Q42 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

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

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

Solution

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 $\quad\left[\frac{x}{3}\right]=k$ $\because \quad[x]=3 k$ $\because \quad x \in[3 k, 3 k+1)$ $\{x \in \mathbf{R} / 3 k \leq x < 3 k+1, k \in \mathbf{Z}\}$

Hence, the correct answer is option: 'D'.

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