If $a \in R$ 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

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

$(1, 2)$

$(- 2, - 1)$

$(- \infty, - 2) \cup (2, \infty)$

answer is A.

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Detailed Solution

Here, $a \in R$ and equation is

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

Let $t = x - [x]$

$∴ - 3 t^{2} + 2 t + a^{2} = 0$

$\begin{array}{ll} \Rightarrow t = \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} \\ ∵ t = x - [x] = {X} \\ ∴ 0 \leq t \leq 1 \\ \Rightarrow 0 \leq \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} \leq 1 \end{array}$ (fractional part)

Taking positive sign,

$[{x} > 0]$

$\begin{array}{l} ∴ 0 \leq \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} < 1 \\ \Rightarrow \sqrt{1 + 3 a^{2}} < 2 \\ \Rightarrow 1 + 3 a^{2} < 4 \\ \Rightarrow a^{2} - 1 < 0 \\ \Rightarrow (a + 1) (a - 1) < 0 \\ ∴ a \in (- 1, 1) \end{array}$