The range of the function $f(x) = \frac{x}{{1 + \left| x \right|}},\,x \in R,$ is

Show that the Modulus Function $f : R \rightarrow R$ given by $(x)=|x|$, is neither one - one nor onto, where $|x|$ is $x$, if $x$ is positive or $0$ and $| X |$ is $- x$, if $x$ is negative.

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 -

Let $f:(1,3) \rightarrow \mathrm{R}$ be a function defined by

$f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ where $[\mathrm{x}]$ denotes the greatest

integer $\leq \mathrm{x} .$ Then the range of $f$ is

Let  $a,b,c\; \in R.$ If $f\left( x \right) = a{x^2} + bx + c$ is such that $a + b + c = 3$ and $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) + xy,$ $\forall x,y \in R,$ then $\mathop \sum \limits_{n = 1}^{10} f\left( n \right)$ is equal to : 

Show that none of the operations given above has identity.