If $f(x) = \frac{{{x^2} - 1}}{{{x^2} + 1}}$, for every real numbers. then the minimum value of $f$

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

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

If $f(x) = \frac{1}{{\sqrt {x + 2\sqrt {2x - 4} } }} + \frac{1}{{\sqrt {x - 2\sqrt {2x - 4} } }}$ for $x > 2$, then $f(11) = $

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 -

Show that the function $f: R_* \rightarrow R_*$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $R_*$ is the set of all non-zero real numbers. Is the result true, if the domain $R_*$ is replaced by $N$ with co-domain being same as $R _*$ ?