Question

Period of $f\left(x\right)=\left\{x\right\}+\{x+\frac{1}{3}\}+\{x+\frac{2}{3}\}$ is equal to (where $\{.\}$ denotes fraction part function )

• A. $1$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $1$

$\left\{x\right\}+\{x+\frac{1}{3}\}+\{x+\frac{2}{3}\}=x+x+\frac{1}{3}+x+\frac{2}{3}-\left(\right[x]+[x+\frac{1}{3}]+[x+\frac{2}{3}\left]\right)$,

where [.] denotes GIF(Greatest Integer Function)

This is because, $x=\left\{x\right\}+\left[x\right]$ .

The period is 1 because

$f(x+1)=x+1+x+1+\frac{1}{3}+x+1+\frac{2}{3}-\left(\right[x+1]+[x+1+\frac{1}{3}]+[x+1+\frac{2}{3}\left]\right)=x+1+x+1+\frac{1}{3}+x+1+\frac{2}{3}-\left(\right[x]+1+[x+\frac{1}{3}]+1+[x+\frac{2}{3}]+1)=x+x+\frac{1}{3}+x+\frac{2}{3}-\left(\right[x]+[x+\frac{1}{3}]+[x+\frac{2}{3}\left]\right)=\left\{x\right\}+\{x+\frac{1}{3}\}+\{x+\frac{2}{3}\}$