solve $\frac{{1 - \left| x \right|}}{{2 - \left| x \right|}} \ge 0$
$R$
$\left[ { - 1\,,\,2} \right)\,\, \cup \,\left( {2\,,\,\infty } \right)\,$
$\left[ { - 1\,,\,1} \right]\,\, \cup \,\left( {2\,,\,\infty } \right)\,$
$\left( { - \infty ,\, - 2} \right)\,\, \cup \,\,[ - 1,\,1]\,\, \cup \,\,(2,\infty )$
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 $x$.
Domain of the function $f(x) = {\sin ^{ - 1}}(1 + 3x + 2{x^2})$ is
The domain of the function $f(x){ = ^{16 - x}}{\kern 1pt} {C_{2x - 1}}{ + ^{20 - 3x}}{\kern 1pt} {P_{4x - 5}}$, where the symbols have their usual meanings, is the set
The range of $f(x) = \cos (x/3)$ is
If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
{3 + x;\,\,\,\,\,x \geqslant 0} \\
{2 - 3x;\,\,\,\,\,x < 0}
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -