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)
- (-∞,-2) ∪ (6, ∞)
- (-∞,-2) ∪ (5, ∞)
- (-∞,-3) ∪ (6, ∞)
- (-∞,-3) ∪ (5, ∞)
The Correct Option is A
Solution and Explanation
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:
\( x \in (-\infty, -2) \cup [6, \infty) \).
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