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(A) The function $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$ is discontinuous where either $[\frac{x}{2} + 3]$ or $[\sqrt{x}]$ is discontinuous.$1$. The t

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$17$

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(A) The function $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$ is discontinuous where either $[\frac{x}{2} + 3]$ or $[\sqrt{x}]$ is discontinuous.$1$. The term $[\frac{x}{2} + 3]$ is discontinuous when $\frac{x}{2} + 3$ is an integer.For $x \in [0, 8]$,$\frac{x}{2} + 3$ takes values in $[3, 7]$.Thus,$\frac{x}{2} + 3 \in \{3, 4, 5, 6, 7\}$.Solving for $x$: $\frac{x}{2} \in \{0, 1, 2, 3, 4\} \implies x \in \{0, 2, 4, 6, 8\}$.$2$. The term $[\sqrt{x}]$ is discontinuous when $\sqrt{x}$ is an integer.For $x \in [0, 8]$,$\sqrt{x} \in [0, \sqrt{8}] \approx [0, 2.82]$.Thus,$\sqrt{x} \in \{0, 1, 2\}$.Solving for $x$: $x \in \{0, 1, 4\}$.$3$. The set $S$ of points of discontinuity in $[0, 8]$ is the union of these points:$S = \{0, 1, 2, 4, 6, 8\}$.However,we must check if the jumps cancel out at any point.At $x=0$: $f(0) = [3] - [0] = 3$. $\lim_{x 	o 0^+} f(x) = [3] - [0] = 3$. Continuous.At $x=4$: $f(4) = [2+3] - [2] = 5 - 2 = 3$.$\lim_{x 	o 4^-} f(x) = [4.99] - [1.99] = 4 - 1 = 3$.$\lim_{x 	o 4^+} f(x) = [5.00...] - [2.00...] = 5 - 2 = 3$. Continuous.Thus,the points of discontinuity are $S = \{1, 2, 6, 8\}$.The sum of elements in $S$ is $1 + 2 + 6 + 8 = 17$.

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to:

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