If `a in R` and the equation `-3(x-[x])^2+2(x-[x])+a^2=0` (where [x] denotes the greatest integer `le x`) has no integral solution, then all possible values of a lie in the interval: (1) (-2,-1) (2) `(oo,-2) uu (2,oo)` (3) `(-1,0) uu (0,1)` (4) (1,2)

Options

A. `(-1,0) cup (0,1)`
B. `(1,2)`
C. `(-2,-1)`
D. `(-oo,-2)cup (2,oo)`

Explanation

To solve the problem, we need to analyze the given equation and determine the conditions under which it has no integral solutions. The equation is:

\[ -3(x - [x])^2 + 2(x - [x]) + a^2 = 0 \]

Where \([x]\) denotes the greatest integer less than or equal to \(x\). Let's denote \(t = x - [x]\), which means \(t\) is the fractional part of \(x\) and lies in the interval \([0, 1)\).

### Step 1: Rewrite the equation in terms of \(t\)

Substituting \(t\) into the equation gives:

\[ -3t^2 + 2t + a^2 = 0 \]

### Step 2: Rearranging the equation

We can rearrange this into a standard quadratic form:

\[ 3t^2 - 2t - a^2 = 0 \]

### Step 3: Determine the discriminant

For the quadratic equation \(3t^2 - 2t - a^2 = 0\) to have no real solutions, the discriminant must be less than zero. The discriminant \(D\) of a quadratic equation \(at^2 + bt + c = 0\) is given by:

\[ D = b^2 - 4ac \]

Here, \(a = 3\), \(b = -2\), and \(c = -a^2\). Thus, the discriminant becomes:

\[ D = (-2)^2 - 4 \cdot 3 \cdot (-a^2) = 4 + 12a^2 \]

### Step 4: Set the discriminant less than zero

We want:

\[ 4 + 12a^2 < 0 \]

However, since \(4\) is positive and \(12a^2\) is non-negative (as \(a^2 \geq 0\)), the expression \(4 + 12a^2\) cannot be less than zero for any real value of \(a\). Therefore, we must analyze the conditions under which the quadratic does not intersect the \(t\) axis in the interval \([0, 1)\).

### Step 5: Finding the range of \(a\)

To ensure that the quadratic does not have any integral solutions in the interval \([0, 1)\), we need to find the values of \(a\) such that the vertex of the parabola lies outside the interval \([0, 1)\).

The vertex \(t_v\) of the quadratic \(3t^2 - 2t - a^2\) is given by:

\[ t_v = -\frac{b}{2a} = \frac{2}{2 \cdot 3} = \frac{1}{3} \]

### Step 6: Analyze the conditions for \(a\)

We need to ensure that the maximum value of the quadratic (which occurs at the endpoints of the interval) does not equal zero or is negative. The maximum value occurs at \(t = 0\) and \(t = 1\):

1. At \(t = 0\):
\[ f(0) = -a^2 < 0 \implies a^2 > 0 \implies a \neq 0 \]

2. At \(t = 1\):
\[ f(1) = 3(1)^2 - 2(1) - a^2 = 3 - 2 - a^2 = 1 - a^2 < 0 \implies a^2 > 1 \implies |a| > 1 \]

Thus, \(a\) must lie in the intervals:

\[ (-\infty, -1) \cup (1, \infty) \]

### Final Answer

The possible values of \(a\) lie in the interval:

\[ (-\infty, -2) \cup (2, \infty) \]

Thus, the correct option is:

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