If $f(x) = \begin{cases} x, & \text{when } x \text{ is rational} \\ 0, & \text{when } x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} 0, & \text{when } x \text{ is rational} \\ x, & \text{when } x \text{ is irrational} \end{cases}$,then $(f - g)$ is:

Let the functions $f$ and $g$ be $f: [0, \frac{\pi}{2}] \rightarrow R$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow R$ given by $g(x) = \cos x$,where $R$ is the set of real numbers. Consider the following statements:
Statement $(I)$: $f$ and $g$ are one-one.
Statement $(II)$: $f+g$ is one-one.
Which of the following is correct?

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)

Prove that the function $f: R \rightarrow R$,defined by $f(x)=2x$,is one-one and onto.

If $P(S)$ denotes the set of all subsets of a given set $S$,then the number of one-to-one functions from the set $S = \{ 1, 2, 3 \}$ to the set $P(S)$ is

If $R$ denotes the set of all real numbers,then the function $f: R \rightarrow R$ defined by $f(x)=|x|$ is