Set of all values of $x$ satisfying
$\frac{{{x^4} - 4{x^3} + 3{x^2}}}{{({x^2} - 4)({x^2} - 7x + 10)}} \ge 0$
$\left( { - \infty ,\, - 2} \right)\,\, \cup \,\,[1,\,3]\,\, \cup \,\,(5,\infty )$
$\left( { - \infty ,\, - 2} \right)\,\,\, \cup \,\,\,[0,\,1]\,\,\, \cup \,\,\left( {2,3} \right]\,\, \cup \,\,(5,\infty )$
$\left( { - \infty ,\, - 2} \right)\,\,\, \cup \,\,\,[1,\,3]\,\,\, \cup \,\,(5,\infty )\,\, \cup \,\,\{ 0\} $
$\left( { - \infty ,\, - 2} \right)\,\,\, \cup \,\,\,[1,\,2]\,\,\, \cup \,\,\left( {2,3} \right]\,\,\, \cup \,\,(5,\infty )\,\, \cup \,\,\{ 0\} $
If $f\left( x \right) = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} - 1$ , $x \in R$ , then the equation $f(x) = 0$ has
If $f:R \to R$ and $g:R \to R$ are given by $f(x) = \;|x|$ and $g(x) = \;|x|$ for each $x \in R$, then $\{ x \in R\;:g(f(x)) \le f(g(x))\} = $
Let $f :R \to R$ be defined by $f(x)\,\, = \,\,\frac{x}{{1 + {x^2}}},\,x\, \in \,R.$ Then the range of $f$ is
The domain of the function $f(x)=\frac{1}{\sqrt{[x]^2-3[x]-10}}$ is (where $[x]$ denotes the greatest integer less than or equal to $x$ )
If $x \in [0, 1]$, then the number of solution $(s)$ of the equation $2[cos^{-1}x] + 6[sgn(sinx)] = 3$ is (where $[.]$ denotes greatest integer function and sgn $(x)$ denotes signum function of $x$)-