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Absolutely Greatest II (Posted on 2014-02-20) Difficulty: 3 of 5
Given that x is a real number, and:
               {x}
|x-1| - 1 =  ---------
              |x-1|
where, [x] is the greatest integer less than or equal to x, and {x}= x-[x]

Find the largest possible value of x.

*** |x| denotes the absolute value of x.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Quadratic equation (spoiler) Comment 1 of 1
Let x = n + f, where n is an integer and 0 <= f <1

First, let's hope that the maximum x is >= 1

If so, then (n+f-2) = f/(n+f-1)

Multiplying by (n+f-1) and collecting terms gives:

n^2 + n(2f-3) + (f^2-4f+2) = 0

Solving for n has two roots, the larger of which is

n = (3 - 2f + sqrt(1-8f))/2

This is maximized when f = 0, and fortunately this makes n = 2, which is an integer.  Since this is greater than 1, there are no other cases that need to be considered

Hope I didn't make any math mistakes.

Final answer: x = 2

Checking:

a) |2-1| - 1 = 0/|2-1|, so that works!

b) If x is between 2 and 3, say 2 + f, then by substitution we get 
   f = f/(1+f), which implies f is 0
   
c) If x is between 3 and 4, say 3 + f, then by substitution we get 
   1+f = f/(2+f)
   But LHS >= 1 and RHS <1
   And this only gets worse as X gets bigger.

   So x = 2 checks out.

Edited on February 20, 2014, 11:44 pm
  Posted by Steve Herman on 2014-02-20 23:14:39

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