If $a, b$ be two fixed positive integers such that $f(a + x) = b + {[{b^3} + 1 - 3{b^2}f(x) + 3b{\{ f(x)\} ^2} - {\{ f(x)\} ^3}]^{\frac{1}{3}}}$ for all real $x$, then $f(x)$ is a periodic function with period
$a$
$2a$
$b$
$2b$
Let $[x]$ denote the greatest integer $\leq x$, where $x \in R$. If the domain of the real valued function $\mathrm{f}(\mathrm{x})=\sqrt{\frac{[\mathrm{x}] \mid-2}{\sqrt{[\mathrm{x}] \mid-3}}}$ is $(-\infty, \mathrm{a}) \cup[\mathrm{b}, \mathrm{c}) \cup[4, \infty), \mathrm{a}\,<\,\mathrm{b}\,<\,\mathrm{c}$, then the value of $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is:
The function $f(x) =$ ${x^{\frac{1}{{\ln \,x}}}}$
The domain of definition of the function $y(x)$ given by ${2^x} + {2^y} = 2$ is
Let the sets $A$ and $B$ denote the domain and range respectively of the function $f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$ where $\lceil x \rceil$ denotes the smallest integer greater than or equal to $x$. Then among the statements
$( S 1): A \cap B =(1, \infty)-N$ and
$( S 2): A \cup B=(1, \infty)$
Suppose $f$ is a function satisfying $f ( x + y )= f ( x )+ f ( y )$ for all $x , y \in N$ and $f (1)=\frac{1}{5}$. If $\sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to $...............$.