Question

The set of all real values of $x$ such that \[ f(x) = \frac{[x] - 1}{\sqrt{[x]^2 - [x] - 6}} \] is a real valued function is

• A. $[1, \infty)$
• B. $(-\infty, -2) \cup [4, \infty)$
• C. $[-1, 3)$
• D. $[-1, 2) \cup [4, \infty)$

Hint:

When dealing with functions involving greatest integer functions inside radicals, analyze the domain by considering integer values stepwise and ensure the radicand is non-negative and denominator non-zero.

Answer 1

The Correct Option is D

Given the function \[ f(x) = \frac{[x] - 1}{\sqrt{[x]^2 - [x] - 6}} \] where $[x]$ denotes the greatest integer function (floor function). For $f(x)$ to be real valued, the expression under the square root must be non-negative and the denominator must not be zero. Set $n = [x]$, then: \[ n^2 - n - 6 \geq 0 \] Factorizing: \[ (n - 3)(n + 2) \geq 0 \] This inequality holds when \[ n \leq -2 \quad \text{or} \quad n \geq 3. \] But the denominator cannot be zero: \[ n^2 - n - 6 \neq 0 \implies n \neq 3, -2. \] Therefore, valid integer values of $n$ are: \[ n \leq -3 \quad \text{or} \quad n \geq 4. \] Since $n = [x]$, for \[ n \leq -3 \Rightarrow x<-2, \] and \[ n \geq 4 \Rightarrow x \geq 4. \] Also, the numerator must be defined, so values where $[x] = 1$ are allowed, but we must check if the denominator is defined at $[x] = 1$: \[ 1^2 - 1 - 6 = 1 - 1 - 6 = -6<0, \] so it is not valid. Finally, the solution set for $x$ becomes: \[ [-1, 2) \cup [4, \infty) \] considering the floor values and the domain restrictions. Hence, the domain of $x$ such that $f(x)$ is real valued is $[-1, 2) \cup [4, \infty)$.