If $f: R \rightarrow C$ is defined by $f(x)=e^{2 i x}$ for $x \in R$,then $f$ is (where $C$ denotes the set of all complex numbers)

If $f: R \rightarrow R$ is defined by $f(x) = x^2 + 3x + 4$,then the function $f$ is . . . . . . .

The function $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$ defined as $f(x) = \frac{x}{1 + x^2}$ is

Let $A=\{1,2,3\}, \,B=\{4,5,6,7\}$ and let $f=\{(1,4),\,(2,5),\,(3,6)\}$ be a function from $A$ to $B$. Show that $f$ is one-one.

Given that for any $n \in N$, there exist an odd integer $q$ and a non-negative integer $r$ such that $n$ can be written uniquely as $n = q \times 2^r$. Let $f: N \rightarrow N \times N$ be a function defined by $f(n) = \left(r+1, \frac{q+1}{2}\right)$. Then,

Let $X$ and $Y$ be subsets of $R$,the set of all real numbers. The function $f:X \to Y$ defined by $f(x) = x^2$ for $x \in X$ is one-one but not onto if (Here $R^+$ is the set of all positive real numbers):