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

Options

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

Answer

C

Explanation

Let {x} = x - [x] be the fractional part of x. Then, `0 le {x} lt 1 and {x} = 0` when x is an integer.
If the equation `-3(x-[x])^(2)+ 2(x-[x]) + a^(2) = 0` has no integral solution, then roots of the equation `-3{x}^(2) + 2{x} + a^(2) = 0` lie in the interval (0, 1). This is possible if
(i) `4 + 12 a^(2) ge 0" "["Discriminant" ge 0]`
(ii) `(-2)/(-6) gt 0" "[because -(b)/(2a) gt 0]`
(iii) `-3 + 2 |a^(2) lt 0" "[because f(0) lt 0, f(1) lt 0]`
These three suggest that `a in (-1, 1)`. But, a = 0 gives {x} = 0 i.e. x is an integer. So, a `ne` 0. Hence, a `ne (-1, 0) uu (0, 1)`.