Let [x] denote the greatest integer leq x,whe | vedclass

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(C) For the function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ to be defined,the expression inside the square root must be non-negative and the denomina

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$-2$

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(C) For the function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ to be defined,the expression inside the square root must be non-negative and the denominator must be non-zero.$\frac{|[x]|-2}{|[x]|-3} \geq 0$ and $|[x]|-3 
eq 0$.Let $Y = [x]$. We solve $\frac{|Y|-2}{|Y|-3} \geq 0$.The critical points are $|Y| = 2$ and $|Y| = 3$,which means $Y \in \{-3, -2, 2, 3\}$.Testing intervals for $|Y|$:$1$. If $|Y|  0$ (True). This corresponds to $-2 $2$. If $2 \leq |Y| $3$. If $|Y| > 3$,then $\frac{+}{+} > 0$ (True). This corresponds to $Y > 3$ or $Y Combining these,the domain is $(-\infty, -3) \cup [-1, 2) \cup [4, \infty)$.Comparing with $(-\infty, a) \cup [b, c) \cup [4, \infty)$,we get $a = -3$,$b = -1$,and $c = 2$.Thus,$a+b+c = -3 + (-1) + 2 = -2$.

Let $[x]$ denote the greatest integer $\leq x$,where $x \in \mathbb{R}$. If the domain of the real-valued function $f(x) = \sqrt{\frac{|[x]|-2}{|[x]|-3}}$ is $(-\infty, a) \cup [b, c) \cup [4, \infty)$,where $a < b < c$,then the value of $a+b+c$ is:

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