If the function $f: R \rightarrow R$ is defined by $f(x) = (x^{2} + 1)^{35}, \forall x \in R,$ then $f$ is

Let $R$ denote the set of all real numbers and $R^{+}$ denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$,define $f: A \rightarrow B$ by $f(x) = x^2$ for $x \in A$. Match the items in Column-$I$ with the items in Column-$II$.

Column-$I$	Column-$II$
$A$. $f$ is one-one and onto,if	$1$. $A = R^{+}, B = R$
$B$. $f$ is one-one but not onto,if	$2$. $A = B = R$
$C$. $f$ is onto but not one-one,if	$3$. $A = R, B = R^{+}$
$D$. $f$ is neither one-one nor onto,if	$4$. $A = B = R^{+}$

Statement-$I$: Let $f : R \rightarrow R$ be a function such that $f(x) = x^3 + x^2 + 3x + \sin x$. Then $f$ is a one-one function.
Statement-$II$: $f(x)$ is a decreasing function.

If a real-valued function $f:[a, \infty) \rightarrow [b, \infty)$ defined by $f(x) = 2x^2 - 3x + 5$ is a bijection,then $3a + 2b =$

If a function $f: R-\{l\} \to R-\{m\}$ defined by $f(x) = \frac{x+3}{x-2}$ is a bijection,then $3l - 2m =$

The number of bijective functions $f : \{1, 3, 5, 7, \ldots, 99\} \rightarrow \{2, 4, 6, 8, \ldots, 100\}$ such that $f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)$ is: