Which one of the following best represent the graph of $y = \frac{|x-x^2|}{x^2-x}$ ? 

Show that function $f : R \rightarrow\{ x \in R :-1< x <1\}$ defined by $f ( x )=\frac{x}{1+|x|^{\prime}} x \in R$ is one-one and onto function.

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 ${a_2},{a_3} \in R$ such that $\left| {{a_2} - {a_3}} \right| = 6$ and $f\left( x \right) = \left| {\begin{array}{*{20}{c}}
1&{{a_3}}&{{a_2}}\\
1&{{a_3}}&{2{a_2} - x}\\
1&{2{a_3} - x}&{{a_2}}
\end{array}} \right|,x \in R.$ Then the greatest value of $f(x)$ is

Domain of $f (x)$ = $\sqrt {{{\log }_2}\left( {\frac{{10x - 4}}{{4 - {x^2}}}} \right) - 1} $ , is

Let $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5,6\}$. Then the number of functions $f: A \rightarrow B$ satisfying $f(1)+f(2)=f(4)-1$ is equal to