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(D) Given $f(x) = |x|$ and $g(x) = [x]$.We need to find the set of $x \in R$ such that $g(f(x)) \leq f(g(x))$.Substituting the functions,we get $[|x|]

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(D) Given $f(x) = |x|$ and $g(x) = [x]$.We need to find the set of $x \in R$ such that $g(f(x)) \leq f(g(x))$.Substituting the functions,we get $[|x|] \leq |[x]|$.Case $1$: If $x \geq 0$,then $|x| = x$ and $[x] \geq 0$. The inequality becomes $[x] \leq |[x]|$. Since $[x]$ is an integer and $|[x]| = [x]$ for $[x] \geq 0$,this holds true for all $x \geq 0$.Case $2$: If $x If $x$ is an integer,$x = -n$ $(n > 0)$,then $[|x|] = [n] = n$ and $|[x]| = |-n| = n$. Thus $n \leq n$,which is true.If $x$ is not an integer,let $x = -n - \delta$ $(n \geq 0, 0 The inequality becomes $n \leq n + 1$,which is true.Therefore,the inequality holds for all $x \in R$.

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are given by $f(x)=|x|$ and $g(x)=[x]$ for each $x \in R$,then $\{x \in R: g(f(x)) \leq f(g(x))\}$ is equal to

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