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The period of $f(x)=x-[x]$, if it is periodic, is
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Verified Answer
The correct answer is:
$1$
Let $f(x)$ be periodic with period $T$.
Then, $f(x+T)=f(x)$ for all $x \in R$
$\Rightarrow x+T-[x+T]=x-[x]$, for all $x \in R$
$\Rightarrow x+T-x=[x+T]-[x]$
$\Rightarrow[x+T]-[x]=T$ for all $x \in R \Rightarrow T=1,2,3,4, \ldots \ldots .$.
The smallest value of $T$ satisfying
$f(x+T)=f(x)$ for all $x \in R$ is 1 .
Hence $f(x)=x-[x]$ has period 1 .
Then, $f(x+T)=f(x)$ for all $x \in R$
$\Rightarrow x+T-[x+T]=x-[x]$, for all $x \in R$
$\Rightarrow x+T-x=[x+T]-[x]$
$\Rightarrow[x+T]-[x]=T$ for all $x \in R \Rightarrow T=1,2,3,4, \ldots \ldots .$.
The smallest value of $T$ satisfying
$f(x+T)=f(x)$ for all $x \in R$ is 1 .
Hence $f(x)=x-[x]$ has period 1 .
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