If $a \in \mathbb{R}$ and the equation $-3(x-[x])^{2}+2(x-[x])+a^{2}=0$ (where $[x]$ denotes the greatest integer $\leq x$) has no integral solution, then all possible values of $a$ lie in the interval:

If a ∈ R and the equation -3(x-[x])²+2(x-[x])+a²=0 (where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval:

$-3(x-[x])^{2}+2(x-[x])+a^{2}=0$

$[x] = \text{greatest integer} \le x, \quad a \in \mathbb{R}$

$\text{To find: } a \text{ such that no integral } x \text{ satisfies the equation}$

$t = x - [x] = \{x\} \in [0,1)$

$-3t^2 + 2t + a^2 = 0$

$a^2 = 3t^2 - 2t$

$f(t) = 3t^2 - 2t$

$x \in \mathbb{Z} \iff t = x - [x] = 0$

$t = 0:\ a^2 = f(0) = 0 \Rightarrow a = 0$

$1.\ \text{range of } f(t) \text{ on } [0,1)$

$2.\ \text{which } a^2 \text{ give a non-integer solution}$

$3.\ \text{exclude the } a \text{ giving an integral solution}$

$f(t) = 3t^2 - 2t, \quad t \in [0,1)$

$f'(t) = 6t - 2 = 0 \Rightarrow t = \frac{1}{3}$

$f\!\left(\frac{1}{3}\right) = -\frac{1}{3}$

$f(0)=0, \quad f(1^-)\to 1$

$f(t) \in \left[-\frac{1}{3},\,1\right)$

$f(0)=0, \quad f\!\left(\frac{2}{3}\right)=0$

$a^2 = f(t) \ \Rightarrow\ a^2 \in [0,1)$

$\Rightarrow\ |a| < 1$

$|a| = 1 \Rightarrow a^2 = 1 \Rightarrow t = 1 \notin [0,1)$

$a = 0 \Rightarrow t = 0 \text{ (integral solution)}$

$\therefore\ a \neq 0$

$a^2=\tfrac{1}{2}:\ t=\frac{2+\sqrt{10}}{6}\approx 0.86 \in (0,1)$

$a \in (-1,0)\cup(0,1)$

$\boxed{a \in (-1,0)\cup(0,1)} \quad \text{option (3)}$