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(D) Given $f(x) = \frac{\{x\}}{1+[x]^2}$.$I.$ For any $x \in R$,let $n = [x]$,then $x = n + \{x\}$ where $0 \le \{x\} < 1$. Thus $f(x) = \frac{\{x\}}{

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None of $I, II$ and $III$

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(D) Given $f(x) = \frac{\{x\}}{1+[x]^2}$.$I.$ For any $x \in R$,let $n = [x]$,then $x = n + \{x\}$ where $0 \le \{x\} $II.$ At $x = n$ (an integer),$\lim_{x 	o n^-} f(x) = \lim_{x 	o n^-} \frac{x-[x]}{1+[x]^2} = \frac{n-(n-1)}{1+(n-1)^2} = \frac{1}{1+(n-1)^2}$,while $f(n) = \frac{n-n}{1+n^2} = 0$. Since the limit does not equal the function value at integers,$f$ is discontinuous at all integers. So,statement $II$ is false.$III.$ Note that $f(0) = \frac{0}{1+0^2} = 0$ and $f(1) = \frac{1-1}{1+1^2} = 0$. Since $f(0) = f(1)$ but $0 
eq 1$,$f$ is not one-one. So,statement $III$ is false.Therefore,none of the statements are true.

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