Understanding the Problem

We are given the equation: $-3{(x-[x\left]\right)}^2+2(x-[x\left]\right)+a^2=0$, where $\left[x\right]$ denotes the greatest integer less than or equal to x (the floor function). We are told the equation has no integral solution (i.e., no solution where x is an integer). We need to find all possible values of the real number $a$ for which this is true.

Key Concept: Fractional Part Function

Let's define a new variable. For any real number $x$, we can write it as $x=\left[x\right]+\left\{x\right\}$, where $\left\{x\right\}$ is the fractional part of x. By definition, $0\le \left\{x\right\}<1$.

Notice that $x-\left[x\right]=\left\{x\right\}$. Let's denote this fractional part by $t$. So, $t=\left\{x\right\}$ and $0\le t<1$.

Substituting into our original equation, we get a new equation in terms of $t$: $-3t^2+2t+a^2=0$, where $0\le t<1$.

Step 1: Analyze the Condition "No Integral Solution"

The problem states the equation has no integral solution. Let's see what an integral solution ($x\in \mathbb{Z}$) implies.

If $x$ is an integer, then its fractional part $t=\left\{x\right\}=0$.

Let's substitute $t=0$ into our transformed equation to see when an integer x would be a solution: $-3(0{)}^2+2(0)+a^2=0$ $\Rightarrow a^2=0$ $\Rightarrow a=0$

This is a crucial finding: An integral solution exists if and only if $a=0$. Therefore, the condition that the equation has no integral solution is equivalent to the simple condition: $a\ne 0$.

Step 2: Re-examining the Question in Context

The question asks for the interval of $a$ where there is no integral solution. From Step 1, we found this happens for all $a\ne 0$. However, this seems too broad and doesn't match any of the given options directly. This suggests we might have misinterpreted the problem's phrasing.

Let's read the problem again carefully: "...has no integral solution". This could be interpreted in two ways:

1. Interpretation A (What we did): There exists no solution $x$ that is an integer.
2. Interpretation B (Likely intended): The equation has no solution at all (i.e., for any real $x$), and in addition, we are given that this case of having no solution occurs specifically when the parameter $a$ takes values such that an integral solution is also not possible. The options hint that this is the intended, more complex meaning.

The presence of specific intervals like (-1,0) U (0,1) in the options strongly suggests that Interpretation B is correct. The problem is asking: For which values of $a$ does the equation have no real solution (for $x$), given that it also has no integral solution? Since we already know $a=0$ gives an integral solution, we must have $a\ne 0$ for the "no integral solution" part of the condition to hold. Now let's find when the equation has no real solution $x$ (and hence no fractional part $t$ in [0,1) that satisfies it).

Step 3: Finding When the Equation Has No Real Solution

We have the quadratic in $t$: $-3t^2+2t+a^2=0$, where $0\le t<1$.

Let's rewrite this quadratic in standard form: $3t^2-2t-a^2=0$

For this quadratic to have no real solution $t$ in the domain $[0,1)$, the quadratic itself must either have no real roots at all, or its real roots must lie completely outside the interval $[0,1)$.

Let $f\left(t\right)=3t^2-2t-a^2$.

The discriminant $D$ of this quadratic is: $D={(-2)}^2-4\cdot 3\cdot (-a^2)=4$