Question

Domain of the function f(x) = \(\frac{1}{\sqrt{[x]^2-[x]-6}}\) where [x] is greatest integer ≤ x is

• A. (-∞, -2) ∪ [4, ∞]
• B. (-∞, -2) ∪ [3, ∞]
• C. [-∞, -2) ∪ [4, ∞]
• D. [-∞, -2] ∪ [3, ∞)

Answer 1

The Correct Option is A

The given function is \(f(x) = \frac{1}{\sqrt{[x]^2-[x]-6}}\), where \([x]\) is the greatest integer less than or equal to x.

For the function to be defined, the expression inside the square root must be positive, i.e., \( [x]^2 - [x] - 6 > 0 \).

Let \(y = [x]\). Then, we have \(y^2 - y - 6 > 0\).

Factoring the quadratic expression, we get \((y-3)(y+2) > 0\).

This inequality holds when \(y < -2\) or \(y > 3\).

So, we have two cases:

Case 1: \([x] < -2\). Since \([x]\) is an integer, this means \([x] \leq -3\). Thus, \(x < -2\). Therefore, \(x \in (-\infty, -2)\).

Case 2: \([x] > 3\). Since \([x]\) is an integer, this means \([x] \geq 4\). Thus, \(x \geq 4\). Therefore, \(x \in [4, \infty)\).

Combining the two cases, the domain of the function is \((-\infty, -2) \cup [4, \infty)\).

Answer 2

The function involves a square root and for it to be defined, the expression under the square root must be non-negative. Hence, we need: \[ [x]^2 - [x] - 6 \geq 0 \] Solving the inequality, we find that \( [x] \) must satisfy the condition \( [x] \leq -2 \) or \( [x] \geq 4 \).

Therefore, the domain of \( f(x) \) is \( (-\infty, -2) \cup [4, \infty) \), which is option (A).