Let $R$ be the set of real numbers and $f: R \rightarrow R$ be defined by $f(x)=\frac{\{x\}}{1+[x]^2}$, where $[x]$ is the greatest integer less than or equal to $x$, and $\left\{x{\}}=x-[x]\right.$. Which of the following statements are true?

$I.$ The range of $f$ is a closed interval.

$II.$ $f$ is continuous on $R$.

$III.$ $f$ is one-one on $R$

Let ${f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)\;,x \in R$ and $k \ge 1$, then ${f_4}\left( x \right) - {f_6}\left( x \right)$ is equal to

Let $f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)$. If $f^{a+1}(x)=f\left(f^{n}(x)\right)$ for all $n \in N$, then $f^{\prime}(6)+f(7)$ is equal to

Let $f(x) = {(x + 1)^2} - 1,\;\;(x \ge - 1)$. Then the set $S = \{ x:f(x) = {f^{ - 1}}(x)\} $ is

Let $f(x)=2 x^{2}-x-1$ and $S =\{n \in Z :|f(n)| \leq 800\}$ . Then value of $\sum_{n \in S} f(n)$ is . . . .  .

If a function $f(x)$ is such that $f\left( {x + \frac{1}{x}} \right) = {x^2} + \frac{1}{{{x^2}}};$ then  $(fof )$ $\sqrt {11} )$ =