If the function $f\,:\,R - \,\{ 1, - 1\}  \to A$ defined by $f\,(x)\, = \frac{{{x^2}}}{{1 - {x^2}}},$ is surjective, then $A$ is equal to

Consider a function $f:\left[0, \frac{\pi}{2}\right]$ $ \rightarrow$ $R$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] $ $\rightarrow$ $R$ given by $g(x)=\cos x .$ Show that $f$ and $g$ are one-one, but $f\,+\,g$ is not one-one.

If $f:\left\{ {1,2,3,4} \right\} \to \left\{ {1,2,3,4} \right\}$ and $y=f(x)$ be a function such that $\left| {f\left( \alpha  \right) - \alpha } \right| \leqslant 1$,for $\alpha  \in \left\{ {1,2,3,4} \right\}$ then total number of such functions are

For $x\,\, \in \,R\,,x\, \ne \,0,$ let ${f_0}(x) = \frac{1}{{1 - x}}$ and ${f_{n + 1}}(x) = {f_0}({f_n}(x)),$ $n\, = 0,1,2,....$  Then the value of ${f_{100}}(3) + {f_1}\left( {\frac{2}{3}} \right) + {f_2}\left( {\frac{3}{2}} \right)$ is equal to

Let, $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ and $A=\{x \in R: f(x)=1\} .$ Then, $A$ has

$f : R \to R$ is defined as

$f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2} + 2mx - 1\,,}&{x \leq 0}\\
{mx - 1\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x > 0}
\end{array}} \right.$

 If $f (x)$ is one-one then the set of values of $'m'$ is