If `a in R` and the equation `-3(x-[x])^2+""2""(x-[x])""+a^2=""0` (where [x] denotes the greatest integer x) has no integral solution, then all possible values of a lie in the interval (1) `(-1,""0)uu(0,""1)` (2) (1, 2) (3) `(-2,-1)` (4) `(-oo,-2)uu(2,oo)`

Options

A. `(-2,-1)`
B. `(-oo,-2)uu(2,oo)`
C. `(-1,0)uu(0,1)`
D. `(1,2)`

Answer

B

Explanation

`:'x-[x]={x}` [fractional part of `x`]
For no integral solution `{x}!=0`
`:.a!=0`……….i
The given equation can be written as
`3{x}^(2)-2{x}-a^(2)=0`
`implies{x}=(2+sqrt((4+12a^(2)))/6=(1+sqrt((1+3a^(2)))/3[:'0lt{x}lt1]`
`implies0lt(1+sqrt((1+3a^(2)))/3lt1impliessqrt(1+3a^(2)))lt2`
`impliesa^(2)lt1=-1ltalt1` ..........ii
From Eqs i and ii we get
`a epsilon(-1,0)uu(0,1)`