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Is |x| + |x -1| = 1?

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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink] New post 18 Aug 2013, 05:15
btg9788 wrote:
Shouldn't it be mandatory that X is an integer? Or that is assumed implicitly?


No you should not assume it to be a integer, if it is not mentioned. The problem here is the range is such that even if you pick a non integer value, you will get the answer as 1.

After combining we have 0<=x<=1

For example if x=0.33

|0.3|+|1-0.3|=.3+.7=1 . Hope it is clear
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink] New post 29 Sep 2013, 12:20
Bunuel wrote:
samark wrote:
Bunuel,

I am confused here..
"B. 0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true."

I am confused that how first x is +ive and second one -ve...after we take condition 0<=x<=1?
Pls, explain.

Thanks!


We know that for |x|:
When x\leq{0}, then |x|=-x;
When x\geq{0}, then |x|=x.

We have |x| + |x -1| = 1.

Now for the range: 0\leq{x}\leq{1} --> |x|=x (as x in given range is positive) and |x-1|=-(x-1)=-x+1 (as expression x-1 in the given range is negative, to check this try some x from this range, let x=-0.5 then x-1=0.5-1=-0.5=negative). So |x| + |x -1| = 1 in this range becomes: x-x+1=1 --> 1=1, which is true. That means that for ANY value from the range 0\leq{x}\leq{1}, equation |x| + |x -1| = 1 holds true.

Hope it's clear.



Bunuel, I have a question on the part in red. Shouldn't it actually be:

We know that for |x|:
When x<{0}, then |x|=-x; (I have changed the "less than or equal to" to only "less than")
When x\geq{0}, then |x|=x.

Because we should consider 2 cases -
a) greater than or equal to zero
AND
b) less than zero. [Not less than or equal to zero]

In the part B of your solution we are also considering the case where x=1, right? If this is the case, how can |x-1| be -x +1? At x=1, I am guessing |x-1| = x-1.
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink] New post 29 Sep 2013, 12:24
Expert's post
emailmkarthik wrote:
Bunuel wrote:
samark wrote:
Bunuel,

I am confused here..
"B. 0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true."

I am confused that how first x is +ive and second one -ve...after we take condition 0<=x<=1?
Pls, explain.

Thanks!


We know that for |x|:
When x\leq{0}, then |x|=-x;
When x\geq{0}, then |x|=x.

We have |x| + |x -1| = 1.

Now for the range: 0\leq{x}\leq{1} --> |x|=x (as x in given range is positive) and |x-1|=-(x-1)=-x+1 (as expression x-1 in the given range is negative, to check this try some x from this range, let x=-0.5 then x-1=0.5-1=-0.5=negative). So |x| + |x -1| = 1 in this range becomes: x-x+1=1 --> 1=1, which is true. That means that for ANY value from the range 0\leq{x}\leq{1}, equation |x| + |x -1| = 1 holds true.

Hope it's clear.



Bunuel, I have a question on the part in red. Shouldn't it actually be:

We know that for |x|:
When x<{0}, then |x|=-x; (I have changed the "less than or equal to" to only "less than")
When x\geq{0}, then |x|=x.

Because we should consider 2 cases -
a) greater than or equal to zero
AND
b) less than zero. [Not less than or equal to zero]

In the part B of your solution we are also considering the case where x=1, right? If this is the case, how can |x-1| be -x +1? At x=1, I am guessing |x-1| = x-1.


No, it works with = sign as well: |0|=0=-0.

If x=1, then |x-1|=0 and -x+1=0 too.
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink] New post 29 Sep 2013, 20:21
I didn't know this. Thanks for clarifying, Bunuel!
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink] New post 04 Jun 2014, 04:40
wow! That is a lot of discussion. My solution will baffle you all.
Basically he is asking if the sum of the distance between 0 and x & x and 1 equal to one.

<-------><-------->
-----------------0---------x----------1-------------


I.E x must lie between 0 and 1
That condition is only satisfied when we combine the two. Hence answer is C.
Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1   [#permalink] 04 Jun 2014, 04:40
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