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the number of solution of |x-1| ={x} where {} represents fractional function,

the number of solution of |x-1| ={x} where {} represents fractional function,

Grade:12th pass

2 Answers

Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
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Samyak Jain
333 Points
5 years ago
Method 1 :  Draw graphs of |x – 1| and {x} and note values of x where intersection of two graphs occur.
Method 2: By algebra.
|x – 1| = {x}                            …...............(1)
Case 1 : Let x \geq 1 & x \in Z    \therefore {x} = 0
We can write (1) as x – 1 = 0   or   x = 1
Taking intersection, we get x = 1                  .…..........(2)
Case 2 : Let x > 1 & x \notin Z 
(1) becomes x – 1 = {x}  or  [x] + {x} – 1 = {x}
i.e., [x] = 1    \therefore  1 \leq x
Taking intersection, we get 1
Case 3 : 0 \leq x < 1  & x \in Z 
So (1) is   1 – x = 0  i.e. x = 1
Taking intersection, we get \in \phi                 …............(4)
Case 4 : 0 \notin Z
So (1) can be written as 1 – x = {x}  or  1 – [x] – {x} = {x}
{x} = (1 – [x] ) / 2    We know that for 0 \leq x \leq 1, [x] = 0
\therefore {x} = (1 – 0) / 2 = ½  or  x = 1 / 2               …............(5)
Case 5 :  –1 \leq x \therefore 1 – x = [x}   or   [x] = 1  – 2{x}
For –1 \leq x
So, 1  – 2{x} = –1  or  {x} = 1, which is not true.
\in \phi                                                            …...........(6)
Similarly, for negative x there is no solution of (1).
Taking union of (2), (3), (4), (5) & (6), we get \in [1,2) \cup 1/2 ,
which are the solutions of |x – 1| = {x}.
Hence, there are infinitely many solutions of |x – 1| = {x} .

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