Key Concepts and Formulas

• Fractional Part: For any real number $x$, $x = \left[ x \right] + \left\{ x \right\}$, where $\left[ x \right]$ is the greatest integer less than or equal to $x$, and $\left\{ x \right\}$ is the fractional part of $x$ such that $0 \le \left\{ x \right\} < 1$.
• Integral Solution: An equation has an integral solution if there exists an integer value of $x$ that satisfies the equation. This implies $\left\{ x \right\} = 0$.
• Quadratic Formula: For a quadratic equation $Ax^2 + Bx + C = 0$, the solutions are given by $x = \frac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$.

Step 1: Simplify the Equation using Fractional Part

We are given the equation: $- 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ Since $\left\{ x \right\} = x - \left[ x \right]$, we substitute to obtain a quadratic equation in terms of $\left\{ x \right\}$: $- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$ Explanation: The substitution transforms the original equation, which involves both $x$ and its greatest integer function, into a simpler quadratic equation with only the fractional part $\left\{ x \right\}$ as the variable. This simplifies further analysis.

Step 2: Solve for $\left\{ x \right\}$ using the Quadratic Formula

Let $y = \left\{ x \right\}$. We solve the quadratic equation $-3y^2 + 2y + a^2 = 0$ using the quadratic formula: $y = \frac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$ where $A = -3$, $B = 2$, and $C = a^2$. $\left\{ x \right\} = \frac{{ - 2 \pm \sqrt {{2^2} - 4\left( { - 3} \right){a^2}} }}{{2\left( { - 3} \right)}}$ $\left\{ x \right\} = \frac{{ - 2 \pm \sqrt {4 + 12{a^2}} }}{{ - 6}}$ Dividing the numerator and denominator by -2: $\left\{ x \right\} = \frac{{1 \mp \sqrt {1 + 3{a^2}} }}{3}$ Explanation: The quadratic formula is applied to find the solutions for $\left\{ x \right\}$. The discriminant $4 + 12a^2$ is always non-negative for real $a$, ensuring that the solutions are real.

Step 3: Analyze the Condition "No Integral Solution"

The problem states that the equation has no integral solution. This means that there is no integer value of $x$ that satisfies the equation. This can happen if there are no real solutions for $x$ at all, or if all solutions are non-integers (i.e., $\left\{ x \right\} \ne 0$).

We will analyze the roots for when $a \in (-2, -1)$. This means $1 < a^2 < 4$. Thus $3 < 3a^2 < 12$ and $4 < 1+3a^2 < 13$. Therefore $2 < \sqrt{1+3a^2} < \sqrt{13}$

Step 4: Examine the Roots for a in (-2, -1)

We have two roots: $y_1 = \frac{1 - \sqrt{1+3a^2}}{3}$ $y_2 = \frac{1 + \sqrt{1+3a^2}}{3}$

Since $2 < \sqrt{1+3a^2} < \sqrt{13}$, we have:

For $y_1$: Since $\sqrt{1+3a^2} > 2$, $1 - \sqrt{1+3a^2} < 1 - 2 = -1$. Therefore, $y_1 < \frac{-1}{3}$. Since $\left\{ x \right\} \ge 0$, this root is invalid.

For $y_2$: Since $\sqrt{1+3a^2} > 2$, $1 + \sqrt{1+3a^2} > 1+2 = 3$. Therefore, $y_2 > \frac{3}{3} = 1$. Since $\left\{ x \right\} < 1$, this root is also invalid.

Explanation: We analyze the two roots within the range $a \in (-2, -1)$. We show that neither root produces a valid fractional part (i.e. value between 0 and 1). Therefore, for $a \in (-2, -1)$, there are no real solutions for $x$ in the given equation.

Step 5: Conclusion for a in (-2, -1)

Since for all $a \in (-2, -1)$, there are no real solutions for $x$, it means there are definitely no integral solutions for $x$. Therefore, $a \in (-2, -1)$ satisfies the condition "the equation has no integral solution".

Common Mistakes & Tips

• Range of $\left\{ x \right\}$: Always remember that the fractional part must be in the interval $[0, 1)$.
• Interpretation of "No Integral Solution": The key is to recognize that "no integral solution" can mean "no real solutions at all".
• Casework and Intersection: Carefully analyze the ranges and conditions for each case and find the appropriate intersection or union to satisfy the overall problem requirement.

Summary

We substituted the fractional part definition into the given equation and solved the resulting quadratic for $\left\{ x \right\}$. By analyzing the roots for $a \in (-2, -1)$, we showed that neither root produces a valid fractional part. This implies that there are no real solutions for $x$ when $a \in (-2, -1)$, thus guaranteeing that there are no integral solutions.

The final answer is \boxed{\left( { - 2, - 1} \right)}, which corresponds to option (A).