If $(x, y) \in R$ and $x, y \neq 0$,and the function $f(x, y) = \frac{x}{y}$,then this function is a/an:

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

For $0 \leq x \leq 1$,what type of function is $f(x) = |x| + |x - 1|$?

The function $f: R \rightarrow R$ is defined by $f(x)=3^{-x}$. Observe the following statements about it:
$I$. $f$ is one-one
$II$. $f$ is onto
$III$. $f$ is a decreasing function
Out of these, the true statements are:

Let $A = \{1, 2, 3, \ldots, 7\}$ and let $P(A)$ denote the power set of $A$. If the number of functions $f: A \rightarrow P(A)$ such that $a \in f(a)$ for all $a \in A$ is $m^n$,where $m, n \in N$ and $m$ is the least possible value,then $m + n$ is equal to . . . . . . .

Let $f : R \to R$ be defined by $f(x) = \frac{|x| - 1}{|x| + 1}$. Then $f$ is