Let $A= \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$. The correct statement is

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.

Let $f(x)=a x^{2}+b x+c$ be such that $f(1)=3, f(-2)$ $=\lambda$ and $f (3)=4$. If $f (0)+ f (1)+ f (-2)+ f (3)=14$, then $\lambda$ is equal to$...$

Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{\,{x^3} - {x^2} + 10x - 5\,\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \le 1\,\,\,\,\,\,\,\,\,\,\,\,}\\
{ - 2x + {{\log }_2}({b^2} - 2),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, > 1\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}} \right.$ the set of values of $b$ for which $f(x)$ has greatest value at $x = 1$ is given by 

If $f(x)$ be a polynomial function satisfying $f(x).f (\frac{1}{x}) = f(x) + f (\frac{1}{x})$  and $f(4) = 65$ then value of $f(6)$ is

If $f(x) = \frac{x}{{x - 1}}$, then $\frac{{f(a)}}{{f(a + 1)}} = $