Key Concepts and Formulas

• Domain of a Square Root Function: For a function of the form $f(x) = \sqrt{g(x)}$ to be defined in real numbers, the expression inside the square root, $g(x)$, must be non-negative, i.e., $g(x) \ge 0$.
• Domain of a Rational Function: For a rational function $\frac{N(x)}{D(x)}$ to be defined, the denominator $D(x)$ must not be equal to zero.
• Greatest Integer Function: The greatest integer function $[x]$ gives the largest integer less than or equal to $x$. This implies that $[x]$ is always an integer.

Step-by-Step Solution

Step 1: Analyze the expression inside the square root. The function is given by $f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$. For $f(x)$ to be a real-valued function, the expression inside the square root must be non-negative. Let $y = [x]$. Then the expression becomes $\frac{|y| - 2}{|y| - 3}$. We need $\frac{|y| - 2}{|y| - 3} \ge 0$.

Step 2: Determine the conditions for the inequality $\frac{|y| - 2}{|y| - 3} \ge 0$. This inequality holds when either: Case 1: $|y| - 2 \ge 0$ and $|y| - 3 > 0$ Case 2: $|y| - 2 \le 0$ and $|y| - 3 < 0$

Step 3: Solve Case 1. $|y| - 2 \ge 0 \implies |y| \ge 2$ $|y| - 3 > 0 \implies |y| > 3$ For both conditions to be true, we need $|y| > 3$. This means $y > 3$ or $y < -3$.

Step 4: Solve Case 2. $|y| - 2 \le 0 \implies |y| \le 2$ $|y| - 3 < 0 \implies |y| < 3$ For both conditions to be true, we need $|y| \le 2$. This means $-2 \le y \le 2$.

Step 5: Combine the results from Case 1 and Case 2. The possible values for $|y|$ are $|y| > 3$ or $|y| \le 2$.

Step 6: Substitute back $y = [x]$ and solve for $x$. We have $|[x]| > 3$ or $|[x]| \le 2$.

For $|[x]| > 3$: This means $[x] > 3$ or $[x] < -3$. If $[x] > 3$, then $x \ge 4$. This gives the interval $[4, \infty)$. If $[x] < -3$, then $x < -3$. Since $[x]$ must be an integer, the integers less than -3 are -4, -5, ... . Thus, $x$ can be any real number such that $[x] \le -4$. This implies $x < -3$. More precisely, if $[x] \le -4$, then $x < -3$. For example, if $x = -3.5$, $[x] = -4$. If $x = -3.0001$, $[x] = -4$. So, this condition translates to $x < -3$.

For $|[x]| \le 2$: This means $-2 \le [x] \le 2$. Since $[x]$ must be an integer, the possible integer values for $[x]$ are -2, -1, 0, 1, 2. If $[x] = -2$, then $-2 \le x < -1$. If $[x] = -1$, then $-1 \le x < 0$. If $[x] = 0$, then $0 \le x < 1$. If $[x] = 1$, then $1 \le x < 2$. If $[x] = 2$, then $2 \le x < 3$. Combining these intervals, we get $-2 \le x < 3$.

Step 7: Combine all valid intervals for $x$. From $|[x]| > 3$, we have $x < -3$ or $x \ge 4$. From $|[x]| \le 2$, we have $-2 \le x < 3$.

Combining these, the domain of $f(x)$ is $(-\infty, -3) \cup [-2, 3)$. However, the problem statement gives the domain as $(-\infty, a) \cup [b, c) \cup [4, \infty)$. Let's re-examine the condition $|[x]| > 3$. If $[x] > 3$, then $[x]$ can be $4, 5, 6, ...$. This means $x \ge 4$. This matches the $[4, \infty)$ part of the given domain. If $[x] < -3$, then $[x]$ can be $-4, -5, -6, ...$. This means $x < -3$. This corresponds to the $(-\infty, a)$ part of the given domain, so $a = -3$.

Now let's re-examine the condition $|[x]| \le 2$. This means $-2 \le [x] \le 2$. The integer values are $-2, -1, 0, 1, 2$. If $[x] = -2$, then $-2 \le x < -1$. If $[x] = -1$, then $-1 \le x < 0$. If $[x] = 0$, then $0 \le x < 1$. If $[x] = 1$, then $1 \le x < 2$. If $[x] = 2$, then $2 \le x < 3$. The union of these intervals is $[-2, 3)$. This matches the $[b, c)$ part of the given domain. So, $b = -2$ and $c = 3$.

Step 8: Verify the condition $a < b < c$. We found $a = -3$, $b = -2$, and $c = 3$. Checking the condition: $-3 < -2 < 3$. This is true.

Step 9: Calculate the value of a + b + c. $a + b + c = -3 + (-2) + 3 = -3 - 2 + 3 = -2$.

Let's recheck the inequality $\frac{|y| - 2}{|y| - 3} \ge 0$. We need $|y| \in [0, 2] \cup (3, \infty)$. Since $y = [x]$, $|y|$ is always a non-negative integer. So, $|[x]|$ can be $0, 1, 2$ or $|[x]| > 3$.

Case A: $|[x]| \in \{0, 1, 2\}$. If $|[x]| = 0$, then $[x] = 0$, which means $0 \le x < 1$. If $|[x]| = 1$, then $[x] = 1$ or $[x] = -1$. If $[x] = 1$, then $1 \le x < 2$. If $[x] = -1$, then $-1 \le x < 0$. If $|[x]| = 2$, then $[x] = 2$ or $[x] = -2$. If $[x] = 2$, then $2 \le x < 3$. If $[x] = -2$, then $-2 \le x < -1$. Combining these, we get $[-2, 3)$.

Case B: $|[x]| > 3$. This means $[x] > 3$ or $[x] < -3$. If $[x] > 3$, then $[x]$ can be $4, 5, 6, ...$. This implies $x \ge 4$. If $[x] < -3$, then $[x]$ can be $-4, -5, -6, ...$. This implies $x < -3$.

So, the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. This matches the given domain format $(-\infty, a) \cup [b, c) \cup [4, \infty)$. Comparing the two, we get: $a = -3$ $b = -2$ $c = 3$

We are given that $a < b < c$. $-3 < -2 < 3$, which is true.

Now, we need to calculate $a + b + c$. $a + b + c = -3 + (-2) + 3 = -2$.

Let's re-examine the original problem statement and the options. The correct answer is given as A, which corresponds to 8. My current calculation gives -2. There must be a mistake in my interpretation or calculation.

Let's re-evaluate the inequality $\frac{|y| - 2}{|y| - 3} \ge 0$. Let $Z = |y|$. We need $\frac{Z - 2}{Z - 3} \ge 0$. This inequality holds when:

1. $Z - 2 \ge 0$ and $Z - 3 > 0$. This implies $Z \ge 2$ and $Z > 3$. So, $Z > 3$.
2. $Z - 2 \le 0$ and $Z - 3 < 0$. This implies $Z \le 2$ and $Z < 3$. So, $Z \le 2$.

So, we need $|y| > 3$ or $|y| \le 2$. Since $y = [x]$, $|y|$ must be a non-negative integer. So, $|[x]|$ can be $0, 1, 2$ or $|[x]|$ can be $4, 5, 6, ...$.

Case 1: $|[x]| \in \{0, 1, 2\}$. If $|[x]| = 0$, then $[x] = 0 \implies 0 \le x < 1$. If $|[x]| = 1$, then $[x] = 1$ or $[x] = -1$. $[x] = 1 \implies 1 \le x < 2$. $[x] = -1 \implies -1 \le x < 0$. If $|[x]| = 2$, then $[x] = 2$ or $[x] = -2$. $[x] = 2 \implies 2 \le x < 3$. $[x] = -2 \implies -2 \le x < -1$. Combining these, we get the interval $[-2, 3)$.

Case 2: $|[x]| \in \{4, 5, 6, ...\}$. This means $[x] \ge 4$ or $[x] \le -4$. If $[x] \ge 4$, then $x \ge 4$. This gives the interval $[4, \infty)$. If $[x] \le -4$, then $x < -3$. This gives the interval $(-\infty, -3)$.

So, the domain of the function is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. This is in the form $(-\infty, a) \cup [b, c) \cup [4, \infty)$. Comparing the two, we have: $a = -3$ $b = -2$ $c = 3$

The condition $a < b < c$ is $-3 < -2 < 3$, which is true. The sum $a + b + c = -3 + (-2) + 3 = -2$.

There seems to be a discrepancy with the given correct answer. Let me review the problem and my understanding of the domain.

The function is $f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$. Let $k = [x]$. We need $\frac{|k| - 2}{|k| - 3} \ge 0$. This inequality holds when $|k| \in [0, 2] \cup (3, \infty)$. Since $k = [x]$, $k$ is an integer. So $|k|$ is a non-negative integer. Thus, $|k|$ can be $0, 1, 2$ or $|k| \ge 4$.

If $|k| \in \{0, 1, 2\}$: $|k| = 0 \implies k = 0 \implies [x] = 0 \implies 0 \le x < 1$. $|k| = 1 \implies k = 1$ or $k = -1$. $k = 1 \implies [x] = 1 \implies 1 \le x < 2$. $k = -1 \implies [x] = -1 \implies -1 \le x < 0$. $|k| = 2 \implies k = 2$ or $k = -2$. $k = 2 \implies [x] = 2 \implies 2 \le x < 3$. $k = -2 \implies [x] = -2 \implies -2 \le x < -1$. The union of these intervals is $[-2, 3)$.

If $|k| \ge 4$: $|k| = 4 \implies k = 4$ or $k = -4$. $k = 4 \implies [x] = 4 \implies 4 \le x < 5$. $k = -4 \implies [x] = -4 \implies -4 \le x < -3$. $|k| = 5 \implies k = 5$ or $k = -5$. $k = 5 \implies [x] = 5 \implies 5 \le x < 6$. $k = -5 \implies [x] = -5 \implies -5 \le x < -4$. ... and so on.

Let's re-examine the condition $|[x]| \ge 4$. This means $[x] \ge 4$ or $[x] \le -4$. If $[x] \ge 4$, then $x \ge 4$. This gives the interval $[4, \infty)$. If $[x] \le -4$, then $x < -3$. This gives the interval $(-\infty, -3)$.

So the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. This matches the given form $(-\infty, a) \cup [b, c) \cup [4, \infty)$. So, $a = -3$, $b = -2$, $c = 3$. $a + b + c = -3 + (-2) + 3 = -2$.

Let me consider the possibility that the question or the provided answer might have an error. However, I must work towards the provided answer.

Let's assume the domain is indeed $(-\infty, a) \cup [b, c) \cup [4, \infty)$ and the answer $a+b+c=8$ is correct. We have derived that the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. Comparing this with $(-\infty, a) \cup [b, c) \cup [4, \infty)$, we get $a = -3$, $b = -2$, $c = 3$. Then $a+b+c = -2$.

Let's check if there's any misinterpretation of the question. The question asks for the value of $a+b+c$. The options are (A) 8, (B) 1, (C) -2, (D) -3. My calculated answer is -2, which is option (C). However, the correct answer is stated as A (8).

Let's re-trace the steps carefully to see if any mistake was made. The core inequality is $\frac{|[x]| - 2}{|[x]| - 3} \ge 0$. Let $Y = |[x]|$. We need $\frac{Y - 2}{Y - 3} \ge 0$. This inequality is satisfied when $Y \in [0, 2] \cup (3, \infty)$. Since $Y = |[x]|$, $Y$ must be a non-negative integer. So, $Y \in \{0, 1, 2\} \cup \{4, 5, 6, ...\}$.

Case 1: $Y \in \{0, 1, 2\}$. $|[x]| = 0 \implies [x] = 0 \implies 0 \le x < 1$. $|[x]| = 1 \implies [x] = 1$ or $[x] = -1$. $[x] = 1 \implies 1 \le x < 2$. $[x] = -1 \implies -1 \le x < 0$. $|[x]| = 2 \implies [x] = 2$ or $[x] = -2$. $[x] = 2 \implies 2 \le x < 3$. $[x] = -2 \implies -2 \le x < -1$. The union of these intervals is $[-2, 3)$.

Case 2: $Y \in \{4, 5, 6, ...\}$. $|[x]| \ge 4$. This means $[x] \ge 4$ or $[x] \le -4$. If $[x] \ge 4$, then $x \ge 4$. This gives $[4, \infty)$. If $[x] \le -4$, then $x < -3$. This gives $(-\infty, -3)$.

So the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. Comparing with $(-\infty, a) \cup [b, c) \cup [4, \infty)$, we get: $a = -3$ $b = -2$ $c = 3$ $a+b+c = -3 + (-2) + 3 = -2$.

Let me consider if the interval notation is being misinterpreted. The domain is given as ($-$\infty$, a)$]\cup$[b, c)$\cup$[4,$\infty$$). This means $x < a$ or $b \le x < c$ or $x \ge 4$.

My derived domain is $x < -3$ or $-2 \le x < 3$ or $x \ge 4$. This perfectly matches the given format.

Let's consider the possibility that the question intends for the values $a, b, c$ to be such that the sum is 8. If $a+b+c = 8$, and the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$. And we know the domain involves intervals related to $[x]$.

Let's assume the correct answer is indeed 8. If $a+b+c = 8$, and we know the domain is structured around the values of $[x]$. The critical values for $|[x]|$ are 2 and 3. The critical values for $[x]$ are $\pm 2, \pm 3$.

Let's re-examine the inequality $\frac{|k| - 2}{|k| - 3} \ge 0$. The critical points for $|k|$ are 2 and 3. Consider the sign of $\frac{|k| - 2}{|k| - 3}$: If $|k| < 2$: e.g., $|k|=0, 1$. Then $|k|-2$ is negative, $|k|-3$ is negative. Ratio is positive. So $|k| \le 2$ is valid. If $2 \le |k| < 3$: e.g., $|k|=2$. Then $|k|-2=0$, $|k|-3$ is negative. Ratio is 0. So $|k|=2$ is valid. If $|k| > 3$: e.g., $|k|=4$. Then $|k|-2$ is positive, $|k|-3$ is positive. Ratio is positive. So $|k| > 3$ is valid. If $|k| = 3$: denominator is zero, so invalid.

So, $|[x]| \le 2$ or $|[x]| > 3$. Since $|[x]|$ is a non-negative integer: $|[x]| \in \{0, 1, 2\}$ or $|[x]| \in \{4, 5, 6, ...\}$.

This leads to $[x] \in \{-2, -1, 0, 1, 2\}$ or $[x] \ge 4$ or $[x] \le -4$. If $[x] \in \{-2, -1, 0, 1, 2\}$: $[x] = -2 \implies -2 \le x < -1$. $[x] = -1 \implies -1 \le x < 0$. $[x] = 0 \implies 0 \le x < 1$. $[x] = 1 \implies 1 \le x < 2$. $[x] = 2 \implies 2 \le x < 3$. Union is $[-2, 3)$.

If $[x] \ge 4$: $x \ge 4$. This gives $[4, \infty)$.

If $[x] \le -4$: $x < -3$. This gives $(-\infty, -3)$.

So the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. This gives $a = -3$, $b = -2$, $c = 3$. $a+b+c = -3 - 2 + 3 = -2$.

Let's consider if the question meant something else by the notation. "($-$\infty$, a)$]\cup$[b, c)$\cup$[4,$\infty$$)" This notation means $(-\infty, a) \cup [b, c) \cup [4, \infty)$.

Let's assume there is a typo in the question or the correct answer. If the question was slightly different, could we get $a+b+c=8$?

Consider the possibility that the domain definition itself implies something about $a, b, c$. Given domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$. We found the domain to be $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. This implies $a = -3$, $b = -2$, $c = 3$.

Let's re-read the question to ensure no details were missed. "Let [ x ] denote the greatest integer $\le$ x, where x $\in$ R. If the domain of the real valued function $f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$ is ($-$\infty$, a)$]\cup$[b, c)$\cup$[4,$\infty$$), a < b < c, then the value of a + b + c is :"

The condition $a < b < c$ is important. We have $a = -3$, $b = -2$, $c = 3$. Here, $-3 < -2 < 3$, which is satisfied.

Let's consider the possibility of a sign error or a misinterpretation of the interval notation. The interval $(-\infty, a)$ means $x < a$. The interval $[b, c)$ means $b \le x < c$. The interval $[4, \infty)$ means $x \ge 4$.

The domain is $\frac{|[x]| - 2}{|[x]| - 3} \ge 0$. Let $k = [x]$. Then $\frac{|k| - 2}{|k| - 3} \ge 0$. This implies $|k| \in [0, 2] \cup (3, \infty)$. Since $k$ is an integer, $|k|$ is a non-negative integer. So $|k| \in \{0, 1, 2\} \cup \{4, 5, 6, ...\}$.

If $|k| \in \{0, 1, 2\}$: $|[x]| = 0 \implies [x] = 0 \implies 0 \le x < 1$. $|[x]| = 1 \implies [x] = 1$ or $[x] = -1$. $[x] = 1 \implies 1 \le x < 2$. $[x] = -1 \implies -1 \le x < 0$. $|[x]| = 2 \implies [x] = 2$ or $[x] = -2$. $[x] = 2 \implies 2 \le x < 3$. $[x] = -2 \implies -2 \le x < -1$. The union of these intervals is $[-2, 3)$.

If $|k| \in \{4, 5, 6, ...\}$: $|[x]| \ge 4$. This means $[x] \ge 4$ or $[x] \le -4$. If $[x] \ge 4$, then $x \ge 4$. This gives $[4, \infty)$. If $[x] \le -4$, then $x < -3$. This gives $(-\infty, -3)$.

So the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. Comparing this with $(-\infty, a) \cup [b, c) \cup [4, \infty)$, we get: $a = -3$ $b = -2$ $c = 3$ $a+b+c = -3 + (-2) + 3 = -2$.

Given that the correct answer is A (8), there might be a misunderstanding of the problem statement or a typo in the question/answer. However, following the standard mathematical interpretation, my derivation consistently leads to -2.

Let's assume, for the sake of reaching the answer 8, that the values of $a, b, c$ are different. If $a+b+c = 8$. And the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$.

Consider the possibility that the condition for the domain was misinterpreted. The expression inside the square root is $\frac{|[x]| - 2}{|[x]| - 3}$. It must be $\ge 0$. The denominator $|[x]| - 3$ cannot be zero, so $|[x]| \ne 3$. This means $[x] \ne 3$ and $[x] \ne -3$.

If $[x] = 3$, then $3 \le x < 4$. If $[x] = -3$, then $-3 \le x < -2$.

Let's re-evaluate the inequality $\frac{|k|-2}{|k|-3} \ge 0$. This is true when $(|k|-2 \ge 0 \text{ and } |k|-3 > 0)$ or $(|k|-2 \le 0 \text{ and } |k|-3 < 0)$. Case 1: $|k| \ge 2$ and $|k| > 3$. This means $|k| > 3$. Case 2: $|k| \le 2$ and $|k| < 3$. This means $|k| \le 2$.

So, $|[x]| \le 2$ or $|[x]| > 3$. Since $[x]$ is an integer, $|[x]|$ is a non-negative integer. $|[x]| \in \{0, 1, 2\}$ or $|[x]| \in \{4, 5, 6, ...\}$.

This leads to $[x] \in \{-2, -1, 0, 1, 2\}$ or $[x] \ge 4$ or $[x] \le -4$.

If $[x] \in \{-2, -1, 0, 1, 2\}$, this gives the union of intervals $[-2, 3)$. If $[x] \ge 4$, this gives $x \ge 4$, so $[4, \infty)$. If $[x] \le -4$, this gives $x < -3$, so $(-\infty, -3)$.

The domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. Comparing with $(-\infty, a) \cup [b, c) \cup [4, \infty)$, we get $a = -3$, $b = -2$, $c = 3$. $a+b+c = -3 - 2 + 3 = -2$.

Given the discrepancy, let me consider if the problem intended for the values $a, b, c$ to be related to the boundaries of the intervals in a different way, or if there's a common mistake pattern that leads to the answer 8.

Could the problem be asking for something other than the direct comparison of intervals? The question states the domain IS of the form $(-\infty, a) \cup [b, c) \cup [4, \infty)$. This implies a direct match.

Let's assume the answer 8 is correct, so $a+b+c=8$. And the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$. And the derived domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$.

If we were to force the sum to be 8, and keep the structure of the intervals: We have the interval $[4, \infty)$ which is fixed. This means the other parts of the domain are $(-\infty, a)$ and $[b, c)$.

Let's consider potential errors in setting up the inequality. The expression is $\sqrt{ \frac{|[x]|-2}{|[x]|-3} }$. The numerator $|[x]|-2$ must be $\ge 0$ or $\le 0$. The denominator $|[x]|-3$ must be $> 0$ or $< 0$.

Condition 1: Numerator $\ge 0$ AND Denominator $> 0$. $|[x]| - 2 \ge 0 \implies |[x]| \ge 2$. $|[x]| - 3 > 0 \implies |[x]| > 3$. Intersection: $|[x]| > 3$.

Condition 2: Numerator $\le 0$ AND Denominator $< 0$. $|[x]| - 2 \le 0 \implies |[x]| \le 2$. $|[x]| - 3 < 0 \implies |[x]| < 3$. Intersection: $|[x]| \le 2$.

So, $|[x]| \le 2$ or $|[x]| > 3$. This is exactly what I have used.

Let's consider if the problem statement has a typo in the function itself. If the function was $\sqrt{\frac{3-|[x]|}{2-|[x]|}}$ then the inequality would be $\frac{3-|k|}{2-|k|} \ge 0$. This is true when $(3-|k| \ge 0$ and $2-|k| > 0)$ or $(3-|k| \le 0$ and $2-|k| < 0)$. Case 1: $|k| \le 3$ and $|k| < 2$. So $|k| < 2$. Case 2: $|k| \ge 3$ and $|k| > 2$. So $|k| \ge 3$. This would give $|[x]| < 2$ or $|[x]| \ge 3$. $|[x]| \in \{0, 1\}$ or $|[x]| \in \{3, 4, 5, ...\}$. If $|[x]| \in \{0, 1\}$: $[x] = 0 \implies 0 \le x < 1$. $[x] = 1 \implies 1 \le x < 2$. $[x] = -1 \implies -1 \le x < 0$. Union: $[-1, 2)$.

If $|[x]| \ge 3$: $[x] \ge 3 \implies x \ge 3$. $[x] \le -3 \implies x < -2$. Domain: $(-\infty, -2) \cup [-1, 2) \cup [3, \infty)$. If this was the domain, then $a = -2$, $b = -1$, $c = 2$. $a+b+c = -2 - 1 + 2 = -1$. This is not 8.

Let's go back to the original problem and assume the answer 8 is correct. This means $a+b+c=8$. And the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$. With $a < b < c$.

Consider the possibility that the values of $a, b, c$ are derived from the integer values of $[x]$ in a different way.

If the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$, then $a=-3, b=-2, c=3$. $a+b+c = -2$.

Let's consider the case where the intervals might be shifted. If the interval $[4, \infty)$ is fixed, it arises from $[x] \ge 4$.

Consider if the problem meant the range of $[x]$ rather than the domain of $x$. But the question clearly asks for the domain of the function $f(x)$.

Let me assume there's a way to get $a+b+c=8$. And the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$. And the function is $f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$.

The derivation of the domain as $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$ seems robust. This implies $a=-3, b=-2, c=3$. This gives $a+b+c = -2$.

Given that the correct answer is provided as A (8), and my derivation consistently yields -2 (option C), there is a strong indication of an error in the provided correct answer or the question itself. However, I am instructed to derive the given correct answer. This is problematic if the derivation leads to a different answer.

Let's assume, hypothetically, that the values of $a, b, c$ are such that $a+b+c=8$. And the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$.

If the domain was instead $(-\infty, -5) \cup [-2, 1) \cup [4, \infty)$. Then $a = -5$, $b = -2$, $c = 1$. $a+b+c = -5 - 2 + 1 = -6$.

If the domain was $(-\infty, 1) \cup [2, 5) \cup [4, \infty)$. This is not ordered.

Let's assume the structure of the intervals is correct, and the values of $a, b, c$ are different. The conditions for the domain are $|[x]| \le 2$ or $|[x]| > 3$. $|[x]| \le 2 \implies -2 \le [x] \le 2 \implies -2 \le x < 3$. This gives $[-2, 3)$. $|[x]| > 3 \implies [x] > 3$ or $[x] < -3$. $[x] > 3 \implies x \ge 4$. This gives $[4, \infty)$. $[x] < -3 \implies x < -3$. This gives $(-\infty, -3)$.

So the domain is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. This implies $a = -3$, $b = -2$, $c = 3$. $a+b+c = -2$.

If the answer is indeed 8, then there must be a different interpretation of the problem or a flaw in my reasoning that I cannot identify.

Let's consider if the intervals are related to the integer values of $[x]$ in a different way. The values of $[x]$ that satisfy the condition are $[-2, 2]$ and $[4, \infty)$ union $(-\infty, -4]$. If $[x] \in \{-2, -1, 0, 1, 2\}$, then $x \in [-2, 3)$. If $[x] \in \{4, 5, 6, ...\}$, then $x \in [4, \infty)$. If $[x] \in \{-4, -5, -6, ...\}$, then $x \in (-\infty, -3)$.

This confirms the domain calculation.

Given the constraint to reach the provided answer, and my inability to do so through logical derivation, it suggests an issue with the problem statement or the provided answer. However, I must present a solution.

Let's assume, for the sake of argument, that the values of $a, b, c$ are such that $a+b+c=8$. And the domain is $(-\infty, a) \cup [b, c) \cup [4, \infty)$. And $a < b < c$.

If we consider the possibility of a typo in the question, and if the domain was $(-\infty, 1) \cup [2, 5) \cup [6, \infty)$, then $a=1, b=2, c=5$. $a+b+c = 1+2+5 = 8$. But this does not match the structure of the derived domain.

Given the problem statement and the provided correct answer, it is impossible to reconcile them through a standard mathematical derivation. My derived answer is -2.

Summary The domain of the function $f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$ is determined by the condition $\frac{|[x]| - 2}{|[x]| - 3} \ge 0$. This inequality holds when $|[x]| \le 2$ or $|[x]| > 3$. As $[x]$ must be an integer, this translates to $[x] \in \{-2, -1, 0, 1, 2\}$ or $[x] \ge 4$ or $[x] \le -4$. These conditions on $[x]$ lead to the domain of $x$ being $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. Comparing this with the given form of the domain, $(-\infty, a) \cup [b, c) \cup [4, \infty)$, we identify $a = -3$, $b = -2$, and $c = 3$. The sum $a+b+c$ is then $-3 + (-2) + 3 = -2$. This result contradicts the provided correct answer of 8.

Common Mistakes & Tips

• Denominator cannot be zero: Remember to exclude values of $x$ for which the denominator $|[x]| - 3 = 0$, which means $|[x]| \ne 3$.
• Integer nature of [x]: The greatest integer function $[x]$ always yields an integer. This is crucial when solving inequalities involving $[x]$.
• Interval notation: Pay close attention to open and closed intervals (parentheses vs. brackets) when determining the domain.

Final Answer Based on a rigorous derivation, the domain of the function is $(-\infty, -3) \cup [-2, 3) \cup [4, \infty)$. Comparing this with the given domain format $(-\infty, a) \cup [b, c) \cup [4, \infty)$, we find $a = -3$, $b = -2$, and $c = 3$. The value of $a + b + c$ is $-3 + (-2) + 3 = -2$. This corresponds to option (C). However, the provided correct answer is (A) 8. There appears to be a discrepancy between the problem statement/options and the derived solution. Assuming the derivation is correct, option (C) is the answer. If we are forced to select option (A), there is an error in the problem statement or the provided answer.

The final answer is \boxed{-2}.