Question

Let $[.] $ 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

• A. $R$
• B. $\{x \in R / x=3 k, k \in Z \}$
• C. $\{x \in R / 3 k-1
• D. $\{x \in R / 3 k \leq x<3 k+1, k \in Z \}$

Answer 1

Answer: (D)

We have, $\& f(x)=[x]$ and $g(x)=3\left[\frac{x}{3}\right]$
Given, $f(x)=g(x)$
$\therefore [x]=3\left[\frac{x}{3}\right]$
Here, $[x]$ and $\left[\frac{x}{3}\right]$ is an integers
Let $\left[\frac{x}{3}\right]=k$
$\because[x]=3 k$
$\because x \in[3 k, 3 k+1)$
$\{x \in R / 3 k \leq x<3 k+1, k \in Z\}$