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

The domain of the function f(x) = $${1 \over {\sqrt {\left| x \right| - x} }}$$ is

Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge - 1$$

Statement - 1 : The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.

Statement - 2 : $$f$$ is a bijection.

For real x, let f(x) = x³ + 5x + 1, then