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Step-by-Step Solution
Step 1: Understand the Function
We are given the function
$$f(x) = \sqrt{\frac{\lvert [x] \rvert - 2}{\lvert [x] \rvert - 3}},$$
where $[x]$ denotes the greatest integer less than or equal to $x$. We need to find the value of $a + b + c$ if the domain of $f(x)$ is
$$(-\infty,\, a)\,\cup\,[b,\, c)\,\cup\,[4,\, \infty), \quad a < b < c.$$
Step 2: Determine the Condition for the Domain
The expression inside the square root must be non-negative:
$$\frac{\lvert [x] \rvert - 2}{\lvert [x] \rvert - 3} \ge 0.$$
For a fraction to be non-negative, the numerator and denominator must have the same sign (both non-negative or both non-positive) and the denominator must not be zero.
Step 3: Analyze Cases Based on the Sign of Numerator and Denominator
Case I: Numerator ≥ 0 and Denominator > 0
1. Numerator ≥ 0:
$$\lvert [x] \rvert - 2 \ge 0 \quad \Longrightarrow \quad \lvert [x] \rvert \ge 2.$$
2. Denominator > 0:
$$\lvert [x] \rvert - 3 > 0 \quad \Longrightarrow \quad \lvert [x] \rvert > 3.$$
Thus, $\lvert [x] \rvert \ge 2$ must simultaneously satisfy $\lvert [x] \rvert > 3$. The stricter condition here is
$$\lvert [x] \rvert > 3.$$
Since $[x]$ is an integer, $\lvert [x] \rvert > 3$ implies $[x] \le -4$ or $[x] \ge 4.$
Corresponding $x$ ranges would be:
$x < -3$ when $[x] \le -4,$
$x \ge 4$ when $[x] \ge 4.$
Hence, from this case, $x \in (-\infty, -3) \cup [4, \infty).$
Case II: Numerator ≤ 0 and Denominator < 0
1. Numerator ≤ 0:
$$\lvert [x] \rvert - 2 \le 0 \quad \Longrightarrow \quad \lvert [x] \rvert \le 2.$$
2. Denominator < 0:
$$\lvert [x] \rvert - 3 < 0 \quad \Longrightarrow \quad \lvert [x] \rvert < 3.$$
Thus we combine:
$$\lvert [x] \rvert \le 2 \quad \text{and} \quad \lvert [x] \rvert < 3.$$
This essentially means $\lvert [x] \rvert \le 2.$ Hence $[x]$ can be $-2, -1, 0, 1, 2.$
Corresponding real $x$ values (because $[x]$ is the greatest integer function) will be:
$$x \in [-2, 3).$$
Therefore, from this case, $x \in [-2, 3).$
Step 4: Combine the Results to Find the Domain
Combining both cases:
$$(-\infty, -3) \cup [-2, 3) \cup [4, \infty).$$
We see that the domain of $f(x)$ is given in the form
$$(-\infty,\, a)\,\cup\,[b,\, c)\,\cup\,[4,\, \infty), \quad a < b < c.$$
Comparing intervals:
$a = -3,$
$b = -2,$
$c = 3.$
Step 5: Find the Required Sum a + b + c
$$a + b + c = (-3) + (-2) + 3 = -2.$$
Conclusion:
The value of $a + b + c$ is $-2$.
