$\begin{array}{l} I f a \in R a n d t h e e q u a t i o n - 3 {(x - [x])}^{2} + 2 (x - [x]) + a^{2} = 0 (w h e r e [x] d e n o t e s t h e g r e a t e s t int e g e r < x) \\ h a s n o int e g r a l s o l u t i o n, t h e n a l l p o s s i b l e v a l u e s o f a l i e i n t h e int e r v a l \end{array}$

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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

$\begin{array}{l} G i v e n E q u a t i o n - 3 {(x - [x])}^{2} + 2 (x - [x]) + a^{2} = 0 \\ L e t x - [x] = y \Rightarrow 0 < y < 1 \\ \Rightarrow 3 y^{2} - 2 y - a^{2} = 0 \\ y = \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} \Rightarrow \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} \\ ∴ 0 < \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} < 1 \\ L e t \frac{1 \pm \sqrt{1 + 3 a^{2}}}{3} < 1 \Rightarrow \sqrt{1 + 3 a^{2}} < 2 \\ \Rightarrow 1 + 3 a^{2} < 4 \\ \Rightarrow 3 a^{2} - 3 < 0 \\ \Rightarrow a^{2} - 1 < 0 \\ \Rightarrow (a + 1) (a - 1) < 0 \\ ∴ a \in (- 1, 1) \\ f o r a = 0, y = 0 i f i s n o t p o s s i b l e \\ ∴ a \in (- 1, 0) \cup (0, 1) \end{array}$