Let $f :R \to R$ be defined by $f(x)\,\, = \,\,\frac{x}{{1 + {x^2}}},\,x\, \in \,R.$ Then the range of $f$ is

Minimum integral value of $\alpha$ for which graph of $f(x) = ||x -2| -\alpha|-5$ has exactly four $x-$intercepts-

Let $f(x)$ and $g(x)$ be two functions given by $f\left( x \right) = \frac{{2\sin \pi x}}{x}$ and $g\left( x \right) = f\left( {1 – x} \right) + f\left( x \right).$ If $g\left( x \right) = kf(\frac{x}{2})f\left( {\frac{{1 – x}}{2}} \right)$,then the value of $k$ is

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by

$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$. If the composition of $f, \underbrace{(f \circ f \circ f \circ \ldots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$, then the value of $\sqrt{3 \alpha+1}$ is equal to………………..

If $f\left( x \right) + 2f\left( {\frac{1}{x}} \right) = 3x,x \ne 0$ and $S = \left\{ {x \in R:f\left( x \right) = f\left( { – x} \right)} \right\}$;then $S :$