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 $R$

If $f(x) = \cos (\log x)$, then the value of $f(x).f(4) - \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$

If $x \in [0, 1]$, then the number of solution $(s)$ of the equation $2[cos^{-1}x] + 6[sgn(sinx)] = 3$ is (where $[.]$ denotes greatest integer function and sgn $(x)$ denotes signum function of $x$)-

If the domain of the function $f(x)=\log _e$ $\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$ is $(\alpha, \beta]$, then the value of $5 \beta-4 \alpha$ is equal to

If $f(x) = \log \left[ {\frac{{1 + x}}{{1 - x}}} \right]$, then $f\left[ {\frac{{2x}}{{1 + {x^2}}}} \right]$ is equal to

Domain of the function $f(x) = \sqrt {2 - {{\sec }^{ - 1}}x} $ is