Question

The domain of the function f(x) = \(\frac{1}{\sqrt{[x]^2-3[x]-10}}\) is (where [x] denotes the greatest integer less than or equal to x)

• A. (-∞,-2) ∪ (6, ∞)
• B. (-∞,-2) ∪ (5, ∞)
• C. (-∞,-3) ∪ (6, ∞)
• D. (-∞,-3) ∪ (5, ∞)

Answer 1

The Correct Option is A

Given Function: 

\[ F(x) = \frac{1}{\lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10}. \]

To ensure \( F(x) \) is defined, the denominator must be positive:

\[ \lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10 > 0. \]

Step 1: Solve the Inequality

Factorize the quadratic expression:

\[ \lfloor x \rfloor^2 - 3\lfloor x \rfloor - 10 = (\lfloor x \rfloor + 2)(\lfloor x \rfloor - 5). \]

The inequality becomes:

\[ (\lfloor x \rfloor + 2)(\lfloor x \rfloor - 5) > 0. \]

Step 2: Analyze the Intervals

The roots of the quadratic are \( \lfloor x \rfloor = -2 \) and \( \lfloor x \rfloor = 5 \). Using a sign chart:

Interval	Sign of \( (\lfloor x \rfloor + 2)(\lfloor x \rfloor - 5) \)
\( (-\infty, -2) \)	+
\( (-2, 5) \)	-
\( (5, \infty) \)	+

The inequality is satisfied in the intervals:

\[ \lfloor x \rfloor < -2 \quad \text{or} \quad \lfloor x \rfloor > 5. \]

Step 3: Refine the Solution

Since \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \), the solution must be refined to:

\[ \lfloor x \rfloor \leq -3 \quad \text{or} \quad \lfloor x \rfloor \geq 6. \]

Step 4: Write the Solution for \( x \)

The corresponding intervals for \( x \) are:

\[ x \in (-\infty, -2) \cup [6, \infty). \]

Final Answer:

\( x \in (-\infty, -2) \cup [6, \infty) \).