Let $A = \left\{ {{x_1},{x_2},{x_3},…..,{x_7}} \right\}$ and $B = \left\{ {{y_1},{y_2},{y_3}} \right\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f:A \to B$ which are onto, if there exist exactly three elements $x$ in $A$ such that $f(x) = {y_2}$ , is equal to

The graph of $y = f(x)$ is shown then number of solutions of the equation $f(f(x)) =2$ is

If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$  and $g\left( x \right) = {-x^2} – 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval

If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne – 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$

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