The function $f(x) = \;|px - q|\; + r|x|,\;x \in ( - \infty ,\;\infty )$, where $p > 0,\;q > 0,\;r > 0$ assumes its minimum value only at one point, if

Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Domain of $f (x)$ 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

$( S 2): A \cup B=(1, \infty)$

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.

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

If the domain of the function $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is $R-(\alpha, \beta)$ then $12 \alpha \beta$ is equal to :