Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Domain of $f (x)$ is
$(-1, 0) \cup (0, \infty )$
$R - { 0 }$
$(-\infty , -1) \cup (0, \infty )$
$(0, \infty )$
Range of the function
$f(x) = \sqrt {\left| {{{\sin }^{ - 1}}\left| {\sin x} \right|} \right| - {{\cos }^{ - 1}}\left| {\cos x} \right|} $ is
Prove that the Greatest Integer Function $f: R \rightarrow R ,$ given by $f(x)=[x]$, is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.
Let $f(x) = {(x + 1)^2} - 1,\;\;(x \ge - 1)$. Then the set $S = \{ x:f(x) = {f^{ - 1}}(x)\} $ is
Greatest value of the function, $f(x) = - 1 + \frac{2}{{{2^x}^2 + 1}}$ is
Let $f$ be a function defined on the set of all positive integers such that $f(x y)=f(x)+f(y)$ for all positive integers $x, y$. If $f(12)=24$ and $f(8)=15$. The value of $f(48)$ is