Domain of the definition of function 

$f(x) = \sqrt {\frac{{4 - {x^2}}}{{\left[ x \right] + 2}}} $ is      $($ where $[.] \rightarrow G.I.F.)$

The range of the function,

$\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)$

$-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))$ is :

Show that function $f : R \rightarrow\{ x \in R :-1< x <1\}$ defined by $f ( x )=\frac{x}{1+|x|^{\prime}} x \in R$ is one-one and onto function.

If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
  {3 + x;\,\,\,\,\,x \geqslant 0} \\ 
  {2 - 3x;\,\,\,\,\,x < 0} 
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -

Range of $f(x) = \;[x]\; - x$ is

Let $f : R \rightarrow R$ be a function defined by $f ( x )=$ $\log _{\sqrt{m}}\{\sqrt{2}(\sin x-\cos x)+m-2\}$, for some $m$, such that the range of $f$ is $[0,2]$. Then the value of $m$ is $............$