If a ∈ R and the equation -3(x - [x])² + 2(x - [x]) + a² = 0 (where [x] denotes the greatest integer

Q: Wait, why can't we just use the vertex formula for the range of the quadratic?

That is a fantastic instinct! The vertex formula gives the global minimum of the parabola, which is indeed at $f = 1/3$. However, we are not looking for the range of the function over all real numbers. We are constrained by the domain $f \in (0, 1)$. We must check the values at the boundaries of this interval, not just the vertex, to see how the function behaves as it approaches the limits of our domain.

Q: Why do we exclude $a^2 = 0$ in the final step?

This is the trap! If $a^2 = 0$, then $g(f) = 3f^2 - 2f = 0$. This equation holds when $f = 0$ or $f = 2/3$. If $f = 0$, then $x$ is an integer, which violates the problem's condition that there are 'no integral solutions'. Therefore, we must exclude the case where $f=0$, which forces us to exclude $a^2 = 0$ to be safe.

Q: Why is the fractional part $f$ strictly between $0$ and $1$?

The fractional part $f = x - [x]$ is defined such that $0 \le f < 1$. If $x$ is an integer, $f$ is exactly $0$. Since the problem explicitly states there are no integral solutions, $x$ cannot be an integer, which means $f$ cannot be $0$. Thus, $f$ is strictly greater than $0$ and less than $1$.

Question

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:

Options:

(A) $(-2, -1)$

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

(C) $(-1, 0) \cup (0, 1)$

(D) $(1, 2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The fractional part function, denoted as $\{x\} = x - [x]$, represents the decimal remainder of a number. By definition, $\{x\}$ is constrained to the interval $[0, 1)$.

However, the problem explicitly states there are **no integral solutions**. If $x$ were an integer, $\{x\}$ would be $0$. Since integral solutions are forbidden, $\{x\}$ cannot be $0$.

Thus, we define $f = \{x\}$, where $f \in (0, 1)$. This shift in the domain is the critical constraint for our analysis.

### Transforming the Equation

Substituting $f$ into the given equation $-3\{x\}^2 + 2\{x\} + a^2 = 0$, we obtain a quadratic equation in terms of $f$:

$$-3f^2 + 2f + a^2 = 0$$

To isolate the parameter $a$, we rearrange the terms:

$$a^2 = 3f^2 - 2f$$

Let $g(f) = 3f^2 - 2f$. We must determine the range of this function as $f$ varies within the open interval $(0, 1)$.

### The Dance of the Parabola

The function $g(f) = 3f^2 - 2f$ represents an upward-opening parabola. To find its vertex, we calculate the derivative:

$$g'(f) = 6f - 2$$

Setting the derivative to zero, we find the critical point at $f = \frac{1}{3}$. This is the location of the parabola's minimum value.

Evaluating the function at this vertex:

$$g\left(\frac{1}{3}ight) = 3\left(\frac{1}{3}ight)^2 - 2\left(\frac{1}{3}ight) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$$

As $f 	o 0$, $g(f) 	o 0$. As $f 	o 1$, $g(f) 	o 3(1)^2 - 2(1) = 1$. Therefore, the range of $g(f)$ for $f \in (0, 1)$ is $[-\frac{1}{3}, 1)$.

### The Final Constraint

Since $a^2 = g(f)$, we must satisfy $a^2 \in [-\frac{1}{3}, 1)$. Because $a$ is a real number, $a^2$ must be non-negative, leading to $a^2 \in [0, 1)$.

We must exclude cases where $f=0$ to satisfy the "no integral solution" condition. If $a^2 = 0$, then $3f^2 - 2f = 0$, which implies $f(3f - 2) = 0$. This yields $f = 0$ or $f = \frac{2}{3}$.

Since $f 
eq 0$, we must exclude $a^2 = 0$. This leaves us with the strict inequality:

$$0 < a^2 < 1$$

Taking the square root, we find $0 < |a| < 1$. The final set of values for $a$ is **$a \in (-1, 0) \cup (0, 1)$**.

Animated Solution for Mathematics - Quadratic Equations: If $a \in R$ and the equation $- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0$ (where $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of $a$ lie in the interval:

Select Answer:

Visualized Solution

The Fractional Part Function

Condition for No Integral Solution

Isolating $a^{2}$

Analyzing the Quadratic $g (f)$

Minimum Value of $g (f)$

Boundary Values of $g (f)$

Range of $g (f)$

Applying the Constraint $a^{2} \geq 0$

The Trap: Checking $a^{2} = 0$

Final Interval for $a$

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

Transforming the Equation

The Dance of the Parabola

The Final Constraint

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Visualized Solution

The Fractional Part Function

Condition for No Integral Solution

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Analyzing the Quadratic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f)

Minimum Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f)

Boundary Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f)

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f)

Applying the Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2≥0

The Trap: Checking .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2=0

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