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If f(x) and [x] denote respectively the fractional and integeral parts of a real number x, then the number of solution of the euation 4{x}=x+[x] , is

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`:'x=[x]+{x}` ……i
Then given equation reduces to
`4{x}=[x]+{x}|[x]`
`implies{x}=2/3[x]`…….ii
`:'0le{x}lt1=0le2/3[x]lt1` or `0le[x]lt3/2`
`:.[x]=0,1`
From Eq. iii `{x}=0,2/3`
From Eq. (i) `x=0,1+2/3` i.e. `x=0,5/3`
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