Question

If domain of $f\left(x\right)$ is $[0,1],$ then domain of $f\left(\right\{x{\}}^3+1),$ (where $\{.\}$ represents fractional part function) is

• A. $\{x:x=k,k\in Q\}$
• B. $\{x:k\le x<k+\frac{1}{2},k\in I\}$
• C. $\{x:x=k,k\in I\}$
• D. $(-\infty ,0)$

Answer 1

The correct option is C $\{x:x=k,k\in I\}$

If domain of $f\left(x\right)$ is $[0,1],$ then for $f\left(\right\{x{\}}^3+1)$ to be defined,
$\Rightarrow 0\le \{x{\}}^3+1\le 1$
$\Rightarrow -1\le \{x{\}}^3\le 0$
$\Rightarrow x\in I$ ( $\because \left\{x\right\}=0$ is the only possiblity )