Let $a > 1$ and $0 < b < 1$. If $f: R \rightarrow [0, 1]$ is defined by $f(x) = \begin{cases} a^x, & -\infty < x < 0 \\ b^x, & 0 \leq x < \infty \end{cases}$,then $f(x)$ is

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

If $f: N \rightarrow R$ is defined by $f(1)=-1$ and $f(n+1)=3f(n)+2$ for $n \geq 1$, then $f$ is

On the set of integers $Z$,define $f: Z \rightarrow Z$ as $f(n) = \begin{cases} \frac{n}{2}, & n \text{ is even} \\ 0, & n \text{ is odd} \end{cases}$. Then $f$ is:

Consider the following statements:
Statement-$I$ : $A$ function $f: A \rightarrow B$ is said to be one-one if and only if $f(x) \neq f(y) \Rightarrow x \neq y$.
Statement-$II$ : $A$ relation $f: A \rightarrow B$ is said to be a function if $x \neq y \Rightarrow f(x) \neq f(y)$.
Then which one of the following is true?

Let $A = \{x_1, x_2, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$ be two sets containing seven and three distinct elements respectively. The total number of onto functions $f : A \to B$ such that there exist exactly three elements $x$ in $A$ with $f(x) = y_2$ is equal to: