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Let $[\cdot]$ denote greatest integer function. If $f(x)=[x]$ and $g(x)=3\left[\frac{x}{3}\right]$, then the set of all real $x$ such that $f(x)=g(x)$ is
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Verified Answer
The correct answer is:
$\{x \in R / 3 k \leq x < 3 k+1, k \in Z\}$
We have,
$f(x)=[x] \text { and } g(x)=3\left[\frac{x}{3}\right]$
Given, $\quad f(x)=g(x)$
$\therefore \quad[x]=3\left[\frac{x}{3}\right]$
Here, $[x]$ and $\left[\frac{x}{3}\right]$ is an integers
Let $\quad\left[\frac{x}{3}\right]=k$
$\because \quad[x]=3 k$
$\because \quad x \in[3 k, 3 k+1)$
$\{x \in \mathbf{R} / 3 k \leq x < 3 k+1, k \in \mathbf{Z}\}$
$f(x)=[x] \text { and } g(x)=3\left[\frac{x}{3}\right]$
Given, $\quad f(x)=g(x)$
$\therefore \quad[x]=3\left[\frac{x}{3}\right]$
Here, $[x]$ and $\left[\frac{x}{3}\right]$ is an integers
Let $\quad\left[\frac{x}{3}\right]=k$
$\because \quad[x]=3 k$
$\because \quad x \in[3 k, 3 k+1)$
$\{x \in \mathbf{R} / 3 k \leq x < 3 k+1, k \in \mathbf{Z}\}$
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