If $y = f(x) = \frac{{ax + b}}{{cx - a}}$, then $x$ is equal to

Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$  $\forall $  $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$  $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.

If $f(x) = 2\sin x$, $g(x) = {\cos ^2}x$, then $(f + g)\left( {\frac{\pi }{3}} \right) = $

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 : 

Domain of function $f(x) = {\sin ^{ - 1}}5x$ is

Range of ${\sin ^{ - 1\,}}\left( {\frac{{1 + {x^2}}}{{2 + {x^2}}}} \right)$ is