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

Options

A. 1
B. 2
C. 3
D. infinitely many

Answer

B

Explanation

We known that x=[x]+{x}
`:. 4{x}=x+[x]`
`implies r{x}=[x]+{x}+[x]`
`implies 3{X}=2[x]" "`….(i)
But , ` 0 le{x} lt 1 `
`:. 0 le 3{X} lt 3 `
`implies 0 le 2[x] lt 3" " `[Using (i)]
`implies 0 le[x] lt (3)/(2) implies [x]=0,1`
Now, `[x]=0 implies {x}=0" "` [Putting [x]=0 in (i)]
`[x]=1 implies {x}=(2)/(3)" "`[Putting [x]=1 in (i) ]
`:. x=0,(5)/(3)" "[ :' x =[x]+{x}]`