|x| = -x..
The LHS is positive because it is absolute mod, so RHS will also be positive..
RHS=-x... now -x will become positive only when x is -ive

Bunuel · Answer

Similar question to practice: https://gmatclub.com/forum/is-x-134652.html

shashankism · Answer

Given : Nothing 
DS : (|x| - 1)/(x - 1) =?

Option 1: x < 0 , 
So, |x| = -x 
(|x| - 1)/(x - 1) = (-x-1)/(x-1) 
 NOT SUFFICIENT

Option 2: |x| = -x 
(|x| - 1)/(x - 1) = (-x-1)/(x-1) 
 NOT SUFFICIENT 
 
Answer E

Sidbendale1 · Answer

chetan2u  what is mod of 0? Is it zero or undefined?
If its undefined, why do we say that |x| >=0. I think i am getting confused between 2 things.
Please advise
Thanks in advance
Siddharth

chetan2u · Answer

Siddharth,

|0|=0, it is not undefined..
Modulus just removes the negative sign

JessicaSheoran · Answer

Firstly this statement isn't correct: "1) if x is 0 or +ive, ans is 1, irrespective of value of x.." 
If x is equal to 1 then ans is 0. 
Secondly shouldn't 0 be the only solution for this: |x| = -x since |x| can never be negative.

revamoghe · Answer

chetan2u  Didn't understand at all. Why are the statements insufficient?

chetan2u · Answer

If x is positive, then |x|=x, and hence the numerator and denominator will be the same. Value will be 1.

If x is negative say -a, then (|x|-1)/(x-1)=(|-a|-1)/(-a-1)=(a-1)/-(a+1)=(1-a)/(a+1). 
Thus answer will depend on value of a, that is the value of x, which is not given by any of the statements

CEdward · Answer

Maybe I missed it, but I concur with Jessica on this one.

For statement 2, we can set x = 0 which would give us a unique value for the expression. No other number would do that as chetah explains. So why isn't this statement sufficient?

chetan2u · Answer

1) The statement that “ if x is 0 or +ive,..” is perfectly fine. x cannot be 1, since then the denominator x-1 will be come 0, making the term undefined. 
2) All values of x as a -ive number will fit in. We are not saying |x| is negative, but that x is negative. -x, that is negative of negative will also become positive. 
For example 
Let x=-2, then |x|=|-2|=2, and -x=-(-2)=2. Thus |x|=-x for all values of x where  x\leq 0

Vaishali2004 · Answer

Doesn't option B imply that x=0 ,and therefore, B is sufficient?

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(|x| - 1)/(x - 1) =?

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(|x| - 1)/(x - 1) =?

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