The function $f(x) = \frac{{{{\sec }^{ - 1}}x}}{{\sqrt {x - [x]} }},$ where $[.]$ denotes the greatest integer less than or equal to $x$  is defined for all  $x$  belonging to

Let $\sum\limits_{k = 1}^{10} {f\,(a\, + \,k)} \, = \,16\,({2^{10}}\, - \,1),$ where the function $f$ satisfies $f(x + y) = f(x) f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $‘ a '$ is

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

Which pair $(s)$ of function $(s)$ is/are equal ?

where $\{x\}$ and $[x]$ denotes the fractional part $\&$ integral part functions.

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 $f(x)$ is a quadratic expression such that $f(1) + f (2)\, = 0$ , and $-1$ is a root of $f(x)\, = 0$, then the other root of $f(x)\, = 0$ is