1.

If a in R and the equation

Answer»

<P>`(-1,0)UU(0,1)`
`(1,2)`
`(-2,-1)`
`(-oo,-2)uu(2,oo)`

Solution :Put `t=xx-[x]=[X]lt1` and then solve it.
Give, `a in R ` and EQUATION is
`-3{x-[x]}^(2)+2{x-[x]}+a^(2)=0`
Let `=x-[x],`then equation is
`-3t^(2)+2t+a^(2)=0`
`impliest=(a+-sqrt(1+3a^(2)))/(3)`
`because t=x-[x]={X}""["fractional part"]`
`because 0letle1`
`0le(1+-sqrt(1+3a^(2)))/(3)le1`
Taking positive SIGN, we GET
`0le (1+sqrt(1+3a^(2)))/(3)lt1""p[because{x}gt0]`
`impliessqrt(1+3a^(2))lt2impliesa+3a^(2)lt4`
`impliesa^(2)-1lt0implies(a+1)(a-1)lt0`
`therefore a in (-1,1)` for no integer solution of a, we consider `(-1,0)ii(0,1)`


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