Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} 2x; & x > 3 \\ x^2; & 1 < x \leq 3 \\ 3x; & x \leq 1 \end{cases}$. Then $f(-1) + f(2) + f(4)$ is

For the mapping $f: R-\{1\} \rightarrow R-\{2\}$, given by $f(x)=\frac{2x}{x-1}$, which of the following is correct?

The function $f: C \rightarrow C$ defined by $f(x) = \frac{ax + b}{cx + d}$ for $x \in C$, where $bd \neq 0$, reduces to a constant function if:

If $f: R \rightarrow R$ is defined by $f(x)=x-[x]-\frac{1}{2}$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then $\{x \in R: f(x)=\frac{1}{2}\}$ is equal to :

The function $f : R \rightarrow (-1, 1)$ defined by $f(x) = \frac{e^x - 1}{e^x + 1}$ is:

Let $f:[0,1] \rightarrow [-1,1]$ and $g:[-1,1] \rightarrow [0,2]$ be two functions such that $g$ is injective and $g \circ f: [0,1] \rightarrow [0,2]$ is surjective. Then,