The function f(x) = [|x|] - |[x]| where [x] d | vedclass

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(D) Let $f(x) = [|x|] - |[x]|$.Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$,so $|[x]| = [x]$. Thus,$f(x) = [x] - [x] = 0$.Case $2$: If $x <

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(D) Let $f(x) = [|x|] - |[x]|$.Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$,so $|[x]| = [x]$. Thus,$f(x) = [x] - [x] = 0$.Case $2$: If $x If $f = 0$,then $x = -n$ (an integer). Then $f(x) = [|-n|] - |[-n]| = [n] - |-n| = n - n = 0$.If $0 $|x| = n + f$. $[|x|] = [n + f] = n$. So,$f(x) = n - (n + 1) = -1$.Thus,$f(x) = 0$ for $x \geq 0$ and $x \in \mathbb{Z}$,and $f(x) = -1$ for $x \in \mathbb{R} \setminus \mathbb{Z}$ where $x Since the function is constant on intervals but jumps at integers for $x The range is $\{0, -1\}$,which is a finite set. Therefore,all given statements are correct.

The function $f(x) = [|x|] - |[x]|$ where $[x]$ denotes the greatest integer function:

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