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Problem Of The Week 36

with 9 comments


Find the value of the expression ……….|||||||x-2|+2|-2|+2|……….

Written by Implex

October 1, 2008 at 11:12 am

9 Responses

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  1. That would be x.

    Celebrating Life

    October 1, 2008 at 11:25 am

  2. hmm. A re look at the question, I’m in a quandary.
    Depends on where the series ends and on the sign of x.
    When x>0 and when it ends with +2 it is x+2
    When x>0 and when it ends with -2 it is x
    When x<0 and the series end with -2, it is |x|+2
    when x<0 and the series ends with +2, it is |x|+4

    Celebrating Life

    October 1, 2008 at 11:38 am

  3. aah!Again a silly mistake .
    When x>0 and when the series ends in +2 it is x
    When x>0 and ends with -2, it’s x-2
    Superb effort implex. Keep ’em coming!

    Celebrating Life

    October 1, 2008 at 11:52 am

  4. It can also have two additional values, 0 and 2 when x=2.

    riyaziq

    October 1, 2008 at 12:10 pm

  5. x or x-2
    4-x ,2-x

    Varun

    October 1, 2008 at 1:26 pm

  6. @Riyaziq
    All the cases fall into only those four categories. Even for 2.

    Celebrating Life

    October 1, 2008 at 1:52 pm

  7. None of the solutions have been sastisfactory
    Let |x-2|=y y>=0

    after every even number of mods we will again get y ( as y+2 will be positive as y is so )

    now if the series ends at +2
    answer will be y+2 means |x-2|+2

    if the series ends at -2
    we will get y or |x-2|

    The solution ends here, there is no need to do anything else
    all the solutions will come in the ambit of this !

    Cheers !

    outtimed

    October 1, 2008 at 2:30 pm

  8. for x>=2 and
    even no. of mods…ans=’x’
    odd no. of mods…ans=’x-2′

    for x<2 and
    even no. of mods…ans=’2-x’
    odd no. of mods…ans=’4-x’

    wolverine

    October 1, 2008 at 5:32 pm

  9. i think my soln…coincides with varun…[;)]

    wolverine

    October 1, 2008 at 5:33 pm


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