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

Let $x-[x]=t$x−[x]=t

$$\therefore -3{t}^{2}+2t+{a}^{2}=0$$∴−3t2+2t+a2=0

As $x$x is not an integer, $t\neq 0$t=0 $\therefore a\neq 0$∴a=0

For a root of $x$x to exist at least one of the roots of $t$t should be between $0$0 and $1$1.

Roots $=\frac{-2\pm \sqrt{4-(4)(-3){a}^{2}}}{-6}$=−6−2±4−(4)(−3)a2​​

$$=\frac{-2\pm \sqrt{4+12{a}^{2}}}{-6}$$=−6−2±4+12a2​​

$$=\frac{-1\pm \sqrt{1+3{a}^{2}}}{-3}=\frac{1\mp \sqrt{1+3{a}^{2}}}{3}$$=−3−1±1+3a2​​=31∓1+3a2​​

As we observe, one root is surely less than zero,

i.e., $\frac{1-\sqrt{1+3{a}^{2}}}{3}$31−1+3a2​​

$\therefore$∴ For a solution to exist,

$$\frac{1+\sqrt{1+3{a}^{2}}}{3}<1$$31+1+3a2​​<1

$$\therefore 1+\sqrt{1+3{a}^{2}}<3$$∴1+1+3a2​<3

$$\therefore \sqrt{1+3{a}^{2}}<2$$∴1+3a2​<2

$$\therefore 1+3{a}^{2}<4$$∴1+3a2<4

$$\therefore 3{a}^{2}<3$$∴3a2<3

$$\therefore {a}^{2}<1$$∴a2<1

$\therefore a\in (-1,0)\cup (0,1)$∴a∈(−1,0)∪(0,1). $[\because a\neq 0]$[∵a=0]