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(D) We are given $f(x) = [x]$ and $g(x) = 3[\frac{x}{3}]$.Setting $f(x) = g(x)$, we have $[x] = 3[\frac{x}{3}]$.Let $[\frac{x}{3}] = k$, where $k \in

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$\{x \in R : 3k \leq x < 3k + 1, k \in Z\}$

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(D) We are given $f(x) = [x]$ and $g(x) = 3[\frac{x}{3}]$.Setting $f(x) = g(x)$, we have $[x] = 3[\frac{x}{3}]$.Let $[\frac{x}{3}] = k$, where $k \in Z$.Then, by the definition of the greatest integer function, $k \leq \frac{x}{3} Also, we have $[x] = 3k$.By the definition of the greatest integer function, $[x] = 3k$ implies $3k \leq x Since $3k \leq x Thus, the set of all such real $x$ is $\{x \in R : 3k \leq x < 3k + 1, k \in Z\}$.

Let $[\cdot]$ denote the greatest integer function. If $f(x) = [x]$ and $g(x) = 3[\frac{x}{3}]$, then the set of all real $x$ such that $f(x) = g(x)$ is

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